Class 9

HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना

Haryana State Board HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 4 परमाणु की संरचना

HBSE 9th Class Science परमाणु की संरचना Intext Questions and Answers
(पृष्ठ संख्या-53)
प्रश्न 1.
केनाल किरणें क्या हैं?
उत्तर:
ई० गोल्डस्टीन ने 1886 ई० में धनावेशित विकिरण की खोज की, जिन्हें केनाल किरणें कहा जाता है।

प्रश्न 2.
यदि किसी परमाणु में एक इलेक्ट्रॉन और एक प्रोटॉन है, तो इसमें कोई आवेश होगा या नहीं?
उत्तर:
यदि किसी परमाणु में एक इलेक्ट्रॉन व एक प्रोटॉन हो तो उस पर कोई आवेश नहीं होगा क्योंकि प्रोटॉन और इलेक्ट्रॉन परस्पर आवेशों को संतुलित करते हैं।

(पृष्ठ संख्या -56)

प्रश्न 1.
परमाणु उदासीन है, इस तथ्य को टॉमसन के मॉडल के आधार पर स्पष्ट कीजिए।
उत्तर:
टॉमसन के मॉडल अनुसार परमाणु धन आवेशित गोले का बना होता है तथा इलेक्ट्रॉन उसमें धंसे होते हैं। इस प्रकार ऋणात्मक और धनात्मक आवेश परिमाण समान होने के कारण परमाणु वैद्युतीय रूप से उदासीन होता है।
HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना img-1

प्रश्न 2.
रदरफोर्ड के परमाणु मॉडल के अनुसार, परमाणु के नाभिक में कौन-सा अवपरमाणुक कण विद्यमान है?
उत्तर:
रदरफोर्ड के परमाणु मॉडल के अनुसार, परमाणु के नाभिक में प्रोटॉन अवपरमाणुक कण विद्यमान है क्योंकि वह अल्फा कण को विक्षेपित करता है।

प्रश्न 3.
तीन कक्षाओं वाले बोर के परमाणु मॉडल का चित्र बनाइए।
उत्तर
HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना img-2

प्रश्न 4.
क्या अल्फा कणों का प्रकीर्णन प्रयोग सोने के अतिरिक्त दूसरी धातु की पन्नी से संभव होगा?
उत्तर:
हाँ, अल्फा कणों का प्रकीर्णन प्रयोग सोने के अतिरिक्त किसी भी दूसरी धातु की पन्नी से संभव होगा।

HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना

(पृष्ठ संख्या – 56)

प्रश्न 1.
परमाणु के तीन अवपरमाणुक कणों के नाम लिखें।
उत्तर:
परमाणु के तीन अवपरमाणुक कण इलेक्ट्रॉन, प्रोटॉन व न्यूटॉन हैं।

प्रश्न 2.
हीलियम परमाणु का परमाणु द्रव्यमान 4u है और उसके नाभिक में दो प्रोटॉन होते हैं। इसमें कितने न्यूट्रॉन होंगे?
उत्तर:
हीलियम परमाणु का परमाणु द्रव्यमान = 4u
हीलियम के परमाणु नाभिक में उपस्थित प्रोटॉन = 2
हीलियम के परमाणु नाभिक में उपस्थित न्यूट्रॉन = परमाणु द्रव्यमान – प्रोटॉन
= 4 – 2 = 2

(पृष्ठ संख्या-57)

प्रश्न 1.
कार्बन और सोडियम के परमाणुओं के लिए इलेक्ट्रॉन-वितरण लिखिए।
उत्तर:
1. कार्बन
द्रव्यमान संख्या = 12
परमाणु संख्या = 6
इसलिए प्रोटॉनों की संख्या = 6
इलेक्ट्रॉनों की संख्या = 6
न्यूट्रॉनों की संख्या = 12 – 6 = 6
इलेक्ट्रॉन वितरण = K = 2, L=4
HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना img-3

2. सोडियम
द्रव्यमान संख्या = 23
परमाणु संख्या = 11
इसलिए प्रोटॉनों की संख्या = 11
इलेक्ट्रॉनों की संख्या = 11
न्यूट्रॉनों की संख्या = 23 – 11 = 12
इलेक्ट्रॉन वितरण = K = 2
L = 8
M = 1
HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना img-4

HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना

प्रश्न 2.
अगर किसी परमाणु का K और L कोश भरा है, तो उस परमाणु में इलेक्ट्रॉनों की संख्या क्या होगी? उत्तर
कोश K = 2 इलेक्ट्रॉन
कोश L = 8 इलेक्ट्रॉन
अतः परमाणु में कुल इलेक्ट्रॉन = 2 + 8 = 10 इलेक्ट्रॉन .

(पृष्ठ संख्या-58)

प्रश्न 1.
क्लोरीन, सल्फर और मैग्नीशियम की परमाणु संख्या से आप इनकी संयोजकता कैसे प्राप्त करेंगे?
उत्तर:
(1) क्लोरीन की परमाणु संख्या 17 है इसलिए इसका इलेक्ट्रॉन वितरण निम्न होगा
K = 2 इलेक्ट्रॉन
L = 8 इलेक्ट्रॉन
M = 7 इलेक्ट्रॉन
इस प्रकार क्लोरीन को अपना अष्टक पूरा करने के लिए (8-7) = 1 इलेक्ट्रॉन की आवश्यकता है। अतः क्लोरीन की संयोजकता 1 है।

(2) सल्फर की परमाणु संख्या 16 है इसलिए इसका इलेक्ट्रॉन वितरण निम्न होगा
K = 2 इलेक्ट्रॉन
L = 8 इलेक्ट्रॉन
M = 6 इलेक्ट्रॉन
इस प्रकार सल्फर को अपना अष्टक पूरा करने के लिए (8 – 6) = 2 इलेक्ट्रॉनों की आवश्यकता है। अतः सल्फर की संयोजकता 2 है।

(3) मैग्नीशियम की परमाणु संख्या 12 है। इसलिए इसका इलेक्ट्रॉन वितरण निम्न होगा
K = 2 इलेक्ट्रॉन
L = 8 इलेक्ट्रॉन
M = 2 इलेक्ट्रॉन
इस प्रकार मैग्नीशियम को अपना बाह्य अष्टक पूरा करने के लिए 2 इलेक्ट्रॉनों को त्यागना आसान है। अतः मैग्नीशियम की संयोजकता 2 है।

HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना

(पृष्ठ संख्या – 59)

प्रश्न 1.
यदि किसी परमाणु में इलेक्ट्रॉनों की संख्या 8 है और प्रोटॉनों की संख्या भी 8 है तब,
(a) परमाणु की परमाणुक संख्या क्या है?
(b) परमाणु का क्या आवेश है?
उत्तर:
परमाणु में इलेक्ट्रॉनों की संख्या = 8
परमाणु में प्रोटॉनों की संख्या = 8
परमाणु में प्रोटॉनों की संख्या = इलेक्ट्रॉनों की संख्या
= 8
(a) परमाणु की परमाणुक संख्या = परमाणु में प्रोटॉनों तथा इलेक्ट्रॉन की संख्या
= 8 + 8 = 16

(b) इलेक्ट्रॉन वितरण = K = 2, L = 6
अतः परमाणु के बाह्यतम कोश को पूरा करने के लिए 2 इलेक्ट्रॉनों की आवश्यकता है। इसलिए आवेश -2 है।

प्रश्न 2.
पाठ्यपुस्तक की सारणी 4.1 की सहायता से ऑक्सीजन और सल्फर-परमाणु की द्रव्यमान संख्या ज्ञात कीजिए।
उत्तर:
दी गई सारणी अनुसार
(1) ऑक्सीजन में परमाणु संख्या = 8
ऑक्सीजन में प्रोटॉनों की संख्या = 8
ऑक्सीजन में न्यूट्रॉनों की संख्या = 8
द्रव्यमान संख्या = प्रोटॉनों की संख्या + न्यूट्रॉनों की संख्या
= 8 + 8 = 16

(2) सल्फर में परमाणु संख्या = 16
सल्फर में प्रोटॉनों की संख्या = 16
सल्फर में न्यूट्रॉनों की संख्या = 16
द्रव्यमान संख्या = प्रोटॉनों की संख्या + न्यूट्रॉनों की संख्या
= 16 + 16 = 32

(पृष्ठ संख्या – 60)

प्रश्न 1.
चिह्न H, D और T के लिए प्रत्येक में पाए जाने वाले तीन अवपरमाणुक कणों को सारणीबद्ध कीजिए।
उत्तर:
(1) चिह्न H प्रोटियम का प्रतीक है अर्थात ,\({ }_1 \mathrm{H}^1\)
परमाणु संख्या = 1
द्रव्यमान संख्या = 1
इलेक्ट्रॉनों की संख्या = 1
प्रोट्रॉनों की संख्या = 1
न्यूट्रॉनों की संख्या = 1 – 1 = 0

(2) चिह्न D ड्यूटीरियम का प्रतीक है अर्थात \({ }_1 \mathrm{H}^2\)
परमाणु संख्या = 1
द्रव्यमान संख्या = 2
इलेक्ट्रॉनों की संख्या = 1
प्रोट्रॉनों की संख्या = 1
न्यूट्रॉनों की संख्या = 2 – 1 = 1

(3) चिह्न T ट्राइटियम का प्रतीक है अर्थात \({ }_1 \mathrm{H}^3\)
परमाणु संख्या = 1
द्रव्यमान संख्या = 3
इलेक्ट्रॉनों की संख्या = 1
प्रोट्रॉनों की संख्या = 1
न्यूट्रॉनों की संख्या = 3 – 1 = 2

HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना

प्रश्न 2.
समस्थानिक और समभारिक के किसी एक युग्म का इलेक्ट्रॉनिक विन्यास लिखिए।
उत्तर:
(i) समस्थानिक युग्म क्लोरीन \({ }_{17} \mathrm{Cl}^{35}\) व \({ }_{17} \mathrm{Cl}^{37}\) का इलेक्ट्रॉनिक विन्यास निम्नलिखित होगा
HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना img-5

(ii) समभारिक युग्म कैल्शियम तथा आर्गन का इलेक्ट्रॉनिक विन्यास निम्न होगा
(1) कैल्शियम \({ }_{20} \mathrm{Cl}^{40}\)
e = 20
P = 20
N = 40 – 20 – 20
HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना img-6

(2) आर्गन 18Ar40
e = 18
P = 18
N = 40 – 18 = 22
HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना img-7

HBSE 9th Class Science परमाणु की संरचना Textbook Questions and Answers

प्रश्न 1.
इलेक्ट्रॉन, प्रोटॉन और न्यूट्रॉन के गुणों की तुलना कीजिए।
उत्तर:

गुणइलेक्ट्रॉनप्रोटॉनन्यूट्रॉन
1. आवेशइन पर इकाई \(\left(-1.6 \times 10^{-19} \mathrm{C}\right)\) ऋण आवेश होता है।इन पर इकाई \(\left(+1.6 \times 10^{-19} \mathrm{C}\right)\) धन आवेश होता है।इन पर कोई आवेश नहीं होता।
2. द्रव्यमानइनका द्रव्यमान हाइड्रोजन के परमाणु के द्रव्यमान का \(\frac{1}{2000}\) वां भाग होता है।इनका द्रव्यमान हाइड्रोजन के परमाणु के द्रव्यमान के \(\left(1.674 \times 10^{-24} \mathrm{~g}\right)\) समान होता है ।इनका द्रव्यमान भी हाइड्रोजन के परमाणु के द्रव्यमान के \(\left(1.674 \times 10^{-24} \mathrm{~g}\right)\) समान होता है।
3. स्थितिये नाभिक के चारों ओर निश्चित वृत्ताकार कोशों में चक्कर लगाते हैं।ये नाभिक में स्थित होते हैं।ये कण भी नाभिक में स्थित होते हैं।
4. संकेतइन्हें e द्वारा प्रदशित किया जाता है।इन्हें p द्वारा प्रदर्शित किया जाता है।इन्हें \(\mathrm{n}\) द्वारा प्रदर्शित किया जाता है।

प्रश्न 2.
जे.जे. टॉमसन के परमाणु मॉडल की क्या सीमाएँ हैं?
उत्तर:
जे.जे. टॉमसन के परमाणु मॉडल ने परमाणु के उदासीन होने की व्याख्या की जिसके अनुसार परमाणु धन आवेशित गोले का बना होता है और इलेक्ट्रॉन उसमें धंसे होते हैं, परंतु यह मॉडल न्यूट्रॉनों के बारे में तथा इलेक्ट्रॉनों की स्थिति के बारे में कुछ भी नहीं बताता जो दूसरे वैज्ञानिकों द्वारा बाद में स्पष्ट किया गया।

HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना

प्रश्न 3.
रदरफोर्ड के परमाणु मॉडल की क्या सीमाएँ हैं?
उत्तर:
अरनेस्ट रदरफोर्ड के परमाणु मॉडल की मुख्य सीमा यह थी कि यह परमाणु की स्थिरता का वर्णन नहीं करता। उसके मॉडल अनुसार, इलेक्ट्रॉन नाभिक के चारों ओर वृत्तीय पथ में चक्कर काटते हैं जिसका परिणाम उनकी त्वरित गति होता है। अतः यह ऊर्जा को विकरित करके वृत्तीय पथ के नाभिक आकार को छोटा कर देगा। परिणामस्वरूप इलेक्ट्रॉन नाभिक में गिर जाएँगे और परमाणु को अस्थिर कर देंगे, परंतु वास्तव में परमाणु अस्थिर नहीं है। चित्र 4.8 : रदरफोर्ड के परमाणु मॉडल की कमी
HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना img-8

प्रश्न 4.
बोर के परमाणु मॉडल की व्याख्या कीजिए।
उत्तर:
नील्स बोर के परमाणु मॉडल के अनुसार
(1) इलेक्ट्रॉन केवल कुछ निश्चित कक्षाओं में ही चक्कर लगा सकते हैं, जिन्हें इलेक्ट्रॉन की विविक्त कक्षा कहते हैं।
(2) जब इलेक्ट्रॉन इस विविक्त कक्षा में चक्कर लगाते हैं, तो उनकी ऊर्जा का विकिरण नहीं होता। इन कक्षाओं (या कोशों) को ऊर्जा-स्तर कहते हैं। चित्र में एक परमाणु के ऊर्जा स्तर को दिखाया गया है।
HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना img-9

प्रश्न 5.
इस अध्याय में दिए गए सभी परमाणु मॉडलों की तुलना कीजिए।
उत्तर:
इस अध्याय में दिए गए विभिन्न परमाणु मॉडलों की तुलना निम्नलिखित प्रकार से हैं-
1. टॉमसन के परमाणु मॉडल अनुसार
(1) परमाणु धन आवेशित गोले का बना होता है और इलेक्ट्रॉन उसमें धंसे होते हैं।
(2) ऋणात्मक और धनात्मक आवेश परिणाम में समान होते हैं। इसलिए परमाणु वैद्युतीय रूप से उदासीन होते हैं।
HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना img-10

2. रदरफोर्ड के परमाणु मॉडल अनुसार
(1) परमाणु का केंद्र धनावेशित होता है जिसे नाभिक कहा जाता है। एक परमाणु का लगभग संपूर्ण द्रव्यमान नाभिक में होता है।
(2) इलेक्ट्रॉन नाभिक के चारों ओर निश्चित कक्षाओं में चक्कर लगाते हैं।
(3) नाभिक का आकार परमाणु के आकार की तुलना में काफी कम होता है।

3. बोर मॉडल के अनुसार
(1) इलेक्ट्रॉन केवल कुछ निश्चित कक्षाओं में ही चक्कर लगा सकते हैं, जिन्हें इलेक्ट्रॉन की विविक्त कक्षा कहते हैं।
(2) जब इलेक्ट्रॉन इस विविक्त कक्षा में चक्कर लगाते हैं, तो उनकी ऊर्जा का विकिरण नहीं होता है। इन कक्षाओं को ऊर्जा स्तर कहते हैं।

HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना

प्रश्न 6.
पहले अठारह तत्त्वों के विभिन्न कक्षों में इलेक्ट्रॉन वितरण के नियम को लिखिए।
उत्तर:
पहले अठारह तत्त्वों के विभिन्न कक्षों में इलेक्ट्रॉन वितरण के लिए बोर और बरी द्वारा निम्न नियम प्रस्तुत किए गए
(1) किसी कक्षा में उपस्थित अधिकतम इलेक्ट्रॉनों की संख्या को सूत्र 2n2 से दर्शाया जाता है, जहाँ ‘n’ कक्ष की संख्या या ऊर्जा स्तर है। इसलिए इलेक्ट्रॉनों की अधिकतम संख्या पहले कक्ष या K कोश में (2 x 12) = 2 होगी, दूसरे कक्ष या L सेल में (2 x 22) = 8 होगी, तीसरे कक्ष या M सेल में होगी = (2 x 32)= 18, चौथे कक्ष या N सेल में होगी = (2 x 42)= 32.

(2) सबसे बाहरी कोश में इलेक्ट्रॉनों की अधिकतम संख्या 8 हो सकती है।
HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना img-11

(3) किसी परमाणु के अगले कोश में इलेक्ट्रॉन तब तक स्थान नहीं लेते हैं जब तक कि उससे पहले वाले भीतरी कोश पूर्ण रूप से भर नहीं जाता। इससे स्पष्ट होता है कि कक्षाएँ क्रमानुसार भरती हैं।

प्रश्न 7.
सिलिकॉन और ऑक्सीजन का उदाहरण देते हुए संयोजकता की परिभाषा दीजिए।
उत्तर:
परमाणु के बाह्यतम कक्ष में इलेक्ट्रॉनों के अष्टक बनाने के लिए जितनी संख्या में इलेक्ट्रॉनों की साझेदारी या स्थानांतरण होता है, वह उस तत्त्व की संयोजकता कहलाती है। जैसे

(1) सिलिकॉन \(({ }_{14} \mathrm{Si}^{28})\) में इलेक्ट्रॉनों का वितरण निम्न होगा
K = 2
L = 8
M = 4
अतः यह अपना अष्टक पूरा करने के लिए 4 इलेक्ट्रॉन स्थानांतरित कर सकता है। इसलिए इसकी संयोजकता 4 है।

(2) ऑक्सीजन (\(({ }_{8} \mathrm{Si}^{16})\)) में इलेक्ट्रॉनों का वितरण निम्न होगा
K = 2
L = 6
अतः इसे अपना अष्टक पूरा करने के लिए (8 – 6) = 2 इलेक्ट्रॉनों की साझेदारी करनी होगी। इसलिए इसकी संयोजकता 2 है।

HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना

प्रश्न 8.
उदाहरण के साथ व्याख्या कीजिए-परमाणु संख्या, द्रव्यमान संख्या, समस्थानिक और समभारिक समस्थानिकों के कोई दो उपयोग लिखिए।
उत्तर:
परमाणु संख्या-परमाणु के नाभिक में उपस्थित कुल प्रोटॉनों की संख्या उसकी परमाणु संख्या (Z) कहलाती है; जैसे ऑक्सीजन की परमाणु संख्या 8 व कार्बन की परमाणु संख्या 6 है।

द्रव्यमान संख्या – परमाणु के नाभिक में उपस्थित प्रोटॉनों और न्यूट्रॉनों की कुल संख्या के योग को द्रव्यमान संख्या (A) कहते हैं। जैसे ऑक्सीजन की द्रव्यमान संख्या 16u तथा कार्बन की द्रव्यमान संख्या 12u है।

समस्थानिक – एक ही तत्त्व के ऐसे परमाणु जिनकी परमाणु संख्या समान हो, परंतु द्रव्यमान संख्या भिन्न हो तत्त्व के समस्थानिक कहलाते हैं। जैसे हाइड्रोजन के तीन समस्थानिक प्रोटियम \(({ }_{1} \mathrm{H}^{1})\), ड्यूटीरियम \(({ }_{1} \mathrm{H}^{2})\) व ट्राइटियम \(({ }_{1} \mathrm{H}^{3})\) तथा कार्बन के दो समस्थानिक \(({ }_{6} \mathrm{C}^{12})\) व \(({ }_{6} \mathrm{C}^{14})\) हैं।

समभारिक – समभारिक वे परमाणु होते हैं जिनकी द्रव्यमान संख्या समान लेकिन परमाणु संख्या भिन्न-भिन्न होती है; जैसे कैल्शियम \(({ }_{20} \mathrm{Ca}^{40})\) तथा आर्गन \(({ }_{18} \mathrm{Ar}^{40})\) समभारिक हैं।

समस्थानिकों के उपयोग
(1) यूरेनियम के एक समस्थानिक (U-235) का उपयोग परमाणु भट्टी (reactor) में ईंधन के रूप में होता है।
(2) कैंसर के उपचार में कोबाल्ट के समस्थानिक का उपयोग होता है।
(3) पेंघा रोग के इलाज में आयोडीन के समस्थानिक का उपयोग होता है।

प्रश्न 9.
Na+ के पूरी तरह से भरे हुए Ka L कोश होते हैं व्याख्या कीजिए। उत्तर-सोडियम परमाणु की परमाणु संख्या 11 होती है। इसलिए सोडियम परमाणु में इलेक्ट्रॉनों की संख्या = 11 अतः सोडियम परमाणु का इलेक्ट्रॉनिक वितरण होगा
K = 2
L = 8
M = 1
यह सोडियम परमाणु स्थाई संरचना प्राप्त करने के लिए एक इलेक्ट्रॉन त्याग कर Na (सोडियम आयन) बन जाएगा जिस कारण इसका इलेक्ट्रॉनिक वितरण K = 2 तथा L = 8 होगा अर्थात K और L कोश पूर्ण भरे होंगे।

प्रश्न 10.
अगर ब्रोमीन परमाणु दो समस्थानिकों \({ }_{35}^{79} \mathrm{Br}\)(49.7%) तथा \({ }_{35}^{81} \mathrm{Br}\) (50.3%)] के रूप में है, तो ब्रोमीन परमाणु के औसत परमाणु द्रव्यमान की गणना कीजिए।
उत्तर:
ब्रोमीन परमाणु दो समस्थानिकों \({ }_{35} \mathrm{Br}^{79}\) और \({ }_{35} \mathrm{Br}^{81}\) के रूपों में पाया जाता है जिनका द्रव्यमान क्रमशः 49.7% व 50.3% है।
इसलिए ब्रोमीन का औसत परमाणु द्रव्यमान होगा
\(\left(79 \times \frac{49.7}{100}+81 \times \frac{50.3}{100}\right)\)
\(\left(\frac{79 \times 497}{1000}+\frac{81 \times 503}{1000}\right)\)
\(\left(\frac{39263}{1000}+\frac{40743}{1000}\right)\) = 39.263 + 40.743 = 80.006 = 80u

HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना

प्रश्न 11.
एक तत्त्व X का परमाणु द्रव्यमान 16.2u है तो इसके किसी एक नमूने में समस्थानिक \({ }_{8}^{16} \mathrm{X}\) और \({ }_{8}^{18} \mathrm{X}\) का प्रतिशत क्या होगा?
हल
माना समस्थानिक \({ }_{8} \mathrm{X}^{16}\) का प्रतिशत = x
तो \({ }_{8} \mathrm{X}^{18}\) का प्रतिशत = 100 – x
तत्त्व X का औसत परमाणु द्रव्यमान = 16.2u
\(\frac{16 \times x}{100}+\frac{18 \times(100-x)}{100}\) = 16.2
दोनों ओर 100 से गुणा करने पर
16x + 1800 – 18x = 1620
या -2x = 1620 – 1800
या -2x = -180
या x = \(\frac{-180}{-2}\) = 90
अतः नमूने (सैंपल) में \({ }_8 \mathrm{X}^{16}\) का प्रतिशत = 90%
तथा \({ }_8 \mathrm{X}^{18}\) का प्रतिशत = 100 — 90 = 10%

प्रश्न 12.
यदि तत्त्व का Z= 3 हो तो तत्त्व की संयोजकता क्या होगी? तत्त्व का नाम भी लिखिए।
हल
तत्त्व की परमाणु संख्या (Z) = 3
तत्त्व में इलेक्ट्रॉनों की संख्या = 3
इलेक्ट्रॉन वितरण
K = 2
L = 1
अतः तत्त्व के बाह्यतम कोश में 1 इलेक्ट्रॉन है। यह स्थाई विन्यास प्राप्त करने के लिए एक इलेक्ट्रॉन का त्याग कर सकता है। इसलिए तत्त्व की संयोजकता 1 है। अतः तत्त्व लीथियम (Li) है।

प्रश्न 13.
दो परमाणु स्पीशीज़ के केंद्रकों का संघटन नीचे दिया गया है
HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना img-12
X और Y की द्रव्यमान संख्या ज्ञात कीजिए। इन दोनों स्पीशीज़ में क्या संबंध है?
हल–परमाणु x के लिए
प्रोटॉनों की संख्या = 6
न्यूट्रॉनों की संख्या = 6
द्रव्यमान संख्या = प्रोटॉनों की संख्या + न्यूट्रॉनों की संख्या
= 6+6
= 12
परमाणु Y के लिए
प्रोटॉनों की संख्या = 6
न्यूट्रॉनों की संख्या = 8
द्रव्यमान संख्या = प्रोटॉनों की संख्या + न्यूट्रॉनों की संख्या
= 6+ 8
= 14
ये दोनों स्पीशीज \({ }_6 X^{12}\) और \({ }_6 Y^{14}\) समस्थानिक हैं।

HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना

प्रश्न 14.
निम्नलिखित वक्तव्यों में गलत के लिए F और सही के लिए T लिखें।
(a) जे.जे. टॉमसन ने यह प्रस्तावित किया था कि परमाणु के केंद्रक में केवल न्यूक्लीयॉन्स होते हैं।
(b) एक इलेक्ट्रॉन और प्रोटॉन मिलकर न्यूट्रॉन का निर्माण करते हैं इसलिए यह अनावेशित होता है।
(c) इलेक्ट्रॉन का द्रव्यमान प्रोटॉन से लगभग \(\frac{1}{2000}\) गुना होता है।
(d) आयोडीन के समस्थानिक का इस्तेमाल टिंक्चर आयोडीन बनाने में होता है। इसका उपयोग दवा के रूप में होता है।
उत्तर:
(a) F,
(b) F
(c) T,
(d) F

नोट-प्रश्न संख्या 15, 16 और 17 में सही के सामने (√ ) का चिह्न और गलत के सामने (X) का चिह्न लगाइए।

प्रश्न 15.
रदरफोर्ड का अल्फा कण प्रकीर्णन प्रयोग किसकी खोज के लिए उत्तरदायी था?
(a) परमाणु केंद्रक
(b) इलेक्ट्रॉन
(c) प्रोटॉन
(d) न्यूट्रॉन
उत्तर:
(a) परमाणु केंद्रक (√),
(b) इलेक्ट्रॉन (x),
(c) प्रोटॉन (x),
(d) न्यूट्रॉन (x)

प्रश्न 16.
एक तत्त्व के समस्थानिक में होते हैं
(a) समान भौतिक गुण
(b) भिन्न रासायनिक गुण
(c) न्यूट्रॉनों की अलग-अलग संख्या
(d) भिन्न परमाणु संख्या
उत्तर:
(a) समान भौतिक गुण (x),
(b) भिन्न रासायनिक गुण (x),
(c) न्यूट्रॉनों की अलग-अलग संख्या (√),
(d) भिन्न परमाणु संख्या (x)।

HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना

प्रश्न 17.
Cl आयन में संयोजक इलेक्ट्रॉनों की संख्या है
(a) 16
(b) 8
(c) 17
(d) 18
उत्तर:
(a) 16 (x)।
(b) 8 (√)।
(c) 17 (x)।
(d) 18 (x)।

प्रश्न 18.
सोडियम का सही इलेक्ट्रॉनिक विन्यास निम्न में कौन-सा है?
(a) 2, 8
(b) 8, 2, 1
(c) 2, 1, 8
(d) 2, 8, 1
उत्तर:
सोडियम का सही इलेक्ट्रॉनिक विन्यास
(d) 2, 8, 1 है।

HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना

प्रश्न 19.
निम्नलिखित सारणी को पूरा कीजिए

परमाणु

संख्या

द्रव्यमान

संख्या

न्यूट्रॉनों की

संख्या

प्रोटॉनों की

संख्या

इलेक्ट्रॉनों

की संख्या

परमाणु

स्पीशीज़

910
1632सल्फर
2412
21
1011

उत्तर

परमाणु

संख्या

द्रव्यमान

संख्या

न्यूट्रॉनों की

संख्या

प्रोटॉनों की

संख्या

इलेक्ट्रॉनों

की संख्या

परमाणु

स्पीशीज़

9191099फ्लोरीन
1632161616सल्फर
1224121212मैग्नीशियम
12111डूयूटीरियम
11011हाइड्रोजन (प्रोटियम)

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HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु

Haryana State Board HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 3 परमाणु एवं अणु

HBSE 9th Class Science परमाणु एवं अणु Intext Questions and Answers
(पृष्ठ संख्या-36)
प्रश्न 1.
एक अभिक्रिया में 5.3g सोडियम कार्बोनेट एवं 6.0g एसीटिक अम्ल अभिकृतं होते हैं। 2.2g कार्बन … डाइऑक्साइड, 8.2g सोडियम एसीटिक एवं 0.9g जल उत्पाद के रूप में प्राप्त होते हैं। इस अभिक्रिया द्वारा दिखाइए कि यह परीक्षण द्रव्यमान संरक्षण के नियम के अनुरूप है।
सोडियम कार्बोनेट + एसीटिक अम्ल → सोडियम एसीटिक + कार्बन डाइऑक्साइड + जल
हल-यहाँ पर
अभिकारकों का कुल द्रव्यमान = (सोडियम कार्बोनेट + एसीटिक अम्ल) का द्रव्यमान
= 5.3g + 6.0g = 11.3g
उत्पादों का कुल द्रव्यमान = (कार्बन डाइऑक्साइड + सोडियम एसीटिक + जल) का द्रव्यमान
= 2.2g + 8.2g + 0.9g
= 11.3g
क्योंकि अभिकारकों का कुल द्रव्यमान उत्पादों के कुल द्रव्यमान के समान है इसलिए यह परीक्षण द्रव्यमान संरक्षण नियम के अनुरूप है। उत्तर

प्रश्न 2.
हाइड्रोजन एवं ऑक्सीजन द्रव्यमान के अनुसार 1:8 के अनुपात में संयोग करके जल निर्मित करते हैं। 3g हाइड्रोजन गैस के साथ पूर्ण रूप से संयोग करने के लिए कितने ऑक्सीजन गैस के द्रव्यमान की आवश्यकता होगी?
हल-यहाँ पर
हाइड्रोजन का द्रव्यमान : ऑक्सीजन का द्रव्यमान = 1:8
अतः 3g हाइड्रोजन गैस के साथ पूर्ण रूप से संयोग ।
करने के लिए आवश्यक ऑक्सीजन का द्रव्यमान = 3g x 8 = 24g उत्तर

प्रश्न 3.
डॉल्टन के परमाणु सिद्धांत का कौन-सा अभिग्रहीत द्रव्यमान के संरक्षण के नियम का परिणाम है?
उत्तर:
डॉल्टन के परमाणु सिद्धांत का अभिग्रहीत है कि परमाणु अविभाज्य सूक्ष्मतम कण होते हैं जो रासायनिक अभिक्रिया में न तो सृजित होते हैं, न ही उनका विनाश होता है। यह द्रव्यमान के संरक्षण के नियम का परिणाम है।

प्रश्न 4.
डॉल्टन के परमाणु सिद्धांत का कौन-सा अभिग्रहीत निश्चित अनुपात के नियम की व्याख्या करता है?
उत्तर:
डॉल्टन के परमाणु सिद्धांत का अभिग्रहीत, किसी भी यौगिक में परमाणुओं की सापेक्ष संख्या एवं प्रकार निश्चित होते हैं। यह निश्चित अनुपात के नियम की व्याख्या करता है।

HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु

(पृष्ठ संख्या-40)
प्रश्न 1.
परमाणु द्रव्यमान इकाई को परिभाषित कीजिए।
उत्तर:
कार्बन-12 समस्थानिक के एक परमाणु द्रव्यमान के 1/12वें भाग को मानक परमाणु द्रव्यमान इकाई के रूप में लेते हैं। कार्बन-12 समस्थानिक के एक परमाणु द्रव्यमान के सापेक्ष सभी तत्त्वों के परमाणु द्रव्यमान प्राप्त किए गए।

प्रश्न 2.
एक परमाणु को आँखों द्वारा देखना क्यों संभव नहीं होता है?
उत्तर:
परमाणु का आकार बहुत छोटा होने के कारण इसे आँखों द्वारा नहीं देखा जा सकता। इसका आकार इतना छोटा होता है कि इसकी त्रिज्या को नेनोमीटर (nm) में मापा जाता है। जहाँ 1nm = 10-9 m के बराबर होता है। अब हम आधुनिक तकनीकों की सहायता से तत्त्वों की सतहों के आवर्धित प्रतिबिंबों को दिखा सकते हैं, जिनमें उपस्थित परमाणु स्पष्ट दिखाई देते हैं।
HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु img-1

(पृष्ठ संख्या-44)

प्रश्न 1.
निम्न के सूत्र लिखिए-
(i) सोडियम ऑक्साइड,
(ii) ऐलुमिनियम क्लोराइड,
(iii) सोडियम सल्फाइड,
(iv) मैग्नीशियम हाइड्रॉक्साइड।
उत्तर:
(i) सोडियम ऑक्साइड का सूत्र
HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु img-2
अतः सोडियम ऑक्साइड का सूत्र Na2o है।

(ii) ऐलुमिनियम क्लोराइड का सूत्र
HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु img-3
अतः ऐलुमिनियम क्लोराइड का सूत्र AICl3, है।

(iii) सोडियम सल्फाइड का सूत्र
HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु img-4
अतः सोडियम सल्फाइड का सूत्र Na2s है।

(iv) मैग्नीशियम हाइड्रॉक्साइड
HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु img-5
अतः मैग्नीशियम हाइड्रॉक्साइड का सूत्र Mg(OH)2 है।

HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु

प्रश्न 2.
निम्नलिखित सूत्रों द्वारा प्रदर्शित यौगिकों के नाम लिखिए
(i) AI2 (SO)3
(ii) CaCl2
(iii) K2SO4
(iv) KNO3
(v) Caco3
उत्तर

रासायनिक सूत्रयौगिक का नाम
(i) AI2 (SO)3ऐलुमिनियम सल्फेट
(ii) CaCl2कैल्शियम क्लोराइड
(iii) K2SO4पोटैशियम सल्फेट
(iv) KNO3पोटैशियम नाइट्रेट
(v) CaCO3कैल्शियम कार्बोनेट

प्रश्न 3.
रासायनिक सूत्र का क्या तात्पर्य है?
उत्तर:
किसी यौगिक का रासायनिक सूत्र उसके संघटक का प्रतीकात्मक निरूपण होता है।

प्रश्न 4.
निम्न में कितने परमाणु विद्यमान हैं?
(1) H2S अणु एवं
(ii) \(\mathrm{PO}_4^{3-}\) आयन
उत्तर:
(i) H2S अणु में परमाणुओं की संख्या = 2 + 1 = 3
(ii) \(\mathrm{PO}_4^{3-}\) आयन में परमाणुओं की संख्या = 1 + 4 = 5

(पृष्ठ संख्या-46)

प्रश्न 1.
निम्न यौगिकों के आण्विक द्रव्यमान का परिकलन कीजिए
H2, 02, Cl2, CO2, CH4, C2H6, C2H4, NH3, एवं CH3OH
हल-
(1) हाइड्रोजन का परमाणु द्रव्यमान = 1u.
H2, में हाइड्रोजन के दो परमाणु हैं।
H2, का आण्विक द्रव्यमान = 2 x 1 = 2u

(2) ऑक्सीजन का परमाणु द्रव्यमान = 16u
O2 में ऑक्सीजन के दो परमाणु हैं।
O2 का आण्विक द्रव्यमान = 2 x 16 = 32u

(3) क्लोरीन का परमाणु द्रव्यमान = 35.5u .
Cl2, में क्लोरीन के दो परमाणु हैं।
CI2, का आण्विक द्रव्यमान = 2 x 35.5 = 71u

(4) कार्बन का परमाणु द्रव्यमान = 12u
ऑक्सीजन का परमाणु द्रव्यमान = 16u अतः
CO2, जिसमें एक परमाणु कार्बन एवं दो परमाणु ऑक्सीजन होते हैं।
CO2, का आण्विक द्रव्यमान = 1 x 12 + 2 x 16
= 12 + 32 = 44u

(5) कार्बन का परमाणु द्रव्यमान = 12u
हाइड्रोजन का परमाणु द्रव्यमान = 1u
अतः CH4. जिसमें एक परमाणु कार्बन तथा चार परमाणु हाइड्रोजन होते हैं।
CH4, का आण्विक द्रव्यमान = 1 x 12 + 4 x 1 = 12 + 4 = 16u

(6) कार्बन का परमाणु द्रव्यमान = 12u
हाइड्रोजन का परमाणु द्रव्यमान = lu
अतः C2H6, जिसमें दो परमाणु कार्बन तथा छः परमाणु हाइड्रोजन के होते हैं।
C2H6 का आण्विक द्रव्यमान = 2 x 12 + 6 x 1
= 24+ 6 = 30u

(7) कार्बन का परमाणु द्रव्यमान = 12u
हाइड्रोजन का परमाणु द्रव्यमान = 1u अतः
C2H4, जिसमें दो परमाणु कार्बन तथा चार परमाणु हाइड्रोजन होते हैं।
C2H4, का आण्विक द्रव्यमान = 2 x 12 +4 x 1
= 24+4 = 28u

(8) नाइट्रोजन का परमाणु द्रव्यमान = 14u
हाइड्रोजन का परमाणु द्रव्यमान = lu
अतः NH3, जिसमें एक परमाणु नाइट्रोजन तथा तीन परमाणु हाइड्रोजन होते हैं।
NH3, का आण्विक द्रव्यमान = 1 x 14 +3 x 1
= 14+3 = 17u

(9) कार्बन का परमाणु द्रव्यमान = 12u
हाइड्रोजन का परमाणु द्रव्यमान = 1u
ऑक्सीजन का परमाणु द्रव्यमान = 16u
अतः CH3OH जिसमें एक परमाणु कार्बन, चार परमाणु हाइड्रोजन व एक परमाणु ऑक्सीजन हैं।
CH3OH का आण्विक द्रव्यमान = 1 x 12 + 4 x 1 + 16 x 1
= 12 + 4 + 16
= 32u

HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु

प्रश्न 2.
निम्न यौगिकों के सूत्र इकाई द्रव्यमान का परिकलन कीजिएZnO, Na20 एवं K2CO3 दिया गया है
Zn का परमाणु द्रव्यमान = 65u
Na का परमाणु द्रव्यमान = 23u
K का परमाणु द्रव्यमान = 39u
C का परमाणु द्रव्यमान = 12u एवं
O का परमाणु द्रव्यमान = 16u है।
उत्तर:
(1) ZnO का सूत्र इकाई द्रव्यमान = 1 x 65u + 1 x 16u
= 65u + 16u
81u

(2) Na2O का सूत्र इकाई द्रव्यमान = 2 x 23u + 1 x 16u
= 46u + 16u
= 62u.

