Class 9

AP State Syllabus AP Board 9th Class Physical Science Important Questions Chapter 9 Floating Bodies

AP State Syllabus 9th Class Physical Science Important Questions 9th Lesson Floating Bodies

9th Class Physical Science 9th Lesson Floating Bodies 1 Mark Important Questions and Answers

Question 1.
Scientifically what is density?
Answer:
Density is defined as mass per unit volume.

Question 2.
Define relative density.
Answer:
Relative density is the ratio of density of an object to the density of water.

Question 3.
What instrument is used to determine the purity of milk?
Answer:
Purity of milk can be determined by using lactometer.

Question 4.
What instrument is used to know the relative density of liquids?
Answer:
Relative density of liquids can be determined by using hydrometer.

Question 5.
What is atmospheric pressure?
Answer:
Air in our surroundings exerts pressure on the surface of earth called atmospheric pressure.
Atmospheric pressure P₀ = ρhg

Question 6.
State Archimedes’ principle.
Answer:
Archimedes’ principle states that when a body is immersed in a fluid it experiences an upward force of buoyancy equal to the weight of fluid displaced by the immersed portion of the body.

Question 7.
State Pascal’s principle.
Answer:
Pascal’s principle states that external pressure applied to an enclosed body of fluid is transmitted equally in all directions throughout the fluid.

Question 8.
What is the principle involved in working of hydraulic lift?
Hydraulic lift works using Pascal’s law.

Question 9.
In what direction does the buoyant force on an object immersed in a liquid act?
Answer:
Buoyant force on an object acts upward direction.

9th Class Physical Science 8th Lesson Gravitation 2 Marks Important Questions and Answers

Question 1.
Define density and give its units.
Answer:
Density is defined as the mass per unit volume.

Units : Unit for density is gm/cm³ (or) kg/m³.

Question 2.
Define relative density and give its units.
Answer:
Relative density is the ratio of density of an object to the density of water.

Units : Relative density has no units.

Question 3.
Why does the wooden block float on water even though it is heavier than a marble?
Answer:
The relative density of wooden block is less than the relative density of water whereas relative density of marble (glass) is more than that of water. Hence wooden block floats on water and marble sinks in water.

Question 4.
What do we mean by ‘heavy’, what do we mean by ‘light’?
Answer:
‘Heavy’ and ‘light’ can be decided based on the density of the object. Among the objects of equal volume, the one which weighs more can be treated as ‘heavy’ and the other one can be treated as ‘light’.

Question 5.
Relative density of silver is 10.8. What is density of silver in S.I. unit?
Answer:
Relative density of silver = 10.8

Density of water in S.I. system = 0³ kg/m³.
Density of silver = Relative density × Density of water = 10.8 × 10³ kg/m³.

Question 6.
Why does a block of plastic released under come up to the surface of water?
Answer:
Density of plastic is less than water. As the density of water is more than plastic, water exerts upward force which is known as buoyant force. So, the block of plastic come up to the surface of water.

Question 7.
The volume of 50 g. of a substance is 20 cm³. If the density of water is 1 g/cm³, will the substance float or sink?
Answer:

The density of substance is more than water so the substance sinks in the water.

Question 8.
An iron nail sinks in water. While a large ship floats on the surface of water. Give reason.
Answer:
The weight of water displaced by iron needle is less than its own weight, so it sinks. On the other hand, the ship displaces water more than its own weight, so it floats on water.

Question 9.
The density of gold is 19 g/cm³. Find the volume of 95 g of fold.
Answer:

Question 10.
A floating boat displaces water weighing 6000 newtons.
a) What is the buoyant force on the boat?
b) What is the weight of the boat?
Answer:
a) By Archimedes’ principles, the buoyant force acting on the object by it.
So, the buoyant force on the boat = 6000 newtons.

b) According to the principle of floating
Weight of object = Weight of liquid displaced by it
⇒ Weight of the boat = 6000 newtons

Question 11.
The volume of 50 g. of a substance is 20 cm . If the density of water is 1 g cm^-3, will the substance float or sink?
Answer:

2. Density of water = 1 g cm^-3.
As the density of substance is greater than that of water, so, the substance will sink in water.

Question 12.
Write any two differences between density and relative density.
Answer:

9th Class Physical Science 8th Lesson Gravitation 4 Marks Important Questions and Answers

Question 1.
Derive an expression for atmospheric pressure.
Answer:
1) All objects on the surface of the earth are subject to constant atmospheric pressure.

2) We know that the pressure is thrust (force) per unit area.

Question 2.
How can you measure the atmospheric pressure?
Answer:
1) Atmospheric pressure can be calculated with the help of the height of mercury column.

2) The weight of the mercury column in the tube is equal to the force applied on it by the mercury in the bowl due to atmospheric pressure.
∴ Weight of mercury column (W)
= Mass of mercury (m) × g
= Volume × density × g
= Cross sectional area of the tube × height of the tube × density × g
= Ahρg

Let P₀ be the atmospheric pressure.
Force on the column due to atmospheric pressure = P₀ A
Then Ahρg = P₀ A ⇒ P₀ = hρg
Here ‘ρ’ and ‘g’ are constants. So the height of mercury column depends on the atmospheric pressure.
We know height of mercury column h = 76 cm = 76 × 10^-2 m
Density of mercury (ρ) = 13.6 gm/cc = 13.6 × 10³ kg/m³3
Acceleration due to gravity (g) = 9.8 m/sec2
P₀ = hρg = 76 × 10^-2 × 13.6 × 10³ × 9.8 = 1.01 × 10⁵ kg m/m²s²
1 kg m\sec² = 1 Newton
P₀ = 1.01 × 10⁵ N/m²
This value is called atmospheric pressure.
1 Atmosphere = 1.01 × 10⁵ N/m² = 1.01 × 10⁵ Pascal

Question 3.
Calculate the pressure at a depth ‘h’ in a liquid.
Answer:

1) Let us consider a container which contains a liquid in it of density ‘ p

2) Consider cylindrical column of height ‘h’ from the surface of the liquid of cross sectional area “A”.

3) The volume of the liquid column V = Ah
Mass = Volume x density
⇒ m = Ahρ
Weight W = mg = Ahρg
From Newton’s law the net force on the liquid column is zero, because it is at rest.

4) The forces acting on the water column are
a) Weight (W), vertically down
b) Force on top surface due to atmospheric pressure (P₀ A) acting vertically down.
c) Force on the bottom surface of the column due to static pressure of liquid (PA), acting vertically up.

5) From Newton’s law we get

PA = P₀A + W ⇒ PA = P₀A + hρgA =⇒ PA = A (P₀ + hρg)
∴ P = P₀ + hρg
P = Pressure at depth h’ from the surface of the liquid
P₀ = Atmospheric pressure
∴ The pressure inside the liquid at a constant depth is constant.

Question 4.
Calculate the pressure difference at different levels of depth in fluids.
Answer:

1) Let us consider a cylindrical column of liquid of height ‘h’ with cross sectional area
‘A’ and let ‘P’ be the density of the liquid.

2) The pressure P₁ in the liquid at depth h, is
P₁ = P₀ + h₁ρg ………(1)

3) Similarly pressure P₂ in the liquid at depth h₂ is
P₂ = P₀ + h₂ρg …………..(2)

4) From (1) and (2), P₂ – P₁ = h₂ρg – h₁pg
P₂ – p₁ = ρg (h₂ – h₁)

5) From the figure h₂ – h₁ = h
∴ P₂ – p₁ = hρg

6) The pressure difference between two levels in that liquid = hρg

7) Here ρ and g are constants, so the pressure difference increases with an increase in depth.

Question 5.
State Pascal’s principle and explain it with an example.
(OR)
Write Pascal’s law. Draw the diagram of any device works based as Pascals’ law.
Answer:
Pascal’s principle :
Pascal’s principle states that external pressure applied to a enclosed body of fluid is transmitted equally in all directions throughout the fluid volume and the walls of the containing vessel.

Explanation :

1. Look at the figure.
2. Here we have an enclosed volume of fluid in a U-shaped tube.
3. The fluid is enclosed in the tube by two leak-proof pistons in each arm.
4. The ratio of cross-section areas of the right and left tubes is A₂ : A₁ and A₂ > A₁
5. When a force F₁ is applied to the left piston, the excess pressure acting on the fluid volume is F₁/A₁.
6. According to Pascal’s principle, this excess pressure is transmitted equally throughout the fluid volume.
7. The excess pressure in the right-side tube (Area A₂) is also F₁/A₁ and since its area is A₂, the upward force acting on the right piston is F₂ = A₂ × F₁/A₁, which is much larger in magnitude than F₁
8. Thus the application of Pascal’s principle results in a large upward force on the right piston when a small downward force is applied on the left piston.

Question 6.
The volume of a solid of mass 500g is 350 cm³.
a) What will be the density of this solid?
b) What will be the relative density of the solid?
c) Will it float or sink in water?
Answer:

c) Relative density of solid is greater than the relative density of water. So, it will sink in water.

AP State Syllabus AP Board 9th Class Hindi Textbook Solutions Chapter 5 फुटबॉल Textbook Questions and Answers.

AP State Syllabus 9th Class Hindi Solutions Chapter 5 फुटबॉल

9th Class Hindi Chapter 5 फुटबॉल Textbook Questions and Answers

InText Questions (Textbook Page No. 22)

प्रश्न 1.
चित्र में क्या दिखाई दे रहा है?
उत्तर:
चित्र में दो लड़के (टेबुल टेन्निस के खिलाड़ी, हाथ में राकेट (बैट) लेकर खडे हुए हैं।) एक के गले में मेडल है। दूसरे के पीठ पर थैला है।

प्रश्न 2.
वे क्या कर रहे हैं?
उत्तर:
वे दोनों एक दूसरे के कमर पर हाथ रखकर एक दूसरे के कंधे मिलाकर खडे हुए हैं।
(या)
वे दोनों प्यार से गले लग रहे हैं।

प्रश्न 3.
एक – दूसरे से क्या कहना चाहते हैं?
उत्तर:
वे एक – दूसरे से जीत की शुभकामनाएँ कहना चाहते हैं।

अर्थव्राह्यता-प्रतिक्रिया

अ) प्रश्नों के उत्तर सोचकर बताइए।

प्रश्न 1.
‘खेलों से मनोरंजन होता है’ अपने विचार बताइए।
उत्तर:
खेलों से मनोरंजन होता है। बच्चों के मनोरंजन और पसंदी साधनाओं में खेलों का ही अधिक महत्व है। खेलों से शारीरिक विकास के साथ-साथ मानसिक विकास होता है। विद्यार्थी सुबह दस बजे से लेकर शाम को पाँच बजे तक पाठशालाओं में पढ़ाई के कारण मानसिक थकान का अनुभव करते हैं तो खेल उन्हें मनोरंजन देकर पढ़ाई में रुचि उत्पन्न करते हैं।

प्रश्न 2.
अपने मनपसंद खेल के बारे में बताइए।
उत्तर:
मेरा मनपसंद खेल क्रिकेट है । इसमें दो टीम होते हैं। एक – एक टीम में ग्यारह खिलाडी होते हैं। एक एक्स्ट्रा खिलाडी भी होता है। एक टीम बाउलिंग करता तो दूसरा टीम बैटिंग करता है। बैटिंग पूरा होने के बाद एक पक्ष वाले जितने रन बना लेते उनसे एक रन अधिक बनाना दूसरे पक्ष का लक्ष्य होता है । यदि वह टीम उतने रन बनाने में विफल हो तो हार जाएगा ।

आ) पाठ के आधार पर वाक्यों को उचित क्रम दीजिए।

1. जीतने की कोशिश करो। ( 4 )
2. यह प्रेम का खेल है। ( 2 )
3. फुटबॉल सचाई का खेल है। ( 1 )
4. इस बार हारोगे, तो अगली बार जीतोगे। (3)

इ) अनुच्छेद पढ़िए। भाव लिरिवए ।

खेल बच्चों के जीवन का अनोखा अंग है। इससे मानसिक और शारीरिक विकास होता है। खेलों से मानसिक थकान दूर होती है । जीवन परिश्रमी बनता है। शरीर स्वस्थ रहता है। इसलिए खेलों को अपने जीवन में महत्व देना अनिवार्य है।
उत्तर:
खेल बच्चों के जीवन का अंग है। खेलों से मानसिक और शारीरिक विकास होता है। मानसिक थकावट दूर होता है और बच्चे परिश्रमी बनते हैं | स्वस्थ रहते हैं। खेलों को महत्व देना ज़रूरी है।

ई) इन प्रश्नों के उत्तर तीन वाक्यों में दीजिए।

प्रश्न 1.
फुटबॉल खेल में गोल रक्षक का क्या महत्व है?
उत्तर:
फुटबॉल एक खास मैदानी खेल है । इस खेल में गोल रक्षक का महत्वपूर्ण स्थान रहता है। खेल में दोनों पक्षों के खिलाडी गेंद को मारते विपक्षी के बाक्स में भेजने का भरसक प्रयत्न करते हैं। गोल रक्षक बडी सतर्कता से गेंद पर ही नज़र रखकर, गेंद किस ओर से भी आये, अंदर घुसने न देने का प्रयत्न करता है। गोल रक्षक के सामर्थ्य पर ही खेल में जीत या हार संभव हो सकता है |