(3) K2CO3 का सूत्र इकाई द्रव्यमान = 2 x 39u + 1 x 12u + 3 x 16u
= 78u + 12u + 48u = 138u

HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु

(पृष्ठ संख्या-48)

प्रश्न 1.
यदि कार्बन परमाणुओं के एक मोल का द्रव्यमान 12g है तो कार्बन के एक परमाणु का द्रव्यमान क्या होगा?
हल-यहाँ पर
कार्बन का मोलर द्रव्यमान = 12g
1 मोल = 6.022 x 1023 परमाणु
अर्थात् 6.022 x 1023 परमाणु कार्बन का द्रव्यमान = 12g
1 परमाणु कार्बन का द्रव्यमान = \(\frac{12}{6.022 \times 10^{23}} \mathrm{~g}\)
= 1.993 x 10-23g उत्तर

प्रश्न 2.
किसमें अधिक परमाणु होंगे- 100g सोडियम अथवा 100g लोहा (Fe) ? (Na का परमाणु द्रव्यमान = 23u, Fe का परमाणु द्रव्यमान = 56u)
सोडियम का मोलर द्रव्यमान = 23g
1 मोल = 6.022 x 1023 परमाणु
अतः 23g सोडियम में परमाणुओं की संख्या = 6.022 x 1023
1g सोडियम में परमाणओं की संग = \(\frac{6.022 \times 10^{23}}{23}\)
100g सोडियम में परमाणुओं की संख्या = \(\frac{6.022 \times 10^{23}}{23}\) x 100
= 26.18 x 1023
लोहे का मोलर द्रव्यमान = 56g
1 मोल = 6.022 x 1023
अतः 56g लोहे में परमाणुओं की संख्या = 6.022 x 1023
1g लोहे में परमाणुओं की संख्या = \(\frac{6.022 \times 10^{23}}{23}\)
100g लोहे में परमाणओं की संख्या = \(\frac{6.022 \times 10^{23}}{23}\) x 100
= 10.75 x 1023
इस प्रकार 100g सोडियम में 100g लोहे की अपेक्षा अधिक परमाणु होंगे। उत्तर

HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु

HBSE 9th Class Science परमाणु एवं अणु Textbook Questions and Answers

प्रश्न 1.
0.24g ऑक्सीजन एवं बोरॉन युक्त यौगिक के नमूने में विश्लेषण द्वारा यह पाया गया कि उसमें 0.096g बोरॉन एवं 0.144g ऑक्सीजन है। उस यौगिक के प्रतिशत संघटन का भारात्मक रूप में परिकलन कीजिए।
हल
0.24g यौगिक में बोरॉन की मात्रा = 0.096g
1g यौगिक में बोरॉन की = \(\frac{0.096}{0.24}\)g
100g यौगिक में बोरॉन की मात्रा = \(\frac{0.096}{0.24}\) x 100
= 40g
अतः यौगिक में बोरॉन की मात्रा = 40% उत्तर
0.24g यौगिक में ऑक्सीजन की मात्रा = 0.144g
1g यौगिक में ऑक्सीजन की मात्रा = \(\frac{0.144}{0.24}\)g
100g यौगिक में ऑक्सीजन की मात्रा = \(\frac{0.144}{0.24}\) x 100 = 60g
अतः यौगिक में ऑक्सीजन की मात्रा = 60% उत्तर

प्रश्न 2.
3.0g कार्बन, 8.00g ऑक्सीजन में जलकर 11.00g कार्बन डाइऑक्साइड निर्मित करता है। जब 3.00g कार्बन को 50.00g ऑक्सीजन में जलाएंगे तो कितने ग्राम कार्बन डाइऑक्साइड का निर्माण होगा? आपका उत्तर रासायनिक संयोजन के किस नियम पर आधारित होगा?
उत्तर:
3.0g कार्बन, 8.00g ऑक्सीजन में जलकर 11.00g कार्बन डाइऑक्साइड निर्मित करता है तो 3.00g कार्बन को 50.00g ऑक्सीजन में जलाने पर 53.00g कार्बन डाइऑक्साइड का निर्माण होगा जोकि रासायनिक संयोजन के द्रव्यमान संरक्षण नियम पर आधारित है।

प्रश्न 3.
बहुपरमाणुक आयन क्या होते हैं? उदाहरण दीजिए।
उत्तर:
परमाणुओं का वह समूह जो आयन की तरह व्यवहार करते हैं, बहुपरमाणुक आयन कहलाते हैं; जैसेबहुपरमाणुक आयन संकेत अमोनियम

बहुपरमाणुक आयनसंकेत
अमोनियम\(\mathrm{NH}_4^{+}\)
हाइड्रॉक्साइड\(\mathrm{OH}^{-}\)
नाइट्रेट\(\mathrm{NO}_3^{-}\)
हाइड्रोजन कार्बोनेट\(\mathrm{HCO}_3^{-}\)
कार्बोनेट\(\mathrm{CO}_3^{2-}\)
सल्फेट\(\mathrm{SO}_4^{2-}\)
फॉस्फेट\(\mathrm{PO}_4^{3-}\)

HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु

प्रश्न 4.
निम्नलिखित के रासायनिक सूत्र लिखिए
(a) मैग्नीशियम क्लोराइड,
(b) कैल्शियम क्लोराइड,
(c) कॉपर नाइट्रेट,
(d) ऐलुमिनियम क्लोराइड,
(e) कैल्शियम कार्बोनेट।
उत्तर:
(a) मैग्नीशियम क्लोराइड का सूत्र
HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु img-9
अतः मैग्नीशियम क्लोराइड का सूत्र = MgCl2,

(b) कैल्शियम क्लोराइड का सूत्र ।
HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु img-10
अतः कैल्शियम क्लोराइड का सूत्र = CaCl2,

(c) कॉपर नाइट्रेट का सूत्र
HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु img-6
अतः कॉपर नाइट्रेट का सूत्र = Cu (NO3)2,

(d) ऐलुमिनियम क्लोराइड का सूत्र
HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु img-7
अतः ऐलुमिनियम क्लोराइड का सूत्र = AlCl3,

(e) कैल्शियम कार्बोनेट का सूत्र ।
HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु img-11
अतः कैल्शियम कार्बोनेट का सूत्र = Caco3,

प्रश्न 5.
निम्नलिखित यौगिकों में विद्यमान तत्त्वों का नाम दीजिए
(a) बुझा हुआ चूना
(b) हाइड्रोजन ब्रोमाइड
(c) बेकिंग पाउडर (खाने वाला सोडा)
(d) पोटैशियम सल्फेट
उत्तर:
(a) बुझा हुआ चूना = Ca (OH)2,
अतः बुझे हुए चूने में विद्यमान तत्त्व कैल्शियम (Ca), ऑक्सीजन (O) व हाइड्रोजन (H) हैं।

(b) हाइड्रोजन ब्रोमाइड = HBr
अतः हाइड्रोजन ब्रोमाइड में विद्यमान तत्त्व हाइड्रोजन (H) व ब्रोमीन (Br) हैं।

(c) बेकिंग पाउडर (खाने वाला सोडा) = NaHCO,
अतः बेकिंग पाउडर में विद्यमान तत्त्व सोडियम (Na), हाइड्रोजन (H), कार्बन (C) तथा ऑक्सीजन (O) हैं।

(d) पोटैशियम सल्फेट = K2SO4.
अतः पोटैशियम सल्फेट में विद्यमान तत्त्व पोटैशियम (K), सल्फर (S) तथा ऑक्सीजन (O) हैं।

प्रश्न 6.
निम्नलिखित पदार्थों के मोलर द्रव्यमान का परिकलन कीजिए
(a) एथाइन, C2H2
(b) सल्फर अणु, S8
(c) फॉस्फोरस अणु, P4 (फॉस्फोरस का परमाणु द्रव्यमान = 31)
(d) हाइड्रोक्लोरिक अम्ल, HCI
(e) नाइट्रिक अम्ल, HNO3
हल-हम
जानते हैं कि C = 12, H = 1, S = 32, P = 31, CI = 35.5, N = 14, 0 = 16

(a) एथाइन (C2H2) का मोलर द्रव्यमान = 2 x 12 + 2 x 1
= 24 +2
= 26g उत्तर

(b) सल्फर अणु (S8) का मोलर द्रव्यमान = 8 x 32 = 256g उत्तर

(c) फॉस्फोरस अणु (P4) का मोलर द्रव्यमान = 4 x 31 = 124g उत्तर

(d) हाइड्रोक्लोरिक अम्ल (HCI) का मोलर द्रव्यमान = 1 x 1 + 1 x 35.5
= 1+ 35.5 = 36.5g उत्तर

(e) नाइट्रिक अम्ल (HNO3) का मोलर द्रव्यमान = 1 x 1 + 1 x 14 + 3 x 16
= 1 + 14 + 48
= 63g उत्तर

HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु

प्रश्न 7.
निम्न का द्रव्यमान क्या होगा
(a) 1 मोल नाइट्रोजन परमाणु?
(b) 4 मोल ऐलुमिनियम परमाणु (ऐलुमिनियम का परमाणु द्रव्यमान = 27)?
(c) 10 मोल सोडियम सल्फाइट (Na2so3) ?
हल
(a) 1 मोल नाइट्रोजन परमाणु का द्रव्यमान = 14g उत्तर
(b) 1 मोल ऐलुमिनियम परमाणु का द्रव्यमान = 27g
4 मोल ऐलुमिनियम परमाणु का द्रव्यमान = 4 x 27 = 108g उत्तर
(c) 1 मोल सोडियम सल्फाइट (Na,So,)
का द्रव्यमान = 2 x 23 + 1 x 32+3 x 16
= 46 + 32+ 48
= 126g उत्तर
10 मोल सोडियम सल्फाइट (Na2so3) का द्रव्यमान = 10 x 126g
= 1260g उत्तर

प्रश्न 8.
मोल में परिवर्तित कीजिए
(a) 12g ऑक्सीजन गैस,
(b) 20g जल,
(c) 22g कार्बन डाइऑक्साइड।
हल
(a) हम जानते हैं कि 1 मोल ऑक्सीजन (O2) = 2 x 16 = 32g
अतः 32g ऑक्सीजन = 1 मोल
1g ऑक्सीजन = 1 मोल
12g ऑक्सीजन = 12 = 0.375 मोल उत्तर

(b) हम जानते हैं कि 1 मोल जल (H2O) = (2 x 1 + 1 x 16)g
= (2 + 16)g
= 18g
अतः 18g जल = 1 मोल
1g जल = \(\frac { 1 }{ 18 }\) मोल
20g जल = \(\frac { 1 }{ 18 }\) x 20 = 1.11 मोल उत्तर

(c) हम जानते हैं कि 1 मोल कार्बन डाइऑक्साइड (CO2) = (1 x 12 + 2 x 16)g
= (12 + 32)g = 44g
अतः 44g कार्बन डाइऑक्साइड = 1 मोल
1g कार्बन डाइऑक्साइड = \(\frac { 1 }{ 44 }\) मोल
22g कार्बन डाइऑक्साइड = \(\frac { 1 }{ 44 }\) x 22 = 0.5 मोल उत्तर

प्रश्न 9.
निम्न का द्रव्यमान क्या होगा
(a) 0.2 मोल ऑक्सीजन परमाणु?
(b) 0.5 मोल जल अणु?
हल
(a) 1 मोल ऑक्सीजन परमाणु का द्रव्यमान = 16g
0.2 मोल ऑक्सीजन परमाणु का द्रव्यमान = 16 x 0.2g
= 3.2g उत्तर

(b) 1 मोल जल अणु (H2O) का द्रव्यमान = (2 x 1 + 1 x 16)g
= (2 + 16)g = 18g
0.5 मोल जल अणु का द्रव्यमान = 0.5 x 18 = 9.0g उत्तर प्रश्न

प्रश्न 10.
16g ठोस सल्फर में सल्फर (S8) के अणुओं की संख्या का परिकलन कीजिए।
हल
सल्फर S8 का अणु द्रव्यमान = 8 x 32 = 256g
1 मोल = 6.022 x 1023 अणु
अतः 256g सल्फर में अणुओं की संख्या = 6.022 x 1023
1g सल्फर में अणुओं की संख्या = \(\frac{6.022 \times 10^{23}}{256}\)
16g सल्फर में अणुओं की संख्या = \(\frac{6.022 \times 10^{23}}{256}\) x 16
= 3.76 x 1022
अतः 16g ठोस सल्फर में 3.76 x 1022 अणु होंगे।

HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु

प्रश्न 11.
0.051g ऐलुमिनियम ऑक्साइड (AI2O3) में ऐलुमिनियम आयन की संख्या का परिकलन कीजिए।
(संकेतः किसी आयन का द्रव्यमान उतना ही होता है जितना कि उसी तत्त्व के परमाणु का द्रव्यमान होता है। ऐलुमिनियम का परमाणु द्रव्यमान = 27u है।)
हल:
1 मोल ऐलुमिनियम ऑक्साइड (AI2O3) का द्रव्यमान = (2 x 27 + 3 x 16)g
= (54+ 48)g = 102g
102g ऐलुमिनियम ऑक्साइड = 1 मोल
0.051g ऐलुमिनियम ऑक्साइड = \(\frac { 1 }{ 102 }\) x 0.051
= 5 x 10-4 मोल
1 मोल AI2O3 में A1 आयनों की संख्या = 2 x 6.022 x 1023
5 x 10-4 मोल AI2O3 में आयनों की संख्या = 2 x 6.022 x 1023 x 5 x 10-4
= 6.022 x 1020 उत्तर

HBSE 9th Class Science Solutions Chapter 3 परमाणु एवं अणु Read More »

HBSE 9th Class Science Solutions Chapter 2 क्या हमारे आस-पास के पदार्थ शुद्ध हैं

Haryana State Board HBSE 9th Class Science Solutions Chapter 2 क्या हमारे आस-पास के पदार्थ शुद्ध हैं Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 2 क्या हमारे आस-पास के पदार्थ शुद्ध हैं

HBSE 9th Class Science क्या हमारे आस-पास के पदार्थ शुद्ध हैं Intext Questions and Answers
(पृष्ठ संख्या-16)
प्रश्न 1.
पदार्थ से आप क्या समझते हैं?
उत्तर:
पदार्थ वे होते हैं जो एक ही प्रकार के कणों से मिलकर बने होते हैं तथा उस पदार्थ में विद्यमान सभी कण समान रासायनिक प्रकृति के होते हैं; जैसे सोना, ताँबा आदि।

प्रश्न 2.
समांगी और विषमांगी मिश्रणों में अंतर बताएँ।
उत्तर:
समांगी मिश्रण-ऐसे मिश्रण जिनका संघटन एक समान होता है; उन्हें समांगी मिश्रण कहते हैं। विषमांगी मिश्रण-ऐसे मिश्रण जिनका संघटन असमान होता है, उन्हें विषमांगी मिश्रण कहते हैं।

(पृष्ठ संख्या – 20)

प्रश्न 1.
उदाहरण के साथ समांगी एवं विषमांगी मिश्रणों में विभेद कीजिए।
उत्तर:
समांगी मिश्रण-ऐसे मिश्रण जिनका संघटन एक समान होता है, उन्हें समांगी मिश्रण कहते हैं; जैसे जल में चीनी का मिश्रण, जल में नमक का मिश्रण, जल और ऐल्कोहॉल का मिश्रण। विषमांगी मिश्रण-ऐसे मिश्रण जिनका संघटन असमान होता है, उन्हें विषमांगी मिश्रण कहते हैं; जैसे बालू और नमक का मिश्रण, नमक और चीनी का मिश्रण।

HBSE 9th Class Science Solutions Chapter 2 क्या हमारे आस-पास के पदार्थ शुद्ध हैं

प्रश्न 2.
विलयन, निलंबन और कोलाइड एक-दूसरे से किस प्रकार भिन्न हैं?
उत्तर:
विलयन, निलंबन व कोलाइड में अंतर निम्नलिखित हैं

विलयननिलंबनकोलाइड
1. यह विलयन समांगी तथा पारदर्शी होता है; जैसे पानी में नमक का विलयन।1. यह विलयन विषमांगी तथा अपारदर्शी होता है; जैसे गंदला पानी, पेंट।1. यह विलयन समांगी परंतु अल्प पारदर्शी होता है; जैसे दूध, रक्त, स्याही, टूथपेस्ट।
2. इसमें विलेय के कणों का आकार मी० तक होता है।2. इसमें विलेय के कणों का आकार $10^{-7}$ मी० या इससे अधिक होता है।2. इसमें विलेय के कणों का आकार $10^{-9}$ से $10^{-7}$ मी० तक होता है अर्थात् विलयन के कणों से बड़ा आकार होता है।
3. इसमें विलेय के कणों को सूक्ष्मदर्शी यंत्र द्वारा भी नहीं देखा जा सकता है।3. इसमें विलेय कणों को नंगी आंखों से भी देखा जा सकता है।3. इसमें विलेय के कण एक शक्तिशाली सूक्ष्मदर्शी यंत्र द्वारा देखे जा सकते हैं।
4. इसमें विलेय कणों को छानकर पृथक् नहीं किया जा सकता।4. इसके कणों को छानकर पृथक् किया जा सकता है।4. इसके कणों को छानकर पृथक् नहीं किया जा सकता है।
5. अपने छोटे आकार के कारण विलयन के कण गुजर रही प्रकाश किरण को फैलाते नहीं हैं। इसलिए विलयन में प्रकाश का मार्ग दिखाई नहीं देता।5. निलंबित कण प्रकाश की किरण को फैला देते हैं जिससे उसका मार्ग दृष्टिगोचर हो जाता है।5. कोलाइड कण इतने बड़े होते हैं कि प्रकाश की किरण को फैलाते हैं तथा उसके मार्ग को दृश्य बनाते हैं।

प्रश्न 3.
एक संतृप्त विलयन बनाने के लिए 36g सोडियम क्लोराइड को 100g जल में 293K पर घोला जाता है। इस तापमान पर इसकी सांद्रता प्राप्त करें।
हल-293K ताप पर..
विलेय पदार्थ का द्रव्यमान (सोडियम क्लोराइड) = 36g
विलायक का द्रव्यमान (जल) = 100g
विलयन का द्रव्यमान = विलेय पदार्थ का द्रव्यमान + विलायक का द्रव्यमान
= 36g + 100g = 136g
विलयन की सांद्रता = HBSE 9th Class Science Solutions Chapter 2 क्या हमारे आस-पास के पदार्थ शुद्ध हैं img-1
\(\frac{36}{10}\) x 100 = 26.47%

(पृष्ठ संख्या – 26)

प्रश्न 1.
पेट्रोल और मिट्टी का तेल (Kerosene Oil) जो कि आपस में घुलनशील हैं, के मिश्रण को आप कैसे पृथक् करेंगे? पेट्रोल तथा मिट्टी के तेल के क्वथनांकों में 25°C से अधिक का अंतराल है।
उत्तर:
पेट्रोल और किरोसीन तेल जोकि आपस में घुलनशील हैं, के मिश्रण को आसवन विधि द्वारा अलग किया जाता है। यह विधि इस तथ्य पर आधारित है कि विभिन्न अवयवों का क्वथनांक भिन्न-भिन्न होता है क्योंकि पेट्रोल और किरोसीन तेल के क्वथनांकों में 25°C से अधिक का अन्तर है इसलिए निम्न क्वथनांक वाला द्रव पहले और उच्च क्वथनांक वाला द्रव कुछ अन्तराल बाद आसवित होकर अलग-अलग हो जाएँगे।

HBSE 9th Class Science Solutions Chapter 2 क्या हमारे आस-पास के पदार्थ शुद्ध हैं

प्रश्न 2.
पृथक् करने की सामान्य विधियों के नाम दें

  • दही से मक्खन,
  • समुद्री जल से नमक,
  • नमक से कपूर।

उत्तर:

  • दही से मक्खन अपकेंद्रण विधि द्वारा पृथक् किया जाता है।
  • समुद्री जल से नमक वाष्पीकरण विधि द्वारा पृथक् किया जाता है।
  • नमक से कपूर ऊर्ध्वपातन विधि द्वारा पृथक् किया जाता है।

प्रश्न 3.
क्रिस्टलीकरण विधि से किस प्रकार के मिश्रणों को पृथक् किया जा सकता है?
उत्तर:
क्रिस्टलीकरण वह विधि है जिसके द्वारा क्रिस्टल के रूप में शुद्ध ठोस को विलयन से अलग किया जा सकता है; जैसे समुद्र से नमक प्राप्त करना। + पाठ्य-पुस्तक के

(पृष्ठ संख्या-27)

प्रश्न 1.
निम्न को रासायनिक और भौतिक परिवर्तनों में वर्गीकृत करें
पेड़ों को काटना,
मक्खन का एक बर्तन में पिघलना,
अलमारी में जंग लगना,
जल का उबलकर वाष्प बनना,
विद्युत तरंग का जल में प्रवाहित होना तथा उसका हाइड्रोजन और ऑक्सीजन गैसों में विघटित होना,
जल में साधारण नमक का घुलना,
फलों से सलाद बनाना तथा
लकड़ी और कागज़ का जलना।
उत्तर:
भौतिक परिवर्तन-पेड़ों को काटना, मक्खन का एक बर्तन में पिघलना, जल का उबलकर वाष्प बनना, जल में साधारण नमक का घुलना, फलों से सलाद बनाना। रासायनिक परिवर्तन-अलमारी में जंग लगना, विद्युत तरंग का जल में प्रवाहित होना तथा उसका हाइड्रोजन और ऑक्सीजन गैसों में विघटित होना, लकड़ी और कागज़ का जलना।

प्रश्न 2.
अपने आस-पास की चीज़ों को शुद्ध पदार्थों या मिश्रण से अलग करने का प्रयत्न करें।
उत्तर:
शुद्ध पदार्थ-लोहा, सोना, चाँदी, ताँबा, ऐलुमिनियम, चीनी, नमक आदि। मिश्रण-समुद्र का जल, खनिज, मिट्टी, वायु, पेय पदार्थ आदि।

HBSE 9th Class Science Solutions Chapter 2 क्या हमारे आस-पास के पदार्थ शुद्ध हैं

HBSE 9th Class Science क्या हमारे आस-पास के पदार्थ शुद्ध हैं Textbook Questions and Answers

प्रश्न 1.
निम्नलिखित को पृथक् करने के लिए आप किन विधियों को अपनाएंगे?
(a) सोडियम क्लोराइड को जल के विलयन से पृथक् करने में।
(b) अमोनियम क्लोराइड को सोडियम क्लोराइड और अमोनियम क्लोराइड के मिश्रण से पृथक् करने में।
(c) धातु के छोटे टुकड़े को कार के इंजन ऑयल से पृथक् करने में।
(d) दही से मक्खन निकालने के लिए।
(e) जल से तेल निकालने के लिए।
(f) चाय से चाय की पत्तियों को पृथक् करने में।
(g) बालू से लोहे की पिनों को पृथक् करने में।
(h) भूसे से गेहूँ के दानों को पृथक् करने में।
(i) पानी में तैरते हुए महीन मिट्टी के कण को पानी से अलग करने के लिए।
(j) पुष्प की पंखुड़ियों के निचोड़ से विभिन्न रंजकों को पृथक् करने में।
उत्तर:
(a) सोडियम क्लोराइड को जल के विलयन से पृथक् करने में वाष्पीकरण विधि का प्रयोग किया जाता है।

(b) अमोनियम क्लोराइड को सोडियम क्लोराइड और अमोनियम क्लोराइड के मिश्रण से पृथक् करने में ऊर्ध्वपातन विधि का प्रयोग किया जाता है क्योंकि अमोनियम क्लोराइड ऊर्ध्वपातित पदार्थ है।

(c) धातु के छोटे टुकड़े को कार के इंजन ऑयल से पृथक् करने में छानना विधि का प्रयोग किया जाता है।

(d) दही से मक्खन निकालने के लिए अपकेंद्रण विधि का प्रयोग किया जाता है।

(e) जल से तेल निकालने के लिए पृथक्करण कीप का प्रयोग किया जाता है क्योंकि जल और तेल अघुलनशील द्रव हैं।

(f) चाय की पत्तियों से चाय को पृथक् करने के लिए छानना विधि का प्रयोग किया जाता है। छानने के लिए चाय-छलनी का प्रयोग किया जाता है।

(g) बालू से लोहे की पिनों को पृथक् करने में चुंबकीय पृथक्करण विधि का प्रयोग किया जाता है क्योंकि लोहे की पिन चुंबक की ओर आकर्षित हो जाती है।

(h) भूसे से गेहूँ के दाने को पृथक करने के लिए फटकना विधि का प्रयोग किया जाता है क्योंकि फटकना विधि से भूसा हल्का होने के कारण वायु से दूर चला जाता है और गेहूँ के दाने भारी होने के कारण सीधे जमीन पर गिरते हैं।

(i) पानी में तैरते हुए महीन मिट्टी के कणों को भारण विधि द्वारा अलग किया जा सकता है क्योंकि मिट्टी के कण फिटकरी द्वारा भारी होकर नीचे बैठ जाते हैं।

(j) पुष्प की पंखुड़ियों के निचोड़ से विभिन्न रंजकों को पृथक् करने में क्रोमैटोग्राफी विधि का प्रयोग किया जाता है।

HBSE 9th Class Science Solutions Chapter 2 क्या हमारे आस-पास के पदार्थ शुद्ध हैं

प्रश्न 2.
चाय तैयार करने के लिए आप किन-किन चरणों का प्रयोग करेंगे। विलयन, विलायक, विलेय, घुलना, घुलनशील, अघुलनशील, घुलेय (फिल्ट्रेट) तथा अवशेष शब्दों का प्रयोग करें।
उत्तर:
चाय तैयार करने के लिए दिए गए शब्दों के प्रयोग से हम निम्नलिखित प्रकार से चाय बनाएँगे

  1. विलायक-चाय बनाने वाले बर्तन में पानी को विलायक के रूप में लीजिए तथा बर्नर पर रखिए।
  2. विलेय-पानी में चीनी को विलेय के रूप में डालिए।
  3. विलयन-पानी व चीनी का मिश्रण विलयन बन जाएगा।
  4. घुलना-पानी में चीनी घुलकर विलयन बनाएगी।
  5. घुलनशील-पानी में चीनी घुलनशील होने के कारण घुल जाती है तथा उबालने के बाद दूध में भी घुलनशील पदार्थ है।
  6. अघुलनशील-पानी व चीनी के मिश्रण में चायपत्ती की पत्तियाँ अघुलनशील पदार्थ के रूप में डालिए तथा उबालिए।
  7. घुलेय (फिल्ट्रेट) और अवशेष-चाय की पत्तियों के उबलने के बाद चाय को छानने योग्य छलनी का प्रयोग करके छानिए। घुलेय (फिल्ट्रेट) पदार्थ चाय को पीने के लिए प्रयोग कीजिए तथा छलनी में बचे अवशेष को फेंक दीजिए।

प्रश्न 3.
प्रज्ञा ने तीन अलग-अलग पदार्थों की घुलनशीलताओं को विभिन्न तापमान पर जाँचा और नीचे दिए गए आँकड़ों को प्राप्त किया। प्राप्त हुए परिणामों को 100g जल में विलेय पदार्थ की मात्रा, जो संतृप्त विलयन बनाने हेतु पर्याप्त है, निम्नलिखित तालिका में दर्शाया गया है

विलेय पदार्थतापमान केल्विन
283293313333353
पोटैशियम नाइट्रेट213262106167
सोडियम क्लोराइड3636363737
पोटैशियम क्लोराइड3535404654
अमोनियम क्लोराइड2437415566

(a) 50g जल में 313K पर पोटैशियम नाइट्रेट के संतृप्त विलयन को प्राप्त करने हेतु कितने ग्राम पोटैशियम नाइट्रेट की आवश्यकता होगी?

(b) प्रज्ञा 353K पर पोटैशियम क्लोराइड का एक संतृप्त विलयन तैयार करती है और विलयन को कमरे के तापमान पर ठंडा होने के लिए छोड़ देती है। जब विलयन ठंडा होगा तो वह क्या अवलोकित करेगी? स्पष्ट करें।

(c) 293K पर प्रत्येक लवण की घुलनशीलता का परिकलन करें। इस तापमान पर कौन-सा लवण सबसे अधिक घुलनशील होगा?

(d) तापमान में परिवर्तन से लवण की घुलनशीलता पर क्या प्रभाव पड़ता है?
उत्तर:
(a) प्रश्नानुसार
100 ग्राम जल में 313K पर पोटैशियम नाइट्रेट के संतृप्त विलयन के लिए आवश्यक पोटैशियम नाइट्रेट = 62 ग्राम
1 ग्राम जल में 313K पर पोटैशियम नाइट्रेट के संतृप्त विलयन के लिए आवश्यक पोटैशियम नाइट्रेट = \(\frac{62}{100}\) 100 ग्राम
50 ग्राम जल में 313K पर पोटैशियम नाइट्रेट के संतृप्त विलयन के लिए आवश्यक पोटैशियम नाइट्रेट = \(\frac{62}{100}\) x 50 ग्राम
= 31 ग्राम

(b) प्रज्ञा 353K पर पोटैशियम क्लोराइड का संतृप्त विलयन प्राप्त करती है और विलयन को कमरे के तापमान (293K) पर ठंडा होने के लिए छोड़ देती है तो जब विलयन ठंडा होगा तो वह अति संतृप्त विलयन होगा क्योंकि कमरे के तापमान पर उसमें संतृप्तता से (54-35)19 ग्राम अधिक पोटेशियम.क्लोराइड होगा।

(c) 293K तापमान पर 100 ग्राम जल में पोटैशियम नाइट्रेट, सोडियम क्लोराइड, पोटैशियम क्लोराइड व अमोनियम क्लोराइड की घुलनशीलता क्रमशः 32 ग्राम, 36 ग्राम, 35 ग्राम व 37 ग्राम है। अतः इस तापमान पर अमोनियम क्लोराइड लवण की घुलनशीलता सबसे अधिक है।

(d) तापमान में परिवर्तन से लवण की घुलनशीलता पर धनात्मक प्रभाव पड़ता है अर्थात् तापमान को बढ़ाने पर लवण की घुलनशीलता बढ़ती है।

HBSE 9th Class Science Solutions Chapter 2 क्या हमारे आस-पास के पदार्थ शुद्ध हैं

प्रश्न 4.
निम्नलिखित की उदाहरण सहित व्याख्या करें
(a) संतृप्त विलयन,
(b) शुद्ध पदार्थ,
(c) कोलाइड,
(d) निलंबन।
उत्तर:
(a) संतृप्त विलयन-दिए गए निश्चित तापमान पर यदि विलयन में ओर अधिक विलेय पदार्थ नहीं घुलता तो उसे संतृप्त विलयन कहते हैं अर्थात् दिए गए तापमान पर किसी विलयन में जब उसकी क्षमता के अनुसार जितना अधिकतम क्लेिय घुल जाता है तब उसे संतृप्त विलयन कहते हैं। उदाहरणतया एक बीकर में 50 मि०ली० जल लें अब इसमें थोड़ा-थोड़ा करके नमक डालें और घोलें। जब नमक घुलना बंद हो जाए तो वह विलयन संतृप्त विलयन कहलाएगा।

(b) शुद्ध पदार्थ एक ही प्रकार के कणों से बने पदार्थ को शुद्ध पदार्थ कहा जाता है; जैसे लोहा, सोना, चाँदी, चीनी, जल आदि।

(c) कोलॉइड-कोलॉइड वह विषमांगी मिश्रण होता है जिसके कणों का आकार 1 mm से 100 mm के बीच होता है। ये कण आंखों से नहीं देखे जा सकते तथा प्रकाश की किरणों को फैला देते हैं; जैसे दूध, शेविंग क्रीम, टूथपेस्ट, जेली, फेस क्रीम आदि।

(d) निलंबन-निलंबन एक विषमांगी मिश्रण होता है जिसमें विलेय पदार्थ के कण घुलते नहीं हैं, बल्कि माध्यम की समष्टि में निलंबित रहते हैं। निलंबित कण 100 nm (10-‘m) से बड़े होते हैं तथा आंखों से देखे जा सकते हैं, जैसे नदी का गंदला पानी; मोटे चूने के पत्थर व जल का मिश्रण आदि।

प्रश्न 5.
निम्नलिखित में से प्रत्येक को समांगी और विषमांगी मिश्रणों में वर्गीकृत करेंसोडा जल, लकड़ी, बर्फ, वायु, मिट्टी, सिरका, छनी हुई चाय। उत्तर-समांगी मिश्रण-सोडा जल, बर्फ, सिरका, छनी हुई चाय। विषमांगी मिश्रण-लकड़ी, वायु, मिट्टी। प्रश्न 6. आप किस प्रकार पुष्टि करेंगे कि दिया हुआ रंगहीन द्रव शुद्ध जल है?
उत्तर:
हम दिए गए रंगहीन द्रव का क्वथनांक ज्ञात करेंगे यदि वह क्वथनांक 373K आता है तो हम पुष्टि करेंगे कि दिया गया रंगहीन द्रव शुद्ध जल है, परंतु यदि नहीं आता तो यह शुद्ध जल नहीं है।

HBSE 9th Class Science Solutions Chapter 2 क्या हमारे आस-पास के पदार्थ शुद्ध हैं

प्रश्न 7.
निम्नलिखित में से कौन-सी वस्तुएँ शुद्ध पदार्थ हैं?
(a) बर्फ,
(b) दूध,
(c) लोहा,
(d) हाइड्रोक्लोरिक अम्ल,
(e) कैल्शियम ऑक्साइड, पारा,
(g) ईंट,
(h) लकड़ी,
(i) वायु
उत्तर:
दी गई वस्तुओं में से निम्नलिखित वस्तुएँ शुद्ध पदार्थ हैं-
(a) बर्फ,
(c) लोहा,
(d) हाइड्रोक्लोरिक अम्ल,
(e) कैल्शियम ऑक्साइड, पारा।

प्रश्न 8.
निम्नलिखित मिश्रणों में से विलयन की पहचान करें
(a) मिट्टी,
(b) समुद्री जल,
(c) वायु,
(d) कोयला,
(e) सोडा जल।
उत्तर:
(e) सोडा जल एक विलयन है।

प्रश्न 9.
निम्नलिखित में से कौन टिनडल प्रभाव को प्रदर्शित करेगा?
(a) नमक का घोल,
(b) दूध,
(c) कॉपर सल्फेट का विलयन,
(a) स्टार्च विलयन।
उत्तर:
(b) दूध टिनडल प्रभाव को दर्शाता है।

प्रश्न 10.
निम्नलिखित को तत्त्व, यौगिक और मिश्रण में वर्गीकृत करें
(a) सोडियम,
(b) मिट्टी,
(c) चीनी का घोल,
(d) चाँदी,
(e) कैल्शियम कार्बोनेट,
(f) टिन,
(g) सिलिकॉन,
(h) कोयला,
(i) वायु,
(j) साबुन,
(k) मिथेन,
(l) कार्बन डाइऑक्साइड,
(m) रक्त।
उत्तर:
तत्त्व-
(a) सोडियम,
(d) चाँदी,
(f) टिन,
(g) सिलिकॉन।
यौगिक-
(e) कैल्शियम कार्बोनेट,
(j) साबुन,
(k) मिथेन,
(l) कार्बन डाइऑक्साइड।
मिश्रण-(b) मिट्टी, (c) चीनी का घोल, (h) कोयला, (i) वायु, (m) रक्त।

प्रश्न 11.
निम्नलिखित में से कौन-कौन से परिवर्तन रासायनिक हैं?
(a) पौधों की वृद्धि,
(b) लोहे में जंग लगना,
(c) लोहे के चूर्ण और बालू को मिलाना,
(d) खाना पकाना,
(e) भोजन का पाचन,
(f) जल से बर्फ बनना,
(g) मोमबत्ती का जलना।
उत्तर:
रासायनिक परिवर्तन निम्नलिखित हैं
(a) पौधों की वृद्धि,
(b) लोहे में जंग लगना,
(d) खाना पकाना,
(e) भोजन का पाचन,
(f) मोमबत्ती का जलना।

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HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ

Haryana State Board HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ

HBSE 9th Class Science हमारे आस-पास के पदार्थ Intext Questions and Answers

(पृष्ठ संख्या -4)
प्रश्न 1.
निम्नलिखित में से कौन-से पदार्थ हैं कुर्सी, वायु, स्नेह, गंध, घृणा, बादाम, विचार, शीत, नींबू पानी, इत्र की सुगंध।
उत्तर:
कुर्सी, वायु, बादाम व नींबू पानी, पदार्थ हैं क्योंकि ये स्थान घेरते हैं तथा द्रव्यमान रखते हैं।

प्रश्न 2.
निम्नलिखित प्रेक्षण के कारण बताएँ गर्मा-गरम खाने की गंध कई मीटर दूर से ही आपके पास पहुँच जाती है लेकिन ठडे खाने की महक लेने के लिए आपको उसके पास जाना पड़ता है।
उत्तर:
क्योंकि तापमान बढ़ने से विसरण तेज हो जाता है। इसी कारण गर्मा-गरम खाने की गंध कई मीटर से ही हमारे पास पहुँच जाती है, जबकि ठंडे खाने की महक लेने के लिए हमें उसके पास जाना पड़ता है।

HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ

प्रश्न 3.
स्वीमिंग पूल में गोताखोर पानी काट पाता है। इससे पदार्थ का कौन-सा गुण प्रेक्षित होता है?
उत्तर:
जल के कणों के बीच दूरी अपेक्षाकृत अधिक होने के कारण इसमें संपीडय का गुण पाया जाता है इसी कारण स्वीमिंग पूल में गोताखोर पानी (जल) को काट पाता है।

प्रश्न 4.
पदार्थ के कणों की क्या विशेषताएँ होती हैं? ।
उत्तर:
पदार्थ के कणों की निम्नलिखित विशेषताएँ होती हैं

  1. पदार्थ के कणों के बीच रिक्त स्थान होता है।
  2. पदार्थ के कण निरंतर गतिशील होते हैं अर्थात् उनमें गतिज ऊर्जा होती है।
  3. तापमान बढ़ने से पदार्थ के कणों की गतिज ऊर्जा बढ़ जाती है।
  4. पदार्थ के कण अपने आप अंतः मिश्रित हो जाते हैं।
  5. पदार्थ के कण एक-दूसरे को आकर्षित करते हैं।

(पृष्ठ संख्या -6)

HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ

प्रश्न 1.
किसी तत्त्व के द्रव्यमान प्रति इकाई आयतन को घनत्व कहते हैं। (घनत्व = द्रव्यमान/आयतन)। बढ़ते हुए घनत्व के क्रम में निम्नलिखित को व्यवस्थित करें-वायु, चिमनी का धुआँ, शहद, जल, चॉक, रुई और लोहा।
उत्तर:
दिए गए तत्त्वों को घनत्व बढ़ने के साथ आरोही क्रम में लिखने पर निम्नलिखित क्रम प्राप्त होगाचिमनी का धुआँ, वायु, रुई, चॉक, जल, शहद व लोहा।

प्रश्न 2.
(a) पदार्थ की विभिन्न अवस्थाओं के गुणों में होने वाले अंतर को सारणीबद्ध कीजिए।
(b) निम्नलिखित पर टिप्पणी कीजिए-दृढ़ता, संपीडयता, तरलता, बर्तन में गैस का भरना, आकार, गतिज ऊर्जा एवं घनत्व।
उत्तर:
(a) पदार्थ की विभिन्न अवस्थाओं के गुणों में निम्नलिखित अंतर पाए जाते हैं

क्रमांक सं.गुणठोसद्रवगैस
1.आकारइनका आकार निश्चित होता है।इनका आकार निश्चितइनका आकार निश्चित
2.आयतनइनका आयतन निश्चित होता है।नहीं होता। ये बर्तन केनहीं होता।
3.कठोरताये कठोर होते हैं।आकार के अनुसार अपनाइनका आयतन निश्चित
4.ढेर लगानाइनको ढेर के रूप में इकट्ठा कियाआकार बदल लेते हैं।नहीं होता।
5.संपीडितजा सकता है।इनका आयतन भीइनमें यह गुण नहीं होता।
क्षमताइन्हें संपीडित नहीं किया जानिश्चित होता है।यह भी बह जाती है।
6.कणों कीसकता।ये कठोर नहीं होते।इन्हें संपीडित किया जा
स्थितिइनमें कण एक-दूसरे के बहुतये बह जाते हैं।सकता है।

(b) दृढ़ता-पदार्थ के कणों के बीच लगने वाला आकर्षण बल पदार्थ की दृढ़ता निश्चित करता है। ठोसों में आकर्षण बल अधिक होने के कारण दृढ़ता अधिक होती है। द्रवों में उससे कम तथा गैसों में सबसे कम दृढ़ता होती है।

संपीडयता – किसी पदार्थ पर बल लगाकर उसके कणों की बीच की दूरी को कम करना संपीडयता कहलाता है। गैसों में संपीडयता का गुण पाया जाता है।

तरलता – जिन पदार्थों में बहने का गुण पाया जाता है, उन्हें तरल पदार्थ कहते हैं। द्रवों में तरलता का गुण पाया जाता है।

बर्तन में गैस का भरना-किसी बर्तन में गैस कणों में अंतराणुक बल क्षीण होने के कारण गैस सारे उपलब्ध स्थान को घेर लेती है। अर्थात् गैसों का आयतन निश्चित नहीं होता। इन्हें बर्तन में उच्च दाब पर भरा जाता है।

आकार-ठोस के कण अधिक अंतराणुक बल के साथ जुड़े होने के कारण ठोस को निश्चित आकार प्रदान करते हैं, जबकि द्रवों में यह अंतराणुक बल कम होता है जिस कारण इनका आकार निश्चित नहीं होता। गैसों में भी ऐसा होता है।

गतिज ऊर्जा-कणों की गति के कारण उत्पन्न ऊर्जा को गतिज ऊर्जा कहा जाता है। तापमान बढ़ाने से कणों की गतिज ऊर्जा बढ़ जाती है। ठोस में अधिक गतिज ऊर्जा नहीं होती, द्रव में कुछ गतिज ऊर्जा होती है जबकि गैसों में उच्च गतिज ऊर्जा होती है।

घनत्व-किसी तत्त्व के द्रव्यमान प्रति इकाई आयतन को घनत्व कहते हैं अर्थात् घनत्व = द्रव्यमान । ठोस पदार्थों का घनत्व उच्च होता है, द्रव में निम्न जबकि गैसों में नगण्य होता है।

HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ

प्रश्न 3.
कारण बताएँ
(a) गैस पूरी तरह उस बर्तन को भर देती है, जिसमें इसे रखते हैं।
(b) गैस बर्तन की दीवारों पर दबाव डालती है।
(c) लकड़ी की मेज़ ठोस कहलाती है।
(d) हवा में हम आसानी से अपना हाथ चला सकते हैं, लेकिन एक ठोस लकड़ी के टुकड़े से हाथ चलाने के लिए हमें कराटे में दक्ष होना पड़ेगा।
उत्तर:
(a) गैस का आयतन निश्चित न होने के कारण गैस पूरी तरह उस बर्तन को भर देती है जिसमें इसे रखा जाता है।
(b) गैसीय अवस्था में कणों की गति अनियमित और अत्यधिक तीव्र होती है। इस अनियमित गति के कारण गैस के कण आपस में एवं बर्तन की दीवारों से टकराते हैं। बर्तन की दीवार पर गैस कणों द्वारा प्रति इकाई क्षेत्र पर लगे बल के कारण गैस का दबाव बनता है।
(c) लकड़ी की मेज का निश्चित आकार, निश्चित आयतन तथा असंपीडय होने के कारण ठोस कहलाती है।
(d) हवा के कणों के बीच अधिक दूरी होने के कारण हवा में हम हाथ को आसानी से चला सकते हैं, जबकि ठोस लकड़ी के टुकड़े के कणों के बीच की दूरी कम होने के कारण हमें हाथ चलाने के लिए कराटे में दक्ष होना पड़ेगा।

HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ

प्रश्न 4.
सामान्यतया ठोस पदार्थों की अपेक्षा द्रवों का घनत्व कम होता है, लेकिन आपने बर्फ के टुकड़े को जल में तैरते हुए देखा होगा। पता लगाइए, ऐसा क्यों होता है?
उत्तर:
सामान्यतया ठोस पदार्थों की अपेक्षा द्रवों का घनत्व कम होता है परंतु जल का घनत्व 4°C पर अधिकतम होता है। जब इसे 4°C से नीचे ठंडा किया जाता है तो 0°C पर बर्फ जम जाती है जिसका घनत्व जल के घनत्व से कम होता है इसी कारण बर्फ जल में तैरती है।

HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ

(पृष्ठ संख्या-9)

प्रश्न 1.
निम्नलिखित तापमान को सेल्सियस में बदलें
(a) 300K
(b) 573K
उत्तर:
(a) 300K = (300 – 273)°C = 27°C
(b) 573K = (573 – 273)°C = 300°C

प्रश्न 2.
निम्नलिखित तापमान पर जल की भौतिक अवस्था क्या होगी?
(a) 250°C
(b) 100°C
उत्तर:
(a) 250°C तापमान पर समस्त जल भाप बनकर उड़ जाएगा।
(b) 100°C पर जल उबलना शुरू कर देगा क्योंकि यह जल का क्वथनांक है।

प्रश्न 3.
किसी भी पदार्थ की अवस्था परिवर्तन के दौरान तापमान स्थिर क्यों रहता है?
उत्तर:
किसी भी पदार्थ की अवस्था परिवर्तन के दौरान तापमान स्थिर रहता है क्योंकि अवस्था परिवर्तन के समय पदार्थ को दी जाने वाली समस्त ऊष्मा कणों के पारस्परिक आकर्षण बल को वशीभूत करके पदार्थ की अवस्था को बदलने में उपयोग हो जाती है।

HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ

प्रश्न 4.
वायुमंडलीय गैसों को द्रव में परिवर्तन करने के लिए कोई विधि सुझाइए।
उत्तर:
वायुमंडलीय गैसों को किसी बंद बर्तन में बंद करके तथा दाब बढ़ाकर व तापमान घटाकर द्रव में परिवर्तित किया जा सकता है।

(पृष्ठ संख्या-11)

प्रश्न 1.
गर्म, शुष्क दिन में कूलर कमरे को अधिक ठंडा क्यों करता है?
उत्तर:
गर्म, शुष्क दिन में आर्द्रता कम होने के कारण वायु शुष्क होती है जब यह वायु कूलर के पंखे द्वारा खींची जाती है तो कूलर में मैट पर टपकता हुआ जल वाष्पीकृत होकर गर्म व शुष्क वायु की गर्मी को अवशोषित कर लेता है जिसके परिणामस्वरूप वायु ठंडी हो जाती है। यह ठंडी वायु कमरे को ठंडा कर देती है।

प्रश्न 2.
गर्मियों में घड़े का जल ठंडा क्यों होता है?
उत्तर:
घड़े में छोटे-छोटे छिद्र होते हैं जिनके माध्यम से गर्मियों में जल रिसकर वाष्पीकृत होता रहता है तथा वाष्पन के लिए वह ऊष्मा जल से ही लेता है जिस कारण घड़े में रखा जल ठंडा हो जाता है।

HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ

प्रश्न 3.
एसीटोन/पेट्रोल या इत्र डालने पर हमारी हथेली ठंडी क्यों हो जाती है?
उत्तर:
एसीटोन पेट्रोल या इत्र को हथेली पर डालने से इसके कण हथेली से ऊर्जा प्राप्त कर वाष्पीकृत हो जाते हैं जिससे हथेली ठंडी हो जाती है।

प्रश्न 4.
कप की अपेक्षा प्लेट से हम गर्म दूध या चाय जल्दी क्यों पी लेते हैं?
उत्तर:
प्लेट की सतह का क्षेत्रफल कप की सतह की अपेक्षा अधिक होता है जिस कारण प्लेट की सतह से वाष्पीकरण अधिक होने के कारण गर्म दूध या चाय जल्दी ठंडी हो जाती है तथा ठंडी होने के कारण जल्दी पी जाती है।

प्रश्न 5.
गर्मियों में हमें किस तरह के कपड़े पहनने चाहिएँ?
उत्तर:
गर्मियों में हमें सूती कपड़े पहनने चाहिएँ क्योंकि गर्मियों में शारीरिक प्रक्रिया के कारण अधिक पसीना आता है, जिससे हमें ठंडक मिलती है जैसा कि हम जानते हैं कि वाष्पीकरण के दौरान द्रव की सतह के कण हमारे शरीर से ऊर्जा प्राप्त करके वाष्प में बदल जाते हैं। वाष्पीकरण की प्रसुप्त ऊष्मा के बराबर ऊष्मीय ऊर्जा हमारे शरीर से अवशोषित हो जाती है, जिससे शरीर शीतल हो जाता है। सूती कपड़े पहनने से जल का अवशोषण अधिक होता है, जिस कारण पसीना इसमें अवशोषित होकर वायुमंडल में आसानी से वाष्पीकृत हो जाता है।

HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ

HBSE 9th Class Science हमारे आस-पास के पदार्थ Textbook Questions and Answers

प्रश्न 1.
निम्नलिखित तापमानों को सेल्सियस इकाई में परिवर्तित करें
(a) 300K
(b) 573K
उत्तर:
(a) 300K = (300 – 273)° = 27°C
(b) 573K = (573 — 273)°C = 300°C

प्रश्न 2.
निम्नलिखित तापमानों को केल्विन इकाई में परिवर्तित करें
(a) 25°C
(b) 373°C
उत्तर:
(a) 25°C = (25 + 273)K 298K
(b) 373°C = (373 + 273)K = 646K

प्रश्न 3.
निम्नलिखित अवलोकनों हेतु कारण लिखें
(a) नैफ्थलीन को रखा रहने देने पर यह समय के साथ कुछ भी ठोस पदार्थ छोड़े बिना अदृश्य हो जाती है।
(b) हमें इत्र की गंध बहुत दूर बैठे हुए भी पहुँच जाती है।
उत्तर:
(a) नैफ्थलीन को रखा रहने देने पर यह समय के साथ कुछ भी ठोस पदार्थ छोड़े बिना अदृश्य हो जाती है क्योंकि नैफ्थलीन एक ऊर्ध्वापातित पदार्थ है जो ठोस से सीधा ही गैसीय अवस्था में परिवर्तित हो जाता है।
(b) हमें इत्र की गंध बहुत दूर बैठे हुए भी पहुँच जाती है क्योंकि इत्र वायु में विसरित होने का गुण रखता है।

HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ

प्रश्न 4.
निम्नलिखित पदार्थों को उनके कणों के बीच बढ़ते आकर्षण के अनुसार व्यवस्थित करें
(a) जल,
(b) चीनी,
(c) ऑक्सीजन।
उत्तर:
कणों के बीच बढ़ते आकर्षण के अनुसार पदार्थों का व्यवस्थित रूप निम्नलिखित है-
ऑक्सीजन < जल < चीनी

प्रश्न 5.
निम्नलिखित तापमानों पर जल की भौतिक अवस्था क्या है
(a) 25°C,
(b) 0°C,
(c) 100°C
उत्तर:
(a) 25°C पर जल द्रव अवस्था में होगा।
(b) 0°C पर जल ठोस (बफ) अवस्था में होगा।
(c) 100°C पर जल क्वथित अवस्था (भाप) में होगा।

HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ

प्रश्न 6.
पुष्टि हेतु कारण दें
(a) जल कमरे के ताप पर द्रव है।
(b) लोहे की अलमारी कमरे के ताप पर ठोस है।
उत्तर:
(a) जल कमरे के ताप पर द्रव है, क्योंकि

  1. इस ताप पर इसके अणुओं के बीच लगने वाला आकर्षण बल सामान्य होता है।
  2. इसके अणुओं की गतिज सामान्य होती है।

(b) लोहे की अलमारी कमरे के ताप पर ठोस है क्योंकि

  1. लोहे के अणुओं के बीच लगने वाला आकर्षण बल काफी अधिक होता है।
  2. लोहे के अणुओं के बीच दूरी नगण्य होती है जिस कारण वे निश्चित दूरी तक घूम सकते हैं।

प्रश्न 7.
273K पर बर्फ को ठंडा करने पर तथा जल को इसी तापमान पर ठंडा करने पर शीतलता का प्रभाव अधिक क्यों होता है?
उत्तर:
273K तापमान पर बर्फ को ठंडा करने पर तथा जल को इसी तापमान पर ठंडा करने पर शीतलता का प्रभाव अधिक होता है, क्योंकि बर्फ में संगलन गुप्त ऊष्मा अधिक होती है।

प्रश्न 8.
उबलते हुए जल अथवा भाप में से जलने की तीव्रता किसमें अधिक महसूस होती है?
उत्तर:
उबलते हुए जल अथवा भाप में से जलने की तीव्रता भाप में अधिक महसूस होती है क्योंकि भाप में अतिरिक्त ऊष्मा होती है जिसे वाष्पन की गुप्त ऊष्मा कहते हैं।

HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ

प्रश्न 9.
निम्नांकित चित्र के लिए A, B, C, D, E तथा F की अवस्था परिवर्तन को नामांकित करें
HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ img-1
उत्तर:
(A) संगलन,
(B) वाष्पन,
(C) संघनन,
(D) जमना,
(E) ऊर्ध्वपातन,
(F) ऊर्ध्वपातन।

HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ Read More »

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Haryana State Board HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Very Short-Answer Type Questions

Question 1.
What are the main components of food?
Answer:
Protein, fat, vitamins, minerals, carbohydrates, etc.

Question 2.
Where from we get food?
Answer:
From plants and animals.

Question 3.
How the crop production increase?
Answer:
By green revolution.

Question 4.
What do you mean by white revolution ?
Answer:
Excess production of milk is called white revolution.

Question 5.
Which practices should be adopted in agriculture and animal husbandry ?
Answer:
Sustained system.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 6.
Write the main sources of carbohydrates.
Answer:
Wheat, maize, rice, barley, millets, etc.

Question 7.
From which products of plants do we get proteins ?
Answer:
Gram, pea, black gram, green gram, pigeon pea, lentil, etc.

Question 8.
Write the name of sources of fat.
Answer:
Soyabean, ground nut, mustard, sesame, linseed, castor, sunflower, etc. ‘

Question 9.
Where from we get vitamins and minerals ?
Answer:
Vegetables, spices, fruits, etc.

Question 10.
Write the names of fodder crops.
Answer:
Berseem, oat, sudan grass, maize, sorghum, etc.

Question 11.
On which factor growth of plants and flowering are dependent ?
Answer:
Growth of plants and flowering are dependent on sunlight.

Question 12.
In rainy season which crops are grown ?
Answer:
Kharif crops.

Question 13.
When rabi crops are grown ?
Answer:
In winter season.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 14.
Write the names of kharif crops.
Answer:
Paddy, maize, cotton, green gram, black gram, soyabean, pigeon pea, etc.

Question 15.
Which are the rabi crops ?
Answer:
Gram, pea, mustard, linseed, etc.

Question 16.
What is called crossing of crops ?
Answer:
Mating or crossing is being done of genetically different plants to improve the desirable characters of the crops, this is called crossing of
crops.

Question 17.
Which is the new technique for improved crops ?
Answer:
Higher yield varieties are prepared through introduction of genes of desirable characters into the – crops.

Question 18.
What is called high yielding ?
Answer:
To increase the production of crop per acre is called high yielding.

Question 19.
What is called biotic resistance ?
Answer:
Resistance from diseases, insects, etc., is called biotic resistance.

Question 20.
Write the names of the conditions for abiotic resistance.
Answer:
Drought, salinity, water logging, heat, cold and frost.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 21.
Is it advantageous or disadvantageous to decrease the maturity period of the crops ?
Answer:
Advantageous.

Question 22.
Why is it useful to grow the dwarf plants for cereals ?
Answer:
Because dwarf plants consume less nutrients.

Question 23.
Why do we need food ?
Answer:
For development, growth and to be healthy.

Question 24.
Where from plants get nutrients ?
Answer:
From air, water and soil.

Question 25.
How many macro-nutrients are in number ?
Answer:
6 (nitrogen, phosphorus, potassium, calcium, magnesium and sulphur).

Question 26.
How many micro-nutrients are in number ?
Answer:
7 (iron, manganese, boron, zinc, copper, molybdenum and chlorine).

Question 27.
How manure is prepared ?
Answer:
Manure is prepared by micro organisms from the decomposition of animal excreta and plant wastes, eradicated weeds, etc. ‘

Question 28.
Which main crops are grown for green manure ?
Answer:
Sun hemp, mung, gur are grown for green manure.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 29.
What are fertilizers ?
Answer:
Which nutritious matters are prepared by artificial methods for plants are called fertilizers.

Question 30.
Are fertilizers cheapest than bio-fertilizers ?
Answer:
No, these are costly.

Question 31.
Write one demerit of fertilizers.
Answer:
It causes water pollution.

Question 32.
Which is the main source of irrigation in India ?
Answer:
Rain.

Question 33.
Which causes much loss to rain-fed crops ?
Answer:
By drought.

Question 34.
How we get underground water ?
Answer:
By digging wells (tubewells).

Question 35.
Write one modern technique to increase the availability of water.
Answer:
By constructing small check dams.

Question 36.
What is called mixed cropping ?
Answer:
Growing of two or more crops simultaneously on the same field is called mixed cropping.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 37.
Write an example of mixed cropping.
Answer:
Wheat + gram, groundnut + sunflower.

Question 38.
What is called intercropping ? ,
Answer:
Growing of two or more crops simultaneously on the same field in a definite pattern is called intercropping.

Question 39.
Write an example of intercropping.
Answer:
Soyabean + maize (one row soyabean + one row maize). .

Question 40.
What is called crop rotation ?
Answer:
The growing of different crops on a piece of land in a pre-planned succession is known as crop rotation.

Question 41.
Write the names of weeds.
Answer:
Xanthium (gokhroo), Parthenium (gajar ghas), Cyperinus rotundus (motha), Chenopodium (bathua).

Question 42.
What are called insects ?
Answer:
The insects which cause loss to crops are called insects. %

Question 43.
What are called chemical which destroy weeds ?
Answer:
Weedicides (Weedkiller).

Question 44.
Write the biotic factors cause loss to foodgrains.
Answer:
Insects, rodents, fungi, bacteria, etc.

Question 45.
Write the names of abiotic factors cause degradation to foodgrains.
Answer:
Lack of appropriate temperature and moisture.

Question 46.
What is called animal husbandry ?
Answer:
Management of livestock is called animal husbandry. ”

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 47.
Who are called draught animals ?
Answer:
Which ones are used for farm labour are called draught animals.

Question 48.
Write the names of two exotic or foreign breeds of cow.
Answer:
Jersey, brown swiss.

Question 49.
Write the names of two local breeds of cow.
Answer:
Red sindhi, sahiwal.

Question 50.
Which are called dairy animals ?
Answer:
Milk-producing female animals are called milch animals (dairy animals).

Question 51.
Write two names of external parasites of animals.
Answer:
Mites, lice.

Question 52.
Write the name of internal parasite of animals which effect liver ? –
Answer:
Liver-fluke.

Question 53.
What is called poultry farming ?
Answer:
Poultry farming is called to domesticate chickens, ducks, guinea-fowls and turkeys.

Question 54.
What is called layers ?
Answer:
Egg-producing chickens are called layers.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 55.
What is called broiler ?
Answer:
Broilers are called those chickens which are domesticated only for meat.

Question 56.
What is called fishery ?
Answer:
Domestication of fishes, prawns and molluscs for economic importance is called fishery.

Question 57.
Give example of saline water fishes.
Answer:
Cat fish, sardines and shark.

Question 58.
Write names of non-saline water fishes.
Answer:
Catla, rohu, mullets, millitus, pomphrets, tuna, bombayduck, etc.

Question 59.
Write the names of fishes which are cultured in sea-water.
Answer:
Mullets, bhetki, pearl spots, prawns, mussels, etc.

Question 60.
Which fishes are farmed in composite fish culture ?
Answer:
Catla, rohu, mrigal, common carp, etc.

Question 61.
What is the main problem of composite fish culture ?
Answer:
The major problem in composite fish culture is the lack of availability of good quality seed.

Question 62.
What is called bee-keeping ?
Answer:
Bee-keeping is called to get honey and wax.

Question 63.
Write the Indian variety of bee.
Answer:
Apis cerana indica.

Question 64.
Write two varieties of bee.
Answer:
Apis dorsata, Apis florae.

Question 65.
Write the name of Italian bee.
Answer:
Apis mellifera.

Short-Answer Type Questions

Question 1.
What are the functions of food ?
Answer:
All living organisms need food for their survival. Functions of food are as follows :
(1) Food provides us energy.
(2) Food is helpful in maintaining the damaged cells tissues and orgAnswer:
(3) Food is helpful for the body development as well as in reproduction.
(4) Food generates the disease resistance energy which saves us from the diseases.

Question 2.
Write the names of the various sources of food.
Answer:
Various sources of our food are as follows :
1. Plants: The main part of our food such as foodgrains, pulses, spices, oil seeds, fruits, vegetables, sugar, etc., all we get from plants. Plants are autotrophic as well as they provide food to animals.
2. Animals: Animals are dependent for their food on plants but these animals provide to other animals such as milk, meat, eggs, butter, ghee, etc.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 3.
Write the names of main constituents and sources of food.
Answer:
Following are the constituents and sources of food :

Name of ComponentsSources
CarbohydratesFor example; wheat, maize, millet, rice, sorghum, potato, etc.
ProteinFor example; pulses, lentil, black gram, gram, pigeon pea, green gram, etc.
FatFor example; soyabean, groundnut, sesame, castor, mustard, linseed, sunflower, etc.
Vitamins and mineralsFor example; vegetables, fruits, spices, etc.

Question 4.
What do you mean by green revolution ?
Answer:
To generate agricultural production speedily is called green revolution. Green revolution has been successed by using higher quality seeds, pesticides, weedicides, fertilizers, modem techniques of irrigation, modern agricultural implements and modem agricultural equipments. The modem agricultural technique which is based on scientific researches has been reached to the farmers by intense agricultural technology. Consequently, there is a unique change in the field of agriculture which is known as green revolution.

Question 5.
What is the need of improvement in crop varieties ?
Answer:
Improvement in the crop varieties is needed due to following objectives :
(i) For high yielding.
(ii) For good quality.
(iii) To increase disease resistance efficiency in crops.
(iv) For pre-mature and uniform maturity.
(v) To increase photo and thermo insensitivity. .
(vi) For the improvement of desirable characters.
(vii) To increase the comprehensive adaptability in crops.

Question 6.
Which are the main water-air conditions required for various crops ?
Answer:
Water, soil, air and temperature are required for the exact growth and development of crops, but amongst these, two conditions are those by which growth of the crops, maturity of the crops and their life cycle are attached. These conditions are-temperature and photoperiod. Temperature is related to the maturity of the crops and photoperiod is related to the flowering and growth of the plants. Plants manufacture their food in photoperiod (sunlight) by the process of photosynthesis.

Question 7.
What do you know about kharif and rabi crops ?
Answer:
Following are seasonally based two types of crops :
(1) Kharif crops
(2) Rabi crops.
1. Kharif crops:
These crops are grown in summer season, especially June and July. These crops are required excess heat and water to be grown. In the very beginning of winter season, these are harvested.

Examples:
Maize, paddy, sorghum, millet, black gram, green gram, sesame, bottle gourd, squash gourd, ash gourd, lady finger, etc.

2. Rabi crops:
These crops are sown in winter season. These crops require less heat and water. These are harvested in the beginning of summer season.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Examples:
Wheat, pea, gram, mustard, carrot, radish, turnip, cauliflower, berseem, etc.

Question 8.
How varieties of plants can be improved ?
Answer:
Varities of plants can be improved by the following characteristics:
1. Selection of good quality seeds:
We should select those seeds to be sown which produce high yields and are disease, insect, and parasite resistants. Variety of plants will be improved by doing so continuously.

2. Good quality imported plants to be grown:
If local varieties produce less yield and are not to be improved, then good quality and high yielding plants should be brought from other countries, e.g., Tui Chug Native -1 rice and Sonara – 64 wheat are brought from foreign countries.

3. Hybridisation:
In crossing technique, the cross-pollination is done in two varieties which are having desirable genes. All the useful characteristics are present in both of the varieties in the cross which we receive after crossing. Many of the good varieties of fruits and vegetables have been introduced by this technique. G-57 and G-67 have been introduced are the improved varieties of cotton.

4. Byindured mutations:
It is that technique in which high temperature, special rays or by chemical substances such changes are brought in the genes of plants that there are desired characters in the produced genes.

Question 9.
Mainly in how many parts the agricultural techniques are divided ?
Answer:
Mainly agricultural techniques are divided into three parts :
(i) Selection of seeds.
(ii) Management of crops.
(iii) Protection of grown crops in the fields and to save the harvested crop.

Question 10.
What is the contribution of improved varieties in higher yield ?
Answer:
To increase the productivity of fhe crop per acre’is caiied’hig’her yieid.Tiigner yielding varieties of crops are meant for :
(i) They should be disease resistant.
(ii) They should be favourable to climate, land and fertilizers.
(iii) They should be of high yielding capacity.
(iv) Should be of high quality for management.
On the basis of above facts criteria can be called as higher yields.

Question 11.
What are the benefits of short maturity period of crops ?
Answer:
From the time of sowing of crops upto the time of harvesting or crop production is called maturity duration. From economical point of view, maturity duration is best, because short duration also reduces the cost of crop production and both the time and labour of the farmer are saved and crop production is got early.

Question 12.
How is it useful to develop the varieties of wider adaptability ?
Answer:
There are different environmental conditions are found in our country, i.e., some part is hot and in other part is cold climate. In coastal areas, there is similar climate. Hence, in these different climatic areas, different varieties of crops can be grown. So, now efforts are being made that varieties with wider adaptability may be developed because one kind of variety can be grown in wider areas.
By doing so, it will help in stabilising the crop production. One variety can then be possible to grow under different climatic conditions in different areas.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 13.
What do you know about the required nutrients for the growth and development of green plants ?
Answer:
At least thirteen nutrients are required for the general growth and development of green plants, which are called nutrients. Nutrients on the basis of required quantity of plants are of two types :
1. Macro-nutrients:
Such nutrients which are required by the plants in large quantities are called macro-nutrients, e.g., nitrogen, potassium, calcium, magnesium, phosphorus and sulphur. These are six in number.

2. Micro-nutrients:
Such nutrients which are required by the plants in small quantities are called micro-nutrients, e.g., iron, zinc, copper, boron, molybdenum, chlorine, manganese. These are seven in number. Plants get carbon and oxygen from air and hydrogen and oxygen from water in kind of nutrients. Above given 13 nutrients plants get from soil.

Question 14.
What are called fertilizers ? Mainly how many kinds of fertilizers are there ?
Answer:
To maintain the fertile efficiency of land, some particular substances are to be mixed in it. These particular substances, which maintain the efficiency of land are called fertilizers. There are the two types of fertilizers :
1. Manures:
These are prepared by the remains of animals, their excreta, and other waste material. In manures, green manure, compost manure, and dung manure are the main manures.

2. Fertilizers:
Which manures are prepared artificially, called fertilizers. These are mainly of three types : nitrogen, phosphorus and potassium, fertilizers. These are called chemical fertilizers.

Question 15.
What are called fertilizers and manure ? Write their differences.
Answer:
Fertilizers: Fertilizers are those chemical substances which contain required nutrients for plants. These are mainly inorganic compounds.

Manure: It is a natural substance. These are made from the decomposition of wastes of plants and animals. These are mainly organic substances.
Following are the differences between fertilizers and manures :

FertilizersManures
1. These are prepared by natural substances.1. These are prepared by natural substances.
2. These are enriched with nutrients.2. These are less enriched with nutrients.
3. These are not helpful in composition of the soil.3. These make the soil.
4. These are mainly non-carbonic substances.4. These are mainly carbonic substances.
5. These cover large area, so these are uneasy to be store and transport.5. These cover large area, so these are uneasy to be store and transport
6. These are required in less quantities.6. These are required in large quantities.
7. Manure is prepared by the help of micro-organisms.7. Manure is prepared by the help of micro-organisms.
8. These effect at a time.8. These effect gradually.
9. Fertilizers cause soil pollution.9.Manures do not cause soil pollution.

Question 16.
What do you mean by compost manure ?
Answer:
Compost manure is prepared by waste plants, dead animals and animal excreta. To prepare compost manure, first of all a trench is dug and a layer of the animal and plant remains is spread. This layer is covered with soil layer so that the remains could be saved from air and light and it may remain moisten. A layer of plant waste or mixed refuse is spread over soil. Micro-organisms present in the trench decompose the biotic remains and change them in a useful manure which is called compost manure. Compost prepared by using earthworms is called vermi-compost.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 17.
What is called green manure ?
Answer:
Prior to the sowing of the crop seeds, sun hemp, green gram, sorghum are grown. When these are grown upto 1\(\frac {1}{2}\) – 2 ft., then they are mulched by ploughing them into the soil and field is filled with water, so that the green plants may convert into manure after decomposition which is called green manure, which helps to increase the fertile efficiency after mixing in soil with quantity of nitrogen and phosphorus.

Question 18.
Fertilizers is a necessary evil. How ?
Answer:
Beeding the decreased fertility ofthe land, it has become necessary to use the fertilizers in land because manures are not available in the demanded quantity. Use of fertilizers is also harmful because fertilizers are flowed away with the irrigative water and mix in water resources which cause water pollution. Continuous use of fertilizers in an area decreases the fertility of the soil because the organic matter in the soil is not replenished and the life cycle ofthe micro-organisms is hampered. Yielding of good crop production in shortterm by using fertilizers is . possible, but after a period of time they cause loss to the soil. Hence, fertilizers is a necessary evil.

Question 19.
What are the benefits by utilising manure ?
Answer:
By utilising manure following are the benefits:
(i) These provide nutrients to soil.
(ii) These increase the quantity of organic substance in the soil.
(iii) Micro-organisms in the soil get food from organic manures which are helpful to provide nutrients to the plants.
(iv) Organic fertilizers are helpful to stabilize the moisture content in the soil.
(v) Use of manures is beneficial to the fertility of the land. Land does not become barren.

Question 20.
Selection of fertilizers is done due to the requirement of the land and crop while for manures it is not so, why ?
Answer:
Every fertilizer has any of the macro-nutrient in itself. It is necessary to be noted that which nutrient is lack of in the land. It is only to be known after the testing of soil. Secondly, the requirement for the nutrients is not same for every type of crop. For example, cereal crops, i.e., nitrogen is required in large quantity for wheat while bean-plants i.e., pulses require nitrogen in lesser quantity. Hence, selection of fertilizers is done due to the requirement of the crop and land. Secondly, manure contains nutrients in lesser quantity. It contains biotic matters in large quantity. Use of manure in excess quantity causes no loss while unwanted use of fertilizers cause lose rather than profit.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 21.
Write a brief note on organic farming.
Answer:
Organic farming means that only organic matters are used to get agriculture production. In this farming, fertilizers, herbicides, pesticides, etc. are not used rather maximum of organic manures are used to fulfil the loss of nutrients and to increase the fertility of the land. Blue green algae in preparation of fertilizers, neem leaves in grain storage used turmeric as bio-pesticides. Production can be increased by destroying insects, weeds with the use of mixed cropping, intercropping and crop rotation.

Question 22.
What do you meant by irrigation ? Which methods are adopted in our country for irrigation ?
Answer:
To fulfil the requirement of water in different steps for the growth, development and maturity of the crops is called irrigation.
Resources of irrigation: Canals, wells, tubewells, ponds, rivers, rain, fountains, etc.

Question 23.
Why irrigation is done to crops by us ? Explain.
Answer:
Following are the reasons to irrigate crops or plants :
(i) Plants absorb mineral salts from the land in liquid form which is possible only by irrigation. Various fertilizers are also given in liquid form.
(ii) Plants get two nutrients H and O from water which are essential for the growth and development of the plants.
(iii) Water is available only by irrigation for the growth and development of the plants. Plants completes its metabolic activites through water.
(iv) Plants get lost water due to the activity of vaporisation. To maintain the water balance exactly, plants get water only by irrigation.
(v) In dry soil seeds are not germinated.

Question 24.
What are the resources of irrigation ?
Answer:
Resources of irrigation are as under:
1. Tanks: In rocky areas, rain water is collected in tanks is used for irrigation purposes. Irrigation through tank is an old method.
2. Canals: Canals are started from rivers and irrigate the areas on their way.
3. Wells: Wells are dug in land and water is collected from water bearing strata for irrigation.
4. Tubewells: Tubewells are the best source of irrigation. Irrigation can be done by collecting water-bearing stroata. Now a days, fountains are also modem resources of irrigation. Rains are uncertain. Therefore, requirement of water to crops is done by other irrigation resources. In the past time in Punjab irrigation was being done by dug wells, i.e., with Persian wheel.
5. River lift systems: Lift system is used to regularise the flow of water in canals. Water is directly drawn from the rivers lift system due to inadequate water in the reservoirs to irrigate the fields close to rivers.

Question 25.
What measure is taken to increase the groundwater level ?
Answer:
There is decrease in water level by continuous utilising of ground water. Water is collected by constructing check-dams to increase the water level. After mixing with ground water, this water increases the water level. Water culture is also possible by planting trees on free land.

Question 26.
What is called mixed cropping ?
Answer:
Growing of two or more crops simultaneously on the same piece of land is called mixed cropping, for example :
(i) Wheat + gram
(ii) Wheat + mustard
(iii) Groundnut + sunflower
(iv) Cotton + green gram
(v) Barley + gram.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 27.
What are the advantages of mixed farming ?
Answer:
The advantages of mixed farming are as under :
(i) Production expenses are less in mixed cropping because expenses which are to be made on one crop, yields can be got on the same expenses for more than one crop.
(ii) Fertility of the land is stabilised by mixed cropping because generally pulses crops are grown with cereal crops.
(iii) Farmer’s income increases by getting yield of more than one crop at a time.
(iv) Yields are increased when advantage of crop protection reach to all the matured crops in the field.
(v) Insects in the land can be destroyed by mixed cropping.
(vi) Weeds are also destroyed by mixed cropping.
(vii) There is no risk of failure of the crop by this technique. For example; if one crop is lost due to failure of rains, then we can get the next crop.

Question 28.
What is called intercropping ? Also write the examples.
Answer:
Growing of two or more crops simultaneously on the same field in a definite pattern is called
intercropping. In few rows of one crop alternate with a few rows of a second crop are grown,
for example:
(i) Soyabean + maize
(ii) Finger millet + cowpea.
HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources - 1

Question 29.
Write advantages of intercropping system.
Answer:
Following are the advantages of intercropping system :
(i) Different seeds of different crops are not mixed in this system.
(ii) Required, different ratio of fertilizers can be provided to the crops.
(iii) It is easy to spray required pesticides on the crops.
(iv) Harvesting of both of the crops can be done separately.
(v) Yields of both of the crops can be obtained separately.
(vi) By adopting this system, yield can be increased.
(vii) To sow the crops in different rows is possible.

Question 30.
What is crop rotation ? What are its advantages ?
Answer:
The growing of different crops in a field in a pre-planned succession is known as crop rotation. If in a field, one crop is sown repetitively, then there is lack of some particular nutrients in the land by which fertility of the land decreases. For example; cereal crops require nitrogenous elements in large quantities for their growth.

If cereals will be sown repetitively, then nitrogenous elements will running decreased and the land will become barren. But once if we sow cereal crop and secondly we sow some legume crop, then content of nitrogen in land will increase because there are some glands in the roots of legume crops in which some microbes (to which plant observes) live, which change the free nitrogen of air into nitrates. By this, nitrogen exceeds in the land and land remains fertile.
Following are the advantages of crop rotation:
(i) There is no decrease of particular nutrient in the land.
(ii) No need to provide fertilizers time to time.
(iii) Harmful toxins to crops do not germinate.
(iv) Various types of weeds are destroyed by themselves.
(v) Fertility of the land remains stable.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 31.
What is intercropping ? How is it different from mixed cropping ?
Answer:
Intercropping : Growing of two or more crops simultaneously on the same field in a definite pattern is called intercropping. For example; groundnut and sunflower can be grown by the technique of intercropping.
Following are the differences between mixed cropping and intercropping:

Mixed CroppingIntercropping
1. This decreases the loss of crop in the field.1. This increases the productivity of the field.
2. Use of pesticides is difficult.2. Use of insecticides is easy.
3. In this rows are not in proper succession.3. In this rows are in a proper succession.
4. In this seeds are mixed before sowing.4. In this seeds are not mixed before sowing.
5. In this cultivation and threshing of crops are difficult.5. In this cultivation and threshing of crops are being done easily.
6. Yields are obtained in kind of mixture.6. In this crop yields are gathered separately.
7. In this fertilizers are not required.7. In this fertilizers are used as per requirement.

Question 32.
What losses are caused due to repetitively growing of crops of one family ?
Answer:
Following are the losses due to repetitively growing of crops of one family :
(i) Diseases and insect are increased in the field.
(ii) Various types of weeds are increased in the field.
(iii) Particular type of nutrients are decreased in the field.
(iv) Yield of production is decreased.
(v) Threat of the failure of the crop increases.
(vi) Only one yield can be obtained.

Question 33.
What are called weeds ? By giving examples write some traits to eradicate them.
Answer:
Those plants which we do not grow ourselves, but those which are grown along with crop in the field automatically, are called weeds. They cover unwanted space in the field and compete with crop plants for water, soil nutrients and sunlight. Generally, according to the conditions, any plant may be weed, but with the crops of paddy and wheat, commonly growing weeds are : convolvulus (Hirankhuri), Sorghum (Jangli oat), grass, Amaranthus (Chaulai) and Chenopodium (Bathua) etc. Weeds like Amaranthus (Chaulai) is commonly grown with every crop.
HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources - 2
Traits to eradicate: Following are the traits to eradicate weeds :
(i) If the weeds are less in number, then by hand hoeing or with ‘khurpi’ they are uprooted.
(ii) While ploughing, these too can be uprooted.
(iii) If the weeds are large in number, then weedicides or herbicides are sprayed. These chemicals destroy them without causing any loss to the crop. Some weedicides are : 2, 4 – D, M.P.C.A. and butachlore etc.
(iv) In this system, to destroy the weeds, help of insects or micro-organisms is obtained, e.g., coboneal insect destroys opentia weed. Weeds can also be destoryed by adopting mixed cropping, intercropping and crop rotation.

Question 34.
What is called crop protection ? Why is it required ?
Answer:
To protect the crops from microbe diseases, mould diseases, insects and other pests is called crop protection. If the crop protection is not practised, then a great loss is caused to farm yields, e.g., cereals, fruits, vegetables and other products. Therefore, it is very necessary to provide protection to the crops grown in fields from insect, pests and diseases which cause harm to the crops. By doing not so, producer has to bear a huge loss.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 35.
Why weeds are to be uprooted ?
Answer:
The weeds are uprooted from the crops because weeds or unwanted plants get their food, water and nutrients supplied to plants by which crop yield can be decreased. Therefore, for higher yielding, weeds are removed from crops so as to supply the water and essential nutrients to crops in sufficient amount and to get higher yield. ,

Question 36.
In how many ways insect-pests attack to the plants ?
Answer:
Insect-pests attack to the plants in three ways and cause harm, e.g.,
(i) They cut the root, stem and leaf.
(ii) They suck the cell rap from various parts of the plant.
(iii) They bore into stem and fruits and cause harm to them.

Question 37.
Explain the advantages and disadvantages by using pesticides.
Answer:
Advantages: Growth in agriculture yield has increased by using pesticides.
Disadvantages:
(i) By using these, useful micro-organisms are also killed along with harmful insects.
(ii) The environment is being polluted by these pesticides.
(iii) By using these, content of underground nitrate and phosphate increases and water turns contaminated. This water is unable to drink.

Question 38.
Which cautions should be adopted for storage of grains ?
Answer:
Following cautions should be adopted for the safe storage of grains :
(i) Before storage, the foodgrains should be dried in open sun heat and become sure that the moisture content in the grains is not more than 14%. Dry and cool grains should be filled in treated gunny bags or grain silage.
(ii) After cooling the grains, it should be filled in treated bags or grain pots.
(iii) The structure of the storage pot should be like this that at different times, the grains may be checked and harmful insects and micro-organisms of crop may be destroyed by fumigation.
(iv) It is beneficial that storage should be in moistureless, well-ventilated and neat, clean rooms.
(v) The grains should be stored in metal tanks to protect from rats etc.
HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources - 3

Question 39.
How can the Insects controlled in stored foodgrains ?
Answer:
There are many methods to decrease the number of insects and microbes in stored foodgrains.
Control of insects: The following measures should be adopted to control the insects:
1. Sprays: Before storage the grains in godowns, it is more suitable to spray the godown to make free from insects and pests. Malathion and pyrethrum are common examples of insecticides.
2. Fumigation: The fumigation system is favourable to control the number of insects at the time of storage of grains. This is a volalite chemical which is evaporated with in short time and this destroyed the insects without affecting the grains.
Examples: Ethylene dibromide (E.D.B.) and aluminium phosphide.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 40.
Write the definition of animal husbandry.
Answer:
Animal husbandry : That branch of agriculture science is called animal husbandry in which the study of cattle farming, food, breeding, their shelters and protection from diseases is done. To improve the variety of animals is the main task in animal husbandry. The main objective of animal husbandry is to earn higher income and to make animals more useful to man.

Question 41.
What should be provided to bovine animals in fodder ?
Answer:
In the fodder of bovine animals, green fodder, chaff, dry grass, tiptia fodder, bran, peels of gram and oil-cake etc., should be provided. Mostly bovine animals are vegetarian. To some milk-producing bovine animals, the mixture of binula, methi, brassica chuca, gur. etc. is given.

Question 42.
What is called domestication ? Write any four names of domestic animals.
Answer:
When the necessities of animals, e.g., food, shelter, protection, health, etc., are filled by the man and also living in this environment, these animals make the descendantal growth by breeding is called domestication.

Name of domestic animals:
(i) Cow
(ii) Buffalo
(iii) Horse
(iv) Dog.
HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources - 4

Question 43.
By which remedies production of food from animal sources can be increased ?
Answer:
Especially the food products obtained from animals are milk and meat. Production of these food yields can be increased by breeding. For example; indigenous cow gives approximately 6-8 liters of milk everyday while hybrid variety of cow approximately gives 60 liters of milk everyday. By improvement of the animal breed food yield can be increased. Furthermore, by giving balanced diet and neat and clean shelters to animals is advantageous to increase the yield. Yield decreases due to animals. Therefore, with disease-free animals, yield increases.

Question 44.
What are called lactation period and dry period ?
Answer:
Lactation period: In milk producing animals, the period of milk production (after the birth of a calf) is called lactation period.
Dry Period: In milk producing animals, when they do not produce milk, i.e. upto the time of birth of a calf is called dry period.

Question 45.
What are the characteristics of a good animal shed ?
Answer:
In animal husbandry, animal shed is considered very important. Following characteristics are necessary :
(i) The shed should be spacious and well-ventilated.
(ii) The shed should be moisture-free. This should be clean and lighted.
(iii) It is necessary to manage the disposal of animal waste (urine and dung).
(iv) The shed should be protective from summer, hot, storm, rain, etc.
(v) The shed should be pollution-free.
(vi) Proper water-supply facility should be there.
(vii) It should be safe from the predators in poultry farming.