प्रश्न 2.
किसी भी खेल में नेतृत्व भावना की क्या भूमिका होती है?
उत्तर:
क्रिकेट, वालीबॉल, फुटबॉल आदि खेलों में अधिक खिलाडी खेलते हैं। उन खिलाडियों में एक योग्य व्यक्ति को नेता (कप्तान) बनाकर टीम का नेतृत्व करने की ज़िम्मेदारी सौंपी जाती है । तब वह नेता बना व्यक्ति समझदारी, निपुणता से खिलाडियों में एकता की भावना भरते खेलने प्रोत्साहित करता रहता है। खिलाडी की योग्यता के अनुसार विविध स्थान देकर खेल में जीत प्राप्त करने का भरसक प्रयत्न करता है। अपने साथी खिलाडियों में उत्साह भरना, किसी कष्ट समय पर भी उनकी निंदा न करके ज़िम्मेदारी से खेलने की चेतना उनमें भरता है। सारे खिलाडियों को बिना किसी भेदभाव के खेल में अपनी पूरी योग्यता दिखाने का मौका देने का सही प्रयत्न नेता करता है । अपने नेतृत्व के टीम को विजयपथ पर ले जाने का गुरुतर भार नेता पर ही रहता है।

अभिव्यक्ति – सृजनात्मकता

अ) इन प्रश्नों के उत्तर लिरिवए ।

प्रश्न 1.
खेलों से किन गुणों का विकास होता है?
उत्तर:

• खेलों से निम्न लिखित गुणों का विकास होता है।
• विपक्षी के खेल को समझने और सराहने का गुण
• मिलजुलकर रहने का भाव
• अपनी हार और जीत से कुछ सीख
• ताकत और सावधानी का विकास
• टीम स्पिरिट का विकास
• परस्पर प्रेम भावना
• सब मिलकर सोचने की भावना
• सचाई से जीने की प्रेरणा
• खेलों से भाईचारे की भावना

प्रश्न 2.
आपको कौन-सा खेल पसंद है और क्यों?
उत्तर:
मुझे फुटबॉल का खेल पसंद है। क्योंकि फुटबॉल का खेल अकेले एक आदमी का खेल नहीं। यह टीम का खेल है, मिलजुलकर खेलने का खेल है। यह ताकत और सावधानी का खेल है। यह चुस्ती का खेल है। यह नज़र का खेल है। फुटबॉल पर नज़र, साथियों पर नज़र। यह ऐसा खेल है जिसमें एक की ताकत सबकी ताकत है और एक की कमज़ोरी सबकी कमज़ोरी है। यह प्रेम और सचाई का खेल है।

प्रश्न 3.
पढाई के साथ – साथ खेलों की आवश्यकता क्यों होती है?
उत्तर:
पढ़ाई के साथ – साथ खेलों की आवश्यकता इन निम्न लिखित कारणों से होती है।

• स्वस्थ शरीर में स्वस्थ मस्तिष्क का निवास होता है। स्वस्थ शरीर के लिए खेलों की जरूरत है। तभी हमें पढ़ने के विषयों के प्रति संबंध स्थापित होता है।
• खेलों से मानसिक तथा शारीरिक थकावट दूर हो कर पढ़ाई के प्रति रुचि उत्पन्न होता है।
• खेलों से आत्म निर्भरता बढ़ती है इसलिए हम जो भी पढ़े वह आत्मसात हो जाता है।
• खेलों से छात्रों में मनोविनोद होता है। इसलिए फिर वे पढने में जी लगा सकते हैं।
• खेलों के द्वारा छात्रों में अनुशासन तथा सहयोग की भावना बढ़ती है ये दोनों छात्रों की पढ़ाई में सहयोग देते हैं।
• खेलों के द्वारा जीवन में परिश्रमी बनने का भाव उत्पन्न होता है जिससे छात्र खूब परिश्रम करके पढ़ाई में जीत पायेंगे।
• खेल बच्चों के जीवन का अनोखा अंग है।
• छात्र खेलते – कूदते, खेलों के माध्यम से अच्छी तरह पढ़ सकते हैं।
• खेलों से मानसिक तथा शारीरिक विकास होता है जो पढ़ाई में सहायक होता है।

आ) “फुटबॉल’ पाठ का सारांश अपने शब्दों में लिखिए।
(या)
फुटबॉल मिलजुलकर खेलने का खेल है। फुटबॉल खेल के बारे में आप क्या जानते हैं? लिखिए।
उत्तर:
‘फुटबॉल” नामक पाठ में लेखक फुटबॉल खेल के बारे में और उसके प्रति आपने लगाव के बारे में वर्णन करते हैं। जो इस प्रकार है –

फुटबॉल सचाई का खेल है। यह प्रेम का खेल है । इसमें दो पक्ष होते हैं। फुटबॉल प्रायः हमारे सभी स्कूलों में खेला जाता है।

लेखक को वॉलीबाल खेलते-खेलते फुटबॉल भी खेलने का शौक पैदा हुआ | तब से फुटबॉल लेखक का प्रिय खेल हो गया। लेखक दिन में, रात में, मैदान में, खेत में जब समय मिले तब फुटबॉल खेलने लगा। __फुटबॉल अकेला एक आदमी का खेल नहीं। यह टीम का खेल है । मिलजुलकर खेलने का खेल है। इसमें एक की ताकत सबकी ताकत है, एक की कमज़ोरी सबकी कमज़ोरी है।

इस खेल में भी हार और जीत दोनों होते हैं। जीत और हार दोनों से सीख मिलता है। अंत में लेखक आशा करता है कि आज तक तो वह किसी बडी टीम में नहीं खेला लेकिन वह एक न एक दिन देश की किसी
भी बड़ी टीम में खेलेगा।

इ) किसी एक खेल के बारे में लिखिए |
उत्तर:
क्रिकेट :
क्रिकेट बडा ही मनोरंजक खेल है। बच्चे, युवक, वृद्ध, अफ़सर, व्यापारी, अध्यापक तथा आफ़ीसों के बाबू आदि सभी के मन में क्रिकेट मैच देखने की उमंग उठती है।

क्रिकेट खेल में दो दल होते हैं। प्रत्येक दल में ग्यारह खिलाडी खेलते हैं। खेल खेलने के लिए बहुत बड़े मैदान की आवश्यकता होती है। इसके बीच में पिच होती है। इस पिच पर मैटिग (चढ़ाई) बिछा दिया जाता है।

खेल शुरू होने से पहले दोनों दल के एक-एक खिलाडी बारी – बारी से आकर विकटों के सामने खडा होता है और गेंद से हिट लगाकर (पीटकर) रन बनाता है। एक खिलाडी विकटों के पीछे खडा होता है। जिसे विकट कीपर कहते हैं। एक दूसरे खिलाडी गेंद फेंकने वाले के पीछे खडा होता है, इसे रनर कहते हैं। दूसरे दल का एक खिलाडी जिसे बउलर या गेंदबाज़ कहते हैं। विकटों को लक्ष्य बनाकर गेंद इस प्रकार फेंकने का प्रयास करता है कि वह विकटों पर जाकर लगे।

यदि एक दल के सभी खिलाडी खेल लेते हैं तो दूसरे दल के खिलाडी जो कि पहले फील्डिंग करते थे, खेलना आरंभ कर देते हैं। जिस दल के रन अधिक होते हैं, वह दल विजयी माना जाता है ।

ई) ‘हार – जीत खेल का एक हिस्सा है।’ इस पर अपने विचार प्रकट कीजिए।
उत्तर:
मानव के शारीरिक व मानसिक विकास के लिए खेल अत्यंत आवश्यक अंग है। खेलों से मानसिक थकावट दूर होती है। शरीर तंदुरुस्त रहकर, जीवन परिश्रमी बनता है। खेलों को खेल ही समझने चाहिए | खेल ही जीवन नहीं समझना है।

हर खेल में दो पक्ष ज़रूर होते हैं। एक पक्ष जीतता है तो दूसरा हार जाता है। यह तो अनिवार्य है। तभी खेल का फल मालूम होता है। हार-जीत दोनों खेल का एक हिस्सा है। इन दोनों का समान अस्तित्व रहता है। खेल में वैर भाव नहीं रखना चाहिए। एक बार हारेगा तो अगली बार जीतेगा | जीतने की कोशिश करनी है। हार या जीत खिलाडियों की गलतियों, कमजोरियों, लिये जानेवाले निर्णयों आदि अनेक मुख्य विषयों पर आधारित होते हैं। अपनी हार को असमर्थता और विपक्षी की सफलता को समर्थता नहीं समझनी चाहिए।

खेल तो खेल ही है । उसे प्रमे से चाव से खेलना चाहिए। इस बार हार होगी तो विफल न होकर अगली बार ज़रूर जीत होगी, ऐसी भावना रखकर कोशिश करनी है। वही सच्चा खिलाडी का महान गुण है। सच्चा खिलाडी जीते हुए पक्षवालों के खेल को समझने की कोशिश करेगा। उनकी प्रशंसा ज़रूर करेगा। अपनी हार का कारण क्या है? इस पर विचार करेगा। इस तरह अपनी हार या जीत से कुछ न कुछ ज़रूर सीखनेवाला ही सच्चा खिलाडी है।

भाषा की बात

अ) वाक्य पढ़िए और अर्थ के आधार पर वाक्य के भेद समझिए ।

परियोजना कार्य

वर्ग-पहेली में खेलों के नाम छिपे हैं । उन्हें पहचानिए । किसी एक के बारे में जानकारी इकट्ठा कीजिए।

उत्तर:

1. खो-खो
2. फुटबॉल
3. क्रिकेट
4. हॉकी
5. टेनिस

इनमें से क्रिकेट खेल के बारे में अभिव्यक्ति -सृजनात्मकता के अंतर्गत प्रश्न संख्या ‘ई में जानकारी दी गयी है। उसे पढ़िए।

फुटबॉल Summary in English

Football is often played in our schools. Soon after I found a football. I went to play the game. We keep on playing football. As I kept on playing. I grew a liking towards the game football. Besides, it became my favourite game.

Whether it is a day or night, field or playground, when I found time I was there. Football was with me. At first I would play the game all alone. Then it was a toy for me. Later I realised that it was everybody’s thing. It’s a wonderful thing. Two teams play the game. Thousands of spectators watch it.

Football is not a one-man game. It is a game played by teams. It is a game played together. It is a game played with energy and concentration. It is a game played with intellect and agility. It is a game played with attention. Attention on football, attention on friends, attention on opposition, attention on goal, attention on line, attention on time and attention on one’s own. It’s a sort of game in which one’s energy becomes everybody’s energy and one’s weakness becomes everybody’s weakness.

Football is an honest game. It’s a game which enhances love. The player who has hostile feelings is not called so. There are two sides. One wins and one loses. Then why should one have feelings such as anger and enmity? If they lose now, they will win next time. Do sincere effort to win. If the opposition wins, try to understand the way they played. Appreciate them. Do learn from your failure. Do learn from your success.

So far I haven’t played in any big team. But one day or other I will certainly play in a big team belonged to my country. Besides, I would like to play outside also. My wish will certainly be accomplished today or tomorrow.

फुटबॉल Summary in Telugu

ఫుట్ బాల్ తరచుగా మన పాఠశాలలన్నింటిలో ఆడబడుచున్నది. ఫుట్ బాల్ లభించగానే ఆడటానికి నేను వెళ్లిపోయాను. మనం వాలీబాల్ ఆడుతూనే ఉంటాం. ఈ విధంగా ఆడుతూ ఆడుతూ ఫుట్ బాల్ ఆట పట్ల ఇష్టం ఏర్పడినది. అంతేకాదు ఫుట్ బాల్ ఆట నా ప్రియమైన ఆట అయినది.

పగలేమి, రాత్రి ఏమి, పొలమేమిటి, ఆటస్థలమేమిటి, ఎప్పుడు సమయం దొరికితే అప్పుడు నేనున్నాను. నా ఫుట్ బాల్ ఉంది. మొదట్లో నేను ఒంటరిగానే ఫుట్ బాల్ ఆడుకునేవాణ్ణి. అప్పుడు నాకు అది ఒక ఆట బొమ్మ. తర్వాత ఇది అందరి వస్తువని నాకు అర్థమైనది. ఇది అద్భుతమైన వస్తువు. ఫుట్ బాల్ అందరి వస్తువు. రెండు పక్షాలు ఆడతాయి. వేలకొలది ప్రజలు చూస్తారు.

ఫుట్ బాల్ ఆట ఒంటరిగా ఒక మనిషి ఆడే ఆట కాదు. ఇది టీం (కొంతమంది ఆటగాళ్ళు) ఆడే ఆట. కలిసిమెలిసి ఆడే ఆట. ఇది శక్తితో మరియు ఏకాగ్రతతో (జాగరూకతతో) ఆడే ఆట. ఇది తెలివితేటలతో చురుకుదనంతో ఆడే ఆట. ఇది దృష్టితో ఆడే ఆట. ఫుట్ బాల్ మీద దృష్టి, స్నేహితుల మీద దృష్టి, విపక్షం మీద దృష్టి, గోల్ మీద దృష్టి, గీత (లైన్) మీద దృష్టి, సమయంపై దృష్టి మరియు తన మీద దృష్టి. ఇది ఎలాంటి ఆట అంటే ఒకరి శక్తి అందరి శక్తి, ఒకరి బలహీనత అందరి బలహీనత అయ్యే ఆట.