Question 46.
Which are the indigenous milk producing cows in our country ?
Answer:
Indigenous milk producing cows mainly are of three types :
1. Red Sindhi: The cow is red in colour with dark and light red shades and is of medium in size.
2. Sahivval: This breed is superior in comparison to other milk producing cows. This cow is large in size and heavier.
3. Gir: This breed is the main breed of jungles of Gir in Gujarat. These cows are medium in size and fairly good milk yielders. Daugi, Devni, Tharparkar and Hariana are dual purposes cows which are high milk yielders.
HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources - 5

Question 47.
In our country which are the selected exotic breeds of cows for cross breeding ?
Answer:
In our country, following are the exotic breeds of cows which are used in cross-breeding :
1. Jersey: This cow is the main breed of Island of Jersey (U.S.A.).
2. Holstein Friesia: This breed is mainly of Holland.
3. Brown Swiss: This is the dual purpose breed from Switzerland.
HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources - 6

Question 48.
Which are improved breeds of cows developed by NDRI, Karnal ?
Answer:
In our country, improved breeds of dairy cows have been developed at the National Dairy Research Institute (NDRI), Karnal, Haryana. Some successfid cross-breeds are:
1. Karan Swiss: This breed developed as cross between Brown Swiss and Sahiwal.
2. Karan Fries: This has been developed by a cross between Tharparkar and Holstein-Friesian.
3. Frieswal: This breed has been developed by a cross between Holstein-Friesian and Sahiwal. These new cross breeds give two to three times more milk than our indigenous breeds.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 49.
Write the name of high milk yielder breeds of buffalo. Write her milk yielding capacity.
Answer:
Variety ofbuffalo with high yield ofmilk is Murrah which produces milk yield 2500 litres in comparison to indigenous breeds of 1800 litres in her each lactation period. Furthermore, high milk yield varieties are : Jafarabadi, Nagpuri, Surti.
HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources - 7

Question 50.
What is called roughage ? How animals obtain it ?
Answer:
Commonly animal food contains : fibrous, coarse and low nutrient context grass which is called roughage. Animals obtain roughage in kind of green grass, berseem, rijka and hay.

Question 51.
How does animal feeding affect the milk production in dairy animals ?
Answer:
In dairy animals, animal feeding affects the milk production and opposite to it, their feeding depends upon the quantity of milk yield. More nutritious feed is given to high yield animal. Milk yield will decrease if the animal is given low nutritious feed. Milk yielding animals are given fibrous, wheat bran, rice bran, low nutritious grass and green fodder, in kind of concentrates, granular cereals, pulses, gram, finger millet, sesame, oil seed cakes, cotton seeds are given. Milk production can be increased approximately three times by giving nutritious feeding to milk producing animals.

Question 52.
Differentiate between roughage and concentrates in relation to feeding of animals.
Answer: Just like persons of various factors such as age, health and nature of work require different types of foods (from view point of nutrition) in the same way animals also require different types of nutritious feeding for their growth etc. In common feeding of animals, there is fibrous, granular, grass of low nutrients, etc., which is called roughage and having one or more nutrients is called concentrated particle. Examples of concentrated particle feeding : cotton seeds, sesame, oil seed cakes and gram etc.

Question 53.
Write some preventive measures for animal diseases.
Answer:
Preventive measures for animal diseases are as under:
(i) Animals should be kept at neat and clean space.
(ii) Compulsory vaccination may be given.
(iii) Provide balanced feeding to animals.
(iv) Space should be provided for proper cleaning of animal body.
(v) It is compulsory to check the animals properly for diseases.
(vi) Dead animals and animal excreta (urine and dung) should be disposed properly.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 54.
Write the symptoms of sick animal.
Answer:
(i) Animal limps or moves slowly.
(ii) Animal becomes less active and looks tired.
(iii) Skin becomes dry and there is no shining on it.
(iv) He looses appetite (stops eating).
(v) His ears look withered.
(vi) Feels difficulty to respire.
(vii) Animal passes loose or extremely fowled dung.
(viii) Nature of animal becomes irritated.

Question 55.
Write the names of any two parasitic and infectious animal diseases.
Answer:
1. Parasitic diseases: Diseases caused by parasites are called parasitic diseases, e.g., cholera (in hens), diarrhoea (in chickens), anthrax (animals and birds), rinderpest (in animals).

2. Infectious Diseases: Diseases caused by pathogens are called infectious diseases or communicable diseases, e.g., cowpox (in cow and buffalo), salmonellosis (sheep-goat), foot and mouth disease (in bovine animals).

Question 56.
Why poultry farming is done ?
Answer:
Mostly poultry farming is done to produce layers for eggs and broilers for meat. The feathers of the chickens are also useful. Weight of cock approximately is 4 to 5 kg and of hen is 3 to 4 kg.

Question 57.
Why there is difference in the diet of broilers and layers ?
Answer:
Layers are farmed for eggs. Therefore, in the diet of layers, grains of wheat, and dry fish in large quantity are given. While broilers are farmed for meat. In the diet of broilers, protein, fat, vitamins A and K are given in large quantity.

Question 58.
Explain the exotic breeds of chicken which are used in our country.
Answer:
Following are the exotic breeds which are used in our country :
1. White leghorn: It is the popular breed which produces long white eggs. It is most popular breed because it has small body size and therefore, needs less feed for its maintenance.
2. Rhode Island Red: This breed has been developed on the farm of Rhode Island in U.S.A. It is a dual type of breed. It is a fairly good egg layer and also meat yielding (broiler).
HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources - 8

Question 59.
Write the names of indigenous, exotic and cross-breeds of chickens.
Answer:
Following are the explanation of indigenous, exotic and cross-breeds of chickens :
1. Indigenous breeds: Aseel, Busra, Chitagang, Dhagus.
2. Exotic breeds: White leghorn, Rhode Island, Plymouth Rock, Australope.
3. Cross-breeds: HH-260, IBL-80, B-77.
HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources - 9

Question 60.
What preventive measures should be adopted in the management of chicks ?
Answer:
Following are the preventive measures in the management of chicks :
(i) Spacious land is required for chicks. With overcrowding their growth decreases.
(ii) Layers require enough space and light. Light intensity and its duration has positive effect on the egg laying output of the hens.
(iii) Proper feed should be provided to hens. The chick feed should be highly enriched with vitamins A and K. Protein and fat should be in the feed of broiler.
(iv) Chickens require their regular cleanliness.
(v) Vaccination should be done well in time to prevent communicative diseases.

Question 61.
How useful are the new breeds of chicks in comparison to indigenous breeds ? Explain with example.
Answer:
New breeds of chick require less feed and meat yielding in high quantity e.g., the new breed requires only 2 kg feed to lay 12 eggs, while indigenous breed requires 6 kg feed for this.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 62.
Where fishes are found ?
Answer:
Fishes are found in saline and non-saline water resources. Non-saline water resources are ponds lagoons and rivers, while saline water is of ocean. People of coastal areas use saline water fish and other aquatic foods. People of Land area use non-saline water fish in their food. Catla, rohu, mullet, etc. are non-saline water fish and teena, shark and cod are sea water fish. In our country 16 lakh hectare inland space and 7500 km long coast line is sufficient for fish yielding.

Question 63.
How do we take fish as food ?
Answer:
Fish dried in fresh form, making pickies and in tin packing, position are used. Marine fishes are captured in large quantity. Therefore, they are preserved by drying, by making pickles or by other techniques.

Question 64.
What is called fisheries ?
Answer:
Domestication of fishes on a large scale and their management is called fisheries. To get eggs of fish, from chatching eggs to obtain young ones (seeds), domestication of fish in proper number in a proper water resource (living), to provide proper feeding, in tanks with sutficient oxygen and light, proper number of fish seeds are to be domesticated and to obtain the yield from them are included in fisheries.

Question 65.
What is called composite fish culture ? What is the cause for no competition in food in this culture ?
Answer:
In a paddy field or in a single fishpond, a combination of five or six fish species are domesticated. It is called composite fish culture. In this culture, these species do not compete for food because catla is a surface feeder, rohu is a column feeder, mrigal and carp are bottom feeders, while grass corp feed on aquatic weeds. In this way, there is no competition in food and in this culture, fish yielding highly increases.

Question 66.
What is the one problem in composite fish culture ?
Answer:
One problem with such composite fish culture is that many of these fish, breed only during monsoon, because of this, good quality seeds are not available. Therefore, fish culture is impossible.

Question 67.
Write a note on bee-keeping.
Answer:
The practice of rearing honeybees is also called silent rearing. Honeybees are reared in artificial wooden hives to get honey. For reared honeybees, box type beehives are prepared, in which a number of sliding vertical frames are fixed in the honey chambers made of thin wax layer. Reared honeybees make the honey from nectar of flowers and fill in these chambers. Now this honey is abstracted with the help of a machine.
HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources - 10

Question 68.
What are the uses of honeybees ?
Answer:
Following are the uses of honeybees:
(i) Honey is obtained from bees. Honey is a nutritious food and is antibiotic substance.
(ii) Wax is also obtained from bees. Wax is also used in preparation of medicines.
(iii) Honeybees also help in cross-pollination by which yield of plants is also increased. Yielding of flowers increases in particular gardens.
(iv) Bee-keeping is an additional income generating activity of farmers.

Question 69.
Which varieties of honeybee are reared on commercial basis ?
Answer:
Following are the varieties of reared honeybees:
(i) Apis cerana indica (Indian bee)
(ii) Apis dorsata (the rock-bee)
(iii) Apis florea (little bpe)
(iv) Apis mellifera (Italian bee)
All the above facts varieties are reared for the yield of honey and wax. Honeybees also increase the yield of all fields and gardens. Their rearing place is called honey-garden or honeybee-farm.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Essay Type Questions

Question 1.
Which irrigation systems are being adopted in India ?
Answer:
Following are the irrigation systems which are adopted in India :
1. Canal System: In canal system, canal receive water reservoirs or rivers. The main canal is divided into branch canals. From these distributaries carry water to the fields. Irrigation by canals, makes the sufficient supply to the fields.

2. Tank System: Small storage reservoirs are called tanks. Water of catchment areas is collected in tanks. Overflow of water is regulated as per its availability in the tanks. Irrigation by tanks is done only in South India.

3. Wells: In those areas where groundwater is available, wells are constructed there. There are two types of wells : dug wells and tube wells. In a dug well, water is utilised from water bearing strata. Water for irrigation from these wells is lifted by mechanical devices such as by Persian wheel.

4. Tubewells: Water is lifted by diesel or electrical pumps from deeper strata. Irrigation by tubewells is a perfect irrigation system.

5. River valley system: Rain water by rivers is used for irrigation in Karnataka and Western ghats in Kerala. In these areas, steep and narrow riverine valleys make available water to the perennial crops such as coconut, rubber, coffee, tapoca, etc. A single crop of rice is grown on the bottom land.

6. River lift system: River lift system is used to make availability of water in canals due to insufficient reservoir release. By lift system on rivers, water is being carried to the canals.

Question 2.
What is need of improvement to bring in crop varieties ? Explain the objectives of improvement in crop varieties.
Answer:
Need of improvement in crop varieties : Ever increasing human and animal population has put more demands for foodgrains, fodder crops, fibre, oil and sugar etc., but due to limited land resources, this increasing need is only possible by increasing the productivity of crops. Therefore, this increasing demand is fulfilled by crop improvement, i.e. variety improvement.

Objectives of improvement in crop varieties:
(i) Variety improvement is required to get high yielding.
(ii) It is required to get high quality yielding, such as content of protein in pulses, quality of preservation in fruits and vegetables, quality of oil in oil seeds, etc.
(iii) Variety improvement is required to improve the disease resistance capacity.
(iv) Variety improvement is required for early and uniform maturity and to develop characters in long duration crops.
(v) The character of photo and thermo-insensitivity in crops can be developed by crop variety improvement. Development of crop varieties insensitive to these factors help in cultivation of these crops in wider areas.
(vi) Desirable agronomic traits, such as: height in fodder crops, character of excess germination, dwarf character in some paddy crops, character of sweetness in fruits, etc., can be developed by variety improvement.
(vii) By developing varieties of wider adaptability, in different agroclimatic conditions, stability can be brought in crop production.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 3.
What is manure ? How many types are of manure ? Explain in detail.
Answer:
Some particular substances are to be added in land to maintain the continuous fertility of the land. To these particular substances which complete the lack of nutrients in land is called manure. Mainly, following are the two types of manure :
(1) Organic manure
(2) Inorganic manure

1. Organic manure: This manure is prepared by the remains of living organisms and plant waste, waste material like livestock excreta (urine and dung), etc. This manure is of three kinds :
(i) Dung manure: This manure is prepared by the decomposition of animal dung. This manure contains sufficient quantity of nutrients.
(ii) Compost manure: This manure is prepared by plant waste, straw, domestic waste, eraticated articles, etc., by decomposing them in pits.
(iii) Green manure: Some small plants growing in the Field are mulched by ploughing into the soil and after their decomposition which manure is prepared is called green manure.
2. Inorganic manure: These manures are prepared artificially. These are also called fertilizers. These are also of three types :
(!) Nitrogen fertilizers: These are those fertilizers which provide nitrogenous elements to the soil such as urea, ammonia, sulphate, etc.
(if) Phosphate fertilizers: These are those fertilizers which provide phosphorus elements to the soil such as super phosphate; etc.
(Hi) Potassium fertilizers: These are those fertilizers which provide potassium elements to the soil such as potassium sulphate, potassium chlorate, etc.

Question 4.
What do you mean by green revolution ? Write its advantages.
Answer:
Highest yielding of food production in between 1960 and 1980 in our country is known as green revolution. In this session, due to higher crop yield and food production, India became self dependent in the matter of foodstuffs and achieved success in making sufficient safe storage of foodgrains (buffer stock). Therefore, this session is also known as golden time. Green revolution begins with the development of high yield variety (HYV) of wheat in which a new high yield variety was developed by the breed of traditional and maxican variety.
Advantages of Green Revolution:
(i) Our country became self dependent in food production.
(ii) Collected a sufficient safe storage (buffer stock) of foodgrains.
(iii) Easily faced the situations such as draught with sufficient safe storage.
(iv) Green revolution changed the living standard of farmers as their income increased.

Padam Vibhushan Prof. M.S. Swaminathan is considered as the creator of green revolution in India, by this reason Dr. Swaminathan was awarded as the father of Economic Ecology in United Nations Environment Programme. After green revolution, now the second green revolution is expected.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 5.
Write a note on poultry farming.
Answer:
In this farming chickens, ducks, guinea-fowls and turkeys are included but mostly chickens are domesticated. Its both indigenous and exotic varieties are available in India.
1. Indigenous breeds: Aseel and Balsara are the two varieties of our indigenous hens. Following are their characteristics :
(i) These are small in size
(ii) Their growth is less
(iii) These lays small eggs.
(iv) These are healthy,
(v) They are used for meat purposes
(vi) Their natural resistance capacity is high.

2. Exotic breeds:
White leghorn and Rhode Island Red are the main breeds. Following are their characteristics :
(i) They are large in size
(ii) Their growth is early
(iii) They lay eggs in large quantity
(iv) Size of their eggs is large
(v) Disease resistance capacity is less.

3. New breeds: These are obtained by the cross of indigenous and exotic breeds. Main breeds are IL S-82, B-77. Following are their characteristics:
(f) They have both the exotic and indigenous breed characters
(ii) Their production capacity is approx. 200 eggs per bird per year
(iii) They require less feeding
(iv) Meat yield is excess in them.

Feed (ration): Their feed contains, grains, mashed cereals, rice bran and water etc. Due to little belly they take feed many a times in a day. Eggs, meat, manure and wealth are earned by poultry farming.

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

Question 6.
Write a brief note on fish farming.
Answer:
Fish farming:
Fish is an aquatic animal and a rich source of protein. Their production comes under underwater agriculture. In our country, 16 lakh hectare inland water area and 7500 km long coastline is sufficient for fish yielding.

On the basis of living place, fish are of following types :
(i) Non-saline or fresh water fish.
(ii) Saline or marine fish.
(iii) Brackish water fish.
1. Non-saline or freshwater fish: Non-saline fish are found in ponds, rivers, lakes and waterfalls. In these main fish are : catla, rohu, mrigal, silver carp and grass carp etc. In catla, growth is very fast.
2. Marine fish: Fishes are found in large number in sea and saline water and these fish are considered main high valued food. In marine fishes main are: ribon fish, cate fish, sardine, shark, hilsa, fladage fish, etc.
3. Brackish fish: These are found in water collected by waves. Pearlsthot and mullet are its examples. Enemies of fish : Water beetles, water-bugs, frogs, snakes and birds, etc., are the enemies of small and
large fish.
Diseases of fish: Main factors of diseases in fish are parasites and pathogens. IPN (Infectious Pancreatic Necrosis) and VHS (Viral Haemorrhagic Septicemia) are main viral diseases. Huge loss is caused to fish yielding due to water pollution. Small and large scale fish-farms are available to fish farmers throughout the country to increase the fish yield.

Practical Work

Experiment 1.
To make a herbarium of cereals, pulses, oil seeds and identify the seasons of their sowing and harvesting.

Work-Method:
With the help of your parents make a herbarium of different seeds and collect the information from the village farmer about the sowing and harvesting of these seeds.

Name of SeedTime of SowingTime of Harvesting
PaddyJune-JulyOctober
MaizeJune-JulyOctober
Green gramJune-JulyOctober
SoyabeanJune-JulyOctober
WheatOctober-NovemberApril-May
GramOctober-NovemberMarch-April
SarsonOctober-NovemberMarch-April

Quick Review of the Chapter

1. We get food from:
(A) animals
(B) only plants
(C) plants and animals both
(D) grainaries
Answer:
(C) plants and animals both

2. For crops the number of essential nutrients is:
(A) 6
(B) 7
(C) 8
(D) 13
Answer:
(D) 13

3. The number of macro-nutrients is:
(A) 6
(B) 7
(C) 13
(D) 3
Answer:
(A) 6

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

4. The number of micro-nutrients is:
(A) 6
(B) 7
(C) 13
(D) 3
Answer:
(B) 7

5. With decomposition of plants the prepared manure is :
(A) compost fertiliser
(B) vermi-compost
(C) green manure
(D) chemical fertiliser
Answer:
(C) green manure

6. The honey bee prepares honey from:
(A) fruits
(B) nectar
(C) pollens
(D) nectar and pollens both
Answer:
(D) nectar and pollens both

7. Which of following is the dwarf animal ?
(A) goat
(B) sheep
(C) camel
(D) lion
Answer:
(C) camel

8. Sudan is a crop of:
(A) paddy
(B) fodder (sorghum)
(C) vegetable
(D) condiment
Answer:
(B) fodder (sorghum)

9. The crop of kharif is :
(A) soyabean
(B) linseed
(C) mustard
(D) peas
Answer:
(A) soyabean

’10. Which is not a rabi crop ?
(A) linseed
(B) mustard
(C) peas
(D) black gram
Answer:
(D) black gram

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

11. The local breed of broiler is:
(A) frog
(B) tiger
(C) grasshopper
(D) shark
Answer:
(A) frog

12. Italian honey bee is:
(A) Apis serena
(B) Apis dorsetta
(C) Apis malifera
(D) Apis flory
Answer:
(C) Apis malifera

13. Non-saline (freshwater) fish is:
(A) prawn
(B) tuna
(C) mussel
(D) pearl-spot
Answer:
(B) tuna

14. Cause of soil pollution is:
(A) composite fertiliser
(B) dung manure
(C) inorganic fertiliser
(D) vermi-compost
Answer:
(C) inorganic fertiliser

15. In carbonic agriculture which is used is:
(A) organic fertilisers
(B) chemical fertilisers
(C) pesticides
(D) weedicides
Answer:
(A) organic fertilisers

16. The selected exotic breeds of cows for sankar reproduction in our country are
(A) Sahiwal
(B) Karan Fries
(C) Frieswal
(D) Brown Swiss
Answer:
(D) Brown Swiss

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

17. Which is not a breed of buffalo?
(A) Gir
(B) Zafarabadi
(C) Nagpuri
(D) Surti
Answer:
(A) Gir

18. Average weight of a cock used to be between 4-5 kg, whereas the weight ola hen is:
(A) 2-3kg
(B) 3-4kg
(C) 5 – 6 kg
(D) none of these
Answer:
(B) 3 – 4 kg

19. Which is not an exotic breed of cock (rooster)?
(A) IBL-80
(B) Astralop
(C) White leghorn
(D) Plymouth rock
Answer:
(A)IBL-80

20. Fish, surviving on water weed ¡s:
(A) Grasscorp
(B) Corp
(C) Catla
(D) Rohu
Answer:
(A) Grasscorp

21. The superior egg laying breed is:
(A) HH-260
(B) B-77
(C) ILS – 82
(D) White leghorn
Answer:
(A) HR – 260

22. Murrah is the superior breed of:
(A) cow
(B) buffalo
(C) sheep
(D) goat
Answer:
(B) buffalo

23. In Vhich of the following carbon is present? [H.B.
(A) Protein
(B) Carbohydrate
(C) Lipid
(D) All of these
Answer:
(D) All of these

24. The present population of India is:
(A) less than one billion
(B) more than 121 crores
(C) one billion
(D) all of the above
Answer:
(B) more than 121 crores

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

25. The reason of crop production is:
(A) green revolution
(B) white revolution
(C) blue revolution
(D) mass revolution
Answer:
(A) green revolution

26. Substance(s) which provide fodder is/are:
(A) berseem
(B) oats
(C) maize
(D) all of the above
Answer:
(D) all of the above

27. On which factor are flowering and development dependent in plants?
(A) irrigation
(B) seeds
(C) fertilisers
(D) sunlight
Answer:
(D) sunlight

28. In which season are the crops of kharif grown?
(A) in rainy season
(B) in summer season
(C) in spring season
(D) in winter season
Answer:
(A) in rainy season

29. Photoperiod related to:
(A) light of moon
(B) electric light
(C) light of glow-worm
(D) sunlight
Answer:
(D) sunlight

30. Plants get mineral elements from :
(A) air
(B) water
(C) soil
(D) all of the above
Answer:
(C) soil

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

31. Nutritious chemicals produced by artificial method for plants are called :
(A) manure
(B) fertiliser
(C) compost manure
(D) none of the above
Answer:
(B) fertiliser

32. Dwarf breed of plants is good for:
(A) sugarcane
(B) pulses
(C) fodder
(D) cereals
Answer:
(D) cereals

33. How many nutrients are provided to plants by air and water?
(A) 6
(B) 7
(C) 13
(D) 3
Answer:
(D) 3

34. Plants get from water:
(A) nitrogen and oxygen
(B) oxygen and hydrogen
(C) carbon and oxygen
(D) all of the above
Answer:
(B) oxygen and hydrogen

35. Growing two or more crops in definite row patterns is known as .
(A) high cropping
(B) intercropping
(C) crop rotation
(D) hybridisation
Answer:
(B) intercropping

36. Example of intercropping is:
(A) soyabean and maize
(B) wheat and gram
(C) groundnut and sunflower
(D) wheat and potato
Answer:
(A) soyabean and maize

37. Sowing of crops in the field alternatively, is called:
(A) intercropping
(B) mixed cropping
(C) crop rotation
(D) hybridisation
Answer:
(C) crop rotation

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

38. Main weed is:
(A) gokhroo (xanthium)
(B) gajar ghas (parthenium)
(C) motha (cyperinus rotundus)
(D) all of the above
Answer:
(D) all of the above

39. Milk giving animals are called :
(A) draught animals
(B) dairy animals
(C) general animals
(D) all of the above
Answer:
(B) dairy animals

40. The external parasite of animals is:
(A) sparrow
(B) crow
(C) vulture
(D) louse
Answer:
(D) louse

41. To get wax and honey, there is:
(A) poultry farming
(B) bee keeping
(C) animal husbandry
(D) fish farming
Answer:
(B) bee keeping

42. Farming of hen, duck and swan, is called:
(A) poultry farming
(B) bee keeping
(C) animal husbandry
(D) fish farming
Answer:
(A) poultry farming

43. Farming of fish, prawns and molluscs for economic importance, is called:
(A) poultry farming
(B) bee keeping
(C) animal farming
(D) fish farming
Answer:
(D) tish farming

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources

44. Which of the following is not a method of crop improvement?
(A) Improvement in variety of crops
(B) Management of crop produce
(C) Management of crop protection
(D) Excess use of fertilizer
Answer:
(D) Excess use of fertilizer

45. Which of the following is a shell fish ?
(A) Prawn
(B) Mullet
(C) Bhetki
(D) Pearispot
Answer:
(A) Prawn

46. Which of these is enriched in protein:
(A) Black gram
(B) Gram
(C) Peas
(D) All of the above
Answer:
(D) All of the above

HBSE 9th Class Science Important Questions Chapter 15 Improvement in Food Resources Read More »

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources

Haryana State Board HBSE 9th Class Science Important Questions Chapter 14 Natural Resources Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 14 Natural Resources

Very Short-Answer Type Questions

Question 1.
What are natural resources?
Answer:
Useful substances available in nature are called natural resources.

Question 2.
Give examples of natural resources.
Answer:
Air, water, soil, minerals, forests, etc.

Question 3.
What are essential factors for life?
Answer:
Ambient temperature, water, and food.

Question 4.
What is called lithosphere ?
Answer:
The outermost layer of earth is called lithosphere.

Question 5.
How much area of earth is covered with water ?
Answer:
75%.

Question 6.
What is called hydrosphere?
Answer:
The part of earth where water is available is called hydrosphere.

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources

Question 7.
Write the definition of atmosphere.
Answer:
Earth is covered with an envelope of air, this envelope of air is called atmosphere.

Question 8.
What is called biosphere ?
Answer:
Integeration of atmosphere, lithosphere and hydrosphere which makes the life possible, is called biosphere.

Question 9.
Name the inorganic components of biosphere.
Answer:
Air, water, soil.

Question 10.
What is called air ?
Answer:
The mixture of various gases, water vapours, dust particles is called air.

Question 11.
How much percentage of C02 is found on Venus and Mars ?
Answer:
95-97%.

Question 12.
is oxygen is available at Venus and Mars ?
Answer:
No.

Question 13.
How much percentage of C02 is in earth atmosphere ?
Answer:
0.033%.

Question 14.
Is atmosphere there on the surface of moon ?
Answer:
No.

Question 15.
What are called winds ?
Answer:
Blowing air is called wind or wave.

Question 16.
What is called air pollution ?
Answer:
Addition of undesirable substances in air is called air pollution.

Question 17.
Write the names of waste air pollutants.
Answer:
Waste gases (CO2, nitrogen and oxides of sulphur, hydrocarbon, etc.) water vapours, dust particles, etc.

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources

Question 18.
What is called fog ?
Answer:
In winter season, when water is also condensed with air, then it is called fog.

Question 19.
Write the names of the diseases caused by air pollution.
Answer:
Respiratory diseases, cancer, heart diseases, allergy.

Question 20.
Mainly how many kinds of water are ?
Answer:
Two, saline and fresh water (non-saline).

Question 21.
What is called the outermost layer of earth ?
Answer:
Earth-crust.

Question 22.
How is soil formed ?
Answer:
By the breaking of rocks.

Question 23.
What is called humus ?
Answer:
Substance prepared after the decomposition of dead decayed animals is called humus.

Question 24.
What is called soil pollution ?
Answer:
Addition of undesirable substances in soil is called soil pollution.

Question 25.
Who is most helpful in preventing soil erosion ?
Answer:
Trees and plants.

Question 26.
How is underground water obtained ?
Answer:
By digging wells.

Question 27.
How much percentage of nitrogen gas is in the earth atmosphere ?
Answer:
78%

Question 28.
Mainly which gas is responsible for the greenhouse effect ?
Answer:
CO2.

Question 29.
What is the main reason of depletion of Ozone Layer?
Answer:
Chloro Fluoro Carbon (CFC)

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources

Short-Answer Type Questions

Question 1.
What are called natural resources ?
Answer:
Many useful substances are found in nature which are called natural resources, e.g., air, water, soil, forests, minerals and living organisms.

Question 2.
What is called atmosphere ? What is the importance of atmosphere for organisms ?
Answer:
An unseen envelope is there around the earth which is called atmosphere. Atmosphere is extended upto 40 km height. 99% of the total air is found in this region. A little volume of air is found at the height of 1000 km from the earth. Water vapours and dust particles are found only near the surface of earth. Above the height of 10-12 km water vapours are not found. Oxygen, the breath of life is present in atmosphere. Ozone layer protects from ultraviolet radiations of the sun.

Question 3.
Write two uses of oxygen present in atmosphere.
Answer:
Following are the uses of oxygen:
(i) Oxygen is used to get energy by breaking the glucose molecules.
(ii) Oxygen is used in combustion and carbon dioxide is released.

Question 4.
Write the names of methods for the fixation of carbon dioxide.
Answer:
Following are the two methods for the fixation of carbon dioxide:
(i) Formation of glucose by green plants in photosynthesis.
(ii) To form the shells from dissolved carbonates in the water by aquatic animals.

Question 5.
On which factor blowing of winds is dependent in coastal areas ?
Answer:
We know that land gets heated much factor and cool down than water. During daytime the air above the land is heated in sunlight and rises up and there creates a low pressure. The air over sea being at a higher pressure moves to area of low pressure. Thus, during daytime winds blow from the sea to land while during night position becomes reverse.

Question 6.
Besides unequal heating of the winds which are other factors which influence these winds ?
Answer:
Unequally heated winds blow on the earth to unequal heating of the atmosphere, but following are the two more factors :
(i) The rotation of earth around the sun and around its own axis.
(ii) Location of mountains in the path of these winds.

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources

Question 7.
How does rain occur ? What is called hailing ?
Answer:
During daytime, water of different water-bodies and flora and fauna evaporates and goes into air. Heated air carry water vapours with it where due to cooling of air, water vapours condense and become droplets. These droplets become bigger and heavier. These droplets come on to the earth in the form of rain. Sometimes, droplets by cooling down become solid ice which are called hail and the rain of hail is called hailing.

Question 8.
What is called air pollution ? Describe its factors.
Answer:
Air Pollution:
Addition of undesirable particles in air makes it harmful to organisms is called air pollution. Following are the natural factors of air pollution:
(i) Gases and toxic particles released oy tne volcano activities.
(ii) Storms.
(iii) Rising of smoke and gases from the forest fires.
(iv) Releasing of methane gas from marshy lands.
(v) Decomposition of plants and animals and pollen grains etc.

Following are the artificial factors of air pollution:
(i) Smoke of vehicles.
(ii) Burning of fuels.
(iii) Industrial smokes and vaporised substances.
(iv) Metallurgical activities
(v) Extraction of minerals
(vi) Pesticides
(vii) Released radio active radiations at the time of originating atomic energy.

Question 9.
What is smog ? Explain its ill effects.
Answer:
Smog:
Dust and smoke particles on combining with water vapours from smog. Toxic gases from houses, industries and toxic gases with the burning of fuels of vehicles are also added in it. Smoke and dust + Fog and toxic gases → Smog Poisonous gases such as sulphur dioxide (S02) or peroxiacetil nitrate (PAN) are added in smog. The smog does not rise high in the atmosphere but on the ocean it makes a form of cloud. It forms commonly in winter season.

Ill Effects:
Following are the ill effects of smog:
(i) It hampers in transportation because long distance things are not visible.
(ii) Creates burning sensation in eye, nose and throat.
(iii) It is harmful to all these organisms, in whom it can easily enter by respiration. Increases the incidence of asthma and heart disorders.
(iv) It affects the growth and development of plants.

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources

Question 10.
What is called acid rain ?
Answer:
Acidic gases in the atmosphere, such as, sulphur dioxide, nitrogen dioxide, hydrogen chloride, etc. adding with droplets made acids, rain of these acids is called acid rain. Acid rain is harmful to monuments, rocks, as well as to fauna and flora.

Question 11.
What is water pollution ? Give reasons of water pollution.
Answer:
Water pollution:
Addition of undesirable substances in water is called water pollution. Following are the reasons of water pollution :
(i) Release of industrial waste and chemical substances etc.
(ii) Agriculture waste such as fertilizers, pesticides, insecticides, weedicides, etc.
(iii) Domestic waste and sewage excreta, such as, garbage, rubbish, polythene, etc.
(iv) Thermal waste such as release of hot water from thermal plants.
(v) Radioscopy wastes such as waste of radioactive substances.
(vi) Social-religious wastes such as festival wastes, by throwing dead bodies of men and animals in rivers.
(vii) The ejected waste, excreta, dead animals from natural resources such as mining, storms, wind stroms due to such type of disasters.

Question 12.
Soil is a natural resources how ?
Answer:
Soil is the uppermost layer of the earth’s crust which is formed by the breaking up of the rocks. Soil is an important natural resources which is essential for the development of life. Our prime requirements of life such as food, clothing and shelter all are obtained from plants that grow in soil or from animals which are dependent upon these plants. Therefore, soil is a precious natural resources for us.

Question 13.
Write the usefulness of humus.
Answer:
Following is the usefulness of humus:
(i) Mixing with humus soil becomes porous which helps air and water to enter in the soil.
(ii) Required nutrients for plants are found in humus.
(iii) Several organisms get nutrients from humus such as earthworm.
(iv) By mixing humus in sandy soil, water bearing capacity of soil increases.
(v) By mixing humus in sandy soil, soil erosion decreases.

Question 14.
Write four common sources of soil pollution.
Answer:
Following are the four common sources of soil pollution:
(i) Excess use of fertilizers in soil.
(ii) Excess spray of insecticides and pesticides on the soil to protect crops.
(iii) To mix the kitchen waste and other garbage of house.
(iv) Mixing of industrial undesired wastes in soil.

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources

Question 15.
What do you know about water cycle ?
Answer:
Ponds, lakes, rivers and oceans are the different water bodies of water on land. Due to water cycle, water of these water bodies evaporates and make water vapours. These water vapours make clouds. On cooling, condensed clouds make droplets. In the form of rain water falls down again on earth surface. This cycling of water is called ‘water cycle’.

Question 16.
In which substances nitrogen is found ?
Answer:
Nitrogen is mainly found in protein, nucleic acid, DNA, RNA, Vitamins, etc. Vitamins are also found in alkaloids and urea.

Question 17.
How does nitrogen deficiency create in soil ? What measure should be adopted for this kind of soil?
Answer:
Content of nitrogen decreases when the soil is utilized again and again to obtain crop yield. Nitrogen enriched fertilizers are mixed in soil to eradicate this deficiency. Most prevailing fertilizer is ammonium nitrate which is prepared from ammonia gas and nitric acid.

Question 18.
How do green plants prepare starch ?
Answer:
Plants get carbon from the atmosphere in the form of carbon dioxide. By getting energy from the sunlight, plants combine this carbon
HBSE 9th Class Science Important Questions Chapter 14 Natural Resources - 1
These are the carbohydrates and fats which provide energy to the living organisms by which all activities of living beings are completed.

Question 19.
Write two benefits of greenhouse effect.
Answer:
The two benefits of greenhouse effect are following:
(i) The temperature of atmosphere increases by absorption of infrared radiations by greenhouse effect and the life remain smooth in cold climate also.
(ii) The snow covered mountains melt due to greenhouse effect and there is water in rivers throughout the year by this reason.

Question 20.
In which form, the oxygen is found on earth ?
Answer:
21 % oxygen is found in our environment. The oxygen is present in the compound form in carbon dioxide. The oxygen is also found in oxides of metals and silicons, carbonates, sulphates, nitrate and in other minerals. The oxygen is also present in biomolecules like carbohydrates, proteins, nucleic acid and fats.

Question 21.
What will be happen if the ozone layer disappear from atmosphere ?
Answer:
At a height of 15 km above the earth’s surface ozone (O3) layer exists. It grows thicker with the increase in height. It is thickest at the height of 23 km. This layer absorbs the ultraviolet radiation that is present in the sunlight. This ultraviolet light develops skin cancer in the living beings. Therefore, if ozone layer gets disappeared from the atmosphere, the protection envelop will come to an end. The harmful ultraviolet radiation after, reaching to us will produce skin cancer to us and other organisms. Thus, the stratosphere will run with a huge loss. Along with, ultraviolet radiation will become the cause of diseases like glucoma and heat immolation. Also, the division of cells in plants will stop.

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources

Essay Type Questions

Question 1.
How does water-cycle In nature accomplish?
Answer:
HBSE 9th Class Science Important Questions Chapter 14 Natural Resources - 2
Water is continuously flow in environment. This is called hydrological cycle. In water, hydrogen is the basic element that circulates in the form of water compound water constantly. Evaporation from water bodies (ocean, sea, lakes and rivers) and from the skin of the living organisms and it goes on mingling into the air. Alongwith the water absorbed by the plants from the underground too transpires through stomata of leaves and gathers into the atmosphere. Thus, by different in sources the water mingled in atmosphere in form of water vapours. Water vapours present in atmosphere changes in rain drops by condensation at low temperature. These causes raining or hailing and in this way water again comes on earth (in soil and ponds). Thus water cycle in continues in biosphere.

Question 2.
Describe briefly the nitrogen cycle.
Answer:
Proteins are the compounds of nitrogen. They are the important physical part of plants and animals. Due to the deficiency of protein growth of plants and animals stops because protein is responsible for physical growth ¡n living beings.
HBSE 9th Class Science Important Questions Chapter 14 Natural Resources - 3
Animals consume their food from plants. Also, it contains protein. When the plants and animals becomes dead and decay then some saprophytic bacteria and fungi turns them into the ammonia salt. These ammonia-salts are turned into nitrates by nitrifying bacteria. Some of its part is absorbed by the root rains of roots tips of plants which is further converted into protein. Remaining part of the nitrate gets transformed into nitrogen by the micro-organisms (nitrifying bacteria). Thus, nitrogen once again appears in the atmosphere.

The bacteria that stabilise cosmic nitrogen are found in the soil and other root-glands of the legume plants. These bacteria convert the nitrogen in free state into its compounds, which are absorbed by the plants. Some part of atmospheric nitrogen remains fixed into the earth by thunder lightning. Fertilizers, manufactured in the factories are also a fixed form of nitrogen. This nitrogen gets absorbed by the plants and passes through the nitrogen food network. Thus, this nitrogen cycle continues in nature.

Question 3.
How does Carbon cycle gets completed in the biosphere ? Give explanation.
Answer:
Carbon is an important physical element for all living beings. In living beings it is found in the form of carbohydrates, fats, protein and nucleic acids. Its transfer takes place through food chain. In atmosphere, it is found in the form of CO2. In oceans it is found in the form of biocarbon and carbonates. The green plants through photosynthesis process consume CO2and prepare carbohydrates. These carbohydrates then get transferred to animals. Decomposes again set them free into the environment. All living beings produce CO2 during respiration. Burning of fuels consisting ef carbon to increase the quantity of CO2 in the atmosphere. Eruption of volcanoes too is the main source of receiving CO2.

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources

Question 4.
Draw a labelled diagram of Oxygen-cycle in nature.
And.
Oxygen in nature, is nearly the 21 % part of the gaseous constituents in atmosphere. It gets dissolved into the water stores, which is utilised by the aquatic animals in breathing process. In bio-bodies it enters and exits in the form of compounds of CO2 and H2O. In biosphere the main source of providing oxygen are the green plants. Two stages of oxygen cycle are as under:

1. Intrusion of oxygen in the atmosphere: Carbon dioxide reached from different sources is utilised by the green plants to undergo photosynthesis process. In photosynthesis, oxygen is produced as a coproduct.

2. Expulsion of oxygen from the atmosphere: Atmospheric oxygen is utilised by all animals in respiration and they release in form of CO2 in the atmosphere. In same way the oxygen dissolved in the water bodies is utilised by the aquatic creatures and they release CO2 as by product underwater. Atmospheric oxygen is utilised in burning of coal, wood and carbon other fuels and is released in the form of CO2 in the atmosphere. In this way, a balance is created in the atmosphere for intrusion and expulsion of oxygen.
HBSE 9th Class Science Important Questions Chapter 14 Natural Resources - 4

Practical Work

Experiment 1.
To study the conduction current waves.

Procedure:
Fix a candle in a beaker and light it up, not take a glowing incense stick near to the mouth of the beaker and observe it:
(1) On bringing the glowing incense stick near to the mouth of beaker the smoke is pushed back, since hot air escapes out.
(2) On keeping the glowing incense stick above the candle flame the smoke directly rises up because the flame of the candle heats up the upper air quickly and thus, let is escape right above.
(3) While keeping the glowing incense stick around the burning candle at differeht positions the smoke is repelled because the entire air from inside the beaker escapes out on heating up.

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources

Experiment 2.
To study the process of formation of fog.

Procedure:
(1) Take 5-10 ml of water in a plastic bottle and tight-up the cap.
(2) Now, shake it well and let escape little quantity of smoke form a glowing incense stick into the bottle by unscrewing the cap. Now place the cap once again to the mouth of the bottle.
(3) Take the bottle in between both palms and squeeze the bottle hard. After a short while, again squeeze the bottle and make your observation. On the water vapours saturated and with an increase in the pressure, fog gets formed inside the bottle. The fog is disappeared during keeping the bottle for a short time still. The pressure increases by squeezing and keep it in motion. During the collection of water vapour around the smoke particles, the fog is formed.