ఫుట్ బాల్ నిజాయితీ ఆట. ఇది ప్రేమను పెంచే ఆట. వైరభావం ఉన్నవాడు ఆటగాడే కాదు. రెండు పక్షాలుంటాయి. ఒకరు గెలుస్తారు. ఒకరు ఓడతారు. దీనిలో కోపం, వైరం(శతృత్వం) అనే భావాలెందుకు ? ఇప్పుడు ఓడిపోతే ఇంకోసారి గెలుస్తారు. గెలిచే ప్రయత్నం చేయి. ఎదుటివారు (ప్రతిపక్షం) గెలిస్తే వారి ఆటను అర్థం చేసుకో, వారిని పొగుడు. నీ ఓటమి నుండి నేర్చుకో, నీ విజయం నుండి నేర్చుకో.

ఈనాటి వరకు నేనే పెద్ద టీంలోనూ ఆడలేదు. కానీ నేను ఏదో ఒక రోజున నా దేశానికే చెందిన ఏదో ఒక పెద్ద టీంలో తప్పనిసరిగా ఆడతా. అంతేకాదు నేను బయట కూడా ఆడాలని అనుకుంటున్నా. ఈ రోజు కాకపోతే రేపైనా నా కోరిక తప్పనిసరిగా నెరవేరుతుంది.

अर्थग्राहयता – प्रतिक्रिया

निम्नलिखित गद्यांश पढ़कर दिये गये प्रश्नों के उत्तर एक वाक्य में लिखिए।

“फुटबॉल सचाई का खेल है। यह प्रेम का खेल है। इसमें जो वैर – भाव रखे, वह खिलाड़ी नहीं। दो पक्ष हैं, एक जीतेगा और दूसरा हारेगा। इसमें नाराज़गी या वैर – भाव की क्या बात ? इस बार हारोगे तो अगली बार जीतोगे।”
प्रश्न :
1. फुटबॉल किस प्रकार का खेल है?
उत्तर:
फुटबॉल सचाई और प्रेम का खेल है।

2. खेल में वैरभाव रखनेवाला क्या नहीं कहलाता है?
उत्तर:
खेल में वैरभाव रखनेवाला खिलाडी नहीं कहलाता है।

3. फुटबाल में कितने पक्ष खेलते हैं?
उत्तर:
फुटबाल में दो पक्ष खेलते हैं।

4. “हारना” शब्द का विलोम शब्द क्या है?
उत्तर:
हारना x जीतना (‘हारना’ शब्द का विलोम शब्द है “जीतना”।

5. यह गद्यांश किस पाठ से लिया गया है?
उत्तर:
यह गद्यांश “फुटबाल” पाठ से लिया गया है।

निम्न लिखित गद्यांश पढ़कर नीचे दिये गये वैकल्पिक प्रश्नों के उत्तर दीजिए। सही विकल्प से संबंधित अक्षर चुनकर कोष्ठक में रखिए।

“मनुष्य के लिए स्वस्थ रहना अत्यन्त आवश्यक है। तंदुरुस्त रहने से मन प्रसन्न रहता है। मनुष्य दुगुने उत्साह के साथ अपने काम में लग सकता है। इसके विपरीत अस्वस्थ रहने से वह उदास हो जाता है। मन नीरस रहता है। इसलिए कहा गया है कि स्वस्थ तन में स्वस्थ मन और मस्तिष्क का
निवास होता है।”
प्रश्न :
1. मनुष्य के लिए कैसे रहना अत्यंत आवश्यक है?
A) उदास
B) स्वस्थ
C) नीरस
D) दुःखी
उत्तर:
B) स्वस्थ

2. तन्दुरुस्त रहने से मन कैसा रहता है?
A) प्रसन्न
B) नीरस
C) उदास
D) बेचैनी
उत्तर:
A) प्रसन्न

3. मनुष्य अस्वस्थ रहने पर क्या हो जाता है?
A) स्वस्थ
B) चुस्त
C) अच्छा
D) उदास
उत्तर:
D) उदास

4. स्वस्थ तन में किसका निवास होता है?
A) स्वस्थ मन
B) स्वस्थ मस्तिष्क
C) स्वस्थ और मस्तिक
D) कुछ नहीं
उत्तर:
A) स्वस्थ मन

5. “स्वस्थ’ शब्द का विलोम शब्द क्या है?
A) अस्वस्थ
B) प्रसन्न उत्तर
C) सुख
D) उत्साह
उत्तर:
A) अस्वस्थ

निर्देश के अनुसार उत्तर दीजिए।

1. मैं हूँ, मेरा फुटबाल । (वाक्य में सर्वनाम शब्द क्या हैं ?)
A) मैं
B) हूँ
C) फुटबॉल
D) कुछ नहीं
उत्तर:
A) मैं

2. शुद्ध रूप पहचानिए।
A) हजारों लोग देखता है।
B) हजार लोग देखता हैं।
C) हजारों लोग देखते हैं।
D) हजारों लोग देखते है।
उत्तर:
C) हजारों लोग देखते हैं।

3. समय मिलता तो हम फुटबाल खेलते। (संज्ञा शब्द पहचानिए।)
A) मिलता
B) तो
C) हम
D) फुटबाल
उत्तर:
D) फुटबाल

4. सही क्रम वाला वाक्य पहचानिए।
A) प्रायः हमारे सभी स्कूलों में खेला जाता है फुटबॉल।
B) फुटबॉल प्रायः हमारे सभी स्कूलों में खेला जाता है।
C) हमारे सभी स्कूलों में फुटबाल प्रायः खेला जाता है।
D) सभी स्कूलों में खेला जाता है फुटबाल प्रायः।
उत्तर:
B) फुटबॉल प्रायः हमारे सभी स्कूलों में खेला जाता है।

5. पढ़ाई के साथ – साथ खेलों की आवश्यकता है। (रेखांकित शब्द का विलोमशब्द पहचानिए।)
A) निरावश्यकता
B) परावशयकता
C) आवश्यकताहीन
D) अनावश्यकता
उत्तर:
D) अनावश्यकता

6. राम दिन – रात फुटबॉल खेलता रहता है। रेखांकित शब्द का समास पहचानिए।
A) तत्पुरुष समास
B) द्विगु समास
C) द्वंद्व समास
D) अव्ययीभाव समास
उत्तर:
C) द्वंद्व समास

7. खेल में विपक्ष पर नज़र रखना है। रेखांकित शब्द का उपसर्ग पहचानिए।
A) क्ष
B) पक्ष
C) वि
D) विप
उत्तर:
C) वि

8. खिलाड़ी ने फुटबॉल को ज़ोर से मारा। इस वाक्य में क्रिया विशेषण शब्द पहचानिए।
A) मारा
B) खिलाडी
C) जोर से
D) फुटबॉल
उत्तर:
B) खिलाडी

9. फुटबॉल सचाई ………. खेल है। उचित कारक चिह्न पहचानिए|
A) के
B) का
(B )
C) की
D) को
उत्तर:
B) का

10. फुटबॉल अकेले एक आदमी का खेल नहीं है। संज्ञा शब्द को पहचानिए।
A) अकेले
B) एक
C) नहीं है
D) फुटबॉल
उत्तर:
D) फुटबॉल

11. फुटबॉल ताकत और सावधानी का खेल है। (रेखांकित शब्द का अर्थ क्या है?)
A) कमज़ोरी
B) शक्ति
C) तप
D) ताज
उत्तर:
B) शक्ति

अर्थग्राह्यता – प्रतिक्रिया

पठित – गद्यांश

निम्नलिखित गद्यांश पढ़कर दिये गये प्रश्नों के उत्तर एक वाक्य में दीजिए।

1. फुटबॉल सचाई का खेल है। यह प्रेम का खेल है। इसमें जो वैर – भाव रखे, वह खिलाड़ी नहीं। दो पक्ष हैं, एक जीतेगा और दूसरा हारेगा। इसमें नाराज़गी या वैर – भाव की क्या बात? इस बार हारोगे तो अगली बार जीतोगे। जीतने की कोशिश करो। विपक्षी अगर जीतता है तो उसके खेल को समझो और उसकी सराहना करो। अपनी हार से भी सीखो और अपनी जीत से भी सीखो।
प्रश्न :
1. फुटबॉल कैसा खेल है?
उत्तर:
फुटबॉल सचाई का खेल है। यह प्रेम का खेल है।

2. फुटबॉल खेल में कितने पक्ष होते हैं?
उत्तर:
फुटबॉल खेल में दो पक्ष होते हैं।

3. विपक्षी अगर जीतता है तो हमें क्या करना है?
उत्तर:
विपक्षी अगर जीतता है तो उसके खेल को समझना और उसकी सराहना करना चाहिए।

4. हमें किन – किनसे सीखना है?
उत्तर:
हमें हार और जीत दोनों से सीखना है।

5. वैर – भाव रखनेवाला क्या नहीं हो सकता?
उत्तर:
वैर भाव रखने वाला खिलाड़ी नहीं हो सकता।

2. फुटबॉल का खेल अकेले एक आदमी का खेल नहीं। यह टीम का खेल है, मिलजुलकर खेलने का खेल है, यह ताकत और सावधानी का खेल है, यह चुस्ती का खेल है, यह नज़र का खेल है। फुटबॉल पर नज़र, साथियों पर नज़र, विपक्षियों पर नज़र, गोल पर नज़र, लाइन पर नज़र, समय पर नज़र, अपने पर नज़र । यह ऐसा खेल है जिसमें एक की ताकत सबकी ताकत है और एक की कमज़ोरी सबकी कमज़ोरी है।
प्रश्न :
1. टीम का खेल क्या है?
उत्तर:
फुटबॉल टीम का खेल है।

2. ताकत और सावधानी का खेल क्या है?
उत्तर:
ताकत और सावधानी का खेल है “फुटबॉल”।

3. इस उपर्युक्त अनुच्छेद में किस खेल के बारे में बताया गया है?
उत्तर:
इस उपर्युक्त अनुच्छेद में फुटबॉल खेल के बारे में बताया गया है।

4. फुटबॉल खेल में किन – किन पर नज़र रखनी पडती है?
उत्तर:
फुटबॉल खेल में फुटबॉल पर नज़र, साथियों पर नज़र, विपक्षियों पर नज़र, गोल पर नज़र, लाइन पर नज़र, समय पर नज़र और अपने पर नज़र रखनी पडती हैं।

5. उपर्युक्त अनुच्छेद किस पाट से लिया गया है?
उत्तर:
उपर्युक्त अनुच्छेद “फुटबॉल” पाठ से लिया गया है।

3. फुटबॉल प्रायः हमारे सभी स्कूलों में खेला जाता है। फुटबॉल मिला और लेकर खेलने निकल गये। यों भी हम लोग वालीबॉल तो खेलते ही आये हैं। इस तरह खेलते – खेलते फुटबॉल भी खेलने का शौक पैदा हो गया और फुटबॉल मेरा प्रिय खेल बन गया।
प्रश्न :
1. प्रायः सभी स्कूलों में क्या खेला जाता है?
उत्तर:
प्रायः सभी स्कूलों में फुटबॉल खेला जाता है।

2. यों भी हम लोग क्या खेलते ही आये हैं?
उत्तर:
यों भी हम लोग वालीबॉल खेलते ही आये हैं।

3. लेखक का प्रिय खेल क्या बन गया ?
उत्तर:
फुटबॉल लेखक का प्रिय खेल बन गया ।

4. लेखक को क्या मिला ? और क्या खेलने निकल गये ?
उत्तर:
लेखक को फुटबॉल मिला और फुटबॉल खेलने निकल गये।

5. उपर्युक्त गद्यांश किस पाठ से लिया गया है?
उत्तर:
उपर्युक्त गद्यांश ‘फुटबॉल’ पाठ से लिया गया है।

अपठित – गद्यांश

नीचे दिये गये गद्यांश को पढ़कर प्रश्नों के उत्तर चुनकर कोष्ठक में लिखिए।

1. लक्ष्मीबाई का जन्म 16 नवंबर, 1835 ई. को हुआ था। इनके पिता का नाम मोरोपंत ताम्बे और माता का नाम भागीरथी बाई था। लक्ष्मीबाई के बचपन का नाम मनुबाई था। मनु की माता का स्वर्गवास उनके बचपन में ही हो गया। पिता ने ही उनका पालन – पोषण किया। नाना साहेब बचपन में मनु के साथी थे।
प्रश्न :
1. लक्ष्मीबाई का जन्म किस साल में हुआ?
A) 1935
B) 1835
C) 1845
D) 1855
उत्तर:
B) 1835

2. लक्ष्मीबाई के पिता कौन थे?
A) मोरोपंत तांबे
B) नाना साहेब
C) गंगाधर
D) भागीरथी बाई
उत्तर:
A) मोरोपंत तांबे

3. लक्ष्मीबाई की माता का नाम क्या है?
A) जमुना बाई
B) लता बाई
C) भागीरथी बाई
D) मनुबाई
उत्तर:
C) भागीरथी बाई

4. मनुबाई किनके बचपन का नाम था?
A) भागीरथी बाई
B) जमुना बाई
C) राधा बाई
D) लक्ष्मीबाई
उत्तर:
D) लक्ष्मीबाई

5. लक्ष्मीबाई का पालन – पोषण किसने किया?
A) माता
B) पिता
C) भाई
D) चाचा
उत्तर:
B) पिता

2. मुगल सम्राटों में शाहजहाँ बड़ा ही सहृदय, विलासप्रिय एवं कला मर्मज्ञ था। रानी मुमताज महल रूप की रानी थी। अपने गुणों एवं सौंदर्य से उन्होंने शाहजहाँ को वशीभूत कर रखा था। दोनों ओर प्रेम का बन्धन था और इस बन्धन में ही सुख था। किन्तु मृत्यु की क्रूर दृष्टि से किसका सुख देखा गया है? महारानी बीमार हुई और अन्तिम घड़ियाँ निकट आने लगीं। जगत् का मोह भी बढ़ने लगा। मृत्यु की ओर बढ़ती हुई मुमताज़ ने कामना की कि उसकी समाधि असाधारण हो । पति ने आर्द्र हृदय से बिछुड़ती हुई प्रिया को कुछ वचन दिये थे। ताजमहल उसी वचन की पूर्ति का प्रयत्न है।
प्रश्न :
1. रूप की रानी कौन थी?
A) रानी मुंताज महल
B) रानी जुबेदा बेगम
C) रुद्रमदेवी
D) रानी चंडी देवी
उत्तर:
A) रानी मुंताज महल

2. बडा ही सहृदय, विलासप्रिय एवं कलामर्मज्ञ कौन था?
A) जहंगीर
B) औरंगजेब
C) अकबर
D) शाहजहाँ
उत्तर:
D) शाहजहाँ

3. मृत्यु की ओर बढ़ती हुई मुंताज की कामना क्या थी?
A) उसकी समाधि असाधारण हो
C) सुंदर भवन की कामना चाहिए
D) ये सब चाहिए
उत्तर:
A) उसकी समाधि असाधारण हो

4. मुमताज ने शाहजहाँ को कैसे वशीभूत कर रखा?
A) दंड देते हुए
B) अपने सौंदर्य और गुणों से
C) प्यार से
D) इन सबसे
उत्तर:
B) अपने सौंदर्य और गुणों से

5. शाहजहाँ इस वंश का सम्राट था
A) गुलाम
B) तुम्लक
C) संगम
D) मुगल
उत्तर:
D) मुगल

3. आज की शिक्षा प्रणाली में जो सबसे बड़ी खराबी है, उसे अब मैं तीव्रता से महसूस कर रही हूँ। पिछले 18 वर्षों के मेरे अध्ययन काल में मुझे न कभी जंगल से लकड़ी लानी पड़ी थी, न कभी गायें चरानी पड़ी थी। इसलिए मेरा शरीर किसी भी प्रकार के परिश्रम के लिए सर्वथा असमर्थ बन गया था। यात्रा में शरीक होने पर पहले ही दिन मेरा स्वास्थ्य बिगड़ गया । पिछले 18 वर्षों के दरमियान में किसी भी विषय में प्रवीण न हो सकी, यह बात तो थी ही, लेकिन उधर वर्षा के महिलाश्रम की नर्मदा बेन मुझ – जैसी ही नौसिखुआ होते हुए भी बड़ी फुर्ती से विनोबाजी के साथ चल रही थी।
प्रश्न :
1. आज की शिक्षण प्रणाली कैसी है?
A) बहुत अच्छी
B) बडी खराबी
C) सुविधाजनक
D) रटंत
उत्तर:
B) बडी खराबी

2. लेखिका का शरीर किसके लिए असमर्थ बन गया ?
A) किसी भी प्रकार के परिश्रम के लिए
B) बोझ ढोने
C) आराम लेने के लिए
D) इन सबके लिए
उत्तर:
A) किसी भी प्रकार के परिश्रम के लिए

3. यात्रा में शरीक होने पर लेखिका का स्वास्थ्य कब बिगड गया ?
A) दूसरे ही दिन
B) तीसरे ही दिन
C) पहले ही दिन
D) पाँचवे दिन
उत्तर:
C) पहले ही दिन

4. कितने वर्षों के पिछले के दरमियान में लेखिका किसी भी विषय में प्रवीण न बनी?
A) 28
B) 19
C) 38
D) 18
उत्तर:
D) 18

5. बड़ी फुर्ती से विनोबा के साथ कौन चल रही थी?
A) लेखिका
B) कस्तूरी बाई
C) विमला
D) नर्मदा बेन
उत्तर:
D) नर्मदा बेन

4. डॉ. अंबेडकर राजनैतिक आज़ादी के साथ सामाजिक और आर्थिक आज़ादी भी चाहते थे। उनको कमज़ोर वर्ग के प्रति सहानुभूति थी। वे उनके दुखों को दूर करने का प्रयत्न करते थे। दर असल वे पीड़ित मानवता के प्रवक्ता थे। वे सच्चे राष्ट्रप्रेमी और समाज सुधारक थे।
प्रश्न :
1. कमज़ोर वर्ग के प्रति सहानुभूति किसे थी?
A) राजाजी को
B) गाँधीजी को
C) डॉ. अंबेडकर को
D) नानक को
उत्तर:
A) राजाजी को

2. डॉ. अंबेड्कर किसके प्रवक्ता थे?
A) पीडित मानवता के
B) हिंसा के
C) विज्ञान के
D) अशांति के
उत्तर:
A) पीडित मानवता के

3. सच्चे राष्ट्रप्रेमी और समाज सुधारक कौन थे?
A) नेहरू
B) तिलक
C) बोस
D) अंबेड्कर
उत्तर:
D) अंबेड्कर

4. डॉ. अंबेडकर राजनैतिक आज़ादी के साथ – साथ किस आजादी को चाहते थे?
A) धार्मिक
B) नैतिक
C) सामाजिक तथा आर्थिक
D) समानता रूपी
उत्तर:
C) सामाजिक तथा आर्थिक

5. उपर्युक्त गद्यांश के लिए उपर्युक्त शीर्षक निम्न में से क्या होगा?
A) आज़ादी
B) डॉ. अंबेड्कर
C) डॉ. राधाकृष्णन
D) डॉ. मेहता।
उत्तर:
B) डॉ. अंबेड्कर

5. आज के दिन इसी समय मैंने अपने दोस्त कैलाश के साथ किशनसिंह होटल में तीन नबंर की चाय पी थी। किशनसिंह की बनाई चाय के नंबर हुआ करते थे – एक नंबर की चाय हलकी, दो नंबर की मध्यम तेज़ और तीन नंबर की स्पेशल हुआ करती थी।
प्रश्न :
1. किस होटल में चाय पी थी?
A) किशोर सिंह
B) किलाडी सिंह
c) किरण सिंह
D) किशन सिंह
उत्तर:
D) किशन सिंह

2. चाय किसने बनायी?
A) किसान सिंह
B) किशन सिंह
C) किशोर सिंह
D) ये सब
उत्तर:
B) किशन सिंह

3. किशनसिंह की बनाई चाय के कितने नबंर हुआ करते?
A) दो
B) तीन
C) चार
D) पाँच
उत्तर:
B) तीन

4. तीन नंबर की चाय कैसी हुआ करती?
A) मध्यम
B) हलकी
C) स्पेशल
D) तेज़
उत्तर:
C) स्पेशल

5. इस अनुच्छेद में एक दोस्त का नाम आया – वह कौन है?
A) किशनसिंह
B) किशोर
C) कैलाश
D) विनोद
उत्तर:
C) कैलाश

AP State Syllabus AP Board 9th Class Physical Science Important Questions Chapter 11 Sound

AP State Syllabus 9th Class Physical Science Important Questions 11th Lesson Sound

9th Class Physical Science 11th Lesson Sound 1 Mark Important Questions and Answers

Question 1.
Mention two important devices used in SONAR system?
Answer:
Sonar system consists of a transmitter and a detector.

Question 2.
A sound of wavelength 0.6 cm is travelling in air with a velocity of 300 m/s. Is this sound audible?
Answer:
We know that v = nλ

This sound is not in the audible range. Hence this is not audible.

Question 3.
How do we get the sensation of sound?
Answer:
Sound travels in air due to the to and fro motion of the air particles, which act upon the ear and produce the sensation of sound.

Question 4.
On which factor does the pitch of a tuning fork depend?
Answer:
The pitch of the tuning fork depends on the length of the prongs.

Question 5.
How does the sound travel?
Answer:
Sound travels^in the form of waves.

Question 6.
What are longitudinal waves?
Answer:
If the particles of the medium vibrate along the direction of wave, the wave is called a longitudinal wave.

Question 7.
Which wave involves change in the density of the medium?
Answer:
Longitudinal wave involves change in the density of the medium.

Question 8.
What do you call the regions formed in longitudinal wave propagation?
Answer:
The regions formed during the longitudinal wave propagation due to change in density are compressions and rarefactions.

Question 9.
When do we call a wave as transverse wave?
Answer:
If the particles of the medium vibrate perpendicular to the direction of wave, then the wave is called a transverse wave.

Question 10.
What is an echo?
Answer:
A reflection of sound, arriving at the listener in more than 0.1 sec after direct sound is called an echo.

Question 11.
What is reverberation?
Answer:
A reflection of sound, arriving at the listener, in less than 0.1 sec after direct sound is called reverberation.

Question 12.
What is the audible range of an ordinary person?
Answer:
20 Hz to 20 K.Hz.

Question 13.
Give examples to the animals which produce infrasonics.
Answer:
Animals such as elephants and whales produce infrasonics.

Question 14.
Expand SONAR.
Answer:
SONAR stands for Sonographic Navigation and Ranging.

Question 15.
Define a compression and a rarefaction.
Answer:

1. Compressions are the regions where density, as well as pressure of particles, is high.
2. Rarefactions are the regions where the density, as well as pressure of particles, is low.

Question 16.
Define time period of the sound wave.
Answer:
The time taken to complete one oscillation of the density of the medium is called the time period of the sound wave, denoted by T, and unit in S.I. system is second.

Question 17.
Define speed of a sound wave.
Answer:
The distance by which a point on the wave, such as a compression or rarefaction, travels in unit time is called speed of sound wave.

Question 18.
In which of the three media air, water, or iron sound travel the faster at a particular temperature?
Answer:
Sound travels faster in iron because speed of sound is maximum in solids then liquids and least in gases.

Question 19.
What is audiable range of the average human ear?
Answer:
The audiable range of the average human ear is 20 Hz to 20 KHz.

Question 20.
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen why?
Answer:
Speed of sound is less (i.e., 342 m/s) is less than speed of light (i.e., 3 x 10⁸ m/s). So, thunder is heard a few seconds after the flash is seen.

Question 21.
The frequency of source of sound is 100 Hz. How many times does it vibrates in 1 minute?
Answer:
Frequency (υ) = 100 Hz = 100 vibrations/ second.
Number of vibrations in minute = 100 × 60 = 6000 vibrations/ minute.

Question 22.
Which fundamental particles were discovered by
i) Thomson
ii) Goldstein
iii) Chadwick.
Answer:
i) Thomson – Electron
ii) Goldstein – Proton
iii) Chadwick – Neutron

Question 23.
What is a nucleus?
Answer:
The small positively charged central part of an atom is called nucleus.

Question 24.
What is an orbit?
Orbit is the path of the electron around the nucleus.

Question 25.
What will the addition of a neutron to the nucleus of an atom do?
Answer:
It will increase the atomic mass of the atom.

Question 26.
Which constituent particles of the atom determine the following
a) size of the atom
b) change on the nucleus?
Answer:
a) Electrons.
b) Protons.

Question 27.
Magnesium atoms has 12 electrons. Which energy shell is incomplete?
Answer:
M – shell is incomplete.

Question 28.
If K and L shells of an atom are full, then what would be the total number of electrons in the atom?
Answer:
K – 2 ; L – 8.
2 + 8 = 10 is the total number of electrons if K and L – shells are full.

Question 29.
An atom of an element has 7 electrons in its L shell
a) What is its atomic number?
b) State its valency.
Answer:
a) Atomic number = 9 [∴ K : 2 ; L : 7]
b) Valency is ‘I’.

Question 30.
The atomic number and mass number of an element are 11 and 23 respectively. Find the number of neutrons in its nucleus.
Answer:
Number of neutrons (N) = Mass number (A) – Atomic number (Z) = 23 – 11 = 12.

Question 31.
The mass number of chlorine atom is 35 and its atomic number is 17. How will this chlorine atom be represented?
Answer:
\({ }_{17}^{35} \mathrm{Cl}\)

Question 32.
What is an anion?
Answer:
When an atom gains one or more electrons, it becomes negatively charged and is known as anion.

Question 33.
Give one example each of diatomie and triatomic molecules.
Answer:

1. Diatomic molecule – Oxygen (O₂)
2. Triatomic molecule – Ozone (O₃)

Question 34.
If Z = 3, what would be the valency of the element?
Answer:

1. If Z = 3, the distribution of electrons is (2,1).
2. Thus the valency of the Element is 1.

Question 35.
Fluorine atom has 9 electrons and 9 protons. How many energy shells it has?
Answer:
Two energy shells. [∴ K = 2 ; L = V]

Question 36.
Which subatomic particle is not present in an ordinary hydrogen atom?
Answer:
Neutron.

Question 37.
What is the maximum number of electrons which can be accommodated in the
a) innermost shell of an atom?
b) outermost shell of an atom?
Answer:
a) 2
b) 8

Question 38.
What is the usual symbol for a) an electron b) a proton and c) a neutron?
Answer:
a) electron = e^–
b) proton = p⁺
c) neutron = n°

Question 39.
Name the shell of an atom which can accommodate a niaximum of a) 8 electrons b) 18 electrons.
Answer:
a) L-shell

Question 40.
Name the negatively charged particle present in the atones of all the elements.
Answer:
electron (e^–).