Experiment 3.
Making of a hygrometer and note down the time of maximum rainfall in your city and state.

Procedure:
Take a glass funnel with its mouth equal to that of the size of the mouth of the beaker and fit it to the mouth of the beaker, mark the wall of the beaker and draw a scale.

Hygrometry:
As soon as the rain starts keep the hygrometer at such a place, where water may fall down into the funnel without any obstacle. Note down the reading as soon as the rain stops. With the help of this device, note down the quantity of rainfall for a fixed period (months) of everyday. Thus, we can find out the day which received the maximum rainfall. To note down the maximum rainfall of a particular state the quantity of rainfall is noted down of metropolitan cities or big cities and thus average rainfall is known.

Quick Review of the Chapter

1. The gas found in our atmosphere different to the atmosphere of Venus and Mars is :
(A) CO2
(B) O2
(C) Ar
(D) N2
Answer:
(B) O2

2. The cause of air blow s:
(A) pressure difference
(B) temperature difference
(C) density difference
(D) surface difference
Answer:
(A) pressure difference

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources

3. Which of the following pollute the air ?
(A) burning of fossil fuels
(B) movement of vehicles
(C) smoke of factories
(D) all of the above
Answer:
(D) all of the above

4. The main cause (s) of soil erosion is (are)
(A) floods
(B) heavy winds
(C) deforestation
(D) all of the above
Answer:
(D) all of the above

5. Earthworms are helpful ¡n:
(A) making nitrogen
(B) making humas
(C) decreasing the fertily
(D) all of the above
Answer:
(B) making humas

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources

6. Pattern of rainfall decided on which factor?
(A) wind pattern
(B) forest pattern
(C) evaparation pattern
(D) people’s life style pattern
Answer:
(A) wind pattern

7. Soil erosion can be controlled by:
(A) growing grass
(B) anti air
(C) levelling fields
(D) all of the above
Answer:
(D) all of th above

8. The protective cover ¡n our environment (atmosphere) is of:
(A) O2
(B) CO2
(C) O3
(D) N2
Answer:
(C) O3

9. The water vapours when passes through cold regions, condense in the form of water drops which is called:
(A) snow
(B) water
(C) sky
(D) cloud
Answer:
(D) cloud

10. The human activity/activities which increase the quantity of CO2 in atmosphere is/are:
(A) burning of fuels
(B) use of petrol in vehicles
(C) respiration process
(D) all of the above
Answer:
(D) all of the above

11. The increase in atmospheric temperature by the absorption of infrared radiations by CO2 reflected from the surface of earth is called:
(A) ozone hole
(B) greenhouse effect
(C) ultraviolet effect
(D) absorption effect
Answer:
(B) greenhouse effect

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources

12. The outermost layer of earth is called:
(A) hydrosphere
(B) lithosphere
(C) atmosphere
(D) ozonosphere
Answer:
(B) lithosphere

13. The cover of air spread around the earth is called:
(A) hydrosphere
(B) lithosphere
(C) atmosphere
(D) ozonosphere
Answer:
(C) atmosphere

14. The part of earth which contains water is called:
(A) hydrosphere
(B) lithosphere
(C) atmosphere
(D) biosphere
Answer:
(A) hydrosphere

15. The non-living factors of biosphere are:
(A) air
(B) water
(C) soil
(D) all of the above
Answer:
(D) all of the above

16. The air pollutant (s) ¡s (are):
(A) CO
(B) NO2
(C) SO2
(D) all of the above
Answer:
(D) all of the above

17. Gas produced in combustion is:
(A) oxygen
(B) nitrogen
(C) carbon dioxide
(D) helium
Answer:
(C) carbon dioxide

18. What is the formula of ozone?
(A) O2
(B) O3
(C) O4
(D) O5
Answer:
(B) O3

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources

19. Necessary for photo-synthesis:
(A) sunlight
(B) chlorophil
(C) water and carbon dioxide
(D) all of the above
Answer:
(D) all of the above

20. The substance made of decomposition of dead animals ¡s called:
(A) humus
(B) fertiliser
(C) soil
(D) coal
Answer:
(A) humus

21. The mixing of undesirable materials in soil is called:
(A) water pollution
(B) air pollution
(C) soil pollution
(D) noise pollution
Answer:
(C) soil pollution

22. The percentage of nitrogen in our atmosphere is:
(A) 21 %
(B) 78 %
(C) 0.033 %
(D) 0.1 %
Answer:
(B) 78%

23. The change of nitrogen of atmosphere into nitrates and nitrites by bacteria is called:
(A) oxygen fixation
(B) denitrification
(C) nitrogen fixation
(D) ammonification
Answer:
(C) nitrogen fixation

24. Soil ¡s a natural resource which:
(A) provides necessary nutrients to plants
(B) is compulsory for life and development
(C) provides eating materials
(D) all of the above
Answer:
(D) all of the above

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources

25. CO2 absorbs:
(A) visible light
(B) infrared radiations
(C) ultraviolet radiations
(D) none of above
Answer:
(B) infrared radiations

26. Oxides of nitrogen and sulphur are present in:
(A) acid rain
(B) basis rain
(C) normal rain
(D) all of the above
Answer:
(A) acid rain

27. Acid rain results due to:
(A) Nitrogen and sulphur oxides in air
(B) Increase in suspended practices in air
(C) Increase in hydrocarbons
(D) Excess C0 in air
Answer:
(A) Nitrogen and sulphur oxides in air

28. Which of the following ¡s wrong regarding causes of water pollution?
(A) Presence of unwanted substances in water bodies
(B) Decrease in amount of dissolved oxygen in water:
(C) Change in temperature of water
(D) Increased no. of aerobic bacteria
Answer:
(D) Increased no. of aerobic bacteria

29. Green House Effect is caused by:
(A) Green plants
(B) mfra-red rays
(C) UV rays
(D) X-rays
Answer:
(C) UV rays

30. Cause of Ozone depletion is:
(A) CO2
(B) O2
(C) CFC
(D) N2
Answer:
(C) CFC

HBSE 9th Class Science Important Questions Chapter 14 Natural Resources Read More »

HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

Haryana State Board HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

Very Short-Answer Type Questions

Question 1.
What is the meaning of health?
Answer:
The proper shape of the entire organs and systems of the body, their location, and functioning is called as health. Health is the normal position of the ability of physical, mental, and social life.

Question 2.
What is called an ailment or disease?
Answer:
Any kind of disruption developing in good health is called a disease. Or in other words, occurring of any disorder in the physical system is called a disease.

Question 3.
Give the name of a factor that influences health.
Answer:
Personal hygiene.

Question 4.
Give a reason that affects the community’s cleanliness.
Answer:
Failure of disposing of garbage scientifically.

Question 5.
How does food affect our health ?
Answer:
Excess of deficiency of nutritious elements in diet both leave ill effects on our health.

HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

Question 6.
How does our environment harm us ?
Answer:
Unwanted materials found in the environment proved to be the cause of diseases for us.

Question 7.
Is it called health that the body is disease-free ?
Answer:
‘No, even if we are disease-free, we cannot consider ourselves healthy.

Question 8.
How is a disease diagnosed ?
Answer:
From the symptoms of the disease.

Question 9.
What are the symptoms of diseases ?
Answer:
The signs that indicate any disorder in our body, are called as the symptoms of disease.

Question 10.
What do the physicians do to confirm a certain disease ?
Answer:
They refer the laboratory’ test,

Question 11.
How many types of diseases are there (on the basis of time duration) ?
Answer:
Two types: Acute diseases, Chronic diseases.

Question 12.
What are called as acute diseases ?
Answer:
Ailments that are short lived are called as acute diseases.

HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

Question 13.
Write down names of two acute diseases.
Answer:
Cough-cold, headache.

Question 14.
What are called as chronic diseases?
Answer:
Chronic diseases-the sustain for long time or throughout life.

Question 15.
Write down the name of chronic disease.
Answer:
Filariasis (elephantiasis).

Question 16.
Can a person fall ill in the inadequacy of diet?
Answer:
Yes, in the inadequacy diet a person can fall ill.

Question 17.
The antibiotic penicillin blocks the processes that build in bacteria.
Answer:
cell wall.

Question 18.
What are called as communicable diseases ?
Answer:
Diseases that spread through contact are called as communicable diseases.

Question 19.
Write down the names of two communicable diseases.
Answer:
Cholera, jaundice.

Question 20.
What are called as non-communicable diseases ?
Answer:
Diseases that do not spread through infection or pathogens are called as non-communicable diseases.

Question 21.
Write down names of two non-communicable diseases.
Answer:
Goitre and Diabetes.

Question 22.
Write names of two diseases caused due to virus.
Answer:
Cough-cold, Influenza, Polio.

HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

Question 23.
What type of disease is cancer ?
Answer:
Non-communicable disease.

Question 24.
Which diseases are caused due to bacteria ?
Answer:
Cholera and Tuberculosis.

Question 25.
Give two examples of the diseases caused due to fungi.
Answer:
Skin diseases (dermatosis)-Ringworm, eczema etc.

Question 26.
Write names of two diseases caused by protozo
Answer:
Malaria and kala-azar.

Question 27.
What are the names of diseases caused by a worm?
Answer:
Elephantiasis (Filariasis), ascariasis.

Question 28.
Write down the name of the factor of acne or pimple.
Answer:
Staphylococci bacteria.

Question 29.
What is the cause of sleeping sickness?
Answer:
Trypanosoma protozoa.

Question 30.
Which disease is caused due to lachmania protozoan?
Answer:
Kala-azar.

Short-Answer Type Questions

Question 1.
What is health ? Write the advantages of good health.
Answer:
Health is a state of being well enough for a person to function well physically, mentally and socially. The structure and functions of all the body parts of a healthy person are normal. He is free from all types of psychological, mental and social tensions. A person free from only physical diseases is not frilly healthy.
Advantages: Following are the advantages of good health:
(1) The personal life of a healthy person is full of happiness.
(2) A healthy person utilises his potential fully as an individual, in the family, in society and in the country and does not become a burden on anyone.

HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

Question 2.
What are the conditions necessary to maintain good health?
Answer:
The following conditions are necessary to maintain good health:
(1) Personal and family hygiene.
(2) Proper exercise and relaxation.
(3) Balanced diet.
(4) Community hygiene.
(5) Good habits and abstinence from intoxicants/drugs.

Question 3.
How does the environment affect health?
Answer:
The health of all persons is based on the surrounding environment. Physical factors constitute the environment. A person can fulfil his basic requirements only from the environment. This is possible only in a hygienic environment. A polluted environment has a bad effect on our health.

Question 4.
What things should be kept in mind while cleaning OUT homes 2
Answer:
Apart from dusting, cleaning cobwebs and disposing the wastes, we should also regularly keep a check on insects like cockroaches, mosquitoes, houseflies, spiders etc. Apart from this, it is necessary to white wash the house to keep it clean and beautiful. We should also keep our immediate surroundings neat and clean so that we don’t catch a disease.

Question 5.
What is a disease? How does disease occur?
Answer:
Any variation in the normal structure or function of a tissue or organ of a body. Or a change in psychological condition is called a disease. A diseased person does not feel comfortable.

Causes of disease:
The disease can occur because of any of the following three causes:
(1) Diseases caused due to deficiency or accumulation of some nutritive components e.g. Beri-beri, Obesity etc.
(2) Through infection (communicable) e.g., cholera, tuberculosis etc.
(3) Non-communicable diseases, like heart diseases, stones in the kidney or gall bladder etc.

Question 6.
What are non-communicable diseases ? Name any two.
Answer:
The diseases which are not caused by infection (and remain confined to the bearer only) are called non-communicable diseases, e.g. heart diseases, non-functioning of kidney, gall bladder or kidney stones etc.

Question 7.
What are hereditary (congenita!) diseases ? Give examples.
Answer:
The diseases which are carried by the parents to the offspring are called hereditary diseases. One such disease is haemophilia, in which the patient’s blood does not clot easily. Generally, the time taken for the clotting of blood increases from 2-8 minutes to 30 minutes-24 hours. A haemophilic person can die because of a continuous flow of blood even from a minor injury. Another such type of disease in sickle cell anaemia. The patient’s RBCs (Red Blood Cells) have defective haemoglobin, which reduces the oxygen-carrying capacity of the RBCs and the person goes on to become weak.

HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

Question 8.
What are antibiotics? Why is their excessive use harmful?
Answer:
Antibiotics are the drugs which are derived from bacteria and are fatal for other bacteria. They do not harm the protoplasm of humans For example; penicillin, Terramycin, streptomycin etc. These save us by stopping the growth of pathogens or by. destroying them. If we use them excessively, then the useful micro-organisms in our alimentary canal will start getting destroyed which will affect our digestive system. .

Question 9.
What are the simple and easy precautions to remain free from the diseases caused by microorganisms ?
Answer:
The simple and easy precautions are:
(1) To keep the infected person away from a healthy one.
(2) By vaccination.
(3) To boil water before drinking.
(4) By keeping our surroundings neat and clean.
(5) By keeping check (on the growth) of mosquitoes and houseflies.
(6) By wearing properly washed clothes.
(7) By dumping the waste products in covered places. .
(8) By spraying insecticides so as to keep the surroundings free from harmful insects and pathogens.

Question 10.
What is the difference between bacteria and virus ?
Answer:
Given is the difference between bacteria and virus:
Bacteria:
1. They are living organisms.
2. They have a cellular structure.
3. They can be seen under a microscope.
4. Bacteria are bom from the same type of bacteria and multiply on their own.
5. Bacteria can be harmful as well as useful.

Virus:
1. They possess the features of both living as well as non-living.
2. Their structure is not cellular.
3. They can be seen only with the help of an electron microscope.
4. Viruses can be multiplied only inside a jost cell.
5. Viruses are only harmful.

Question 11.
How do antibiotics act against bacterial infection? Why are they not effective against virus? Explain with reason.
Answer:
Bacteria create a cell wall for their protection. But the antibiotic substances hinder the process of creation of the cell soil due to which bacteria cannot protect themselves and the antibiotics destroy them. Pathogens are destroyed by the same process. Because virus do not create any cell wall so, that antibiotics are not able to effect virus.

Question 12.
How does infection through air take place ?
Answer:
When an infected person exhales, sneezes or coughs, the pathogens spread in the air. When another person inhales the same air, these pathogens enter his body. Common cold, pneumonia and tuberculosis spread through this mode only. The occurence of common cold in crowded places is again due to this.

HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

Question 13.
How does infection take place through water ?
Answer:
When the pathogens from the infected person enter the drinking water, then a healthy person also gets infected when he drinks this water. There is more incidence of water-borne diseases in rainy season. Cholera Spreads through this method.

Question 14.
Why are V.D. and gonorrhoea considered a social evil ?
Answer:
They are both sexual diseases. Their pathogens enter a healthy person’s body through cuts, wounds or genitals. In such a condition, these diseases are spread from a woman to a man or from a man to a woman during sexual contact. As these diseases also spread through sexual activity, so these diseases are considered a social evil because the incidence of these diseases is more in people who establish illegitimate sexual relations. Prostitution is the main cause of this. Social awareness can reduce the incidence of these diseases.

Question 15.
What is the sign inflammation ?
Answer:
During infection, immune system recruits many cells to the affected tissue to kill the disease causing microbes, this process is called imflammation. As a part of this process, there are local effects such as swelling, pain and general effects such as fever.

Question 16.
Why is vaccination of children and expectant mothers done ?
Answer:
Polio, whooping cough, measles diphtheria, typhoid and tetanus are some infective diseases common in the children. To prevent these diseases newly bom babies and children must be got vaccinated or immunised. To immunise all the children and expectant mothers under a plan, sufficient quantity of vaccines have been manufactured and distributed in India. If the expectant mothers are treated with the tetanus vaccine before she delivers the baby, the newly bom baby will have the ability of immunity. Therefore, lacs of lives of children can be saved by vaccination at proper time.

Essay Type Questions

Question 1.
How do pathogens spread diseases ?
Answer:
Those pathogens, that spread diseases are called as pathogens. With the entrance of pathogens into the human body and their reaction a man falls ill.

Pathogens can spread diseases in the following ways:
1. Through Contact: Some pathogens spread diseases through direct contact with the patient or the thing used by the patient; for example, if a healthy person puts on the clothes of a patient or uses his bed, the pathogens get entered into his body and make him ill, like pathogens causing skin diseases.

2. Through eatables: Pathogens of typhoid, cholera etc. the remaining food of the patient enter into our body and spread diseases.

3. Through air: Pathogens of common cold, influenza etc. enter the body of a healthy person through breathing.

4. Through vectors: Houseflies, mosquitoes, mice, bugs etc. are some insects that spread diseases; like malaria spreads through female anopheles mosquito bite. Plague is spread by mice.

5. Through water: Pathogen on mixing into drinking water reach to the healthy person and diseases
like cholera, diarrhoea, typhoid etc. surround the human beings. Therefore, specially during rainy season it is suggested to boil drinking water and then cool it down to drink there after, so that the pathogens might destroy.

HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

Question 2.
Give a brief introduction of the infectious diseases.
Answer:
Infectious diseases are caused due to micro-organisms like:
1. Cholera: The cause of qholera is vibriocoli. It occurs due to drinking dirty water or infected eatables. Pathogens of cholera are spread through housefly. The patient suffers from painless water loss of due to acute vomitting and diarrhoea.

2. Tuberculosis: It’s cause is micro bacterium tuberculosis. This disease affects the lungs. The patient suffers from constant fever, blood in sputum, bodyache, diarrhoea and the neck ribs get swollow.

3. Typhoid: It spreads through salmonela bacterium typall. Its main symptoms are fever, headache, bleeding nose, black circles on the skin, reddish blisters on the chest.

4. Cough and Cold: It is caused by rhino works. Constant flowing of liquid through nostrils, redness in the eyes, irritation and headache are its main symptoms.

5. Chickenpox: It is caused by the pathogen berisola joster. Here, big circles on the patients body
get formed, severe backache, high fever and headache are its other symptoms. Its a effect lasts long for 14 days. .

6. Polio: It is an infectious disease among the children. It spreads through the pathogen of polio mylitis.
Its symptoms are fever, vomiting, bodyache, stiffness and difficulty in controlling muscles. The children develop paralysis. .

7. Rabies: It is caused by the Rabies pathogen. It is caused due to the bite of dog, cat, bat, jackal, fox etc. Patient suffering from Rabies scares water (hydrophobia). It affects the nervous system. Due to this the patient meets a painful death.

8. AIDS: It comes under the category of sexually transmitted diseases which is caused due to virus. A person suffering from HIV-AIDS, Immunisation system becomes weaker, with the result the body fails to face even the minor infection. The patient of HIV runs with pneumonia even with minor cough and cold. Presently, AIDS is an incurabled disease.

Question 3.
Which programmes have been adopted the Govt, of control communicable diseases ?
Answer:
Following programme have been adopted by the Govt, of India to control communicable diseases :

1. National Malaria Eradiction Programme:
This started as a control measure in 1953. In 1975, it was converted from control of the disease to total eradication programme. The changed programmed was started in 1977 which was effectively enforced. The programme include inspection of areas affected by malaria educated the people using suitable medicines, spraying of insecticides to kill mosquitoes, identifying the malaria patients, distribution of antimalarial drugs to keep their accounts etc.

HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

Proper Vaccination List:

For Pregnant WomenType of Injection:
As early as possible in pregnancyTetanus-1 injection
One month after of Tetanus-1Tetanus-2 or Booster injection
For Children
1\(\frac {1}{2}\) monthsB.C.G. and D.P.T-1 vaccines and oral dose of Polio-1
2\(\frac {1}{2}\) monthsD.P.T-2 vaccine and oral dose of Polio-2
3\(\frac {1}{2}\) monthsD.P.T.-3 vaccine and oral dose of Polio-3
9 monthsMeasles vaccine
Between 16 and 24 monthsBooster vaccine of D.P.T. and booster oral dose of polio
Between 5 and 6 yearsTyphoid and D.P.T Vaccine
10 yearsTyphoid and T.T. Vaccine
16 yearsTyphoid and T.T. Vaccine

2. National Tuberculosis Eradication Programme:

Following stages are there in this programme:
(1) To identify the person suffering from tuberculosis in the very first stage and giving him treatment.
(2) To prevent the diseases by inoculating the children at an early age by BCG vaccine.
(3) Many TB clinics, district centres and education centres started in country.

3. National Cholera Eradication Programme: It includes cleaning the cholera effected area, making fresh drinking water availability and proper disposal of urban faecol waste. To educate people and for immediate treatment, mobile treatment units have been established for it. To inoculate people with cholera injections has been arranged in these.

4. National Leprosy Eradication Programme : A survey was conducted for thistdiseases in India and many treatment centres have been setup. Houses have been constructed for rehabilitation of the infected persons. Tamilnadu Govt, has opened a Leprosy Education and Research Centres, where there is an advisory committee for this diseases. Sulpha drugs are used for the treatment of the disease.

Practical Work

Experiment 1.
To study the effect of possible calamities on the health of people.

Procedure:
There are two effects of natural calamities on the health. Prepare its table with the help of science-teacher.

Immediate effect:Effect after calamity:
1. When people wounds, there is a diverse effect on health.1. The infectious diseases can spread in its lack of pure drinking water.
2. The health disturbed, if proper treatment is not given.2. The people can fall ill due to lhalnutrition in the lack of sufficient nutrition.
3. The tension is created due to the fear and epidemic.3. Due to lack of proper cleanliness, diseases are spreads from the dirtiness all around.

Experiment 2.
To study the provisions made by local municipal corporation for the supply of clear drinking water.

Procedure:
Prepare a list of provisions made by the local municipal corporation for the supply of clear drinking water.
(1) The cholorination of drinking water is done.
(2) Water storage tank is cleaned regulary.
(3) Tubewells are used to get underground drinking water.
(4) Water treatment plant is setup for canal water.
(5) Water supplies by pipes to avoid pollution.

HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

Experiment 3.
To study the provisions made by local authority to manage garbage.

Procedure:
Prepare a list of provisions related to it from officer related to local conservator.
(1) The garbage is cleaned by landfill.
(2) Organic garbage is treated by roasting.
(3) Covered vehicles are used to throw garbage.
(4) Composte prepared from organic garbage is selling to farmers.
(5) Local water is planned to set up solid waste treatment plant.

Quick Review of the Chapter

1. Cholera is a ………… disease.
(A) bacterial
(B) viral
(C) protozoan
(D) fungal
Answer:
(A) bacterial

2. Malaria spreads by:
(A) mouse
(B) fly
(C) mosquito
(D) bug
Answer:
(C) mosquito

3. Cholera spreads by :
(A) fly
(B) mouse
(C) mosquito
(D) none of these
Answer:
(A) fly

HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

4. Respiratory diseases spread by :
(A) food
(B) water
(C) air
(D) animal
Answer:
(C) air

5. Communicable disease is :
(A) goitre
(B) beri-beri
(C) obesity
(D) tuberculosis
Answer:
(D) tuberculosis

6. Example of chronic disease is :
(A) cough
(B) cold
(C) elephantiasis
(D) headache
Answer:
(C) elephantiasis

7. Acute disease is:
(A) lungs tuberculosis
(B) Dengue fever
(C) blindness
(D) disentary
Answer:
(D) disentary

8. Which of the following disease is caused by virus ?
(A) Anthrax
(B) Dengue fever
(C) Malaria
(D) Cholera
Answer:
(B) Dengue fever

9. The reason of AIDS is :
(A) bacteria
(B) protozoa
(C) virus
(D) fungi
Answer:
(C) virus.

 

10. The prostitution in society is the reason which of the following disease
(A) pneumonia
(B) gonorrhoea
(C) giardiasis
(D) influenza
Answer:
(B) gonorrhoea

11. AIDS can spread by:
(A) kiss
(B) embracing
(C) combined used blade of shave
(D) hug
Answer:
(C) combined used blade of shave

12. AIDS cannot spread by:
(A) sexual relations
(B) blood transfusion
(C) from placenta of mother
(D) by shaking hands
Answer:
(D) by shaking hands

13. Which disease can be caused by attack on lever?
(A) diabetes
(B) malaria
(C) jaundice
(D) goitre
Answer:
(C) jaundice

HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

14. Which disease can be caused by fungi?
(A) kala-azar
(B) cholera
(C) malaria
(D) skin disease
Answer:
(D) skin disease

15. Vomiting is related to :
(A) mind
(B) liver
(C) lungs
(D) kidney
Answer:
(A) mind

16. Which fight against microbes :
(A) RBC
(B) WBC
(C) platelets
(D) lymphocytes
Answer:
(B) WBC

17. Which disease happens first and last times ?
(A) rabies
(B) polio
(C) chicken pox
(D) small pox
Answer:
(D) small pox

18. The main symptom of AIDS is:
(A) the loss in resistance power
(B) fever
(C) arthritus
(D) high cough ‘
Answer:
(A) the loss in resistance power

19. Rabies is a …………… disease.
(A) bacterial
(B) viral
(C) protozoan
(D) fungal
Answer:
(B) viral

20. Before going to fair, there is vaccination of:
(A) tuberculosis
(B) hepatitis
(C) typhoid
(D) cholera
Answer:
(D) cholera

HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

21. Deficiency disease is not:
(A) diarrohea
(B) scurvy
(C) pellagra
(D) xeropthalmia
Answer:
(A) diarrohea

22. Communicable disease is :
(A) intluenza
(B) hemophilia
(C) goitre
(D) anaemia
Answer:
(A) influenza

23. The disease in mother’s placenta is:
(A) polio
(B) AIDS
(C) goitre
(D) scurvy
Answer:
(B) AIDS

24. Vaccination of which of the following is not available?
(A) whooping cough
(B) polio
(C) diphtheria
(D) AIDS
Answer:
(D) AIDS

25. Which disease can be caused by throwing solid waste in open?
(A) elephantiasis
(B) goitre
(C) cholera
(D) AIDS
Answer:
(C) cholera

26. Match the following:

(a) AIDS(i) Staphylococcus aureus
(b) Acne(ii) Leishmanja donovan
(c) Kalazar(iii) Trypanosoma garnbiense
(d) Sleeping sickness(iv) Human Immune Virus

(A) a-iv, b-i,c-ii,d-iii
(B) a-i,b-ii,c-iii,d-iv
(C) a-ii,, b-ic—iii,d-ii,
(D) a-iii. b-Ii, c-i, d’-iv
Answer:
(A) a-iv b-i, c-ii d-III

HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

27. Which disease is caused by viruses?
(A) Common cold
(B) Dengue fever
(C) Influenza
(D) All of these
Answer:
(D) All of these

HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill Read More »

HBSE 9th Class Science Important Questions Chapter 12 Sound

Haryana State Board HBSE 9th Class Science Important Questions Chapter 12 Sound Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 12 Sound

Very Short-Answer Type Questions

Question 1.
What is sound?
Answer:
Sound is a kind of energy that is produced by vibration.

Question 2.
What is the state of the body that is producing sound?
Answer:
Vibrating state.

Question 3.
How does sound travel in air?
Answer:
Sound travels in compressions and rarefactions.

Question 4.
In which the speed of sound is more-solids, liquids or gases ?
Answer:
In solids.

Question 5.
What is the speed of sound in air at 0°C ?
Answer:
331 m/s.

Question 6.
What is the speed of sound in air at 22°C ?
Answer:
344 m/s.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 7.
What is the speed of sound in distilled water at 25°C ?
Answer:
1498 m/s.

Question 8.
In which the speed of sound is 5100 m/s at 20°C ?
(i) steel
(ii) aluminium
(iii) alcohol
(iv) mercury.
Answer:
(ii) aluminium.

Question 9.
What is the velocity of light ?
Answer:
3 x 108 m/s.

Question 10.
Name two forms of energy.
Answer:
(i) Heat energy
(ii) Light energy.

Question 11.
What is medium ?
Answer:
The liquid or material through which sound propagates, is called medium.

Question 12.
What moves forward in sound propagation-particle of medium or disturbance ?
Answer:
Disturbance moves forward.

Question 13.
Why are sound waves called mechanical waves ?
Answer:
Because sound waves are characterise by the particles of medium. Therefore, the sound waves are called mechanical waves.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 14.
What is compression ?
Answer:
The region of high pressure in medium is called compression (C).

Question 15.
What is rarefaction ?
Answer:
The region of low pressure in medium is called rarefaction (R).

Question 16.
What is the meant of 256 written on tuning fork ?
Answer:
It means its frequency is 256 Hz.

Question 17.
What are electromagnetic waves ?
Answer:
Those waves which can travel without any medium are called electromagnetic waves.

Question 18.
Can the sound waves travel in vacuum ?
Answer:
No, sound waves cannot travel in vacuum.

Question 19.
Which type of waves are used by astronauts to talk on moon ?
Ans:
Radio waves.

Question 20.
Define wave velocity.
Answer:
The distance covered by the wave per unit time is called wave velocity.

Question 21.
What is the relation between frequency and time period ?
Answer:
Frequency is the reciprocal of time period. That is v = \(\frac {1}{T}\) .

Question 22.
What is the relation among velocity of wave, wavelength and frequency ?
Answer:
The velocity of wave is equal to the product of frequency and wavelength. That v = v X

Question 23.
How many type of waves can slinky produce ?
Answer:
Slinky can produce three types of waves :
(1) Transverse waves.
(2) Longitudinal waves.
(3) Periodic waves.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 24.
Which physical quantity is propagated by waves ?
Answer:
Energy.

Question 25.
What is the SI unit of frequency ?
Answer:
Hertz.

Question 26.
What is the SI unit of wavelength ?
Answer:
Metre.

Question 27.
What is the SI unit of wave velocity ?
Answer:
Metre/second (m/s)

Question 28.
Define pulse.
Answer:
Waves of shorter duration is called pulse.

Question 29.
What is the effect on the particles of the medium when a wave moves ?
Answer:
When a wave moves, the particles of the medium execute only pendulum motion, and not moves with sound.

Question 30.
What are longitudinal waves ?
Answer:
When the individual particles of the medium move in a direction parallel to the direction of propagation of disturbance are called
longitudinal waves.

Question 31.
How many types are there of waves ?
Answer:
There are three types of waves:
(i) Longitudinal waves
(ii) Transverse waves
(iii) Periodic waves.

Question 32.
Which are longitudinal waves among these ?
(i) waves on pond water
(ii) sound waves
(iii) light waves.
Answer:
(ii) Sound waves.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 33.
What are transverse waves ?
Answer:
When the individual particles of the medium moves in a direction perpendicular to the direction of wave propagation, are called
transverse waves.

Question 34.
The propagation of sound is the example of which wave ?
Answer:
Longitudinal wave.

Question 35.
What types of waves are produced on pond water ?
Answer:
Transverse waves.

Question 36.
How do the longitudinal waves travel ?
Answer:
The longitudinal waves travel in compressions and rarefactions.

Question 37.
What is a trough ?
Answer:
The maximum displacement in negative direction on Y-axis in the graph of periodic waves is called a trough.

Question 38.
Define the pitch.
Answer:
How the brain interprets the frequency of an emitted sound is called the pitch.

Question 39.
Define note.
Answer:
A sound which is produced due to a mixture of several frequencies is called a note.

Question 40.
What are the causes of variations in sounds ?
Answer:
The main cause of variations in sounds are loudness of sound, quality, pitch of sound and frequency.

Question 41.
What is the hearing range of our ears ?
Answer:
20 Hz to 20,000 Hz.

Question 42.
Define ultrasonic sound.
Answer:
Frequencies higher than 20 kHz are called ultrasonic sound or ultrasound.

Question 43.
What is the frequency of the waves produced during earthquake and valcano eruption ?
Answer:
Frequencies less than 20 Hz.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 44.
Write the name of two creatures which produce sounds of frequencies less than 20 Hz.
Answer:
(i) Whale
(ii) Elephant.

Question 45.
Write the name of two creatures which produce ultrasonic sound.
Answer:
(i) Bats
(ii) Dogs.

Question 46.
Which sound helps the bat to locate their way ?
Answer:
Ultrasound.

Question 47.
What is reverberation ?
Answer:
The repeated reflection that results in the persistence of sound is called reverberation.

Question 48.
What is meant by supersonic speed ?
Answer:
When the speed of any object exceeds the speed of sound it is said to be travelling at supersonic speed.

Question 49.
Which types of aircraft produce shock waves ?
Answer:
Supersonic aircraft.

Question 50.
What is the sound called produced by motion of supersonic aircrafts ?
Answer:
Sonic boom.

Question 51.
What is meant by reflection of sound ?
Answer:
When sound strikes on a surface, it changes its direction, it is known as reflection of sound.

Question 52.
When sound produces in deep wells, huge buildings and mountains, we experience a sound after some time besides mean sound, what is that ?
Answer:
Echo.

Question 53.
On which principle the megaphone works ?
Answer:
Megaphone works on the principle of reflection of sound.

Question 54.
What should be the minimum time to have echo between mean sound and echo ?
Answer:
0.1 second or \(\frac {1}{10}\) second.

Question 55.
How much minimum distance should be their to listen echo from a reflecting object ?
Answer:
17.2 metres.

Question 56.
What will be the minimum distance covered by the sound to hear echo ?
Answer:
34.4 metres.

Question 57.
Define sonar.
Answer:
Sonar means Sound Navigation And Ranging. It is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 58.
Where is SONAR used ?
Answer:
It is used to determine the depth of the sea and to locate underwater hill, valleys, submarine, icebergs and sunken ships, etc.

Question 59.
Which type of waves are used in SONAR system ?
Answer:
Ultrasonic waves.

Question 60.
Upto what frequency, the bat can produce ultrasonic waves ?
Answer:
Upto 120 kHz.

Question 61.
By emitting which waves to the painful part of body, we feel very like rubing (malish) ?
Answer:
Ultrasound wave.

Question 62.
Why is it beneficial to use ultrasound instead of X-rays to take the photograph of inner body ?
Answer:
Because ultrasound does not affect tissues and cells.

Question 63.
What is ultrasonography ?
Answer:
The technique which is used to examine the foetus during pregnancy to detect congenial defects and growth abnormalities is called ultrasonography.

Question 64.
Which type of waves are used to locate the eye tumor and to cure that ?
Answer:
Ultrasonic waves.

Question 65.
Which characterstics of ultrasound is used to develop photographic film ?
Answer:
Mixing of immiscible liquids by ultrasound.

Question 66.
Which type of waves are used in breadth therapy ?
Answer:
Ultrasonic waves.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Short Answer Type Questions

Question 1.
Prove that sound is a form of energy.
Answer:
When we speak for long time or loudly, we get tired. We spend our energy during talking. This energy produces vibration in throat. It proves that sound is a form of energy.

Question 2.
How is sound produced ? Justify with example.
Answer:
It is compulsory to vibrate for the production of sound. For example; when we speak, the vocal cord of throat vibrate. When trumphets or guitar play, its wires vibrate. When bell is rung, there is also vibrations. The vibrations in bell can be felt by touching it by hand. In this way, the sound is produced by vibrations.

Question 3.
Define compression and rarefaction. Draw a graph to show the density of air molecules due to sound waves.
Answer:
Compression: A region of high pressure is called compression.
Rarefaction: A region of low pressure is called rarefaction.
The graph between the density of air in unit interval and distance can be drawn.
HBSE 9th Class Science Important Questions Chapter 12 Sound - 1
In fig. the distribution of air atoms produced by sound propagation of sound wave is shown. The graph between density and distance is also shown.

Question 4.
Justify with an example that how the compression and rarefaction formed in vibrating body move ?
Answer:
Experiment:
Take a spiral spring as shown in fig. Press one end of spring slowly by hand and then leave. We see that the compressions and rarefactions are formed in the spring and they propagate from one end to another end. The propagating motion of compressions and rarefactions is called wave motion. In the same way, the compressions and rarefactions produced in air propagate from one layer to another and in this way, the sound travels from one place to another.
HBSE 9th Class Science Important Questions Chapter 12 Sound - 2

Question 5.
Justify the sound system in air by tuning fork.
Answer:
The prongs of tuning fork start vibrating when hit with a rubber pad. The vibrating prong exert pressure of air particles and displaces them. The air particles come in motion from rest and come in original postition after displacing the nearby particles. The compression of high pressure is formed at the region where the air particles come close, and after detaching pressure of vibration, the particles come in their original position. This region of low pressure is called rarefaction. The particles of medium do not move forward itself but disturbance of particle moves forward by compression and rarefaction. This is called sound wave. Air is the most common medium for the propagation of sound.
HBSE 9th Class Science Important Questions Chapter 12 Sound - 3

Question 6.
Can sound make a light spot dance ? Justify it.
Answer:
Take an empty tin with both ends open. Tie stretched membrane of balloon on one end. Take one piece of plane mirror and stick it to the membrane of balloon in such a way that its reflecting surface points upward and the rays coming from a source should fall on the mirror and after reflection the light spot should be on the wall. According to figure, make a loud sound from the open end of tin. You will see the light will dance.

Cause:
Due to vibration of sound, the membrane and the mirror on it vibrate, due to this the image formed on wall seems to be shaking. Stop the sound. The spot will be on fixed place because the vibration of membrane is stopped.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 7.
Define longitudinal and transverse waves.
Answer:
(i) Longitudinal wave : When the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance is called longitudinal wave.
(ii) Transverse wave : When the individual particles of the medium move in a direction perpendicular to the direction of wave propagation, is called transverse wave.

Question 8.
Define crest, trough and amplitude and show them for a transverse wave by graph.
Answer:
The crest, trough and amplitude of a transverse wave are shown in fig.
Crest: Peaks formed during the motion of transverse wave is called crest.
Trough: The depression formed during the motion of transverse wave is called trough.
Amplitude: The maximum displacement of medium particles from their mean position during transmission of waves is called amplitude. The unit of amplitude is metre.
HBSE 9th Class Science Important Questions Chapter 12 Sound - 4

Question 9.
Distinguish between longitudinal and transverse waves.
Answer:
The differences between longitudinal and transverse waves are as follows

Longitudinal wavesTransverse waves
1. The individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance in these waves.1. The individual particles of the medium move in a direction perpendicular to the direction of propagation of the disturbance in these waves.
2. Compression and rarefaction are formed in it which make a wave.2. Crest and trough are formed in it, which make a wave.
3. These waves can be formed in any medium, solid, liquid or gas.3. Solid or liquid medium is necessary for the formation of these waves.
4. These waves cannot be polarised.4. These waves can be polarised.
Longitudinal waves:Transverse waves:

Question 10.
What do you know about these :
(i) Loudness of sound
(ii) Pitch
(iii) Frequency.
Or
Write two differences between the loudness and pitch of sound.
Answer:
(i) Loudness of sound:
The level of the sensation of the sound is called the loudness of sound. This level of sensation depends upon the sharp or light vibrations of source. When vibrations are more, the loudness of sound will be more and when vibrations are light, the loudness of sound will be low.

HBSE 9th Class Science Important Questions Chapter 12 Sound

(ii) Pitch:
That property’of sound by which we can differentiate between the louder and softer sound. This property depends upon the frequency of the source of sound. When the frequency of the source of sound is more, the pitch is also more.

(iii) Frequency:
The number of vibrations completed by any object in one second is called frequency.

Question 11.
What is the difference between the sound of mosquito and lion ?
Answer:
The loudness of the sound of mosquito is less but frequency is more. On the other side, the loudness of the sound of lion is more but frequency is less. So the sound of mosquito is more sharp than that of lion.

Question 12.
Show through graph that how the variations occur in density and pressure when sound wave travels in a medium.
Answer:
HBSE 9th Class Science Important Questions Chapter 12 Sound - 5
On a certain time, both the density’ and pressure of the medium vary up and down from their average value. In fig. a region of high pressure is compression and low pressure is rarefaction. The peak is called crest and valley is called trough of wave.

Question 13.
What is the difference between the sound of violin and flute of an orchestra ?
Answer:
Both the violin and flute travel in same medium and reach to our ear in same time, the speed of both is same, but both sounds are obtained from different characterstics. That is pitch. How the brain interprets the frequency of an emitted sound is called the pitch. Therefore, the sound of violin and flute are heard differently.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 14.
Differentiate between low and high pitch through graph.
Answer:
HBSE 9th Class Science Important Questions Chapter 12 Sound - 6

Question 15.
What is the main difference between soft and loud sound ? Clarify it by graph.
Answer:
The softness and loudness of sound depends upon the amplitude. For instance, knocking the door slowly produces a sound of’low amplitude and on loudly it produces a sound of high amplitude. Less energy is emerged in low amplitude and more energy is emerged in high amplitude. That is why the sound of low amplitude covers less distance while of high amplitude covers more distance
HBSE 9th Class Science Important Questions Chapter 12 Sound - 7

Question 16.
Show by an experiment that speed of sound is more in liquids than in air.
Answer:
Take a balloon full of water and rub its surface slowly. You will hear a sound. Now keep the balloon near your ear and rub it from other side, we will hear more and clear sound. It proves that the speed of sound is more in liquids than in air.
HBSE 9th Class Science Important Questions Chapter 12 Sound - 8

Question 17.
How can we hear the sound of train coming from very far when ear is kept on track ?
Answer:
We know that the speed of sound in iron on normal temperature is 5950 m/s. Therefore, we can hear the sound of train when it is so far by keeping the ear on line which is made of iron girders.