Question 41.
Which part of an atom was discovered by Ruthorford’s alpha particle scattering experiment?
Answer:
Nucleus.

Question 42.
State the relative mass and charge of a proton.
Answer:

1. Relative mass = 14
2. Relative charge = + 1

Question 43.
Name the radioactive isotope which is used in the treatment of cancer.
Answer:
Cobalt – 60.

Question 44.
Which radioactive isotope is used to determine the activity of thyroid gland?
Answer:
Iodine – 131 radioisotope

Question 45.
What name is given to the pair of atoms such as \({ }_{7}^{14} \mathrm{N}\) and \({ }_{7}^{15} \mathrm{N}\)?
Answer:
Isotopes.

Question 46.
Which noble gas has less than 8 electrons in the valence shell of its atom?
Answer:
Helium.

Question 47.
What is the general name of the elements having 8 electrons in the valence of shell of their atoms?
Answer:
Noble gases.

9th Class Physical Science 11th Lesson Sound 2 Marks Important Questions and Answers

Question 1.
How can you say that the sound waves are longitudinal?
Answer:

1. When sound wave passes through air, the layers in the medium are alternately pushed and pulled.
2. Thus the particles of the medium move to and fro along the direction of propagation.
3. Therefore sound waves in air are longitudinal.

Question 2.
What are the characteristics of a sound wave?
Answer:
The characteristics of a sound wave, which play an important role in describing the nature of a wave are

1. Wavelength (λ)
2. Amplitude (A)
3. Frequency (υ)
4. Wave speed (v)

Question 3.
Deduce a relationship between time period and frequency of a sound wave.
Answer:

1. Let the time taken for o oscillations = 1 sec.
2. The time taken for one oscillation = \(\frac{1}{υ}\) sec.
3. But the time taken for one oscillation is called the time period (T) and the number of oscillations per second is called the frequency (υ).
4. Hence frequency and time period are related as T = \(\frac{1}{υ}\) (or) υ = \(\frac{1}{T}\)

Question 4.
What is the speed of sound wave in air, water and iron at 20 °C?
Answer:
The speed of sound in air at 20°C is 343.2 m/s.
The speed of sound in water at 20°C is 1484 m/s.
The speed of sound in iron at 20°C is 5120 m/s.

Question 5.
Define echo and reverberation.
Answer:
Echo :
A reflected sound arriving at the position of listener in more than 0.1 s after the direct sound is called an echo’.

Reverberation :
A reflection of sound, arriving at the listener in less than 0.1 s after the direct sound is called reverberation’.

Question 6.
Explain the working of a megaphone and a horn.
(Or)
How multiple reflections of sound are used in working of mega phone and a horn?
Answer:

In megaphone and a horn, a tube followed by a conical opening reflects sound successively to guide most of the sound waves from the source in the forward direction towards the audience.

Question 7.
Describe the working of a stethoscope.
Answer:

1. Stethoscope is a medical instrument used for listening to sounds produced with in the body, chiefly in heart or lungs.
2. In stethoscopes the sound of the patient’s heartbeat reaches the doctor’s ears by multiple reflection and amplifying the sound.

Question 8.
How the concert hails and cinema halls are designed to use multiple reflections of sound?
Answer:

Generally the ceilings of concert halls, conference halls and cinema halls are designed such that the sound after reflection reaches all corners of the hall as shown in the figure.

In some halls, a curved ceiling is arranged in such a way that the sound after reflecting from the ceiling spreads evenly across the hall.

Question 9.
What are the medical applications of ultrasound?
Answer:
1) Imaging of organs :
Ultrasonics are useful in Electro Cardio Graphy (ECG) to form an image of heart.

Ultrasonography is used in the formation of images of organs such as liver, gall bladder, uterus, etc. to identify the abnormalities and tumors. Ultrasonography is also used to monitor the growth of a foetus inside the mother’s womb.

2) Surgical use of ultrasound :
Ultrasounds are used in cataract removal, breaking up of stones in kidneys, etc. without surgery.

Question 10.
Show the frequencies of various musical notes in a table.
Answer:
In musical terms, the pitch of the note determines the position of the note on the musical scale which is denoted as :

The tuning fork set is prepared based on the above frequencies.

Question 11.
Draw a diagram to show wavelength and amplitude of a wave.
Answer:

Question 12.
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer:

Question 13.
An echo returned in 3s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m/s?
Answer:
Speed of sound (v) = 342 m/s, Time taken to heard echo, t = 3 s.
Distance travelled by sound = v × t = 342 × 3 = 1026 m.

Question 14.
What are the range of frequencies associated with (a) infrasonic sound, b) ultrasonic sound?
Answer:
a) For infrasonic sound frequency is less than 20 Hz.
b) For ultrasonic sound frequency is greater than 20 KHz.

Question 15.
Which characteristic of sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer:
The pitch of voice differs for person to person which inturn depends on frequency. So, the frequency is the characteristic of sound which helps to identify my friends voice.

Question 16.
A person has hearing range from 20 Hz to 20 KHz. What are the typical wavelength of sound waves in air corresponds to these two frequencies? Take speed of sound in air as 344 m/s.
Answer:

Question 17.
A sound wave travels at a speed of 339 m/s. If its wavelength is 1.5 cm. What is the frequency of the wave? Will it be audiable?
Answer:

It is more than 20000 Hz (20 KHz). So the sound is not audiable.

Question 18.
A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound if the distance of an object from the submarine is 3625 m.
Answer:
Distance of an object from the submarine (d) = 3625 m.
The time taken to receive signal (t) = 5s.

Question 19.
Fill in the following blanks in respect of an atom of an element.

Answer:
Mass number = 23
Atomic number =11
Number of electrons =11
Valency = 1

Question 20.
Write the electronic configurations of the following elements.
a) Carbon
b) Neon
c) Chlorine
d) Calcium
Answer:
a) 2, 4
b) 2, 8
c) 2, 8, 7
d) 2, 8, 8, 2

Question 21.
Helium atom has an atomic mass of 4u and two protons in its nucleus. How many neutrons does it have?
Answer:
1. Mass number = No. of protons + No. of neutrons
⇒ 4 = 2 + No. of neutrons
⇒ No. of neutrons = 4 – 2 = 2
2. Thus, the helium atom has 2 neutrons.

Question 22.
Fill in the following blanks.

Answer:
1. Protons : 10
2. Neutrons : 12
3. Electrons : 10
4. Symbol : Ne

Question 23.
Write the three isotopes of carbon and hydrogen.
Answer:

Question 24.
Three different atoms of oxygen are represented as \({ }_{8}^{16} O,{ }_{8}^{17} O, \text { and }{ }_{8}^{18} O\).
a) What do the lower figures and upper figures represent ?
b) Give the nuclear composition of \({ }_{8}^{18} \mathrm{O}\).
Answer:
a) Lower figures → Atomic number
Upper figures → Mass number

b) Composition of \({ }_{8}^{18} \mathrm{O}\) → Protons : 8
Neutrons : 18 – 8 = 10

Question 25.
An atom of an element has 6 electrons in L shell.
a) What is the atomic number of the element?
b) State its valency.
c) Identify the element and write its name. [ K = 2 ; L = 6 Given]
Answer:
a) Atomic number = 2 + 6 = 8
b) Valency = 2
c) Oxygen atom.

Question 26.
State any two similar properties of isotopes.
Answer:

1. Isotopes of an element have same atomic number.
2. Isotopes of an element have similar number of electrons.

Question 27.
If chlorine atom is available in the form of Say, two isotopes \({ }_{17}^{35} \mathrm{Cl}\) (75%) and \({ }_{17}^{35} \mathrm{Cl}\) (25%), calculate the average atomic mass of chlorine atom.
Answer:

Question 28.
Write the name of any two radioactive isotopes.
Answer:
1. Uranium – 235
2. Cobalt – 60

9th Class Physical Science 11th Lesson Sound 4 Marks Important Questions and Answers

Question 1.
“Sound travels in the form of waves”. Justify your argument.
Answer:

1. Sound is a form of energy which travels through the air and reaches our ears to give the sensation of sound.
2. There may be two possible ways by which transfer of energy from the source of sound to our ears take place.
   a) Source of sound produces disturbances in air and they strike our ears.
   b) Some particles are shot off from the source of sound and they reach our ears.
3. If the second explanation is correct, the vibrating body would gradually lose its weight as particles are continuously shot off from it.
4. This is impossible because it would lead to vanishing of the object.
5. Hence the first explanation that the sound travels through disturbances in the form of waves is correct.

Question 2.
Mention the industrial applications of ultrasonic waves.
Answer:
1) Drilling holes and making cuts of desired shapes :
a) Holes can also be drilled using ultrasonic vibrations.
b) Ultrasonic cutting and drilling are very effective for fragile materials like glass, etc.

2) Ultrasonic cleaning :
a) Ultrasonics help in cleaning the parts located in hard-to-reach places.
b) The high frequency ultrasonic vibrations knocks off all dirt and grease particles from the objects.

3) Ultrasonic detection of defects in metals :
The defects in the metallic structures, which are not visible from the outside, can be detected by ultrasonic waves.

Question 3.
Draw the graphical representation of
a) Lower pitch, higher pitch
b) Louder sound, soft sound
Answer:

Question 4.
Draw the graphical representation of sound wave produced by a tuning fork, a violin and a piano playing the same note with equal loudness.
Answer:

Question 5.
Composition of the nuclei of two atomic species X and Y are given as under.

a) Give the mass number of X and Y.
b) What is the relation between the two species ?
Answer:
a) Mass number of X = 6 + 6 = 12
Mass number of Y = 6 + 8 = 14

b) Since X and Y both have atomic numbers as 6 but mass numbers are different.
∴ These are isotopes to each other.

Question 6.
How will you find the valency of Magnesium, Sulphur, and Chlorine?
Answer:
Electronic configuration

1. Valency of Magnesium = 2
2. Valency of Sulphur = 8 – 6 = 2
3. Valency of Chlorine = 8 – 7 = 1

Question 7.
For the symbol H, D and T tabulate three sub-atomic particles found in each of them.
Answer:

Question 8.
Draw a sketch of Bohr’s model of an atom with three shells.
Answer:

Question 9.
Observe the following table and answer the following questions.

a) What is the mass number of X?
b) What is the mass number of Y?
c) What is the relation between X and Y?
d) Which element do they represent?
Answer:
a) Mass number of X = 16 (∵ number protons + number neutrons)
b) Mass number of Y = 17
c) Isotopes
d) Oxygen (¹⁶₈O)

Question 10.
Draw the diagrams of the three isotopes of hydrogen and label the parts.
Answer:

Question 11.
Explain why, Sodium ion, Na⁺ has completely filled K and L shells.
Answer:

1. A sodium ion, Na⁺ has 10 electrons in it.
2. Now, the maximum capacity of K shell is 2 electrons and that of L shell is 8 electrons.
3. Taken together, the maximum capacity of K and L shells is 2 + 8 = 10 electrons.
4. A sodium ion Na⁺ has completely filled K and L shells because its 10 electrons can completely fill up K and L shells.

AP State Syllabus AP Board 9th Class Hindi Textbook Solutions Chapter 8 यक्ष प्रश्न Textbook Questions and Answers.

AP State Syllabus 9th Class Hindi Solutions Chapter 8 यक्ष प्रश्न

9th Class Hindi Chapter 8 यक्ष प्रश्न Textbook Questions and Answers

InText Questions (Textbook Page No. 37)

प्रश्न 1.
ये प्रश्न किसके मन में उत्पन्न हुए थे?
उत्तर:
ये प्रश्न स्वामी विवेकानंद के मन में उत्पन्न हुए थे।

प्रश्न 2.
स्वामी विवेकानंद ने इन प्रश्नों का उत्तर किससे पूछा होगा?
उत्तर:
स्वामी विवेकानंद ने इन प्रश्नों का उत्तर अपने गुरु श्री रामकृष्ण परमहंस से पूछा होगा।

प्रश्न 3.
इनके जीवन से हमें क्या संदेश मिलता है?
उत्तर:
इनके जीवन से हमें यह संदेश मिलता है कि अपने ऊपर विश्वास रखो। सर्व धर्म एक ही हैं। सर्व धर्मों का सार एक ही हैं। देश को प्यार करो। भाईचारे के साथ जीवन जीतो।

अर्थग्राह्यता-प्रतिक्रिया

अ) प्रश्नों के उत्तर सोचकर दीजिए।

प्रश्न 1.
युधिष्ठिर के बारे में बताइए |
उत्तर:
पांडुराज के दो पत्नियाँ थीं – कुंती और माद्री | महर्षि दुर्वास के मंत्र प्रभाव से कुंती के तीन पुत्र और माद्री के दो पुत्र हुए | कुंती का प्रथम पुत्र धर्म देवता यम धर्मराज की कृपा से पैदा हुआ था । वही युधिष्ठिर था। उसे ही धर्मराज, पांडुनंदन नाम से पुकारते थे । युधिष्ठिर तो धैर्यवान, साहसी, निडर, साधु स्वभाव, दया, ममता, मानवता से पूरित महान व्यक्ति था।

अच्छी चालचलन, उत्तम चरित्र, विवेचना शक्ति संपन्न युधिष्ठिर को सब चाहते और आदर करते थे। सचमुच धर्म का दूसरा रूप था युधिष्ठिर, इसीलिए ही उसे धर्मराज कहते थे। यह सचमुच प्रेमालू था । अपने चारों भाइयों के प्रति वह असीम श्रद्धा रखता था । द्यूत क्रीडा में वह निपुण था। अपने अच्छे आचरण और गुणों से वह जीवन भर निष्कलंक बना रहा था । प्रजा की सेवा में तत्परता दिखाकर उनको सदा सुखी रखनेवाला श्रेष्ठ राजा था । नाम सार्थक करनेवाला महान व्यक्ति था युधिष्ठिर |

प्रश्न 2.
युधिष्ठिर की जगह तुम होते तो किसे जीवित करवाना चाहते और क्यों?
उत्तर:
युधिष्ठिर महान चरित्रवान और शील संपन्न व्यक्ति था । धर्म परायण होने के कारण उसने यक्ष के कहने पर नकुल को जीवित करवाना चाहा । अगर उसकी जगह मैं होता तो पराक्रमी, साहसी, धनुर्विद्या प्रवीण, पांडव मध्यम अर्जुन को जीवित करवाना चाहता । क्योंकि अर्जुन महान गुणवाला और तेज संपन्न था । अनेक देवों की कृपा से वर और अस्त्र-शस्त्र प्राप्त किया हुआ महान व्यक्ति था। अपने पराक्रम और सज्जनता से किसी भी तरह वह अपने तीनों भाइयों को जीवित कर सकता है | इसीलिए मैं अर्जुन को जीवित करवाना चाहता।

आ) पाठ पढ़िए । उत्तर दीजिए ।

पाठ के आधार पर वाक्यों को सही क्रम दीजिए |

1. युधिष्ठिर ने नकुल को पानी की तलाश में भेजा । ( 2 )
2. पक्षपात से रहित मेरे प्यारे पुत्र ! तुम्हारे चारों ही भाई जीवित हो । ( 4 )
3. तुम जिस किसी को भी चाहो, वह जीवित हो जाएगा | ( 3 )
4. पांडव ब्राह्मण की व्यथा से प्रभावित हुए | ( 1 )

प्रश्नों के सही उत्तर दीजिए |

रानी सुदेष्णा का भाई कीचक बड़ा ही बलिष्ठ और प्रतापी वीर था । मत्स्य देश की सेना का वही नायक बना हुआ था | उसने अपने कुल के लोगों को साथ लेकर मत्स्याधिपति बूढे . विराटराज की सत्ता पर अप्रत्यक्ष रूप से अधिकार कर लिया था । कीचक की धाक लोगों में ऐसी बनी हुई थी कि लोग कहा करते थे कि मत्स्य देश का राजा तो कीचक ही है ।

1. रानी सुदेष्णा का भाई कौन था ? (आ)
अ) युधिष्ठिर
आ) कीचक
इ) विराटराज
ई) हनुमान
उत्तर:
अ) युधिष्ठिर

2. विराट किस देश के राजा थे?
अ) मगधराज
आ) पांचाल
इ) मत्स्य देश
ई) वैदेह
उत्तर:
इ) मत्स्य देश

3. देश की सेना का नायक कौन बना हुआ था ?
अ) कीचक
आ) विराटराज
इ) भीम
ई) हनुमान
उत्तर:
अ) कीचक

इन प्रश्नों के उत्तर तीन वाक्यों में दीजिए ।

प्रश्न 1.
ब्राह्मण पांडवों के पास आकर कौन-सी व्यथा सुनाने लगा?
उत्तर:
ब्राह्मण पांडवों के पास आकर अपनी व्यथा सुनाने लगा कि एक हिरण उसकी अरणी की लकडी सहित भाग गया । अब उसके पास यज्ञ की अग्नि पैदा करने के लिए दूसरी लकडी नहीं है । इसलिए कृपया उसे वापस दिला दीजिए |

प्रश्न 2.
युधिष्ठिर को सरोवर के पास कौन-सी चेतावनी सुनायी दी?
उत्तर:
युधिष्ठिर को सरोवर के पास यह चेतावनी सुनायी दी कि –

” सावधान ! तुम्हारे भाइयों ने मेरी चेतावनी की ओर ध्यान नहीं दिया । इसलिए उनकी यह दशा हुई । यदि तुम पानी पीना चाहते हो तो पहले मेरे प्रश्नों के उत्तर दो, फिर अपनी प्यास बुझाना ।”

प्रश्न 3.
युधिष्ठिर ने नकुल को जीवित करवाने का अनुरोध क्यों किया?
उत्तर:
माता कुंती का बचा हुआ एक पुत्र युधिष्ठिर हैं । वह चाहता हैं कि माता माद्री का भी एक पुत्र जीवित. रहे । अतः युधिष्ठिर ने नकुल को जीवित करवाने का अनुरोध किया ।

प्रश्न 4.
यक्ष ने युधिष्ठिर के सद्गुणों से मुग्ध होकर कौन-सा वर दिया?
उत्तर:
यक्ष ने युधिष्ठिर के सद्गुणों से मुग्ध होकर वर दिया कि उनके चारों ही भाई जीवित हों ।

अभिव्यक्ति-सृजनात्मकता

अ) इन प्रश्नों के उत्तर लिरिवए ।

प्रश्न 1.
धैर्य ही मनुष्य का साथ देता है। कैसे?
उत्तर:
यह सच है कि धैर्य ही मनुष्य का साथ देता है। धैर्य ही हमें समस्याओं से बाहर निकालता है। धैर्यवान सब कुछ बडे धीरज के साथ कर सकता है। धैर्य से ही हम सफलता पा सकते हैं। धैर्यवान का जीवन ही सच्चे अर्थ में जीवन है। जो धैर्यवान नहीं है। जिसमें धैर्य नहीं है वह हर पल डर से मरता है। इसलिए हम कह सकते हैं कि धैर्य ही मनुष्य का साथ देता है। हमें अकेले किसी स्थान या प्रदेश जाने में धैर्य ही साथ देता है।

प्रश्न 2.
अच्छी संगति से क्या लाभ होते हैं?
उत्तर:
अच्छी संगति से ये लाभ होते हैं –

• अच्छी संगति से हम भी अच्छे बन जायेंगे।
• अच्छी संगति से हमें सुख पहुँचता है।
• अच्छी संगति के कारण ही हमें आध्यात्मिक (परलौकिक) चिंतन मिलता है।
• अच्छी संगति से ही हमें सद्बुद्धि, परोपकारिता, सेवा, भलाई आदि भावनाएँ जागृत होते हैं।
• अच्छी संगति से हम विख्यात बनेंगे।
• अच्छी संगति से हम आसानी से समस्याओं से बाहर आ सकते हैं।
• अच्छी संगति से उत्तम जीवन बिता सकेंगे।
• अच्छी संगति से ही हमें विवेक, विनय, नेक आदि गुण मिलते हैं।
• अच्छी संगति कल्पवृक्ष जैसा है।

प्रश्न 3.
नकुल के लिए जीवन दान माँगकर युधिष्ठिर ने सही किया या गलत? अपने उत्तर का कारण बताइए।
उत्तर:
नकुल के लिए जीवन दान माँगकर युधिष्ठिर ने सही किया। कारण यही है कि युधिष्ठिर जो है वह माता कुंति का बचा हुआ एक मात्र पुत्र है। इसलिए उसने माता माद्री का भी एक पुत्र को जीवित करवाना चाहता था वह था नकुल। इसलिए युधिष्ठिर ने नकुल के लिए जीवन दान माँगकर सही किया।

आ) यक्ष के किन्हीं पाँच प्रश्नों के उत्तर अपने दृष्टिकोण के आधार पर लिखिए ।
उत्तर:
यक्ष के पाँच प्रश्न और उनके उत्तर इस प्रकार हैं –

1. आकाश से भी ऊँचा कौन है ?
उत्तर:
पिता आकाश से भी ऊँचा हैं ।

2. हवा से भी तेज़ चलनेवाला कौन है?
उत्तर:
मन हवा से भी तेज़ चलनेवाला है ।

3. विदेश जानेवाले का कौन साथी होता है?
उत्तर:
विद्या विदेश जानेवाले का साथी है ।

4. मरणासन्न वृद्ध का मित्र कौन होता है?
उत्तर:
मरणासन्न वृद्ध का मित्र दान है ।

5. सब से तुच्छ क्या है ? ।
उत्तर:
सब से तुच्छ चिंता है।

इ) यदि यक्ष के स्थान पर आप होते तो कौन-कौन से प्रश्न पूछते?
उत्तर:
यदि यक्ष के स्थान पर मैं होता तो वे ही प्रश्न पूछता जो यक्ष ने पूछा | जैसे :

• मनुष्य का साथ कौन देता है?
• कौन-सा शास्त्र है, जिसका अध्ययन करके मनुष्य बुद्धिमान बनता है?
• भूमि से भारी चीज़ क्या है?
• आकाश से भी ऊँचा कौन है?
• हवा से भी तेज़ चलनेवाला कौन है?
• विदेश जानेवाले का कौन साथी होता है?
• मरणासन्न वृद्ध का मित्र कौन होता है?
• सुख क्या है?
• सब से तुच्छ क्या है?
• किसके छूट जाने पर मनुष्य सर्व प्रिय बनता है?

ई) ‘यक्ष प्रश्न’ पाठ आपको क्यों अच्छा लगा? अपने शब्दों में लिखिए ।
उत्तर:
“यक्ष प्रश्न सचमुच एक संदेशात्मक और प्रेरणादायक पाठ है। यह पाठ मुझे बहुत अच्छा लगा | क्योंकि हम मानव हैं। समाज में रहते हैं। हमें मानवता, ममता, करुणा, सहयोग, कष्ट सहन आदि अनेक गुणों से रहना है। परमात्मा ने कृपा करके हमें बुद्धि प्रदान की है। इसका उपयोग करके सदा अच्छे काम करने में ही हमें आगे बढना है। आपदाग्रस्त लोगों का दुःख दूर करना हर एक का पवित्र धर्म है। वह काम हमें संपन्न करना है । धर्म के मार्ग में अनेक कठिनाइयों का सामना करना पडता है। निडर होकर धर्म मार्ग पर चलना महानता है। धर्म का पालन करते अपने अस्तित्व को बनाये रखना बड़ा महत्वपूर्ण विषय है। युधिष्ठिर के व्यक्तित्व से इन सभी गुणों का परिचय हमें मिलता है। इसीलिए ‘यक्ष प्रश्न’ पाठ मुझे बहुत अच्छा लगा।

भाषा की बात

अ) इन शब्दों के अर्थ बताइए | वाक्य प्रयोग कीजिए ।
1) तृषा 2) अग्रसर
3) चेतावनी
4) दुस्साहस
5) पक्षपात
उत्तर:
1) तृषा = प्यास, దప్పిక , thrust ; मैं तृषा से व्याकुल हूँ |
2) अग्रसर = मुख्य, ముఖ్యుడు, chief. पांडवों में युधिष्ठिर अग्रसर है।
3) चेतावनी = सावधानी, హెచ్చరిక , warning ; मेरे अध्यापक जी ने मुझे कई बार चेतावनी दी।
4) दुस्साहस = अंधासाहस, దుస్సా హసం, rashness, impudence ; दुस्साहस से कोई प्रयोजन नहीं होता |
5) पक्षपात = तरफ़दारी, పక్షపాత బుద్ధి, partiality ; सब बिना पक्षपात से रहना चाहिए |

आ) पर्यायवाची शब्द पहचानकर रेखा खींचिए ।

उत्तर:

इ) निम्न शब्दों पर ध्यान दीजिए । दो शब्दों के मेल में हुए विकार को समझिए ।

परियोजना कार्य

पंचतंत्र की नीतिप्रद कहानियों में से किसी एक कहानी का संग्रह कीजिए ।
उत्तर:
एकता का बल :
एक जंगल था । वहाँ जंगल में कई जानवर और पक्षी थे । उनमें कबूतरों का झुंड भी था । उसका नायक था बुद्धिमान चित्रग्रीव ।

एक दिन एक शिकारी उस जंगल में आकर जाल बिछाकर चावल छिडक कर झाडों में छिपकर बैठा। कबूतरों का झुंड आसमान में उड़ते नीचे चावल को देखा । एक कबूतर ने उन्हें खाने नीचे उतरने को कहा । नायक बुद्धिमान चित्रग्रीव के कहने पर भी वे नहीं सुने और नीचे उतर गये | वे सब जाल में फँस गये थे । सभी आपस में निंदा करने लगे थे । चित्रग्रीव ने उन्हें एक उपाय बताया कि सभी मिलकर एक बल से जाकर साथ उडें ।

सभी कबूतरों ने वैसा ही किया था । वे एक नदी के किनारे उतरे थे | उस प्रदेश में चित्रग्रीव का मित्र चूहा रहता था । उसने जाल काट दिया तो सब जाल से स्वतंत्रता पाये थे ।

नीति :
एकता का फल और बल बहुत महत्वपूर्ण होता है |

यक्ष प्रश्न Summary in English

This incident happened during the times of Mahabharata. In those days the Pandava princes along with Draupadi were in-cognito exile of 12 years. One day a brahmin came to the Pandavas and complained that a deer had taken away his Arani (The pair of wooden blocks to generate fire by friction) and therefore he was not able to light the fire to perform Vedic rituals. He requested them to bring his Arani back.