Question 18.
Describe an experiment to show the reflection of sound.
Answer:
Take two identical wide tubes of plastic or card. Also take one sheet of plywood or metal. Arrange these according to fig. Keep a clock near the end of one of the tubes. Keep your ear near the other end of tube. Put a thick membrane between two tubes so that sound of watch cannot reach to your ears. Arrange the angle of second tube in such a way that you should be able to hear the sound of clock. You will see that in this situation both the tubes make equal angle to normal. This proves that sound is reflected.
HBSE 9th Class Science Important Questions Chapter 12 Sound - 9

Question 19.
What is meant by reflection of sound ? How is reflection of sound used to measure the depth of sea ?
Answer:
When sound changes its direction after striking a surface, it is called reflection of sound. For example, when the sound is produced near the mouth of well, the wave of mean sound produced enter in the well and after some time they came back after reflecting from the surface of water and heard again like mean sound as echo.

To measure the depth of sea by reflection of sound:
The depth of sea is measured by sonar device. From this instrument, the sound signal is sent to the bottom of sea from the ship. This signal comes back to the instrument after the reflection from the sea bottom. In this way, the time taken to reach the sea bottom and reflection is noted. The depth of sea is measured by the speed of sound in water. Now the depth of sea is calculated by the time and speed of sound in water.
HBSE 9th Class Science Important Questions Chapter 12 Sound - 10

Question 20.
What is the difference between ultrasound and echo ?
Answer:
Ultrasound:
The frequencies more than 20 kHz are called ultrasonic sound. We cannot hear this sound. Some animals like dogs have the capacity to hear ultrasound. The bats have the capacity to produce and listen ultrasound.

Echo:
When sound is produced in deep well, big buildings, mountains, etc., the same sound is heard as mean sound after some time. This is known as echo. We can hear echo only when there is a minimum difference of 1/10 s between the mean sound and reflected sound. The minimum distance covered by the sound should be 34.4 metres to hear the echo. Thus, the object should be at a distance of 17.2 metres or more from the source of sound because the speed of sound at 22°C is 344 m/s. Time taken by sound to reach the obstacle and returning back should be 1/10 seconds. The sensation of sound in our brain lasts for 1/10 seconds.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 21.
What is the audible range of different animals ?
Answer:
The audible range of different animals is different. For example,
(1) Man – 20 Hz to 20,000 Hz.
(2) Dog and babies less than 5 yrs – upto 25 kHz.
(3) Rhino – 5 Hz.
(4) Whale and elephant – Infra sound.
(5) Dolphins, bats and porpoises – sounds of frequency more than 20 kHz.
(6) Moths – The moths can hear the high frequency squeaks of the bat.

Question 22.
What is the reason that we cannot hear the waves produced by earthquake whiie bats or dogs can?
Answer:
Our ear can hear the waves of frequencies from 20 Hz to 20 kHz. This range of frequency is known as audible range. The frequency of the waves produced in earthquake does not lie in this limit. Therefore, we cannot hear these waves, but the bats and dogs are sensitive to these waves and can hear these waves.

Question 23.
Define infrasound waves, audible range and ultrasound.
Answer:
1. Infrasound: Those waves whose frequency is less than 20 Hz are known as infrasound waves. We cannot hear them. i.e. earthquake wave.
2. Audible range : The waves of frequency between 20 Hz and 20 kHz is called audible range.
3. Ultrasound waves : Those waves whose frequency is more than 20 kHz is known as ultrasound waves. We cannot hear them also. Dogs and bats can hear these waves.

Question 24.
Write the uses of ultrasound.
Answer:
The uses of ultrasound are following :
(1) Radar works on this principle.
(2) The milk exposed to them does not get spoilt for long.
(3) Sprouted is more in grain when affected by it.
(4) The cracks can be found in metal sheets melted by it.
(5) The bat find its prey by using this sound.
(6) The depth of sea can be measured by using it.
(7) It is used to wash clothes.
(8) It is used to find the cancer in body parts.
(9) It is used to cut hard substances and to join aluminium.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 25.
A person has been kept in a closed room whose doors and windows are made of glass. When he makes a sound in the room, nobody hear that; but when he knocks the window, the outsider persons hear the knocking. Why ?
Answer:
When a person in a room makes a sound, the waves produced by him is reflected from the glass. This produces weak vibrations in the glass, which is incapable to vibrate the air outside the room. As a result of it sound is not heard outside. Opposite to it, when the glass of window is knocked, the waves produced, reach the outer part of the glass and vibrate the air in its contact. Therefore, the sound of knocking is heard by persons.

Question 26.
What is sonic boom ? How is it harmful ?
Answer:
Jet aircrafts produce shock waves in air. These shocking waves travel with supersonic speed. These waves carry a large amount of energy. The air pressure variation associated with this type of shock waves produces a very sharp and loud sound called the “sonic boom”. The shock waves produced by a supersonic aircraft have enough energy to shatter glass and even damage buildings. Sometime sonic boom produces such a loud noise which is beyond our intolerable limits. This can create trouble in our ears. Remember it, when supersonic aircraft flies, it continuously produces sonic boom around it.

Question 27.
Write a short note on whispering galleries.
Answer:
The whispering gallery of dome of St. Paul’s Cathedral in London is very famous and serves an interesting example of reflection of sound. The gallery is running round the base in the inside of the dome in a circular form. When we whisper near any point of walls, the sound produced is reflected at every place. This can be clearly heard near the walls. But at some distance towards centre, let at point A, this sound can never be heard. One such gallery is in Gol Gumbaz at Bijapur in India.
HBSE 9th Class Science Important Questions Chapter 12 Sound - 11

Question 28.
What is SONAR ? What are its uses ?
Answer:
SONAR: The instrument which produces sound waves and measures the small interval of time of reflected sound waves is known as SONAR.
Uses:
(1) To measure the depth of sea.
(2) To find the distance of groups of fishes.
(3) To locate the submarines and enemy torpedo.
(4) To detect cracks and flaws in metal blocks or sheets.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 29.
How can the bat fly freely in dark by using the principle of sonar ?
Answer:
Bats use the principle of SONAR naturally. Bat can produce ultrasound of frequency 120 kHz or 1.2 x 105 Hz and can hear them. The wavelength of such high frequencies is very less, approximately 0.29 cm. Bat is blind in reality, but in intense dark, it can fly without any difficult while crossing all the obstacles. It uses ultrasound to find its prey. When it flies, emits ultrasound signals of high frequency for small interval of time. These signals strike the surrounding objects and reflect. Thus, the bat calculates the time taken by ultrasonic waves after reflecting from any object. In this way the bat can fly freely in dark.

Question 30.
On what factors the speed of sound depends ?
Answer:
The speed of sound depends on following factors:
1. Density of medium: The speed of sound is proportional to the density of medium. This is, the speed increases as the density increases and decreases with the decrease in density.
2. Temperature: Speed of sound increases with increase in temperature.
3. Wavelength: The speed of sound is proportional to wavelength.
4. State of medium: The speed of sound is maximum in solids, less in liquids and minimum in gases.

Question 31.
What is the audible range ? Write the name and uses of the frequencies more than this range.
Answer:
Audible range:
A person can hear waves of frequency from 20 Hz to 20 kHz. This limit is called audible range.

The sound of frequency more than 20 kHz is called ultrasound. Its applications are as follows :
(1) Radar is made on this principle.
(2) It is used to find the cancer in body parts.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 32.
How can we measure the depth of sea ?
Answer:
An instrument is fitted on the ship to measure the depth of sea which sends ultrasound in sea level. These waves return back after reflecting from sea level and the receiver on ship receives these reflected rays. The time taken (t) by these ultrasonic waves to reach the sea level and return back and the speed of sound waves (v) help to calculate the distance (d) between the ship and sea level.
We know that,
Velocity = HBSE 9th Class Science Important Questions Chapter 12 Sound - 12
or Distance = velocity x time.
= 2 d = v x t
d = \(\frac{\mathrm{v} \times \mathrm{t}}{2}\)
This will be the depth of sea at a certain place.

Essay Type Questions

Question 1.
What is sound ? Prove it by experiment that sound is produced by vibration.
Answer:
Sound:
Sound is a form of energy which produces a sensation of hearing in our ears. It is produced by vibrating bodies. It can be proved by following experiments.

Experiment 1:
Take a tong. Hold it from the curved end and strike it against a surface or a stone. We will see the arms of the tong vibrating. Due to these vibrations a sound is heard. Repeat this process and bring the tong near to the ear. The sound of the tong is heard clearly.
HBSE 9th Class Science Important Questions Chapter 12 Sound - 13

Experiment 2:
Fix two nails on a wooden plank and tie a thin wire tightly between the two nails. Put a triangular piece of wood below the wire. Now pluck the wire and release it. A sound is heard. If we put a small piece ofpaper on the wire and again pluck it and release, the paper wil I fall dowm proving thatthe sound is produced due to vibrations.
HBSE 9th Class Science Important Questions Chapter 12 Sound - 14

Experiment 3.
Take a tuning fork. Strike it against a rubber pad, and make it stand straight on a table. A sound is produced which we can hear. Now bring a ball near to the prongs of the vibrating tuning fork. The prong strikes again and again with the ball. This proves the vibration of the tuning fork. If we bring our finger near to it, we can feel the vibrations. When we touch it the sound coming from the tuning fork will stop. This again proves that the sound is produced due to vibrations.
HBSE 9th Class Science Important Questions Chapter 12 Sound - 15

Question 2.
Describe two types of wave motions.
Answer:
There are two types of wave motions:
(a) Transverse waves:
A wave in which individual particles of medium vibrate about their mean position in a direction perpendicular to the direction of propagation of the wave through the medium, is called transverse wave.

Examples:
(1) Waves produced in a slinky
(2) Waves produced on the surface of the water.

(1) The waves produced in a slinky: Take a slinky, Tie its one end with a book to the wall. Hold its other end and give a jerk. The disturbance produced in the slinky moves from hand to the slinky and the particles of the slinky vibrate up and down from their mean position. This is an example of transverse wave.

(2) The wave produced on the surface of the water:
Throw a stone from a certain height in the still water of a pond then ripples are produced on the surface of the water which moves in all the directions. If we put a leaf on the ripples, the ripples will make the leaf move up and down at the same point. The raised upper part of the wave is called the ‘crest’ and the lower part is called the ‘trough’.

HBSE 9th Class Science Important Questions Chapter 12 Sound

(b) Longitudinal waves:
A wave in which the individual particles of medium vibrate back and forth from their mean position in direction parallel to the direction of propagation of the wave through the medium is called longitudinal wave.

Examples:
(1) Waves produced in a slinky
(2) Waves propagating in air.

(1) Waves produced in a slinky : Longitudinal waves can also be produced with slinky. Tie the one end of slinky with a hook on wall and stretch other end with your hand. Now press some turns in the small part of slinky between your fingers. The compression will reach the other end of slinky.

(2) The waves propagating in air : Take a tuning fork, suspend it forcefully with rubber pad. Its prongs will vibrate left-right. When one of its vibrating prongs moves in right, it press the air molecules due to this the air molecules compressed and the distance between them decreases. This is known as compression when the prong of vibrating fork moves in left, the vacuum is created in right side. To fill this, the air molecules expand to right side, and the distance among these air molecules increases which is known as rarefaction.

Question 3.
What are the characterstics of periodic waves ? Establish relationship among wave velocity, frequency and wavelength.
Answer:
The periodic wave has following characterstics:
1. Amplitude: The magnitude of maximum displacement of vibrating particles from mean position is called amplitude of pendulum.
2. Time period: Time taken by a particle to complete one oscillation is called time period.
3. Frequency: The number of vibrations in a second by a vibrating particle is called frequency. It is denoted by Greek Letter (v). Its unit is Hertz (Hz).
4. Wavelength: The time in which the particles of medium complete one vibration, the distance covered by the wave in that period is called wavelength. It is denoted by X .
5. Velocity: The distance moved by the wave in unit time is called velocity of the wave. It is denoted by v.

The relation among frequency, wavelength and velocity of wave:
According to definition:

The distance covered by the wave T second = λ
The distance covered by the wave in 1 second = \(\frac {λ}{T}\)
But, the distance covered in 1 second = Velocity (v)
∴ v = \(\frac {λ}{T}\) ………….(i)
If time taken in one vibration is T,
Then number of vibrations in T seconds = 1
Numbers of vibrations in 1 second = \(\frac {1}{T}\)
But, number of vibrations in 1 second is called frequency ‘υ’.
υ = \(\frac {1}{T}\)
Putting the value of y in equation (i)
v = \(\frac {λ}{T}\) = λ \(\frac {1}{T}\) [∴ \(\frac {1}{T}\) = v]
y = λv
Thus, Velocity = wavelength x frequency.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 4.
What is meant by reflection of sound ? Write three uses of it.
Answer:
Reflection of Sound: When the sound strikes any surface it changes its direction which is called reflection of sound.
Uses of reflection of sound: The practical uses of reflection of sound are following :

1. Megaphones:
You have seen the use of megaphones to address the crowd in fairs or tourists places. Megaphone is simply a horn shaped instrument. When a person speaks in the megaphone, the sound waves are prevented from spreading out due to repeated and successive reflections from the walls of megaphone tube. For this reason, loudspeakers have horn shapes openings.
HBSE 9th Class Science Important Questions Chapter 12 Sound - 16

2. Ear trumpet or Hearing aid:
People with hard of hearing may need a hearing aid. Hearing aid is an electronic battery operated device. The hearing aid receives sound through a microphone. The microphone converts the sound waves to electrical signals. These electrical signals are amplified by an amplifier. The amplifier electric signals are given to a speaker of the hearing aid. The speaker converts the amplified electrical signal to sound and sends to the ear for clear hearing.
HBSE 9th Class Science Important Questions Chapter 12 Sound - 17

3. Sound drum or boards : The reflection of sound is also possible from curved surfaces. This fact is used in bigger halls to spread the sound in whole hall. Sound drums are used for this purposes. The speaker S is focussed at S sound drum. Sometimes a curved sunboard may be placed behind the stage so that the sound after reflecting from the sound board, spreads evenly across the width of the hall as shown in fig.

Question 5.
Describe the different parts of human ear.
Answer:
There are three main parts of human ear:
(a) External ear : It is that part which is seen from outside. It consists of projecting part known as pinna and the passage auditory canal which is connected to a thin membrane called the eardrum or thin panic membrane.
1. Pinna: It consists of cartilage and muscles. Its lower part is soft and elastic. Pinna collects the sound waves and gives direction to them. It is joined with auditory canal.

HBSE 9th Class Science Important Questions Chapter 12 Sound

2. Auditory Canal: It is a slant about 3-4 cm long. Its skin contains hair and micro glands and liquid comes out of them. This is called ear wax. It prevents the ears from dust particles and insects.

3. Eardrum: There is a thin membrane in the end of auditory canal which is called eardrum. It sends the sound waves to middle ear.

(b) Middle ear: The middle ear is like a air filled room. Middle ear is attached to throat through a canal which is called ‘eustachian canal’. This canal maintains the air pressure in two sides of eardrum. The middle ear contains three tiny bones –
(i) hammer
(ii) anvil
(iii) stirrup.

These three bone are together called ear ossicles.
(c) The Inner Ear: It contains
(i) cochlea
(ii) auditory nerves
(iii) semi-circular canals.

(i) Cochlea: The shape of cochlea is spiral and has two half turns. It contains a liquid which is called ‘andolimph’. It helps in hearing.
(ii) Auditory nerve: It is a small chamber which is connected with cochlea from front and from back with semi circular canals.
(iii) Semi-circular canals: These are three in number and arranged at right angles to-each other in three different planes. The ends of auditory canals swim in ‘andolimph.’

Important Table

StateSubstanceSpeed in m/s
SolidsAluminium6420
Nickel6040
Steel5960
Iron5950
Brass4700
Glass (Flint)3980
LiquidsWater (sea)1531
Water (distilled)1498
Ethanol1207
Methanol1103
GasesHydrogen1284
Helium965
Air346
Oxygen316
Sulphur dioxide213

Numerical Questions

Question 1.
The frequency and wavelength of a sound wave is 2 kHz and 25 cm respectively. How much time will it take to move a distance of 1.5 km ?
Solution:
Here, Frequency v = 2 kHz = 2000 Hz
Wavelength λ = 25cm = 0.25 m
We know that, velocity of wave v= wavelength x frequency v = λv
v = 0.25m x 2000 Hz = 500m/s
Time taken by wave to move a distance of 1.5 km,
t = \(\frac{\mathrm{d}}{\mathrm{v}}=\frac{1.5 \times 1000 \mathrm{~m}}{500 \mathrm{~ms}^{-1}}=\frac{15}{5}\)s = 3s
Thus, sound will take 3 s to travel on distance of 1.5 km.

Question 2.
A person clapped his hands near a cliff and heard the echo after 6 s. What is the distance of the cliff from the person if the speed of the sound, v is taken as 346 ms-1 ?
Solution:
Given, Speed of sound, (v) = 346 ms-1
Time taken for hearing the echo, (t) = 6s
Distance travelled by the sound = v x t = 346 ms-1 x 6s = 2076 m
In 5 s sound has to travel twice the distance between the cliff and the person. Hence, the distance between the cliff and the person = 2076 m/2= 1038 m.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 3.
A ship sends out ultrasound that returns from the seabed and is detected after 3.42 s. If the speed of ultrasound through seawater is 1531 m/s, what is the distance of the seabed from the ship ?
Solution:
Given, time between transmission and detection, (t) = 3.42 s.
Speed of ultrasound in sea water, (v) =1531 m/s
Distance travelled by the ultrasound = 2 x depth of the sea = 2d
Where d is the depth of the sea.
2 d = speed of sound x time = 1531 m/s x 3.42 s = 5236 m
d = 5236 m/s = 2618 m
Thus, the distance of the seabed from the ship is 2618 m or 2.62 km.

Question 4.
Two children are standing at two opposite ends of an iron pipe. One boy hit one end of pipe with stone. Find the ratio of the time taken by the sound to travel through iron and air (Given, speed of sound in air is 344 m/s and in iron 5130 m/s).
Solution:
Here,
Let length of the pipe = L m
Velocity of sound in air (v1) = 344 m/s
Velocity of sound in iron (v2) = 5130 m/s
∴ Time taken by sound in the air (T1) = \(\frac{\mathrm{L}}{\mathrm{v}_1}=\frac{\mathrm{L}}{344}\)s
Time taken by sound in iron (T1) = \(\frac{\mathrm{L}}{\mathrm{v}_2}=\frac{\mathrm{L}}{5130^{\mathrm{s}}}\)
The ratio between two time = \(\frac{T_1}{T_2}=\frac{\frac{\mathrm{L}}{344}}{\frac{\mathrm{L}}{5130}}\)
= \(\frac{\mathrm{L}}{344} \times \frac{5130}{\mathrm{~L}}\)
= \(\frac{14.9}{1}\)
T1 : T2 = 14.9 : 1

Question 5.
A boat at anchor is rocked by waves whose crests are 100 m apart and whose velocity is 20 m/s. How often do the crest reach the boat?
Solution:
Here,
The distance between two consecutive crests of wave = 100 m
Velocity of wave (v) = 20 m/s
Frequency of wave (v) = \(\frac{v}{\lambda}=\frac{20}{100}\) Hz = 0.2 Hz

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 6.
A source of wave produces 40 crests and 40 troughs in 0.4 s. Find the frequency of the wave.
Solution:
Here,
Number of wave crests = Number of wave troughs = 40
time = 0.4 s
Number of crests per second Number of troughs per second =\(\frac{40}{0.4}\) = 100
Thus, the frequency of wave = 100 Hz

Question 7.
Using the SONAR, sound pulses are emitted at the surface. The pulses after bdng reflected from the bottom are detected. If the time interval from the emission to the detection of the sound pulse is 2s, find the depth of the wster. IVelocity of sound ¡n water is 1498 rn/sI
Solution:
The velocity of sound in water (y) = 1498 m/s
Time taken in coming and going pulse = 2 s
Thus, time taken in going pulse = \(\frac{2}{2}\) = 1 s
Depth of water = velocity x time = 1498 x 1 = 1498 metres

Question 8.
A child hears an echo from a cliff 4 seconds after the sound from a powerful cracker is produced. How far away is the cliff from the child?
Solution:
Total time taken by sound in going and returning back to the child = 4 s
The time taken by sound in going to the child = \(\frac{4}{2}\) = 2s
The velocity of sound in air 344 = m/s
Thus, the distance between the child and the cliff = velocity x time = 344 x 2 = 688 m

Question 9.
A stone is dropped into a well 44.1 metres deep. The sound of the splash is heard 3.13 s after the stone is dropped. Find the velocity of sound in air.
Solution:
Here,
Depth of the well (s) = 44.1 m
Suppose time taken by the stone to reach the well = t1 s
And the time taken by the sound to reach on water = t2 s
According to the question t1 + t2 = 3.13
second u = 0, g =9.8 m/s2

We know that s = u t + \(\frac{1}{2}\) gt2
44.1 = 0 x t1 + gt12 (For t1)
or 441 = \(\frac{1}{2}\) x 9.8 x t12
or t12 = \(\frac{44.1 \times 2}{9.8}\) = 9
or t1 = 3s
t2 = 3.13 – t1 = 3.13 – 3 = 0.13 s
Velocity of sound =HBSE 9th Class Science Important Questions Chapter 12 Sound - 12 = \(\frac{44.1}{0.13}\) = \(\frac{441 \times 100}{10 \times 13}\) = 339.2 m/s

Question 10.
A tuning fork produces waves of frequency 512 Hz by vibration, which moves with a velocity of 330 m/s. Find its wavelength.
Solution:
Here,
Frequency (υ) = 512 Hz
Wave velocity (v) = 330 mis
Wavelength (λ) = ?
We know that,
v = υ λ
Or λ = \(\frac{v}{v}=\frac{330}{512}\) m = 0.645 m

Question 11.
What is the speed of that water wave whose wavelength ¡s 20cm and frequency is 20 Hz.
Solution:
Here,
Wavelength (λ) = 20 cm = \(\frac{20}{100}\) m = 0.2m
Frequency (υ) = 20 Hz
Velocity of wave (v) = ?
We know that, v = υ λ = 20 x 0.2 = 4 m/s

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 12.
Find the frequency of that wave whose time period is 0.02 s.
Solution:
Here,
Time period (T) = 0.02 s
Frequency (υ) = ?
We know that
υ = \(\frac{\mathrm{I}}{\mathrm{T}}=\frac{1}{0.02}\) = \(\frac{100}{2}\) = 50 Hz

Question 13.
If the frequency of a tuning fork is 400 Hz and the speed of sound in air is 320 m/s, find the distance covered by the sound when tuning fork completes 20 vibrations.
Solution:
Here,
Wave velocity (v) = 320 mIs
Frequency (υ) = 400 Hz
We know that v = υ λ
or λ = \(\frac{v}{v}=\frac{320}{400}\) = 0.8 m
Thus, the distance covered in I vibration = 0.8 m
The distance covered in 20 vibrations = 20 x 0.8 = 16 m

Question 14.
An object completes 6000 vibrations in one minute. If the speed of sound in air is 330 m/s, find the (a) frequency of vibrations (b) wavelength of produced waves.
Solution:
(a) 1 minute = Number of vibrations in 60 s = 6000
Number of vibrations in I second = \(\frac{6000}{60}\) = 100
Thus, frequency (υ) = 100 Hz

(b) Here, velocity of wave (v) = 330m1s
Frequency (υ) loo Hz
Wavelength (λ) = ?
We know that v = υλ
or λ = \(\frac{\mathrm{v}}{v}=\frac{330}{100}\) = 3.3 m

Question 15.
The velocity of light wave is 3 x 105 km/s and wavelength is 6 x 107 m. Find its frequency.
Solution:
Here,
Velocity of wave (v) = 3 x 105 km/s
3 x 105 x 1000.m/s
3 x 108 m/s
Wavelength (λ) = 6 x 10-7 m
Frequency (υ) = ?
We know that v = υλ
υ = \(\frac{v}{\lambda}=\frac{3 \times 10^8}{6 \times 10^{-7}}\) = 0.5 x 1015 Hz
Thus, frequency is 0.5 x 1015 Hz.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Question 16.
An observer is standing at a distance of 200 metres from a cliff and whistles. If the speed of sound is 332 mIs, when the echo of whistle will be heard?
Solution:
Distance of observer from cliff = 200 m
To hear the echo,
Total distance covered = 200 + 200 400 m
Speed of sound = 332 m/s
Time = HBSE 9th Class Science Important Questions Chapter 12 Sound - 12
\(\frac{400}{332}\) = 12 s

Question 17.
A Sound signal is send from a boat. That signal return after 0.8s by reflecting from the sea bottom. Find the velocity of sound in water if the depth of sea is 600m.
Solution:
Here,
Depth of sea = 600 m
Distance travelled by the signal = 600 + 600 = 1200 m
Time = 0.8s
Velocity of Sound in water = HBSE 9th Class Science Important Questions Chapter 12 Sound - 12
= \(\frac{1200}{0.8}\) m/s
1500 m/s

Question 18.
A sound signal is send from a boat. That signal return after 0.8 s by reflecting from the sea bottom. If the velocity of sound is 1500 mIs, then find the depth of sea.
Solution:
Here,
Velocity of sound in water = 1500 mis
Time taken = 0.8 s
Distance travelled by sound = Velocity of sound x time
1500 x 0.8 = 1200m
Thus Depth of sea latex]\frac{1200}{2}[/latex] = 600 m

Practical Work

Experiment 1.
To study the vibrations produced by tuning fork.

Procedure:
(i) Suspend a small plastic ball by a thread from a support.
(ii) Bring the non-vibrating tuning fork near the ball and observe the situation.
(iii) Now bring the vibrating tuning fork near the ball and observe the situation.
In first situation, the ball does not move far from tuning fork while in second situation, the ball moves far as the tuning fork is taken near to it, because the prong of tuning fork applies the force on ball and pushes it backward.
HBSE 9th Class Science Important Questions Chapter 12 Sound - 18

Experiment 2.
To study the vibrating parts of musical instruments.

Procedure:
Go into the music room of your school and know about the system of producing sound in musical instruments from music teacher.

For example:
(i) In drum: by the vibration of membrane.
(ii) Sitar: by vibration in wire.
(iii) Ektara: by vibration in wire.
(iv) In flute: by air column.
(v) In water wave: by water.

HBSE 9th Class Science Important Questions Chapter 12 Sound

Experiment 3.
To produce longitudinal wave by slinky.

Procedure:
(i) Take a slinky. Ask your friend to hold one end and you hold the other end.
(ii) Give it a sharp push towards your friend. A special motion will be seen in slinky while repeated.
Mark a dot on the slinky and repeat the process again. The dot on the slinky will move back and forth parallel to the direction of the propagation of the disturbance. This is longitudinal wave.

Experiment 4.
To understand the reflection of sound.

Procedure:
Take a hallow cylinder of cardboard opened from both ends. Keep a small clock near the open end of cylinder and keep the wooden block in such a way that the person standing on the other end of the cylinder can hear the sound of tik-tik of clock. Now keep the other wooden block far from the other open end of cylinder. At this time sound reaches the ears after striking from surface. The change in the direction of sound after striking from any surface is called reflection of sound.

Quick Review of the Chapter

1. The sound of a bell is
(A) mechanical energy
(B) sound energy
(C) heat energy
(D) light energy
Answer:
(B) sound energy

2. Sound is produced due to
(A) striking
(B) vibration
(C) beating
(D) rubbing
Answer:
(B) vibration

3. Sound does not travel/transmit in:
(A) solids
(B) liquids
(C) gases
(D) vacuum
Answer:
(D) vacuum

4. Vave is a …………
(A) disturbance
(B) compression
(C) rarefaction
(D) vibration
Answer:
(A) disturbance

5. A region of high pressure is called :
(A) wave
(B) compression
(C) rarefaction
(D) disturbance
Answer:
(B) compression

HBSE 9th Class Science Important Questions Chapter 12 Sound

6. In which of the following the speed of sound maximum?
(A) Dry air
(B) steel
(C) wet air
(D) water
Answer:
(B) steel

7. For sound propagation, which is the most common medium?
(A) water
(B) earth
(C) air
(D) space
Answer:
(C) air

8. Which are waves produced in water?
(A) longitudinal
(B) transverse
(C) partially both (A) and (B)
(D) sometime (A) and sometime (B)
Answer:
(B) transverse

9. The SI unit of wave velocity is:
(A) m
(B) mis
(C) km
(D) kmls
Answer:
(B) m/s

10. Which of the following is represented by λ (lambda)?
(A) Wavelength
(B) Wave velocity
(C) Frequency
(D) Amplitude
Answer:
(A) Wavelength

11. The transfer of takes place in sound propogation.
(A) mass
(B) both (A) and (C)
(C) energy
(D) none of these
Answer:
(C) energy

HBSE 9th Class Science Important Questions Chapter 12 Sound

12. The longitudinal waves are produced in :
(A) solids
(B) liquids
(C) gases
(D) all three
Answer:
(D) all three

13. The transverse waves are produced ¡n :
(A) solid and liquids
(B) liquids and gases
(C) gases and solids
(D) all three
Answer:
(A) solid and liquids

14. Light is wave.
(A) transverse
(B) mechanical
(C) longitudinal
(D) electromagnetic
Answer:
(A) transverse

15. The frequency of a source of sound is loo Hz. How many times does it vibrate in Half an hour?
(A) 1,80,000
(B) 90,000
(C) 60,000 (D) I ,00,000
Answer:
(A) 1,80,000

16. The distance between is called wavelength.
(A) two consecutive compressions
(B) two consecutive rarefactions
(C) a compression and a rarefaction
(D) both (A) and (B)
Answer:
(D) both (A) and (B)

17. The SI unit of wavelength is.
(A) metre
(B) joule
(C) erg
(D) newton
Answer:
(A) metre

18. What is the unit of frequency of a sound wave?
(A) Hertz
(B) Metre
(C) erg
(D) Joule
Answer:
(A) Hertz

HBSE 9th Class Science Important Questions Chapter 12 Sound

19. The SI unit of time period is:
(A) Hertz
(B) Metre
(C) Second
(D) Nm
Answer:
(C) Second

20. The magnitude of the maximum distrubance in the medium on either side of the mean is called
(A) frequency
(B) amplitude
(C) velocity
(D) none of the above
Answer:
(B) amplitude

21. What is represented by AR?
(A) frequency
(B) velocity
(C) amplitude
(D) time period
Answer:
(C) amplitude

22. The sound of single frequency is called
(A) note
(B) tone
(C) noise
(D) none of these
Answer:
(B) tone

23. A sound which is produced due to mixture of several frequencies is called
(A) note
(B) tone
(C) noise
(D) lyrics
Answer:
(A) note

24. The amount of sound passing each second through unit area is called . of sound.
(A) loudness
(B) intensity
(C) pitch
(D) frequency
Answer:
(B) intensity

25. In which medium the speed of sound is maximum at 25°C?
(A) glass
(B) steel
(C) aluminium
(D) nickel
Answer:
(C) aluminium

HBSE 9th Class Science Important Questions Chapter 12 Sound

26. In which liquid the speed of sound is maximum at 25°C?
(A) sea water
(B) distilled water
(C) ethanol
(D) methanol
Answer:
(A) sea water

27. To listen the echo clearly the minimum distance of reflector should be:
(A) 172m
(B) 17.2m
(C) 344 m
(D) 344 m
Answer:
(B) 17.2 m

28. In which gas the speed of sound is maximum at 25°C?
(A) oxygen
(B) air
(C) hydrogen
(D) helium
Answer:
(C) hydrogen

29. The sensation of sound can remain about second in our brain.
(A) 0.1
(B) 0.2
(C) 0.3
(D) 0.4
Answer:
(A) 0.1

30. Sound obsorber substances are installed walls and roofs of building to reduce
(A) echo
(B) sound reflection
(C) reverberation
(D) supersonic sound
Answer:
(C) reverberation

HBSE 9th Class Science Important Questions Chapter 12 Sound

31. The principle of reflection of sound is used:
(A) in megaphone and horns
(B) stethoscope
(C) in concert hail, cinema halls
(D) in all aboe mentioned
Answer:
(D) in all above mentioned

32. What is hearing range of sound in man?
(A) 10-15 Hz
(B) 20-20.000Hz
(C) 30.000 Hz-40,000 Hz
(D) 45,000 Hz-50,000 Hz
Answer:
(B) 20-20,000 Hz

33. How much Hz sound can the children and dogs of age below S hear?
(A) to25kHz
(B) to30kHz
(C) to 35 k Hz
(D) to 45 k Hz
Answer:
(A) to 25 k Hz

34. Sounds of frequencies below are called infrasonic sound.
(A) 20 Hz
(B) 200 Hz
(C) 2000 Hz
(D) 20000 Hz
Answer:
(A) 20 Hz

35. Frequencies higher than is called ultrasonic sound.
(A) 10kHz
(B) 20kHz
(C) 25 k Hz
(D) 30 k Hz
Answer:
(B) 20 k Hz

36. Ultrasonic sound is used :
(A) for cleaning dust, oilness and dirt
(B) to detect cracks in bodies of metal
(C) in ECG and ultrasound
(D) in all above mentioned
Answer:
(D) in all above mentioned

37. SONAR is used to measure the of bodies under sea water.
(A) distance
(B) direction
(C) speed
(D) all above mentioned
Answer:
(D) all above mentioned

38. Which part is lamella of an ear?
(A) outer part
(B) midd’e ear
(C) inner ear
(D) auditory nerve
Answer:
(A) outer part

HBSE 9th Class Science Important Questions Chapter 12 Sound

39. How are the waetength and frequency of a sound wave related to its speed ?
(A) λ= vυ
(B) v = \(\frac {υ}{λ}\)
(C) υ = λ υ2
(D) v = υλ
Answer:
(D) v= vA

40. The number of complete oscillations per unit time is called
(A) Amplitude
(B) Wavelength
(C) Frequency
(D) Time period
Answer:
(C) Frequency

HBSE 9th Class Science Important Questions Chapter 12 Sound Read More »

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Haryana State Board HBSE 9th Class Science Important Questions Chapter 11 Work and Energy Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 11 Work and Energy

Very Short-Answer Type Questions

Question 1.
What is work ?
Answer:
When force is applied on an object and it moves in the direction of force, it is said to be work done.

Question 2.
On which the quantity of work depends ?
Answer:
The quantity of work depends upon the applied force & distance moved.

Question 3.
What is the formula of work ?
Answer:
Work = Force x Displacement.

Question 4.
What is the unit of work ?
Answer:
Newton-metre (Joule).

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Question 5.
What is energy ?
Answer:
The capacity of doing work is called energy.

Question 6.
What are the types of mechanical energy ?
Answer:
Mechanical energy is of two types:
(1) Kinetic energy,
(2) Potential energy.

Question 7.
What is potential energy ?
Answer:
The energy possessed by a body due to its position or change in shape is called potential energy.

Question 8.
Which type of energies are in aircraft flying on height ?
Answer:
Both type of energies-kinetic and potential.

Question 9.
Which type of energy is there in a stone lying on a hill ?
Answer:
Potential energy.

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Question 10.
What is mechanical Energy ?
Answer:
The sum of kinetic energy and potential energy of a body is called mechanical energy.

Question 11.
Which type of energy is there when a bullet is fired from a gun ?
Answer:
Kinetic energy.

Question 12.
Write the formula for kinetic energy.
Answer:
Kinetic energy = \(\frac {1}{2}\) mv2, where m → mass, v → velocity.

Question 13.
Write the formula for potential energy.
Answer:
Potential energy = mgh, where m → mass, g → acceleration due to gravity and h → height.

Question 14.
What is the law of conservation of energy ?
Answer:
According to this law, energy can neither be created nor be destroyed but can only be transferred from one form to another.

Question 15.
What is the effect on energy when work is done ?
Answer:
When work is done, energy decrease in some extent.

Question 16.
What is the work done by a man when he lifts a luggage of mass 10 kg on his head and moves 50 metres on horizontal road ?
Answer:
zero.

Question 17.
From where, our body gets energy ?
Answer:
From food taken by us.

Question 18.
Which type of energy is used by plants ?
Answer:
Solar energy.

Question 19.
Write an important use of wind energy.
Answer:
Wind energy is used to grind the flour by wind mills.

Question 20.
Write the name of any two natural sources of energy.
Answer:
(i) Sun
(ii) Water.

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Question 21.
Write any four types of energy.
Answer:
(i) Mechanical Energy
(ii) Chemical Energy
(iii) Light Energy
(iv) Sound Energy.

Question 22.
What is gravitational potential energy ?
Answer:
The energy stored in a body when work is done against force of gravity is called gravitational potential energy.

Question 23.
What is mean by elastic potential energy ?
Answer:
The energy possessed by an object due to change in its shape is called elastic potential energy.

Question 24.
Why spark is produces when two stones collides with each other ?
Answer:
Because kinetic energy of stone is converted into light energy.

Question 25.
How arrow gets its kinetic energy ?
Answer:
The arrow converts the potential energy from bow into kinetic energy.

Question 26.
Out of energy and acceleration, which is vector ?
Answer:
Acceleration is a vector quantity.

Question 27.
The kinetic energy goes on decreasing when it is thrown vertically upward. What will be the motion at highest point ?
Answer:
Zero.

Question 28.
Write the name of that device which converts electric energy into mechanical energy.
Answer:
Electric motor.

Question 29.
Which energy is converted in other in a hydroelectric power centre ?
Answer:
The kinetic energy of water is converted into electric energy.

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Question 30.
What is power ? What is its unit ?
Answer:
The rate of doing work is power. Its unit is watt.

Question 31.
If the velocity of an object is increased three times, what will be its kinetic energy ?
Answer:
The kinetic energy will increase nine times.

Question 32.
Which type of energy is there in a expanded spring ?
Answer:
Potential energy.

Question 33.
Which type of energy is there in a dam ?
Answer:
Potential energy.

Question 34.
Which type of energy is there in the spring of watch ?
Answer:
Potential energy which gives the kinetic energy to the needles of watch. ,

Question 35.
Which type of energy is there in running train ?
Answer:
Kinetic energy.

Question 36.
What will be potential energy of that body whose height is doubled ?
Answer:
Its potential energy will be doubled.

Question 37.
What is the relation between 1 kilowatt and watt ?
Answer:
1 kilowatt = 1000 watt.

Question 38.
Why the palms become hot when rubbed ?
Answer:
Because kinetic energy of palms is converted into heat energy.

Question 39.
How much energy is required to lift a stone of mass 1 kg to a height of 1 metre ?
Answer:
9.8 Joules [∵ W = mgh = 1 x 9.8 x 1 = 9.8 joules]

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Question 40.
A person and his son are running with uniform velocity. If the mass of person become two times more than of his son, then what will be the ratio between kinetic energy of both ?
Answer:
2 : 1 [ ∵ K.E. = \(\frac {1}{2}\) mv2 and v is equal]

Question 41.
Which type of energy is required to move for a sailboat ?
Answer:
The sailboat moves by the kinetic energy of wind.

Question 42.
What will be the energy a body has a capacity to do a work of 10 joules ?
Answer:
10 Joules.

Question 43.
Write the name of two units of power used in daily life.
Answer:
Kilowatt and megawatt.

Question 44.
How many watts are there in one horse power ?
Answer:
1 horse power = 746 watt.

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Question 45.
What is meant by 1 kilowatt hour energy ?
Answer:
The energy consumed by a source of 1 kilowatt power in 1 hour.

Question 46.
What is the relation between kilowatt hour and joule ?
Answer:
1 kilowatt hour = 36,00,000 joules.

Question 47.
Which type of energy is used when a person climbs stairs ?
Answer:
Muscular energy.

Question 48.
Which type of energy is there in our body ?
Answer:
Chemical energy.

Question 49.
The person requires the least amount of energy in which of the following activity ?
(i) In swimming
(ii) In movement
(iii) In sleeping
(iv) In running foot.
Answer:
(iii) In sleeping.

Question 50.
What is simple pendulum ?
Answer:
Simple pendulum is a metalic bob which is hang by weightless string.

Question 51.
Write the name of energy when simple pendulum is at its extreme point.
Answer:
Potential energy.

Question 52.
Write the name of energy when pendulum is in middle position.
Answer:
Kinetic energy.

Short-Answer Type Questions

Question 1.
What is meant by work ? How is it measured ? What is its unit ?
Answer:
Work done on an object is defined as the magnitude of the force multiplied by the distance moved by the object in the direction of the
applied force.
Thus, Work done = Force x The distance moved in the direction of force
W = F x s
Work is a vector quantity. Unit of work is newton-metre or joule (J).
HBSE 9th Class Science Important Questions Chapter 11 Work and Energy - 1

Question 2.
What is energy ? Write its unit and definition. What are its various types ?
Answer:
The capacity of doing work is energy. Its unit is also joule.
Joule: Work done is said to be one joule when a force of 1 newton acting on an object and the object is displaced through 1 m in the direction of force.