The Pandavas were moved on his request. They went in quest of the deer. After going a little away they found it. They tried to catch it. But the deer disappeared before their very eyes. They became exhausted searching for the dear. They sat under a tree for rest. They felt very thirsty. Dharmaraja (Yudhishtira) sent Nakula to fetch water. Nakula went in search of water and reached a lake. On seeing it, he was overjoyed. He thought that he would first quench has thirst. He got ready to drink water. At the very instant he heard a divine voice. “O Nakula ! Don’t venture to drink water. You have to answer my questions before you do that”. Nakula ignored the warming and drank the water. Soon after it, he fell down dead. Dharmaraja was worried for Nakula’s not coming. He sent Sahadeva, Arjuna and Bheema in turns in search of Nakula. Just same as Nakula, they paid no heed to the warning and fell down dead drinking the lake’s water. Worried over this, Dharmaraja himself went in search of them. Being thirsty, he walked towards the lake quench his thirst. There he found his brothers lying dead. He went forward. He too heard the same words, “Beware! your brothers ignored my warning and so they got this fate. If you want to drink water, first answer my questions. Later you can quench your thirst. “Then Dharmaraja said that he was ready to answer the questions.

Yaksha : Who is the companion of a man?
Dharmaraja : Courage is the companion of a man.
Yaksha : On studying which science will one become virtuous?
Dharmaraja : There is no such a science. By making friends with great persons one will become virtuous. Yaksha : What is heavier than the earth?
Dharmaraja : One’s mother is heavier than the earth.
Yaksha : What is higher than the heavens?
Dharmaraja : Father
Yaksha : What is faster than the wind?
Dharmaraja : Mind
Yaksha : Who is the companion of the person who leaves for foreign countries?
Dharmaraja : Education
Yaksha : Who is the friend of one who is ill and one who is dying?
Dharmaraja : Charity, because it alone follows one after one’s death.
Yaksha : What is pleasure?
Dharmaraja : It is such a thing that depends on one’s character and virtues.
Yaksha : What is the meanest of all?
Dharmaraja : Worry.
Yaksha : What is that which, when renounced, makes one lovable?
Dharmaraja : Pride.

In this manner, Yaksha asked Dharmaraja many other questions. Dharmaraja answered them correctly. Finally Yaksha told him that he would make one of his brothers alive. He asked Dharmaraja whom he would choose.

Dharmaraja thought for a moment and asked Yaksha to make his younger brother Nakula alive. Then Yaksha questioned why he had chosen Nakula instead of Bhima who had the strength of ten thousand elephants.

Dharmaraja replied that among Kunti’s sons he himself was alive. He also wished that one should be alive between Madri’s sons. He pleaded Yaksha to make Nakula alive.

Fully satisfied with the impartial and generous nature, Yaksha gave Dharmaraja a boon that his four brothers would be made alive.

यक्ष प्रश्न Summary in Telugu

ఈ సంఘటన మహాభారతకాలం నాటిది. ఆ సమయంలో పాండవులు ద్రౌపదీ సమేతంగా 12 సం||లు వనవాసం చేస్తూ ఉన్నారు. ఒక రోజున అకస్మాత్తుగా ఒక బ్రాహ్మణుడు పాండవుల వద్దకు వచ్చెను. అతడు తన బాధను ఈ విధంగా తెలియచేయుచుండెను – “ఒక లేడి నా అరణికర్ర (యజ్ఞము చేయుటకు ఉపయోగపడు ఒక కర్రయంత్రం. ఇది నిప్పును రాజేయుటకు ఉపయోగపడును.) తీసుకుని పారిపోయినది, ఇప్పుడు నా వద్ద యజ్ఞపు అగ్నిని రగిలించుటకు వేరొక కర్ర లేదు. దయచేసి నా అరణిని తిరిగి ఇప్పించండి.”

పాండవులు ఆ బ్రాహ్మణుని వ్యధకు ప్రభావితులైరి. అందువలన వారు లేడిని వెదకుటకు బయలుదేరిరి. కొంచెం దూరం వెళ్ళిన తర్వాత వారికి లేడి కన్పించినది. దానిని పట్టుకొనుటకు వారు ముందుకు వెళ్ళిరి. కానీ ఆ లేడి వారి కళ్ళముందే మరలా మాయమైనది. లేడిని వెదకుచూ వారు అలసిపోయిరి. విశ్రాంతి కోసం వారు ఒక చెట్టు క్రింద కూర్చొనిరి. దాహంతో అందరి గొంతులు ఎండిపోయినవి. ధర్మరాజు (యుధిష్ఠిరుడు) నకులుడిని మంచినీటి కోసం పంపెను. నకులుడు నీటిని అన్వేషించుటకు బయలుదేరెను. వెళుతూ వెళుతూ ఒక సరోవరం వద్దకు చేరెను. సరోవరం చూడగానే తన మనస్సు సంతోషించినది. పులకించినది. అతడు ముందుగా నేనెందుకు మంచినీరు త్రాగి దప్పిక తీర్చుకోకూడదు అని తలచెను. అతడు నీరు త్రాగుటకు సన్నద్ధుడయ్యెను. అప్పుడే ఒక మాట వినపడినది, “ఓ నకులా ! నీరు త్రాగే దుస్సాహసం చేయకు. జాగ్రత్త. ముందు నా ప్రశ్నలకు జవాబు ఇమ్ము.” నకులుడు హెచ్చరికను పట్టించుకోకుండా నీరు త్రాగెను. నీరు త్రాగగానే అతడు నేలపై పడిపోయెను. చాలా సేపటి వరకు నకులుడు తిరిగి రాకపోయేసరికి ధర్మరాజును దిగులు బాధించసాగెను. ఆయన అతనిని వెదకుటకు వంతుల వారీగా సహదేవుడు, అర్జునుడు, భీముడిని పంపెను. యక్షుని హెచ్చరికను ఉపేక్షిస్తూ అందరూ ఆ సరోవరం ఒడ్డున దాహం తీర్చుకునే ప్రయత్నం చేసి నకులుని వలెనే భూమిపై పడిపోయిరి. ఎవ్వరూ తిరిగి రాకపోయేసరికి ధర్మరాజు దిగులుతో స్వయంగా తనే వారిని వెదకుతూ బయలుదేరెను. దాహంతో ఉన్న ధర్మరాజు ముందుగా తన దాహాన్ని తీర్చుకోవడానికి సరోవరం వైపు వెళ్ళెను. అక్కడే ధర్మరాజుకు తన సోదరులు నేలపై పడిపోయి ఉండుట కన్పించెను. ఆయన ముందుకు వెళ్ళెను. తనకు కూడా అవే మాటలు వినబడినవి. జాగ్రత్త. మీ సోదరులు నా హెచ్చరికను పట్టించుకోలేదు. అందువలన వారికి ఈ దశ కల్గినది. నీవు నీళ్ళు త్రాగదలచుకొన్నట్లయితే ముందుగా నా ప్రశ్నలకు సమాధానమిమ్ము. తర్వాతే నీ దాహం తీర్చుకో. అప్పుడు ధర్మరాజు “శ్రీమాన్ ! మీరు ప్రశ్నలు అడగండి. నేను వాటికి జవాబు చెప్పుటకు సిద్ధముగా ఉన్నాను.” అనెను.

యక్షుడు : మనిషికి తోడు ఎవరు?
ధర్మరాజు : ధైర్యమే మనిషికి తోడు.
యక్షుడు : ఏ శాస్త్రాన్ని (విద్య) అభ్యసించి మనుష్యుడు బుద్ధిమంతుడవుతాడు?
ధర్మరాజు : అటువంటి శాస్త్రము ఏదీ లేదు. గొప్పవాళ్ళ స్నేహంతోనే మనిషి బుద్ధిమంతుడవుతాడు.
యక్షుడు : భూమి కంటే బరువైన వస్తువేది?
ధర్మరాజు : బిడ్డను కడుపులో (గర్భంలో) దాచుకున్న తల్లి భూమి కంటే బరువైనది.
యక్షుడు : ఆకాశం కంటే కూడా (ఉన్నతమైన) ఎత్తైనవారు ఎవరు ?
ధర్మరాజు : తండ్రి
యక్షుడు : గాలి కంటే కూడా వేగంగా వెళ్ళేది ఎవరు?
ధర్మరాజు : మనస్సు
యక్షుడు : విదేశాలు వెళ్ళేవారికి తోడు ఎవరు?
ధర్మరాజు : విద్య (చదువు)
యక్షుడు : మరణమాసన్నమైన వృద్ధునికి మిత్రుడు ఎవరు?
ధర్మరాజు : దానం, ఎందుకనగా మృత్యువు తర్వాత ఒంటరిగా నడిచే జీవికి తోడు వచ్చేది దానమే.
యక్షుడు : సుఖమంటే ఏమిటి?
ధర్మరాజు : శీలము, మంచి నడవడికపై ఆధారపడేది.
యక్షుడు : అన్నిటికంటే తుచ్ఛమైనది ఏది?
ధర్మరాజు : దిగులు.
యక్షుడు : దీనిని వదిలివేస్తే మానవుడు అందరికీ ప్రియుడగును?
ధర్మరాజు : అహంకారం.

ఇదే విధంగా యక్షుడు ఎన్నో ఇతర ప్రశ్నలు కూడా వేసెను. ధర్మరాజు వాటన్నిటికి సరియగు సమాధానములు చెప్పెను. చివరికి యక్షుడు “ఓ రాజా ! నేను మృతి చెందిన నీ సోదరులలో ఒకరినే జీవింపచేయగలను. నీవు ఎవరిని కోరుకుంటే వారు బ్రతుకుతారు” – అని చెప్పెను.

ధర్మరాజు ఒక క్షణం ఎవరిని బ్రతికించాలా ? అని ఆలోచించి, కొంచెం సేపు ఆగి “నా చిన్న సోదరుడు నకులుడు బ్రతకాలని చెప్పెను.

ధర్మరాజు ఈ విధంగా పలకగానే యక్షుడు, “ఓ ధర్మరాజా ! పదివేల ఏనుగుల బలం కల్గిన భీముణ్ణి వదలి నీవు నకులుణ్ణి ఎందుకు జీవింపకోరితివి? ఇది ఎలా సరియగును ?” అని యక్షుడు ప్రశ్నించెను.

“యక్షరాజా ! నేను నకులుణ్ణి ఎందుకు బ్రతికించమంటున్నానంటే నేను కుంతికి బ్రతికివున్న ఒక కుమారుణ్ణి అవుతాను. మాద్రికి జన్మించిన ఒకరు కూడా బ్రతికివుంటే బాగుంటుందని అలా కోరుకున్నాను. తమరు దయచేసి నకులుణ్ణి బ్రతికించండి” అని ధర్మరాజు యక్షునితో చెప్పెను.

ధర్మరాజు యొక్క ధర్మ స్వభావానికి యక్షుడు ఎంతో సంతృప్తి చెంది “ఓ ధర్మరాజా ! పక్షపాతమెరుగని ఓ నా కుమారా ! నీ నలుగురు సోదరులూ జీవింతురుగాక” – అని వరమిచ్చెను.

अभिव्यक्ति – सृजनात्मकता

निम्नलिखित प्रश्नों के उत्तर छह या आठ पंक्तियों में लिखिए।

1. “यक्ष प्रश्न” कहानी पाट का सारांश अपने शब्दों में लिखिए।
उत्तर:
पांडव वन में पानी की तलाश में घूम रहे थे। युधिष्टिर ने अपने सिवा चारों भाइयों को पानी के लिए एक के बाद एक को भेजा। लेकिन ये चार भाई एक सरोवर के पास मरे पड़े हुए हैं। युधिष्टिर सरोवर में पानी पीने उतरे। तो उसे अदृश्य वाणी से बातें सुनायी दी कि पानी पीने के पहले उनके प्रश्नों का उत्तर अवश्य देना। नहीं तो उसे भी उन के भाइयों के जैसे ही होगा। युधिष्टिर यक्ष के प्रश्नों का ठीक – ठीक उत्तर देता है। प्रसन्न होकर यक्ष उनके चारों भाइयों में से केवल एक को जीवित कर देने का वर देता है तो युधिष्टिर नकुल को जीवित करने को कहते हैं। तो यक्ष कारण जानकर उनके चारों भाइयों को ही जीवित कराते है।

निर्देश के अनुसार उत्तर दीजिए।

1. पांडव हिरण की खोज में निकल पडे। (रेखांकित शब्द का अर्थ पहचानिए।)
A) तलाश
B) इलाज
C) कगार
D) उपेक्षा
उत्तर:
A) तलाश

2. आग पैदा करने के लिए अरणी की नङकी की जरूरत हैं। (रेखांकित शब्द का तत्सम रूप
A) अग
B) अग्नि
C) आगा
D) अगन
उत्तर:
B) अग्नि

3. हिरण उनकी आँखों से पुनः गायब हो गया। ‘आँख’ शब्द कन तत्सम रूप पहचानिए।
A) अक्षि
B) अद्य
C) अष्ट
D) अज्ञ
उत्तर:
A) अक्षि

4. मेरे पास यज्ञ की अग्मि पैदा करने के लिए दूसरी लकङी नहीं हैं। (रेखांकित भब्द का तद्भव रूप पहचानिए।)
A) आग
B) आज
C) आड़
D) आन
उत्तर:
A) आग