Various types of energy:
(i) Mechanical energy
(ii) Heat energy
(iii) Chemical energy
(iv) Electric energy
(v) Nuclear energy’
(vi) Light energy
(vii) Atomic energy
(viii) Magnetic energy
(ix) Sound energy,
(x) Solar energy.

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Question 3.
Justify with example that moving bodies have capacity to do work ?
Answer:
The examples are following :
(i) A moving marble has a capacity to displace that marble which is in rest.
(ii) The blowing wind works by rotating the wheels of windmill.
(iii) The moving water also takes the objects from one place to another.
(iv) The blowing wind is helpful for sailing boat.
Thus, we can say that moving bodies have capacity to do work.

Question 4.
What is meant by potential energy and kinetic energy ? Give two examples for each.
Answer:
Potential energy’: The energy possessed by a body due to its change in position or shape is called the potential energy.
Examples:
(1) The book lying on the table has potential energy due to its position.
(2) A binded watch has potential energy.

Kinetic energy:
The energy possessed by a body due to its motion is called the kinetic energy.

Examples:
(1) Moving air has kinetic energy.
(2) The running water has kinetic energy.

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Question 5.
Is there energy in following substances ? If yes, then what is the type-potential energy or kinetic energy or both ?
(i) The switched off ceiling fan.
(ii) A man climbing up mountain.
(iii) Flying bird.
(iv) The water in a dam.
(v) The stretched bow more than its shape.
(vi) The rubber locket on a table.
(vii) Stretched locket of rubber.
Answer:
Yes, all substances have energy’.
(i) A switched off ceiling fan has potential energy only.
(ii) A man climbing up mountain has both potential and kinetic energy.
(iii) A flying bird has both kinetic and potential energy.
(iv) The water in the dam has potential energy.
(v) A stretched bow has potential energy.
(vi) The rubber locket on the table has potential energy.
(vii) Stretched locket of rubber has potential energy.

Question 6.
Write the difference between potential energy and kinetic energy.
Answer:
The differences between potential energy and kinetic energy are as follows:
Potential Energy (P.E.):
1. The energy possessed by an object due to its position is called potential energy.
2. Potential energy = mgh
3. The potential energy of an object depends upon the height or depth from the surface of earth.

Kinetic Energy (K.E.):
1. The energy possessed by an object due to its motion is called kinetic energy.
2. Kinetic energy = \(\frac {1}{2}\) mv2
3. Kinetic energy depends upon the motion of the object.

Question 7.
Which type of energy is in following ?
(1) Coiled spring of watch.
(2) Running buffallo.
(3) The arrow released from bow.
(4) A table lying on roof.
(5) Compressed air.
(6) A mango on tree.
Answer:
(1) Potential energy
(2) Kinetic energy
(3) Kinetic energy
(4) Potential energy
(5) Potential energy
(6) Potential energy.

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Question 8.
When the capacity of doing work by human body decreases ?
Answer:
The capacity of doing work by human body decreases in following circumstances :
(1) When a person is ill.
(2) When a person grows old.
(3) When a person takes food less than required, his muscular energy decreases.

Question 9.
What is the relation between work and energy ? Write their units.
Answer:
The capacity of doing work is called energy. The magnitude of energy is always equal to that work which has a capacity to do that. Energy is measured in joule and the unit of work is also joule.

For example:
To lift a piece of stone of mass 1 kg upto a height of 1 metre requires 9.8 joules energy. Opposite to it, when stone is dropped from one metre, the stone does the same amount of work (9.8 J). Thus, if an object has an energy of 100 joules. It has a capacity of doing work of 100 joules.

Question 10.
How the energy is transferred in key toy ?
Answer:
There is a spring in key toy or key watch which contracts while binding and stored potential energy in it. After some moments, the spring loses slowly and the toy comes in motion. Thus, potential energy is transferred in kinetic energy.

Question 11.
A ball is thrown upward and it returns back. What is the change in kinetic and potential energy of the ball ? Explain.
Answer:
When a ball is thrown upward its kinetic energy changes continuously into potential energy. At highest point its whole kinetic energy changes into potential energy. When the ball returns back its potential energy changes continuously into kinetic energy and whole of its potential energy changes into kinetic energy near the earth but on reaching the earth, whole of its kinetic energy changes into potential energy.

Question 12.
A boy is standing by holding some weight in his expanded hand. In this stage, there seems no change in the position of both boy and weight. Is he doing any work ? If yes, then justify.
Answer:
Yet, there is no change in the position of boy and weight externally, but internally the boy is working. The muscles of boy are stretched due to weight on palm. Thus, their shape get changes. The heart pumps more blood into muscles. The energy of boy wastes in all these chemical activities and he tired soon. If the boy stand more enough in same position, the external change will be observed as sweating.
HBSE 9th Class Science Important Questions Chapter 11 Work and Energy - 2

Question 13.
How much work is done when an artificial satellite revolve round the earth ? Explain in brief. –
Answer:
The artificial satellite revolves round the earth in almost circular track. The gravitational force between the earth and the satellite works along the line joining the earth and satellite and the satallite moves perpendicular to the direction of force of gravity. Thus, the magnitude of displacement of satellite in the direction of force will be zero and the work done by the earth on the satellite will also be zero.

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Question 14.
How the kinetic energy of wind is utilised ? Write its important uses.
Answer:
The electric energy can be produced by the energy in the blowing air. It has much waste. Thus, wind mill is used in grinding flour and drawing out water from ponds. A small amount of electric energy is produced from wind mill. Those places where the wind blows throughout the years, wind mill are used to grind flour and drawing out water.
Important uses of mechanical energy of wind :
(1) The fields can be irrigated by drawing water from ponds.
(2) It can be used in grinding the flour.
(3) The mechanical energy of wind can be converted into electric energy by joining the various wind wheels together.

Question 15.
What is water energy ? Write its two important uses.
Answer:
The running water has a capacity to do work due to its kinetic energy, which is known as water energy.

Uses of water energy: Following are the uses of water energy :
(1) Water energy is converted into electric energy and it is used in houses and factories.
(2) Water energy is used to grind flour by windmills.
(3) The heavier wooden pieces is taken from one place to another due to water energy.

Question 16.
How water energy is used to produce electricity ?
Answer:
The kinetic energy of running water is stored as potential energy in dam. The stored water is dropped on the larger blades of water turbines. The shaft attached with running turbine rotates the generator or dyanmo quickly. The rotating coil of dyanmo produces electricity which is called ‘hydroelectric power’. The electricity is produced on Bhakhra dam by this method. This is also known as the principle of hydroelectricity.
HBSE 9th Class Science Important Questions Chapter 11 Work and Energy - 3

Question 17.
Why the nail becomes hot when it is hammered in the sheet of wood ?
Answer:
The nail becomes hot when it is hammered in the sheet of wood because the lifted hammer has potential energy due to its position. When the hammer is dropped on the nail, it comes in rest position, but transfers, its whole energy in the nail. As a result of this, the nail gets some kinetic energy and moves inside the wood. When the nail enters completely and we continuous hammering the nail becomes hot because the mechanical energy of the hammer is transferred in nail and changes into heat energy.

Question 18.
What are the benefits of the solar energy ?
Answer:
The benefits of the solar energy are as follows :
(1) The solar energy is used in solar cooker to prepare food,
(2) Solar energy is used to warm the water by solar heater in hotels, hospitals and industries.
(3) Solar energy is used in the production of electricity.
(4) It is used to melt the metals in solar furnances.

Question 19.
Differentiate among work, energy and power. Write the SI unit for each. Or What is Power? Write its S.I. unit.
Answer:
Work:
When force is applied on a body and it moves in the direction of force, it is said to be work done.
Thus, Work (W) = Force (F) x Distance (S)
SI unit of work is joule.

Energy: t he capacity ot doing work is called energy. SI unit of energy is joule.

Power: The rate of doing work is called power.
Thus,
HBSE 9th Class Science Important Questions Chapter 11 Work and Energy - 4
SI unit of power is watt.

Question 20.
What do you mean by transfer of energy ? Write two examples of transfer of energy.
Answer:
According to the law of conservation of the energy, The energy can neither be created nor destroyed. It can be transferred from one form to another. It is called the transfer of energy. Thus, the total magnitude of energy is never changed.

Examples:
(1) Ice on mountains melts and becomes water. This is the transfer of potential energy into kinetic energy.
(2) The electric energy in bulb changes into light energy and glows.

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Numerical Questions

Question 1.
A boy of mass 40 kg is running with a ve’ocity of 0.5 ms. What is his kinetic energy ?
Solution:
Here,
m = 40 k
v = 0.5 ms-1
We know that,
Kinetic Energy (K.E.) = \(\frac {1}{2}\) mv2
= \(\frac {1}{2}\) x 40 x 0.5 x 0.5 = 5 Joule Answer:

Question 2.
An object of mass 15 kg is moving with a uniform velocity (if 4 ms. What is tue kincth energy possessed by the object?
Solution:
Here
Mass of the body (m) = 1 5 kg
Velocity of the body (v) = 4 rn/s
We know that
Kinetic Energy (K.E) = \(\frac {1}{2}\) mv2
= \(\frac {1}{2}\) x 15 x 4(2)2
= \(\frac {1}{2}\) x 15 x 16
= 120 Jule

Question 3.
A stone of mass 100 gram is kept at a height ofS metre. What kind of energy does it possess and find out its value.
Solution:
The stone has potential energy because of height.
Here. m = 100g = \(\frac {100}{1000}\) = 0.1 kg
g = 9.8 ms-2
h = 5m
We know that.
Potential Energy (P.E.) m x g x h = 0.1 x 9.8 x 5 = 4.9 Joule

Question 4.
To what height can an object of mass 1 kg be raised by giving it an energy of 980 joule?
Solution:
Here.
m = 1 kg
g = 9.8 ms-2
h = ?
Potential Energy (RE.) = 980 J
We know that,
Potential Energy (RE.) = m.g.h.
or 980 = 1 x 9.8 x h
or h = \(\frac{980}{1 \times 9.8}\)
= 100 m

Question 5.
What is the increase in kinetic energy of a particle, if its velocity is increased to four times?
Solution:
Suppose the initial velocity of the particle = v
and its final velocity = 4 v
Suppose the mass of the particle = m
∴Initial kinetic energy (K1) of the particle = \(\frac {1}{2}\) mv2 ………….(i)
Final kinetic energy (K2) of the particle = \(\frac {1}{2}\)m (4v)2
= \(\frac {1}{2}\)m x 16 v2 …………..(ii)
From equations (i) and (ii),
\(\frac{\mathrm{K}_1}{\mathrm{~K}_2}=\frac{\frac{1}{2} \mathrm{mv}^2}{\frac{1}{2} \mathrm{~m} \times 16 \mathrm{v}^2}=\frac{1}{16}\)
or K2 = 16K1
Thus, the increase in kinetic energy is 16 times the initial kinetic energy.

Question 6.
The kinetic energy of a ball of mass loo g is 20 joule. Calculate its velocity.
Solution:
Here,
m = 100 g = \(\frac {100}{1000}\)kg = 0.1 kg
v = ?
Kinetic Energy (K.E.) = 20 J
We know that, Kinetic Energy (K.E) = \(\frac {1}{2}\) mv2
Or 20 = \(\frac {1}{2}\) x 0.1 x v2
v2 = \(\frac {20 x 2}{0.1}\) = 400
or v = \(\sqrt{400}\) = 20 ms-1

Question 7.
A coolie lifts an object of mass 20 kg from the earth to 1.5 m and keep It on his head. Calculate the amount of work done by him on the object.
Solution:
Here,
Mass of the object (m) = 20kg
Displacement (s) = 1 .5m
Work done (W) = F x s = mg x s
= 2okg x 10 ms-2 x 15 m
= 300kgms-2 m
= 300Nm= 300J
The amount of work done by the coolie on the object is 30 joules.

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Question 8.
A man of mass 50 kg jumps to a height of 1.2 m. What is its potential energy at the highest point?
Solution:
Here,
m = 50kg
g = 9.8 ms-2
h = 12m
We know that,
Potential Energy (P.E.) m x g x h
= 50 x 9.8 x 1.2
= 588 J
Thus, at the highest point the potential energy of the man is 588 J.

Question 9.
With what velocity should a man of mass 50 kg run so that his kinetic energy becomes equal to 625 J.
SolutIon:
Here,
m = 50kg
v = ?
Kinetic Energy (K.E.) = 625 J
We know that,
Kinetic Energy (K.E) = \(\frac {1}{2}\) mv2
Or 625 = \(\frac {1}{2}\) x 50 x v2
v2 = \(\frac{625 \times 2}{50}\) = 25
or v = \(\sqrt{25}\) 5ms

Question 10.
A man throws a stone of mass 10 kg from a ladder of height 5m. When ¡t reaches the ground, what will be its kinetic energy ? What will be its velocity at the nearest point to the ground?
Solution:
Here,
m = 10kg
h = 5m
v = ?
Kinetic Energy (K.E.) = ?

We know that,
The kinetic energy (K.E.) of the stone = The loss in potential energy due to falling of the stone
= mgh
= 10 kg x 9.8 ms-2 x 5m = 490J ……………(i)
Because kinetic energy = \(\frac {1}{2}\) mv2 .(ii)
From eqns. (i) and (ii),
\(\frac {1}{2}\) mv-2 = 490J
or \(\frac {1}{2}\) x 10kg x v2 = 490J
or v2 =\(\frac{490 \times 2}{10}=\frac{980}{10}\) = 98
or v = \(\sqrt{98}\) = 9.9 ms-1 (approx.)

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Question 11.
A car of mass 1000 kg is moving with a velocity of 30 ms-1. After applying brakes it stops with uniform acceleration at a distance of 50m. Calculate the force and the work done by the brakes?
Solution:
Here,
m = 1000kg
u = 30 ms-1
v = 0 ms-1
s = 50m
F = ?
W = ?
We know that, v2 – u2 = 2as
a = \(\frac{v^2-u^2}{2 s}=\frac{(0)^2-(30)^2}{2 \times 50}=\frac{-900}{100}\) ms-2
= – 9 ms-2 So retardation = 9ms-2
F = m.a.l000kg x 9ms2 = 9000 N
The force applied by the brakes on the car (F) = 9000 N
Work done(W) = F x s = 9000 N x 50 m = 450000 J = 450 kJ

Question 12.
A man of mass 50kg climbs a vertical height of 10m on a mountain in 20 seconds. What ¡s his power ? (Given that g 9.8 ms-2)
Solution:
Here,
F = m x g
= 50 x 9.8 = 490N
Now, W = F x S = 490N x 10m = 4,900 Nm
Time (t) = 20 seconds
Power (P) = \(\frac{\text { Work done }}{\text { Time taken }}=\frac{4,900}{20}\) = 245 watt

Question 13.
A boy of mass 40 kg runs up a staircase of 45 steps in 9 sec. If the height of each step is 15 cm, Calculate his power ? (The value of g is 10 ms2)
Solution:
Weight of the boy (mg) 40kg x 10ms2 = 400N
Total height of the steps (h) = 45 x 15/100m = 675m
Time taken to climb the steps (t) = 9s
We now that
Power (P) = Time taken = \(\frac{\mathrm{mgh}}{t}\) = \(\frac{400 \mathrm{~N} \times 6.75 \mathrm{~m}}{9 \mathrm{~s}}\) = 300 W
The power of the boy is 300 W.

Question 14.
A man of mass 50 kg climbs 30 steps in 30 sec. If height of each step is 20 cm. Then calculate the power to climb the total steps.
Solution:
Mass of the man (m) = 50 kg
Total number of steps = 30
height of each step = 20 cm = \(\frac {20}{100}\) m = 0.2 m
Total height covered by the man = 30 x 0.2 m = 6m
Acceleration due to gravity (g) = 10 ms-2
Thus, work done by the man = m.g.h = 50 x 10 x 6J = 3000J
Time taken to climb the steps = 30 s
HBSE 9th Class Science Important Questions Chapter 11 Work and Energy - 5
= \(\frac{3000 \mathrm{~J}}{30 \mathrm{~s}}\)
= 100 Js-1 = 100 W(watt)

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Question 15.
An electric bulb of 100 Wis used for 2 hours, how much electrical energy is consumed?
Solution:
Power of the bulb = 100W
Time = 2 hours 2 x 3600 = 7200 sec.
Energy consumed = Power x Time
= 100 x 7200Ws = 720000 joule

Question 16.
A rocket has a mass of 3 x 106 kg. Calculate (a) its potential energy (b) its kinetic energy, at a height of 25 km moving with a velocity of 1 km/s. (Take g = 10 ms-2)
Solution:
Here,
Mass of the rocket (m) 3 x 106 kg
Velocity of the rocket (v) = 1 k ms-1 = 1000 ms-1
Height of the rocket (h) = 25 km = 25000 m
Acceleration due to gravity (g) = 10 ms-2

(a) Potential Energy = m.g.h.
3 x 106 x 10 x 25000 J
= 7.5 x 1011

(b) Kinetic Energy = \(\frac {1}{2}\) mv2
= \(\frac {1}{2}\) x 3 x 106 x (1000)2 J
= 7.5 x 1011J

Question 17.
A horse is pulling a cart moving with a velocity of 18 kmh-1 with a force of 300 N. Find out the power of the horse in watt and also in horse power?
Solution:
Here,
Velocity of the cart(v) = 18 kmh-1
= \(\frac{18 \times 1000}{3600}\) ms-1 = 5 ms-1
The force exerted by the horse on cart (F) = 300 N
m.g. = 300N
Work done = m.g.h
= 300 x 5 = 1500J
HBSE 9th Class Science Important Questions Chapter 11 Work and Energy - 5
= \(\frac{15000}{1s}\) = 1500 w
We know that, 746 W = 1 horsepower
1 waft = \(\frac {1}{746}\) horsepower
1500 watt = \(\frac {1}{746}\) x 1500 horsepower
= 2 horsepower

Do Yourself

(1) A car is moving with a uniform velocity of 54 kmh-1 What will be the kinetic energy of a child of mass 40 kg sitting inside it?
Answer:
[4500J]

(2) Calculate the power of that pump which stores the water of loo kg in 25 seconds into a tank kept at a height of 19m. (Let g = 10 ms2)

(3) What will be the energy of an object of mass 10 kg at a height of 40 m from the ground level?
Answer:
[Take g 100 ms2] [40003]

(4) A force of 5 N is acting on a body. The body displaces to a distance of 2m in the direction of the force. Find out the work done on the object.
Answer:
[103)

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

(5) An object of mass 15 kg is moving uniformly with a velocity of 4 ms2. Calculate the kinetic energy of the object.
Answer:
[120J]

(6) What will be the height from the ground level where an object of mass 1 kg will have a potential energy of 1 joule if g= 10 ms2?
Answer:
[0.1 m]

(7) If 4900 joule of energy is utilised in lifting a body of mass 50 kg to a certain height, then what will be the height of the object if g = 9.8 ms2?
Answer:
[10m]

(8) A body of mass 3 kg starts falling down from rest position. After 3 sec., what will be its kinetic energyifgl0m.c2?
Answer:
[1350m]

(9) An object of mass 10 kg is lifted to a height of 6m from the ground level. Calculate the energy present in the object. The value of g is
9.8 ms2.
Answer:
[588 J]

(10) An object of mass 12 kg is kept at a certain height. 1f its potential energy is 480 J; calculate its height from the ground level.
Answer:
[Take g = 10 ms2] [4 m]

(11) A man of mass 50kg takes a packet of mass 20kg upto the terrace of a building of height 14m, what is the work done by the man ?
Answer:
[Take g = 10 ms2] [9604 J]

(12) A vehicle of mass 1800 kg is moving with a velocity of 40 ms. After applying brakes it stops with a uniform acceleration at a distance of 200 m. Calculate the force applied and work done by the brakes.
Answer:
[1440 kJ]

(13) What will be the increase in kinetic energy of a particle if its velocity is doubled?
Answer:
[4 times ]

(14) How much amount of work is done on lifting an object of mass 2 kg to a height of 2m if the value of g is 9.8 ms2?
Answer:
[39.2J1

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

(15) The height of’Kutubminar’ is 70m. What is the work done by a man of mass 50 kg on reaching the top?
Answer:
[Take g 9.8 ms-2] [3430011

(16) A stone of mass 10g is kept at a height of 5m. What kind of energy is present in it and calculate its value?
Answer:
[Take g 9.8 ms-2] [0.49J]

(17) Two girls A and B of mass 400 N each, climbs a rope upto a height of 8m. IfA takes 20 sec. to do this work while B takes 50 sec.,
then calculate the individual power of both the girls.
Answer:
[A 160 W, B = 64W]

(18) Find the energy in KWh consumed inS hours by four devices of power 500 W each.
Answer:
[10Kwh]

Essay Type Questions

Question 1.
Define potential energy. Derive an expression for the potential energy of a body.
Answer:
Potential Energy:
The energy possessed by a body due to its position or change in shape is known as potential energy.

Expression for potential energy:
Consider a body of mass m kept on the.ground level at point ‘A’. it is raised to a point ‘B’ by applying a force F on it. The distance between A and B is ‘h’.
As we know that, the force F acting on the body to lift it, is equal and opposite to the gravitational force
F = \(\frac{G M_e m}{\mathrm{R}_e{ }^2}\) …………(i)

Where, Me = Mass of the earth
Re = Radius of the earth
But, g = \(\frac{G M_e m}{\mathrm{R}_e{ }^2}\) (g = acceleration due to gravity) ………..(ii)
From equations (i) and (ii),
F = mg
The work done in raising the body to a height ‘h’
W = F x S (∵ s = h) = mgh
Because work done is same as the energy. Therefore, at a height ‘h’ the potential energy of the object is,
Potential Energy (P.E.) = mgh
HBSE 9th Class Science Important Questions Chapter 11 Work and Energy - 6

Question 2.
Define kinetic energy. Obtain an expression for it.
Answer:
Kinetic Energy:
The energy possessed by a body due to its motion is called kinetic energy.

Expression:
Let us suppose a body of mass m is dropped from a height h and its velocity becomes V.
on reaching the ground level Then –
Initial velocity of the body (u) = 0
Acceleration (a) = g (acceleration due to gravity)
Final velocity = v
Distance (s) = Height (h)
We know that, v2 – u2 = 2as
v2 – 0 = 2gh
v2 = 2gh
h = \(\frac{v^2}{2 g}\) …………. (i)
As the object comes down, its potential energy decreases and kinetic energy increases. When the body strikes to the ground level, its hoIe potential energy is converted into kinetic energy. According to the law of conservation of energy, Energy can neither be increased nr be decreased.”
Thus, kinetic energy at ground level = Potential energy at height h.
Kineiic Energy = mgh
or K.E. = mg x \(\frac{v^2}{2 g}\) (∵ h = \(\frac{v^2}{2 g}\))
K.E. = \(\frac {1}{2}\) mv2

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Question 3.
Define Law of conservation of energy. Give an example of conservation of energy.
Answer:
Law of conservation of energy-According to this law, energy can neither be created nor be destroyed. ¡t can only be transfèrred from one form to another. When the energy is released in any form, the same amount of energy appears in other forms. Thus, total energy of the universe remains constant. Suppose, an object of mass m starts falling from a height h, on the ground level.
HBSE 9th Class Science Important Questions Chapter 11 Work and Energy - 7

(a)EnergyatpointA, PotentialEnergy = mgh
Kinetic Energy = 0
∴ Total energy = mgh + 0 = mgh ……………(i)

(b)Energy at point B:
Suppose the object falls from A to B at a distance s, then its height from ground level will be (h-S).
∴ Potential energy = mg (h-s)
and v2-u2 = 2gs
v2 = 2gs [∵ u = 0]
Kinetic Energy = \(\frac {1}{2}\) mv2 = \(\frac {1}{2}\)m x 2gs = mgs
Total Energy = mg (h-s) + mgh = mgs – mgs + mgs = mgh …………(2)

(c) Energy at point C; v2-u2 = 2gh
v2 = 2gh [∵u = 0]
Kinetic Energy = \(\frac {1}{2}\) mv2 = \(\frac {1}{2}\) m x 2gh = mgh
and Potential Energy = 0
Total Energy = mgh + 0 = mgh ……………(iii)
in this way we observe from eQuestion (i), (ii) and (iii) that the energy of the object remains constant at every point. There is potential energy in the object at highest point. When it falls litle downs same part of potential energy is converted into kinetic energy and on ground level whole potential energy is converted into kinetic energy.

Practical Work

Experiment 1.
Describe by an experiment that the wark done by a force can either be positive or negative.

Procedure:
A child is pulling a toy car parallel to the ground as shown in figure. The child has exerted a force in the direction of displacement of car. In this situation, the work done will be equal to the product of the force and displacement. In this situation the work done by the force is taken as positive. Consider a situation in which an object is being displaced by the action of forces and we identit one of the forces, F acting Opposite to the direction of the displacement s, thus, the angle between the two directions is 1800.
HBSE 9th Class Science Important Questions Chapter 11 Work and Energy - 8

In such a situation, the work ‘ done by the force F is taken as negative and denoted by the minus sign. The work done by the force is F x (-s) or (-F x s). When the force acts opposite to the direction of the displacement, the work done is negative. When the force acts in the direction ofdisplacernen, the work done is positive.

Experimenf 2.
Describe by an activity in which both positive and negative forces are working.

Procedure:
When an object is lifted up, both positive and negative forces work, because the force applied by us, displaces the object upward and called positive force, while the force of gravity applied by earth pulls the object downward which is called negative force.

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

Experiment 3.
Describe by an experiment that the object at more height has more energy.

Procedure:
(1) Take a heavy ball. Drop it on a thick bed of sand. A wet bed of sand would be better. Drop the ball on the sand bed from height of about 25 cm. The ball creates a depression.
(2) Repeat this activity from heights of 50 cm, I m and 1.5 m.
(3) Ensure that all the depressions are distinctly visible.
(4) Mark the depressions to indicate the height from which the ball was dropped.
(5) Compare their depths.

We will observe that when the ball was dropped from the height of 1.5 m, the depression was the deepest; which proves that the object at more height has more energy which makes the deepest depression.

Expereiient 4.
Describe by an experiment that the kinetic energy of an object increases by increasing its mass.

Procedure:
(1) Set up the apparatus as shown in figure
(2) Place a wooden block of known mass in front of the trolley at a convenient fixed distance.
(3) Place a known mass on the pan so that the trolley starts moving.
(4) The trolley moves forward and hits the wooden block and displaces the block.
(5) Note down the displacement of the block. It is, clear that the work is done on the block by the trolley as the block has gained energy.
(6) Repeat this activity by increasing the mass on the pan.

You will observe that the displacement will increase with the increase in mass, which proves that the kinetic energy increases by increasing the mass.
HBSE 9th Class Science Important Questions Chapter 11 Work and Energy - 9

Quick Review of the Chapter

1. What is required for doing a work?
(A) displacement in object
(B) force should be applied on the object
(C) both (A) and (B)
(D) no displacement in object
Answer:
(C) both (A) and (B)

2. What type of energy is there inside a body kept at height?
(A) potential
(B) kinetic
(C) both (A) and (B)
(D) no relation with energy
Answer:
(A) potential

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

3. The device that converts the electrical energy In heat energy:
(A) Electrical iron
(B) Electrical bulb
(C) Radio
(D) Electrical motor
Answer:
(A) Electrical iron

4. Which type of energy is there in a flying bird?
(A) potential
(B) kinetic
(C) both (A) and (B)
(D) neither of the two
Answer:
(C) both (A) and (B)

5. Rado transfer the electrical enegry in:
(A) light energy
(B) sound energy
(C) mechanical energy
(D) heat energy
Answer:
(B) sound energy

6. Which formula is correct for work?
(A) W = \(\frac {F}{s}\)
(B) W = \(\frac {s}{F}\)
(C) W= Fs
(D) W = \(\frac {1}{Fs}\)
Answer:
(C) W= Fs

7. The unit of force Is:
(A) Newton
(B) Newton metre
(C) Joule
(D) both (B) and (C)
Answer:
(A) Newton

8. If 5n force acts on an object, then displacement is 2m, what will be the work done?
(A) lONm
(B) IOJ
(C) lOJm
(D) both(A)and(B)
Answer:
(D) both (A) and (B)

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

9. A coolie lifts an object of mass iS kg from the earth to 1.5 m keep it on his head. Calculate the
amount of work done by the coolie.
(A) 22.5J
(B) 2.25 J .
(C) 225J
(D) 225N
Answer:
(C) 225 J

10. The potential energy of an object of mass 1kg kept at the height &f 2m will be:
(A) 9.8J
(B) 19.6J
(C) 29.4 J
(D) 39.2 J
Answer:
(B) 19.6 J

11. Whose unit is Newton metre (Nm)?
(A) work
(B) force
(C) acceleration
(D) power
Answer:
(A) work

12. How many watts are in 1 Kilowatt?
(A) 103
(B) 104
(C) 105
(D) 106
Answer:
(A) 103

13. How many joules are in 1 Kilowatt?
(A) 36 lakhs
(B) 36 thousands
(C) 36 hundreds
(D) 36 crores
Answer:
(A) 36 lakhs

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

14. What is another name of newton metre?
(A) Erg
(B) Coulomb
(C) Joule
(D) Hertz
Answer:
(C) Joule

15. ……….. is vector.
(A) potential energy
(B) kinetic energy
(C) work
(D) none of these
Answer:
(D) none of these

16. A man stands for 10 minutes by carrying a weight of 20 kg the work done s ill be:
(A) 200joule
(B) onejoule
(C) zero
(D) 400joule
Answer:
(C) zero

17. The unit of energy is:
(A) Nm
(B) Joule
(C) both (A) and (B)
(D) no unit
Answer:
(C) both (A) and (B)

18. The example of kinetic energy is not:
(A) stone kept on mountain
(B) revolving wheel
(C) blowing air
(D) bullet shot out of a gun
Answer:
(A) stone kept on mountain

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

19. The example of potential energy is not :
(A) an arrow kept in the bow
(B) a stone on a mountain
(C) falling coconut
(D) a bullet in a gun
Answer:
(C) falling coconut

20. An object of 10 kg mass is carried to a height of 6 m. Calculate the energy of the object. (g = 9.8 mc2)
(A) 98 J
(B) 58.8 J
(C) 588 J
(D) 5880 J
Answer:
(C) 588 J

21. The law of conservation of mass is:
(A) all works require energy
(B) energy can be produced
(C) energy can be destroyed
(D) energy neither can be produced nor destroyed
Answer:
(D) energy neither can be produced nor destroyed

22. Which statement is not correct?
(A) Potential energy + kinetic energy constant
(B) Kinetic energy + potential energy = mechanical energy
(C) As much decrease is there in potential energy at a point so is in kinetic energy
(D) As much decrease is there in potential energy so much increase takes place in kinetic energy.
Answer:
(C) As much decrease is there in potential energy at a point so is in kinetic energy

23. Which formula is correct?
(A) P = W x t
(B) P = \(\frac {W}{t}\)
(C) P = \(\frac {t}{W}\)
(D) P = \(\frac{1}{W \times t}\)
Answer:
(B) P = \(\frac {W}{t}\)

24. What is the unit of power?
(A) Joule
(B) Nm
(C) erg
(D) Watt
Answer:
(D) Watt

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

25. How much energy will be consumed by a person of 600 newton weight for climbing a height of 10m?
(A) 60 joule
(B) 600 jouie
(C) ó000joule
(D) 588 joule
Answer:
(C) 6000 joule

26. When will be the minimum amount of potential energy in your body ? When you:
(A) are standing
(B) are sitting in a chair
(C) ire lying on the earth
(D) are sitting on the earth
Answer:
(C) are lying on the earth

27. When ¡s work not done?
(A) on fixing a nail in a wood
(B) on sliding a box on the floor
(C) when there is no constituent of parallel force in the direction of motion
(D) a weight hanging on a peg
Answer:
(C) when there is no constituent of parallel force in the direction of motion

28.
10 seconds are taken to fetch a bucket of 5 kg mass from a well of 10 m depth. The power used in the power will be:
(A) 5ONm
(B) 50)
(C) 50 W
(D) 500 W
Answer:
(C) 50 W

29. How much energy will be consumed by 5 fans of 100 watt each in 4 hours?
(A) 20 kwh
(B) 10 kwh
(C) 2 kwh
(D) 1 kwh
Answer:
(C)2 kwh

30. The rate of doing work is called:
(A) Power
(B) Force
(C) Energy
(D) None of these
Answer:
(A) Power

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy

31. What is the momentum of an object of mass m, moving with a velocity u?
(A) (mυ)2
(B) mυ2
(C) \(\frac {1}{2}\) mυ2
(D) mυ2
Answer:
(C) \(\frac {1}{2}\) mυ2

HBSE 9th Class Science Important Questions Chapter 11 Work and Energy Read More »

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Haryana State Board HBSE 9th Class Science Important Questions Chapter 10 Gravitation Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 10 Gravitation

Very Short-Answer Type Questions

Question 1.
What is meant by gravitation?
Answer:
Everybody on earth attracts each other. Such type force of attraction between bodies is called gravitation.

Question 2.
Who discovered the gravitational force firstly?
Answer:
Sir Isaac Newton.

Question 3.
In which direction the gravitational force between two objects works?
Answer:
The gravitational force between two objects works in the direction of a line joining the centre of their masses.

Question 4.
Which force is responsible for the existence of our solar system?
Answer:
Gravitational force is responsible for our solar system.

Question 5.
What ‘G’ indicates?
Answer:
‘G’ indicates gravitational constant.

Question 6.
What is the value of ‘G’ ? ’
Answer:
VMtte of‘G’ is 6.673 x 1011 Nm2/kg2

Question 7.
How much is the mass of earth ?
Answer:
Mass of earth (Me) = 6 x 1024 kg (approx.).

Question 8.
How much is the radius of earth ?
Answer:
Radius of earth (Re) = 6.4 x 1o6m.

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 9.
At which part of earth, the value of g is maximum and minimum?
Answer:
The value of g is maximum on poles of earth and minimum on the equator. ‘

Question 10.
Write the formula for the force of attraction between two objects of masses m1 and m2 respectively kept at a distance of R.
Answer:
F = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{R}^2}\)

Question 11.
What is acceleration due to gravity (g) ?
Answer:
The acceleration by which earth attracts other object towards its centre is called acceleration due to gravity.

Question 12.
Does the acceleration of an object falling towards earth depend upon its mass ?
Answer:
No, the acceleration of an object falling towards earth does not depend upon its mass.

Question 13.
What will be the force of attraction between two objects if the distance between them becomes twice ?
Answer:
Force of attraction will become one-fourth.

Question 14.
What will be the weight of an object on moon in comparison to earth ?
Answer:
\(\frac {1}{6}\) times.

Question 15.
What is centripetal acceleration ?
Answer:
When an object moves on a circular track’ ah acceleration is produced due to change in velocity, which is called centripetal acceleration.

Question 16.
What is centripetal force ?
Answer:
The force which produces centripetal acceleration is called centripetal force. It is always directed towards centre. That is, F = \(\frac{\mathrm{mv}^2}{\mathrm{r}}\)

Question 17.
What ‘g’ indicates ?
Answer:
‘g’ indicates acceleration due to grayity.

Question 18.
What is the value of ‘g’ near the syrfepe of earth ?
Answer:
The value of ‘g’ is 9.8 m/s2.

Question 19.
On which place, the value of ‘g’ is zero ?
Answer:
The value of‘g’is zero at the centre of earth.

Question 20.
Write the formula for the relation between ‘g’ and ‘G’
Answer:
g = \(\frac{\mathrm{G} \times \mathrm{M}_{\mathrm{e}}}{\mathrm{R}^2}\)

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 21.
What are the factors on which the gravitational force depends ?
Answer:
(i) On the mass of objects.
(ii) On the distance between objects.

Question 22.
What is the distance of moon from earth ?
Answer:
60 times of the radius of earth – 384000 km.

Question 23.
If an object is dropped from height A with zero velocity, then how much distance it will cover in one second ?
Answer:
Distance covered, h = \(\frac {1}{2}\) gt2

Question 24.
Whose value, out of ‘g’ and ‘G’ is constant and whose not ?
Answer:
The value of ‘G’ is constant at every place on earth, but value of ‘g’ is more on poles and less on equators.

Question 25.
How much time does the earth take to move around the sun ?
Answer:
The earth takes 365 \(\frac {1}{4}\) days to move around the sun.

Question 26.
How much time does the moon take to move around the earth ?
Answer:
The moon takes 27 days 8 hours to move around the earth.

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 27.
What is meant by mass ? Write its unit.
Answer:
Mass is the quantity of matter contained in it. Its unit is kilogram.

Question 28.
What is meant by weight ? What is its measurement unit ?
Answer:
The force by which earth attracts an object towards itself is called the weight. Its measurement unit is Newton.

Question 29.
What is meant by weight lessness ?
Answer:
The state of body when gravity of planet (earth) stops acting, the person in spaceship is in the state of weightlessness.

Question 30.
What is meant by centre of mass ?
Answer:
The centre of mass is that point at which whole mass of object is directed.

Question 31.
What is meant by centre of gravity ?
Answer:
The centre of gravity is that point of the object at which the whole force of gravity works.

Question 32.
Where is the centre of gravity of objects having regular shape ?
Answer:
The centre of gravity of die objects having regular shape and equal density lies on geometrical centre.

Question 33.
What will happen if the mass and acceleration of two objects become equal ?
Answer:
If the mass and acceleration of two objects become equal, then they will appear to move towards ‘ each other.

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 34.
What is meant by the force of gravity ?
Answer:
The force between earth and an object is called the force of gravity.

Question 35.
Which balance is used to measure weight ?
Answer:
Spring balance is used to measure weight.

Question 36.
Which balance is used to measure mass ?
Answer:
Physical balance is used to measure mass.

Question 37.
Why is the weight of the object less on the moon in comparison to the earth ?
Answer:
Because the acceleration due to gravity on the ftibon is less than that on the earth.

Question 38.
Two stones of masses 25 kg and 2 kg are dropped at the same time from the top of a building. Which stone will reach first on earth ?
Answer:
The acceleration due to gravity does not depend upon the mass of the object. So both stones will reach on the earth in same time.

Question 39.
There is gravitational force between two objects kept on table. Why they don’t move towards each-other even in the presence of this force ?
Answer:
The gravitational force between two objects is very weak so that they do not move.

Question 40.
The earth attracts an iron pin downward with gravitational force, but the magnet attracts the iron pin upwards against the direction of attraction of earth, why ?
Answer:
Because attractive force applied by magnet on iron pin is more than the gravitational force on pin.

Question 41.
Where the gravitational force will be more-between two objects on earth or between object and earth ?
Answer:
The gravitational force between object and earth is more.

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 42.
Where is the weight of the object zero ?
Answer:
The weight of the object is zero at the centre of earth.

Question 43.
What is the relation among u, v, g and t ?
Answer:
v = u + gt

Question 44.
What is the sign of ‘g’ according to Cartesian Coordinate system ?
Answer:
‘g’ is always taken as negative.

Question 45.
What is meant by projectile range ?
Answer:
The maximum horizontal distance covered by projectile is called projectile range.

Question 46.
What is the angle between velocity and acceleration at the highest point of projectile ?
Answer:
Right angle (90°).

Question 47.
How far is the sun from the earth ?
Answer:
1.5 x 1011 metres.

Question 48.
What is meant by thrust ?
Answer:
The force acting perpendicular to the surface of object is called thrust.

Question 49.
Give an example of thrust in daily life.
Answer:
When we press a pin on drawing board, the force applied by our thumb acts in perpendicular direction on board. Thus, it is thrust.

Question 50.
What is pressure ? What is its SI unit ?
Answer:
The thrust per unit area is called pressure. That is pressure  HBSE 9th Class Science Important Questions Chapter 10 Gravitation - 13 Its SI unit is Pascal (Pa).

Question 51.
What is the SI unit of thrust ? Africa
Answer:
Newton (N).

Question 52.
In regard of which scientist, the SI unit of pressure is Pascal (Pa) ?
Answer:
In honour of scientist Blaise Pascal.

Question 53.
What is the bigger unit of pressure ?
Answer:
Kilopascal (k Pa).

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 54.
Why cutting, tearing and grooving apparatus are always sharp ?
Answer:
The cutting, tearing and grooving apparatus are sharp so that thrust affects minimum area to increase pressure and the object can be cut, tear or groove easily.

Question 55.
When is the pressure of man more -sitting on ground or standing at one place and why ?
Answer:
The pressure of man is more when standing at one place because the area in standing position is less than that in sitting position at one place.

Question 56.
What is the effect of depth on liquid pressure ?
Answer:
The liquid pressure increases with depth.,

Question 57.
Why do the egg floats in salty water ?
Answer:
The density of water increases by adding salt in it, due to that buoyant force of water increases and the egg floats.

Question 58.
In which stage, the object floats ?
Answer:
If the weight of the body is equal to the weight of fluid displaced, then the object floats on the surface of fluid.

Question 59.
Write the name of two devices that work on Archimedes’ principle.
Answer:
(i) Ship and
(ii) submarine.

Question 60.
Which device is used to measure the purity of the milk and what is its principle ?
Answer:
The lactometer is used to measure the purity of the milk which works on Archimedes’ principle.

Question 61.
Write the name and principle of that device which is used to measure the density of liquids.
Answer:
Hydrometer, which works on Archimedes’ principle.

Question 62.
What is the mass of unit volume called ?
Answer:
Density.

Question 63.
What is the SI unit of density ?
Answer:
Kilogram per metre cube (kg m-3).

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Short Answer Type Questions

Question 1.
If a piece of stone is allowed to fall freely from a certain height, then what will be the changes in velocity ?
Answer:
The following are the changes :
(1) The initial velocity of stone is zero and it increases at coming down.
(2) The stone nearest to earth has maximum velocity.
(3) The velocity of stone becomes zero after striking the ground. –

Question 2.
What is gravitation ? In which direction it works ?
Answer:
Everybody of universe attracts other body towards itself. This attraction between two bodies is gravitation. This attraction between two bodies works along the line joining the centres.

Question 3.
Prove that acceleration produced in the object dropped towards the earth does not depends upon its mass.
Answer:
If an object of mass, m is dropped from the centre of earth at a distance of R, then the force exerted by the earth on object,
F = \(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}\)
Where, Me = mass of earth, G = gravitation constant
∴ Acceleration produced in the object = \(\frac {F}{m}\) (F = ma; a = \(\frac {F}{m}\))
= \(\frac{\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}}{\mathrm{~m}}\) = \(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2} \times \frac{1}{\mathrm{~m}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^2}\)
∴ The acceleration produced in an object does not depend upon the mass of dropped object.

Question 4.
What is Newton’s third law of motion ? Is this law applicable on gravitational force ? Explain.
Answer:
According to Newton’s third law, “To every action, there is equal and opposite reaction.” For example, we act on ball when we throw it on wall and ball comes back by the reaction of wall. Thus, an object A exerts a force on object B, then object B is also exerts equal and opposite force on A. This fact is also true for gravitational force. When a stone falls on earth, it happens due to the force of gravity of the earth on stone. In fact, the piece of stone, attracts the earth towards itself with equal force of gravity. But the mass of stone is less than that of the earth so the force exerted by the stone on the earth is less. Therefore the stone seems to attracted towards earth.

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 5.
Discuss the motion of the object thrown upward.
Answer:
When a body is thrown upward, the earth attracts that body continuously towards itself. Therefore, the gravitational force works in opposite direction of motion and the body moves slowly. The speed reduces at a rate of 9.8 ms-2 in every second. At last when the body reaches at highest point, the speed becomes zero. The body again starts falling towards earth with an acceleration of 9.8 ms-2.

Question 6.
Find the unit of ‘G’
Answer:
‘G’ is the ratio of force (F) and \(\frac{M_1 M_2}{R^2}\) Therefore, its unit depends upon the units offorce, mass and distance. If the mass is taken in kilogram, distance in metre and time in second, then the uqit offorce is newton (N).
Unit of G = \(\frac{M_1 M_2}{R^2}\) = Nm2/kg2

Question 7.
Why is G called universal constant ?
Answer:
We know that F = G \(\frac{m_1 m_2}{R^2}\)
‘G’ is constant among diem. The value of ‘G’ does not depend upon the Value of mj, m2 or R. Its value does not depend upon that how, when and where the force is measured. The value of‘G’ remains consant for any two bodies in universe. The ratio of F and \(\frac{m_1 m_2}{R^2}\) is equal to G which is equal for any two bodies. That is why, G is called universal constant.

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 8.
What is the difference between ‘g’ and ‘G’ ?
Answer:
The difference between ‘g’ and ‘G’ is as follows:

g:G:
1. It indicates acceleration due to gravity.1. It indicates gravitational constant.
2. Its value vary from place to place.2. Its value is same everywhere. Therefore, it is called universal constant.
3. Its value on earth is 9.8 ms-2.3. Its value is 6.673 x 10-11 Nm2 kg-2.
4. It is always negative.4. It is always positive.

Question 9.
Prove that the bodies of different masses when dropped together from the same height, strikes in same time (assume, air resistance is negligible).
S0lution:
Suppose two objects of different masses were dropped together from same height h,
In this state s = h, u = 0
According to formula s = ut + \(\frac {1}{2}\) gt2
h = 0 x t + \(\frac {1}{2}\) gt2
Or h = \(\frac {1}{2}\) gt2
Or 2h = gt2
Or gt2 = 2h
t2 = \(\frac {2h}{g}\)
Or t = \(\sqrt{\frac{2 h}{g}}\)
It is clear from the equation that the time taken by objects does not depend upon masses. Thus, both objects fall in same time on earth.

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 10.
What is meant by vertical direction of earth?
Answer:
If we suppose the earth as a ball of uniform density, then its centre of mass will be at its centre. Thus, the force of attraction of the earth on a body works in the direction of its (earth) centre. This is the direction which we call vertical direction.
HBSE 9th Class Science Important Questions Chapter 10 Gravitation - 1

Question 11.
Which are equations related to the freely falling bodies?
Solution:
(i) v = u + gt
(ii) v2 – u2 = 2gs
(iii) s = ut + \(\frac {1}{2}\) gt2
Here,
u = initial velocity
y = final velocity
g = acceleration due to gravity
s = distance or height
t = time

Question 12.
What is meant by weight ? Write the formula for weight.
Answer:
The force of attraction applied by earth on an object is called its weight. if earth produces an acceleration ‘g’ on an object of mass, m then, the force of artraction applied by earth on object, F = mg
But, the force of attraction applied by the earth on object (F) = Weight of object (W) = W = mg

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 13.
Find a formula to compare the weight of a body on earth and on any other heavenly body.
Solution:
Suppose, mass of the object = m
Mass of the earth = Me
Radius of earth = Re
The attractive force applied by earth on object, Fe = \(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}{ }^2}\) ………..(i)
Suppose, mass of hçavenly body = Mm
Radius of heavenly body = Rm
The force of attraction applied by other heavenly body on object
Fm = \(\mathrm{G} \frac{\mathrm{M}_{\mathrm{m}} \times \mathrm{m}}{\mathrm{R}_{\mathrm{m}}^2}\) …………..(ii)

Dividing eq (ii) by eq (1)
\(\frac{\mathrm{F}_{\mathrm{m}}}{\mathrm{F}_{\mathrm{e}}}=\frac{\mathrm{M}_{\mathrm{m}}}{\mathrm{M}_{\mathrm{e}}} \times\left(\frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{m}}}\right)^2\)

Question 14.
Find the relation between acceleration due to gravity (g) and universal gravitational constant (G).
Answer:
Suppose an object of mass m is kept at a distance of R from the centre of the earth. If M is the mass of the earth, then the force applied by earth on that object:
F = \(\mathrm{G} \frac{\mathrm{Mm}}{\mathrm{R}^2}\) ………..(i)
The force by which the earth attracts the object towards its centre, is cal led weight. It is also denoted by’ F’.
F = mass (m) x acceleration due to gravity (g)
or F = mg ……….(ii)
From equations (i) and (ii) mg = \(\mathrm{G} \frac{\mathrm{Mm}}{\mathrm{R}^2}\)
Or g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\)
This is the required relation. The value of’g’ can be found with the help of this.

Question 15.
What is mass ? What is its measurement unit ? WrIte the law to measure mass.
Answer:
Mass: The quantity of matter contained in an object is called its mass. It is a scalar and constant quantity.
The unit of mass is kilogram. It is found by physical balance.

Law to measure mass:
Suppose we have two objects of masses m1 and m2, kept at the same distance from the centre of the earth, then the force of attraction between the earth and m1:
F1 = R2
Here, Me = mass of earth and R = distance of mass m1 from the centre of the earth.
Like this, the force of attraction between the earth and m2
F1 = \(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}_1}{\mathrm{R}^2}\)
Here, Me = mass of earth and R = distance of mass m1 from the centre of the earth.
Like this, the force of attraction between the earth and m2
F2 = \(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}_2}{\mathrm{R}^2}\)
1f both the forces are same, then F1 = F2
\(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}_1}{\mathrm{R}^2}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}_2}{\mathrm{R}^2}\)
∴ m1 = m1
Thus, if two forces are equal then the masses of two objects will be equal and parts of physical balance will be equal.

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 16.
Define projectile and path of projection and give some examples of projection motion.
Answer:
Projectile:
A body thrown in horizontal direction is called projectile.

Path of projection:
The curved path travelled by projectile is called path of projection.

The examples of projection motion are following:
(1) The path of a body dropped from aeroplane, i.e., a ball falling from a building of 80m high is shown in figure.
(2) The path of a bullet fired from gun.
(3) The path of sphere thrown by player.
HBSE 9th Class Science Important Questions Chapter 10 Gravitation - 2

Question 17.
Why do astronauts feel weightlessness in spaceship?
Answer:
When the astronauts sitting in spaceship move around the earth, the gravitational force between the earth and spaceship is balanced by the force used to move around the earth, due to that the astronauts feel weightlessness. The value of acceleration due to gravity becomes zero in spaceship.

Question 18.
Generally, if the cotton ball and stone of same size are allow ed to fall from a certain height, then which one will strike the surface of earth first? What Galello discussed about this fact?
Answer:
Generally, the piece of stone reaches first on the surface of earth in comparison to cotton. But first of all Galelio told that the air resistance becomes zero, cotton ball and piece of stone will reach in same time on the surface of earth with equal acceleration. According to Galelio. “the objects the acceleration of falling on the acceleration of the earth does not depend upon their masses.”
HBSE 9th Class Science Important Questions Chapter 10 Gravitation - 3

Question 19.
How does acceleration due to gravity change at any place (i) due to height, (ii) due to the shape of the earth?
Answer:
In the following way the acceleration due to gravity changes at any place:
(i) Due to height: The value of ‘g’ is less on the mountains situated at some height on earth surface, whereas it ¡s more on plain surface.
∴ \(\frac{W_h}{W}=\frac{g_h}{g}=\frac{R^2}{(R+h)^2}\)
Like this, the value of ‘g’ goes on decreasing by increasing height.

(ii) Due to the shape of earth: The earth is not completely round like a ball, but it is convexed at the equator line and concaved at the polar line. There is a difference of 22 km between the radius of the equator and polar line of the earth. That’s why
the value of’g’ is more on poles and less on equators.
∴ g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\) (Where, G and M are constant.)
∴ g ∝ \(\frac{1}{R^2}\)

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 20.
“The moon falls towards the earth.” What does it mean ? Why does it not fall on the surface of the earth?
Answer:
The moon falls towards the earth means that the earth attracts the moon towards itself due to its gravitational force. But, moon moves around the earth in a circular path and does not fall, because the velocity of the object moving on a circular path changes at every point. The change in velocity or acceleration is due to the change in the direction of motion of the object. The centripetal force which produces this acceleration maintains the body to move in circle, which is always directed towards the centre.
HBSE 9th Class Science Important Questions Chapter 10 Gravitation - 4

Question 21.
The earth attracts the apple, does the apple also attract the earth ? If yes, then why the earth does not move towards the apple?
Answer:
Definitely the apple attracts the earth towards itself, but the mass ofapple is negligible in comparison to earth so the earth does not move towards apple.

Question 22.
What are Kepler’s three laws of motion?
Answer:
Kepler’s is three laws of motion are following:
(i) The orbit of a planet is an ellipse with the sun at one of the foci. As the position of sun is shown by in the figure.
(2) The line joining the planet and the sun sweep equal area in equal intervals of time. Thus, if the time of travel from A to B is same as that from C to D, then the areas OAB and OCD are equal.
(3) The cube of the mean distance of a planet from the sun is proportional to the square of its orbital period, T, or constant. It is important to know that Kepler could not give a theory to explain the motion ofplanets. It was Newton who showed that the cause of the planetory motion is the gravitational force that the sun exerts on them.
HBSE 9th Class Science Important Questions Chapter 10 Gravitation - 5

Question 23.
Why are the tyres of buses or trucks wider than that of cars?
Answer:
The thurst on buses or trucks is more than that on cars. If their tyres are made less wider than the tyres of cars, the pressure on them will increase and they will sink ¡n earth. Therefore, the tyres of buses or trucks are made wider, so that the area will increase and pressure on earth decrease.

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 24.
The camel can walk in desert easily, but ¡t is difficult for horse to walk, why?
Answer:
The area of feet of camel is more than that of horse. So the affected area is more and pressure decreases. The camel can walk easily in desert due to less pressure. But the area of feet of horse is less, and the pressure increases. Due to large pressure, it is difficult to walk by horse in desert.
HBSE 9th Class Science Important Questions Chapter 10 Gravitation - 6

Question 25.
What is buoyant force? [low is it measured?
Answer:
When a body is immersed in a fluid, an upward force works on the body which is called buoyant force. Measurement of Buoyant Force: Take a metal’s sphere. Hang it on spring balance and note its weight. Let its weight is W1 gm. Now put some water in a beaker and take the sphere in the beaker slowly. You will see that the weight of sphere seems decreasing when put in water slowly, which is indicated by spring balance. Note the indicator of spring balance when the solid sphere immersed completely in water.
Let it is W2 gm.
The difference between two weights will be the buoyant force of water.
Buoyant force = weight of sphere in air – weight of sphere in water.
= (W1 – W2)gm
HBSE 9th Class Science Important Questions Chapter 10 Gravitation - 7

Question 26.
What is the reason that, an iron needle sinks in water, but a ship made up of iron floats in water?
Answer:
The iron needle sinks in water, but a ship made of iron floats in water because weight of water displaced by needle is less than the weight of needle. Due to heavy weight, the needle sinks in water. But the weight of water displaced by ship is more than the weight of ship and it floats in water.

Question 27.
Why is it easy to swim in sea water than river?
Answer:
It is easy to swim in sea water than river because the density of sea water is more than the density of river water. Thus, the sea water due to being heavy, applies larger buoyant force on the body of person, duc to this reason it is easy to swim in sea water.

Question 28.
What is density ? What is its SI unit ? What is its importance?
Answer:
The mass per unit volume of a body is called density. Its SI unit is kilogram per metre cube. The density of the substance always remain same in certain circumstances. The density of gold ¡s 19300 kg m-3 and the density of water is 1000 kg m-3.
Importance: The density of the sample of any substance helps in the measurement of its purity.

Question 29.
What is relative density ? What is the condition of sink or float in water on the basis of it?
Answer:
The ratio of density of substance and density of water is called relative density of the body. Thus density of substance
HBSE 9th Class Science Important Questions Chapter 10 Gravitation - 8
∴Relative density is a ratio, so, it has no unit.
The relative density ofgold is 19.3. If the relative density of any solid or liquid is more than I, it will sink in water, if it is less than 1, it will float in water.

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 30.
Why the ice floats on water?
Answer:
The ice floats on water because the density of ice is less than the density of water. when a piece of the ice is put in the water, then its weight is less than the weight of water displaced by it.

Question 31.
Define Archimedes’ principle and explain it. What arc uses of this principle? Or Define Archimedes’ Principle. Write Iwo applications ofArchimedes’ Principle.
Answer:
Statement of Archimedes’ principle: When a body is immersed fully or partially ¡n a fluid, it expirences an upward force that is equal to the weight of the fluid displaced by it.

Explanation:
When a solid is immersed in a fluid, the fluid pushes it upwards by its buoyant force, and its wekht decreases, because buoyant force is equal to the weight of water displaced by the solid. So the decrease in weight of solid when immersed ¡n fluid is equal to the weight of the water displaced.

Uses of Archimedes’ principie in daily life:
(I) It is used in designing of ships and submarines.
(2) The lactometer, which is used to determine the purity of milk, is based on this principle. .
(3) The hydrometer which is used for determining density of liquid, is also based on this principle.

Question 32.
What happens to the gravitational force between two objects If:
(i) the distance between the objects ¡s one-third (1/3)?
(ii) the masses of both the objects are halfed (1/2)?
Answer:
(i) As we know that gravitational force between two objects is inversely proportional the square of the distance between them. So, if the distance between them is then the gravitational force between them will be 9 times.
(ii) As the gravitational force between two objects is proportional to the product of masses of two objects. So, if the masses of two objects are halfed (1/2) then the gravitational force between them will be one tourth (1/4).

Numerical Questions

Question 1.
A bullet is fired with a velocity of 25 ms1 in horizontal direction. How much distant it will fall in first two seconds?
Solution:
Vertical distance is independent to horizontal velocity.
∴ Initial velocity (u) = 0
Acceleration due to gravity (g) = 9.8 ms-2
Time (t) = 2s
Distance (s) = ?
We know that
s = ut + \(\frac {1}{2}\) gt2
= 0 x 2 +\(\frac {1}{2}\) x 9.8 x (2)2
= 19.6 m

Question 2.
Find the weight of an object of mass 5 kg.
Solution:
Mass of the object (m1) = 35 kg
Acceleration due to gravity (g) = 9.8 ms-2
Weight of the object (W) = ?
We know that, W = m x g = 5 x 9.8 = 49 N

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 3.
If the mass of you and your friend is 35 kg, and the distance between you and your friend is 1 m, then what will be the force of attraction between two?
Solution:
Here,
Mass(m1) = 35kg
Ma.ss(m2) = 35kg
Distance(r) = 1 m
Gravitational constant (G) = 6.67 x 10-11 Nm2 kg-2
Force of attraction (F) = ?
We know that, Force of attraction, (F) = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}^2}\)
= \(\frac{6.67 \times 10^{-11} \times 35 \times 35}{(1)^2}\)
= 8.17 x 10-8 N Ans

Question 4.
FInd the weight of an object of mass 10 kg on the surface of that planet whose mass is two times and radius Is three times that of earth.
Solution:
For planet, M = 2Me and R = 3 Re
∴ g1 = \(\frac{\mathrm{G}\left(2 \mathrm{M}_{\mathrm{e}}\right)}{\left(3 \mathrm{R}_{\mathrm{e}}\right)^2}=\frac{2 \mathrm{GM}_{\mathrm{e}}}{9 \mathrm{R}_{\mathrm{e}}^2}\)
= \(\frac {2}{9}\) g
= \(\frac {2}{9}\) x 9.8 = \(\frac {19.6}{9}\)m-2 [∵ g = 9.8 m-2]
∴ Weight of the object of 10 kg on the planet (W) = m x g1
= 10 x \(\frac {19.6}{9}\) = \(\frac {19.6}{9}\) = 21.78 N

Question 5.
The weight of an object is 49 N, what will be its mass?
Solution:
Weight of object (W) = 49 N
Acceleration due to gravity (g) = 9.8 ms-2
Weight = mass x acceleration due to gravity
W = mg
49 = m x 9.8
m = \(\frac {49}{9.8}\) = 5 kg

Question 6.
A car fall off a ledge and drops to the ground in 0.5g. Let g = 10 ms-2
(i) What is the speed of car on striking the ground?
(ii) What ¡s its average speed during \(\frac {1}{2}\) s?
(iii) How high is the ledge from the ground?
Solution:
Time (t) = \(\frac {1}{2}\) s
Initial velocity (u) = 0 ms-1
Acceleration due to gravity (g) = 10 ms-2
Acceleration of car (a) = + 10 ms-2 (downward)
(i) Speed (v) = at
v = 10 ms-2 x \(\frac {1}{2}\) s = 5 ms-1

(ii) Average speed = \(\frac {u + v}{2}\) = \(\frac{\left(0 \mathrm{~ms}^{-1}+5 \mathrm{~ms}^{-1}\right)}{2}\) = 2.5 ms-1

(iii) Distance travelled, s = \(\frac {1}{2}\) at2
= \(\frac {1}{2}\) x 10 ms-2 x (1/2s)2
= \(\frac {1}{2}\) x 10 ms-2x (1/4s)2 = 1.25 m
Thus
(i) its speed on striking the ground = 5 ms-1
(ii) its average speed during the \(\frac {1}{2}\) s = 2.5 ms-1
(iii Height of the ledge from the ground = 1.25 m

Question 7.
A stone ¡s dropped from the edge of the roof:
(i) how long time does It take to fall 4.9 m?
(ii) what is the speed of stone at that time?
(iii) how fast does It move at the end of 1.9 m?
(iv) what is the acceleration after Is and after 2s?
Solution:
(i) Initial velocity of stone (u) = 0
The distance covered by the stone (s) = 4.9 m
g = 9.8 ms-2
Time, t = ?
Now, S = ut + gt2
4.9 = 0 x t +\(\frac {1}{2}\) x 9.8 x t2
Or 4.9 = 0 + 4.9 t2
Or 4.9 t2 = 4.9
Or t2 = \(\frac {4.9}{4.9}\) = 1s

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

(ii) Suppose the speed of stone at that time = v
Now, v = u + gt
v = 0 + 9.8 x 1
v = 0 + 9.8 = 9.8 ms-1

(iii) initial velocity of stone (u) = 0
The distance covered by the stone (s) = 79 m
g = 9.8 mr-2, v = ?

Now, v2 – u2 = 2gs
v2 – (0)2 = 2 x 9.8 x 7.9
v2 – 0 = 154.84
v = 154.84
v = \(\sqrt{154.84}\) = 12.44 ms-1

(iv) Because the body is faLling freely under the action of gravity.
∴ Acceleration of the body after 1 s = 9.8 ms-2
Acceleration of the body after 2 s = 9.8 ms-2

Question 8.
A ball is thrown up vertically upto a height of 100 m, find its initiai speed.
Solution:
Here,
Initial speed (u) = ?
Final speed (v) = 0
Distance (s) = 100m
Acceleration due to gravity (g) = -9.8 ms-2
Weknowthat, v2 – u2 = 2gs
02 – u2 = 2 x (-9.8) x 100
– u2 = – 1960
u = \(\sqrt{1960}\) = 44.2 ms-1 Ans

Question 9.
What will be the height of building Ifa piece of stone takes 4 seconds when dropped freely from the roof of a building?
Solution:
Here, Initial speed (u) = 0
Time(t) = 4s
Height or distance (s) = ?
Acceleration due to gravity (g) = 9.8 ms2
We know that, s = ut + \(\frac {1}{2}\) gt2
= 0 x 4 + \(\frac {1}{2}\) x 9.8 x (4)2
= \(\frac {1}{2}\) x 9.8 x 4 x 4 = 78.4 m

Question 10.
The volume of a 500 g sealed tin (or box) is 350 cm3. What is the density of sealed tin ? Will it float or sink in water? What will be the mass of the water displaced by this tin?
Solution:
Mass of tin (m) = 500 g = \(\frac {500}{1000}\) kg = 0.5 kg
Volume of tin = 350 cm3
= \(\frac{350}{100 \times 100 \times 100}\) m3
= 35 x 10-5 m3
Density of tin = \(\frac{\text { mass }}{\text { volume }}=\frac{0.5}{35 \times 10^{-5}}\) kg m-3
= \(\frac{5 \times 10^5}{35 \times 10}=\frac{10^4}{7}\) kg m3
= 1.429 x 103 kg m-3
We know that the density of water = 1000 kg m-3
= 103 kg m-3

∴ Relative density of tin density of water HBSE 9th Class Science Important Questions Chapter 10 Gravitation - 14 = \(\frac{1.429 \times 10^3}{10^3}\) = 1.429
Because the relative density of tin (is more than 1). Therefore, tin will sink in the water. Thus, mass of the water displaced wilt be less than that of tin.

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 11.
The volume of a piece of iron is 20cm3 and mass ¡s 156 g. What is its density?
Solution:
Mass of the iron = 156 g
Volume of the iron = 20 cm3
HBSE 9th Class Science Important Questions Chapter 10 Gravitation - 15
= \(\frac {156}{20}\) = 7.8g cm-3.

Do Yourself

(1) Write down the expression for the acceleration experienced by a particle on the surface of moon due to the gravitational force on moon. Find the ratio of this acceleration to that experienced by the saine particle on the surface of the earth. if the acceleration due to gravity on earth is 98 ms2 what is the acceleration of a particle on the surface of moon?
Answer:
[(i) a = \(\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}}\) (ii) 0.16 (iii) 1.57 ms-2]

(2) A particle is thrown up in vertical direction with a velocity of 50 ms. What will be its velocity at highest point? What is the maximum height attained by this particle? How much time will it take to reach at highest point?
Answer:
[(i) v = 0 (ii) 125 m (iii) 5 s]

(3) What will be the force due to gravity on the surface of that planet whose mass and radius, both are
half than that of earth ?
Answer:
[Take g = 9.8 ms2 1 [19.6 ms-2]

(4) A force of 20 N is applied on an object of weight 9.8 N. Calculate the mass of the object and the
acceleration produced.
Answer:
[Take g = 9.8 ms-2] [1 kg; 20 ms-2]

(5) The mass of the earth is 6 x 1024 kg and that of the moon is 7.3 x 1o22 kg. 1f the distance between the
earth and the moon is 3.84 x 105 km, calculate the force exerted by the earth on the moon.
Answer:
[19.8 x 1019 N]

(6) Calculate the force of attraction acting on person of mass 35 kg on the surface of earth. The mass of the earth is 6 x 1024 kg and radius is 6.4 x 106m.
Answer:
[342 N]

(7) Find the force of attraction between two objects having mass 50 kg and 10 kg and the distance between them is 1 metre.
Answer:
[3.34 x 10-8 N]

(8) The mass of earth is 6 x kg and mass of moon is 7.4 x 1022 kg. 1f the distance between the earth and the moon is 3.84 x 105 km, find the force exerted by the earth on the moon
Answer:
[G = 6.7 x 10-11 Nm2 kg-2]
[22.01 x 1020N]

(9) An object is thrown up ¡n vertical direction and attains a height of 10 m. Calculate
(i) the velocity with which the object is thrown up and
(ii) time taken by the object to reach the highest point.
[u = 14 ms-1 t = 1 .43s]

(10) A boy dropped a stone from the edge of a rock of 49 m high. After is, he dropped second stone. What was the speed of second stone if both stones reached the eanh in same time ?
[ Take g = 9.8 ms-2]
[12.08 ms-1]

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

(11) By which initial velocity, the ball will be thrown up so that it can go upto a height of 40 m.
Answer:
[28 ms-1] [10 kg]

(12) Find the mass of that object whose weight is 196 N.
Answer:
[Take -9.8 ms-2] (10 kg]

(13) An aeroplane is flying at a height of 1000 m in horizontal direction with a velocity of 100 ms-1. A bomb is dropped from it. Calculate:
(i) the time taken by the bomb to reach earth.
(ii) the velocity before striking the ground.
(iii) the distance of target.
Answer:
[(1) 14.28 s ((ii) 172.4 ms-1 ((iii) 1428.5 m]

(14) The mass of an object ¡s 10 kg. What will be its weight on the earth?
Answer:
[98 N]

(15) The weight of an object is 10 N on weighing. What will be its weight on moon?
Answer:
[98 N]

(16) A block of wood is kept on a table top. The mass of wooden block is 5 kg and its dimcnsions are 40 cm x 20 cm x 10cm. Find the pressure exerted by the wooden block on the table top ¡fit is made to lie on the table top with its sides of dimensions (a) 20 cm x 10 cm and (b) 40 cm x 20 cm.
Answer:
[(a) 2450 Nm-2 (b) 612.5 Nm-2]

(17) The relative density of silver is 10.8 g. The density of water is 1 kgm3. What will be the density
of silver in SI unit?
Answer:
[10.8 x 103 kg m3]

Essay Type Questions

Question 1.
Find the deflnition of universal gravitational constant ‘G’. Write the unit of ‘G’ In SI system.
Answer:
According to Newton’s Universal Law of Gravitation,

F = \(\mathrm{G} \frac{\mathrm{M}_1 \mathrm{M}_2}{\mathrm{R}^2}\)
Or G = \(\frac{\mathrm{FR}^2}{\mathrm{M}_1 \mathrm{M}_2}\)
If M2 = M2 = 1 unit and R = 1 unit, then
G = \(\frac{F \times 1^2}{1 \times 1}=\frac{F \times 1}{1 \times 1}\) = F
∴ Thus, universal gravitational constant is a gravitational force which is placed between the objects of unit mass which are at unit distance from each other.

SI unit of G:
G = \(\frac{\mathrm{FR}^2}{\mathrm{M}_1 \mathrm{M}_2}\)
Because G is the ratio ofF and M1 M2/R2 . Therefore, its unit depends upon the units of force, mass and distance. If mass is taken in kilogram, distance in metres and time in seconds, then unit of force is newton (N). Therefore, unit of G will be Nm-2 kg-2. The value of G found by the experiments on golden balls is:
G = 0.000,000,000,066,734 Nm2 kg2
= 6.673 x 10-11 Nm2 kg2

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 2.
State the universal law of gravitation. Derive the formula for gravitational force between two bodies. Aand B of masses M and m, placed at a distanced.\(1 \frac{1}{2}+2 \frac{1}{2}=4\)
Answer:
Universal Law of Gravitation:
Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the çentres of two objects.
HBSE 9th Class Science Important Questions Chapter 10 Gravitation - 9

Let two objects A and B of masses M and ‘m’ lies at a distance ‘d’ from each other as shown in figure. Let the force of attraction between two objects be F. According to the universal law of gravitation, the force between two objects is directly proportional to the product of their masses, that is,
F ∝ M x m …………(i)
And the force between two objects is inversely proportional to the square of the distance between them, that is,
F ∝ \(\frac{M \times m}{d^2}\) ………..(ii)
Combining eqs. (i) and (ii), we get
F ∝ \(\frac{M \times m}{d^2}\) or F = G \(\frac{M \times m}{d^2}\)
Where G is the constant of proportionality and is called universal gravitational constant.

Question 3.
Compare the weights of body if that are weigh on earth and moon.
Answer:
The value of acceleration due to gravity of each planet (g) depends upon its mass and radius. Due to this reason, weight of an object on moon is \(\frac {1}{6}\) th times than on the earth. Suppose the mass of object is ‘m’e. If the mass of earth is Me and its radius is Re, then weight of the object on earth (Fe).
Fe = \(\frac{\mathrm{GM}_{\mathrm{e} \mathrm{m}}}{\mathrm{R}_{\mathrm{e}}^2}\) ………..(i)
In same way, the mass of the moon is Mm and its radius is Rm then weight of the object on moon (Fm)
Fm = \(\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}^2}\) ………(ii)
Dividing eQuestion (i) by eQuestion (ii) \(\frac{\mathrm{F}_{\mathrm{e}}}{\mathrm{F}_{\mathrm{m}}}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}^2} \times \frac{\mathrm{R}_{\mathrm{m}}^2}{\mathrm{GM}_{\mathrm{m}} \times \mathrm{m}}=\frac{\mathrm{M}_{\mathrm{e}}}{\mathrm{M}_{\mathrm{m}}}\left(\frac{\mathrm{R}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{e}}}\right)^2\)
∴ The mass of the earth is 100 times than that of moon, and radius is 4 times more
Me = 100 Mm and
Re = 4 Rm

By putting the value of Me and Re
\(\frac{\mathrm{F}_{\mathrm{e}}}{\mathrm{F}_{\mathrm{m}}}=\frac{\mathrm{M}_{\mathrm{e}}}{\mathrm{M}_{\mathrm{m}}}\left(\frac{\mathrm{R}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{e}}}\right)^2=\frac{100 \mathrm{M}_{\mathrm{m}}}{\mathrm{M}_{\mathrm{m}}} \times\left(\frac{\mathrm{R}_{\mathrm{m}}}{4 \mathrm{R}_{\mathrm{m}}}\right)^2\)
\(\frac{\mathrm{F}_{\mathrm{e}}}{\mathrm{F}_{\mathrm{m}}}\) = 100 x \(\frac {1}{6}\)
= 6 approx.
Fe = 6 Fm
Thus, it is clear that the mass of the object on the moon will be m, but its weight on the earth is 6 times more in compare to moon.

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

Question 4.
Write the difference between:
(i) Gravitational force and force of gravity
(ii) Centre of mass and centre of gravity.
Answer:
(i) The difference between gravitational force and force of gravity is as follows:

Gravitational ForceForce of Gravity:
1. The gravitational force is the force that arises between any two objects in the universe.1. The force exerted between the earth and the object is called force of gravity.
2. The gravitational force exerted between two objects is proportional to the product of their masses.2. The force of gravity is proportional to the masses of body.
3. The gravitational force exerted between the two objects is inversely proportional to the square of the distance between their centres.3. The force of gravity is proportional to the acceleration due to gravity (g).
Gravitational ForceForce of Gravity:

(ii) Centre of mass and centre of gravity.
Suppose a giant body is made of many particles. Then it is possible for us to define that point where the whole of the mass of the body is concentrated. This point is called ‘centre of mass’. On the surface of earth or nearby it, where force of gravity is constant, centre of mass is only that point where the force of gravity on whole body can be experienced. In this state, the centre of mass is called centre of gravity.

Thus ‘centre of gravity’ of a body is that point, where the force of gravity is acting on the whole body as shown in fig. The centre of gravity of objects having regular size and uniform density lies on their geometrical centre. This can be balanced on a needle at the geometrical centre. For example, the centre of gravity of a spherical ball, or rectangular piece of wood or circular disc of metal is at their centre.
HBSE 9th Class Science Important Questions Chapter 10 Gravitation - 10

Practical Work

Experiment 1.
Describe an activity to understand the motion of moon from earth. Procedure:
(i) Take a piece of thread. Tie a small stone at one end.
(it) Hold the other end of the thread and whirl it round as shown in fig.
(iii) The direction of the motion of stone will be circular motion.
(iv) Now release the thread.
(v) Again, note the direction of motion of the stone.
Before the thread is released, the stone moves in a circular path with a certain speed and changes direction at every point. The change in direction involves change in velocity or acceleration. The force that cause this acceleration and keeps the body moving along the circular path is acting towards the centre. This force is called the centripetal force.
HBSE 9th Class Science Important Questions Chapter 10 Gravitation - 11

Experiment 2.
Explain an activity to understand the meaning of free fall of objects.

Procedure:
(i) Take a stone.
(ii) Throw it upwards.
(iii) It reaches a certain height and then it starts falling down.
We know that the earth attracts objects towards it. This is due to force of gravity. Whenever objects fall towards the earth under this force alone, we say that the objects are in free fall.

Experiment 3.
Prove with experiment that the gravitational force between two objects of general masses is weak.

Procedure:
Take a magnet. Take it near the iron pin kept near the piece of stone. You will see that the iron pin is not attracted towards piece of stone but towards the magnet. It proves that the gravitational force between stone and pin is very weak. So the iron pin is not attracted towards stone. In fact, the gravitational force between piece of stone and iron pin is very weak as compared to magnetic force. So the iron pin attracted towards magnet but not towards piece of stone. It is concluded that the gravitational force between two objects of general masses is weak.

Experiment 4.
Explain with an experiment that why iron sinks into the water while cork floats ?

Procedure:
(i) Take a beaker filled with water.
(ii) Take an iron nail and a cork of equal mass.
(iii) Place them on the surface of water.
(iv) You will observe that the cork floats and the nail sinks.
This is due to the difference in their densities. The density of a substance is defined as the mass per unit volume. The density of cork is less than the density of water. This means that the up thrust of water on the cork is greater than the weight of the cork. So it floats.
HBSE 9th Class Science Important Questions Chapter 10 Gravitation - 12

Quick Review of the Chapter

1. What is the force of attraction called between the two particles of the universe ?
(A) force of gravity
(B) gravitational force
(C) centripetal force
(D) magnetic force
Answer:
(B) gravitational force

2. The force exerted by the earth on an object is called :
(A) force of gravity
(B) gravitational force
(C) centripetal force
(D) magnetic force
Answer:
(A) force of gravity

3. The acceleration which maintains the circular motion always directed towards the centre of circle, that is called :
(A) force of gravity
(B) gravitational force
(C) centripetal force
(D) buoyant force
Answer:
(C) centripetal force

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

4. Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. This law is called :
(A) universal law of gravitation
(B) law of gravity
(C) universal law of centripetal
(D) Archimedes’ principle
Answer:
(A) universal law of gravitation

5. The magnitude of the gravitational force on an object on the surface of earth is :
(A) \(\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^2}\)
(B) \(\frac{\mathrm{GM}}{\mathrm{R}^2}\)
(C) \(\mathrm{GM}_{\mathrm{e}} \mathrm{m}\mathrm{R}_{\mathrm{e}^2}\)
(D) \(\frac{\mathrm{M}_{\mathrm{e} m}}{\mathrm{GR}^2}\)
Answer:
(C) \(\mathrm{GM}_{\mathrm{e}} \mathrm{m}\mathrm{R}_{\mathrm{e}^2}\)

6. The value of gravitational constant is :
(A) different at each place on earth
(B) same at all planets and satellites
(C) vary from place to place
(D) all above statements are false
Answer:
(B) same at all planets and satellites

7. On which part of earth the value of acceleration due to gravity (g) is maximum?
(A) on poles
(B) on equator
(C) on centre
(D) on all of them
Answer:
(A) on poles

8. What is the weight of an object on moon in comparison of earth ?
(A) 6 times
(B) 3 times
(C) \(\frac {1}{6}\) times
(D) \(\frac {1}{3}\) times
Answer:
(C) \(\frac {1}{6}\) times

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

9. The value of acceleration due to gravity (g) is :
(A) same at every place on earth
(B) same at all planets and satellites
(C) vary from place to place
(D) all above statements are false
Answer:
(C) vary from place to place

10. If the weight of a person is 24 N on the earth, then what will be his weight on moon ?
(A) 6N
(B) 8N
(C) 4N
(D) 96 N
Answer:
(C) 4 N

11. The value of acceleration due to gravity (g) is least on which part of earth ?
(A) on poles
(B) on equator
(C) on mountains
(D) on all above
Answer:
(B) on equator

12. The formula of the attractive force (F) between two objects of masses ntj and m2 respectively kept at a distance of R is :
(A) \(\frac{\mathrm{Gm}_1}{\mathrm{~m}_2 \mathrm{R}^2}\)
(B) \(\frac{\mathrm{Gm}_2}{\mathrm{~m}_1 \mathrm{R}^2}\)
(C) \(\frac{\mathrm{R}^2}{\mathrm{Gm}_1 \mathrm{~m}_2}\)
(D) \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{R}^2}\)
Answer:
(D) \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{R}^2}\)

13. The value of acceleration due to gravity (g) on the surface of earth is :
(A) 9.8 ms-2
(B) 4.9 ms-2
(C) 19.6 ms-2
(D) 29.4 ms-2
Answer:
(A) 9.8 ms-2

14. The value of acceleration due to gravity (g) is zero on :
(A) poles
(B) equator
(C) centre
(D) everywhere
Answer:
(C) centre

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

15. The value of acceleration due to gravity (g) at a distance of 12800 km from the centre of earth is:
(A) 1/4 th that of earth
(B) half of the acceleration due to
(C) 4 times that of earth
(D) gravity of earth double that of earth
Answer:
( A) 1/4 th that of earth

16. The distance of the moon from the earth is :
(A) 384000 m
(B) 384000 km
(C) 38400 km
(D) 38400 m
Answer:
(B) 384000 km

17. The quantity of matter present in the body is called :
(A) weight
(B) mass
(C) pressure
(D) thrust
Answer:
(B) mass

18. The weight of an object is zero at:
(A) poles
(B) equator
(C) mountains
(D) centre of earth
Answer:
(D) centre of earth

19. What is the relation among u, y, g and t?
(A) v = u + gt
(B) v + u = gt
(C) v + gt = u
(D) \(\frac {v}{g}\) = u + t
Answer:
(A) v = u + gt

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

20. The distance of sun from earth is approximately:
(A) 1.5 x 1011 m
(B) 1.5 x 1011 km
(C) 1.5 x 1011 cm
(D) 1.5 x 1011 mm
Answer:
(A) 1.5 x 1011 m

21. By which symbol pascal is denoted?
(A) Pa
(B) Ps
(C) Pc
(D) P
Answer:
(A) Pa

22. On what angle a player throws ball with horizontal line so the ball falls at maximum distance?
(A) 900
(B) 0°
(C) 45°
(D) 600
Answer:
(C) 45°

23. The free fall means:
(A) the objects fàlling towards earth
(B) the objects going upward from the earth
(C) the objects moving parallel to earth
(D) none of these
Answer:
(A) the objects falling towards earth

24. SI unit of pressure is:
(A) Newton
(B) kilogram
(C) metre
(D) Pascal
Answer:
(D) Pascal

25. What will be the produced pressure 11300 n force is applied on the area of 20m2?
(A) 6000 Pa
(B) 3OPa
(C) 15 Pa
(D) 7.5 Pa
Answer:
(C) 15 Pa

HBSE 9th Class Science Important Questions Chapter 10 Gravitation

26. It is difficult to hold a school bag having a strap made of a thin and strong string, because thrust affects:
(A) large area
(B) more area
(C) small area
(D) longer area
Answer:
(C) small area

27. When an object is immersed in a liquid, the liquid exerted an upward force, which is called:
(A) buoyant force
(B) gravitational force
(C) frictional force
(D) opposing force
Answer:
(A) buoyant force

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