5. एक दिन अचानक एक ब्राह्मण पांडवों के पास आया। (क्रिया विशेषण शब्द पहचानिए।) पहचानिए।)
A) एक
B) अचानक
C) ब्राह्मण
D) आया
उत्तर:
B) अचानक

अर्थग्राह्यता – प्रतिक्रिया

पठित – पद्यांश

निम्न लिखित गद्यांश पढ़कर दिये गये प्रश्नों के उत्तर एक वाक्य में दीजिए।

1. पांडव ब्राह्मण की व्यथा से प्रभावित हुए। इसलिए वे हिरण की खोज में निकल पड़े। थोड़ी दूर जाने | पर उन्हें हिरण दिखायी दिया। उसे पकड़ने के लिए वे आगे बढ़े परंतु हिरण उनकी आँखों से पुनः गायब हो गया । हिरण की खोज में भटकते हुए वे थक गये और विश्राम के लिए एक वृक्ष के नीचे बैठ गये। प्यास के मारे सभी के कंठ सूख रहे थे ।
प्रश्न :
1. पांडव किनकी व्यथा से प्रभावित हुए?
उत्तर:
पांडव ब्राह्मण की व्यथा से प्रभावित हुए।

2. किसकी खोज में भटकते हुए पांडव थक गये?
उत्तर:
हिरण की खोज में भटकते हुए पांडव थक गये।

3. विश्राम के लिए पांडव कहाँ बैठ गये?
उत्तर:
विश्राम के लिए पांडव एक पेड (वृक्ष) के नीचे बैठ गये।

4. सभी के कंट क्यों सूख रहे थे?
उत्तर:
प्यास के मारे सभी के कंठ सूख रहे थे।

5. थोडी दूर जाने पर उन्हें क्या दिखायी दिया?
उत्तर:
थोडी दूर जाने पर उन्हें हिरण दिखायी दिया।

2. युधिष्ठिर ने कहा – “यक्षराज ! मैंने जो नकुल को जीवित करवाना चाहा, वह सिर्फ इसी कारण कि माता कुंती का बचा हुआ एक पुत्र मैं हूँ, मैं चाहता हूँ कि माता माद्री का भी एक पुत्र जीवित हो उठे। अतः आप कृपा करके नकुल को जीवित कर दें।” युधिष्ठिर के धर्म स्वभाव से यक्ष अत्यंत संतुष्ट हुए और उन्होंने ‘वर’ देते हुए यह कहा, “पक्षपात से रहित मेरे प्यारे पुत्र ! तुम्हारे चारों ही भाई जीवित हों।”
प्रश्न :
1. युधिष्ठिर किसे जीवित करना चाहा?
उत्तर:
युधिष्ठिर नकुल को जीवित करना चाहा।

2. नकुल किसका पुत्र है?
उत्तर:
नकुल माद्री का पुत्र है।

3. युधिष्ठिर के धर्म स्वभाव से कौन अत्यंत संतुष्ट हुये?
उत्तर:
युधिष्ठिर के धर्म स्वभाव से यक्ष अत्यंत संतुष्ट हुए।

4. यक्ष ने क्या वर दिया?
उत्तर:
यक्ष ने यह वर दिया कि युधिष्ठिर के चारों भाई ही जीवित हों।

5. उपर्युक्त गद्यांश किस पाठ से दिया गया है?
उत्तर:
उपर्युक्त गद्यांश ‘यक्ष प्रश्न’ पाठ से दिया गया है।

3. युधिष्ठिर ने नकुल को पानी की तलाश में भेजा। नकुल पानी की खोज में निकला और चलते – चलते.एक सरोवर के पास पहुँचा। सरोवर देखकर उसका मन प्रफुल्लित हो उठा। वह सोचने लगा कि क्यों न मैं पहले अपने प्यास बुझा लूँ। वह पानी पीने के लिए उद्यत ही हुआ था कि सहसा एक आवाज़ गूंज उठी, “ऐ नकुल, पानी पीने का दुस्साहस न करें। सावधान ! पहले मेरे प्रश्नों के उत्तर दो।” नकुल चेतावनी की परवाह न करके पानी पीने लगा।
प्रश्न :
1. पानी की तलाश में युधिष्ठिर ने किसे भेजा?
उत्तर:
पानी की तलाश में युधिष्ठिर ने नकुल को भेजा।

2. नकुल चलते – चलते किसके पास पहुँचा?
उत्तर:
नकुल चलते – चलते सरोवर के पास पहुंचा।

3. सहसा क्या आवाज़ गूंज उठी?
उत्तर:
सहसा यह आवाज़ गूंज उठी कि ” नकुल, पानी पीने का दुस्साहस न करें।”

4. सरोवर देखकर किसका मन प्रफुल्लित हो उठा?
उत्तर:
सरोवर देखकर नकुल का मन प्रफुल्लित हो उठा।

5. सरोवर को देखकर नकुल क्या सोचने लगा?
उत्तर:
सरोवर को देखकर नकल सोचने लगा कि “क्यों ने मैं पहले अपनी प्यास बुझा लूँ?”

अपठित – गद्यांश

• नीचे दिये गये गद्यांश को पढ़कर प्रश्नों के उत्तर चुनकर कोष्ठक में लिखिए।

1. प्रिया, घबराओ नहीं। इस मुद्रिका पर मेरे नाम के जितने अक्षर हैं उतने ही दिनों में मेरे कर्मचारी आकर तुम्हें अंतःपुर में पहुँचाएँगे। मुद्रिका देखते ही मैं भी तुम्हें पहचान सकूँगा। दुष्यंत चला गया और फिर उसने गर्भवती शकुन्तला की सुध नहीं ली। शकुन्तला प्रिय वियोग से व्याकुल होकर अपनी सुधबुध खो गयी। दुर्दैव से एक दिन क्रोधी दुर्वासा का आगमन हुआ और उन्हें आतिथ्य देने में शकुन्तला दत्तचित नहीं हो सकी। क्रोध में आकर दुर्वासा बोले – जिस प्रेमी के ध्यान में पड़कर तुम कर्तव्य भूल गयी हो, वह तुम्हें भूल जाएगा। शकुन्तला को इस कठोर वचन की भी सुध न रही। उस समय कण्व सोमतीर्थ गये हुए थे। उनको तो यह वृत्तान्त बिलकुल ज्ञात न था ।
प्रश्न :
1. प्रिय वियोग से व्याकुल होकर कौन अपनी सुधबुध खो गयी थी?
A) दुष्यंत
B) शकुंतला
C) रेणुका
D) जानकी
उत्तर:
B) शकुंतला

2. शकुंतला का प्रिय कौन है?
A) दुष्यंत
B) विश्वामित्र.
C) कण्व
D) वशिष्ठ
उत्तर:
A) दुष्यंत

3. दुर्देव से एक दिन शकुंतला के पास किनका आगमन हुआ ?
A) विश्वामित्र
B) वशिष्ठ
C) दुर्वास
D) दुष्यंत
उत्तर:
C) दुर्वास

4. उस समय कण्व कहाँ गये थे?
A) तिरुपति
B) उज्जैयनी
C) भद्राचलम
D) सोमतीर्थ
उत्तर:
D) सोमतीर्थ

5. यह वृत्तांत किन्हें बिलकुल ज्ञात न था?
A) दुर्वास
B) कण्व
C) दुष्यंत
D) वशिष्ठ
उत्तर:
D) वशिष्ठ

2. रविवार का दिन था। श्रीमति स्टो गिरिजाघर गयी हुई थी और वहाँ धर्मोपदेश सुन रही थी कि एक साथ उनके मन में पुस्तक प्रारंभ कर देने की प्रेरणा उत्पन्न हुई और उन्होंने पहला अध्याय वहीं पर बैठे – बैठे लिख डाला। फिर उन्होंने वह अध्याय अपने बच्चों को सुनाया जिसे सुनकर बच्चों की आँखों से आँसू टप – टप गिरने लगे। इतने में श्रीमति स्टो के पतिदेव भी आ गये। बच्चों को रोते हुए देखकर वे आश्चर्यचकित रह गये। समझ में नहीं आया कि माजरा क्या है? तब श्रीमित स्टो ने वह अध्याय पति को भी सुनाया और बे भी रोने लगे। इस प्रकार प्रारंभ हुआ इस महत्वपूर्ण ग्रंथ का, जिसने आगे चलकर संसार में अक्षय कीर्ति प्राप्त की, जिसका अनुबाद शीघ्र ही संसार की तेईस भाषाओं में हो गया और जिसकी लाखों कापियाँ जनता के हाथों तक पहुँच गयीं। इस पुस्तक का नाम है – ‘अंकल टाम्स केबिन’ अर्थात् – ‘टाम काका की कुटिया।’
प्रश्न :
1. श्रीमति स्टो कहाँ गयी हुई थी?
A) गिरिजा घर
B) मंदिर
C) मसजिद
D) मदरसा
उत्तर:
A) गिरिजा घर

2. श्रीमति स्टो गिरिजाघर में क्या सुन रही थी?
A) गीतोपदेश
B) खुरान
C) धर्मोपदेश
D) ये सब
उत्तर:
C) धर्मोपदेश

3. उस पुस्तक का अनुवाद संसार के कितनी भाषाओं में हुआ?
A) 23
B) 33
C) 43
D) 53
उत्तर:
A) 23

4. उस पुस्तक का नाम क्या है?
A) मम्मी और डाडी
B) गिरिजाघर में मम्मी
C) अंकल टाम्स केबिन
D) डाडी के केबिन
उत्तर:
C) अंकल टाम्स केबिन

5. श्रीमति स्टो किस दिन गिरिजाघर गई?
A) सोमवार
B) शुक्रवार
C) शनिवार
D) रविवार
उत्तर:
D) रविवार

3. बूढ़ा बोला – ‘बेटा, तुम ठीक कह रहे हो। लेकिन यह पौधा मैं अपने लिए नहीं लगा रहा । एक | दिन वह पौधा बड़ा हो जाएगा। और पेड़ बनाजाएगा। यह अपनी छाया से आने – जानेवाले यात्रियों को आराम देगा। गर्मी और बरसात से उन्हें बचा सकेगा। जब इसमें फल लगेंगे तब शायद मैं इस दुनिया में न रहूँ, लेकिन इससे बहुत से लोग इसके फल खा सकेंगे।
प्रश्न :
1. एक दिन पौधा क्या बन जाएगा?
A) बडा
B) जंगल
C) पेड
D) लता
उत्तर:
C) पेड

2. जब पेड़ को फल लगेंगे तब उन्हें कौन खायेंगे?
A) बूढ़ा
B) बहुत से लोग
C) बेटे
D) ये सब
उत्तर:
B) बहुत से लोग

3. “बेटा, तुम ठीक कह रहे हो। लेकिन यह पौधा मैं अपने लिए नहीं लगा रहा” – इस वाक्य को किसने कहा?
A) बूढ़ा
B) बेटे
C) बेटा
D) स्त्री
उत्तर:
A) बूढ़ा

4. गर्मी और बरसात से यह हमें बचा सकेगा
A) नदी
B) सागर
C) पेड़
D) फल
उत्तर:
C) पेड़

5. उपर्युक्त इस अनुच्छेद में किसके बारे में बताया गया?
A) जानवरों के
B) पक्षियों के
C) पेड़ों के
D) मनुष्यों के
उत्तर:
C) पेड़ों के

AP State Syllabus AP Board 9th Class Hindi Textbook Solutions Chapter 4 प्रकृति की सीख Textbook Questions and Answers.

AP State Syllabus 9th Class Hindi Solutions Chapter 4 प्रकृति की सीख

9th Class Hindi Chapter 4 प्रकृति की सीख Textbook Questions and Answers

InText Questions (Textbook Page No. 18)

प्रश्न 1.
चित्र में क्या दिखाई दे रहा है?
उत्तर:
चित्र में एक नौका सागर को पार कर रही है। इसमें दो आदमी हैं।

प्रश्न 2.
वे क्या कर रहे हैं?
उत्तर:
वे जीवन यापन के लिए मछलियों को पकडते हुए गंभीर सागर को (नाव में से) पार कर रहे हैं।

प्रश्न 3.
इससे क्या संदेश मिलता है?
उत्तर:
इससे (हमें) यह संदेश मिलता है कि जितने भी गंभीर समस्याएँ हों उनकी साहस और कठिन मेहनत से सामना करना चाहिए।

अर्थग्राह्यता-प्रतिक्रिया

अ) प्रश्नों के उत्तर बताइए।

प्रश्न 1.
नदियाँ खेती के लिए किस प्रकार उपयोगी हैं ?
उत्तर:
नदियाँ पहाडों से निकलती हैं । पहाडों में रास्ता बनाकर, जंगलों से होकर मैदान में बहती हैं । प्राणिमात्र के लिए नदियों का पानी बहुत उपयोगी है । खेती के लिए उपजाऊ भूमि तैयार करने का काम नदियाँ करती हैं । नदियों के पानी से ही खेतों की सिंचाई होती है। किसान अनाज के साथ-साथ विविध तरकारियाँ और खाद्यान्न भी उसी पानी से पैदा करते हैं। नदियों पर बाँध बनाकर पानी इकट्ठा किया जाता है । नहरें निकाली जाती है । नहरों द्वारा सिंचाई का काम सुचारू रूप से होता है । हमारे भारत में खेती पूरी तरह नदियों पर निर्भर रहती है । इस तरह खेती के हर काम के लिए नदियाँ बहुत उपयोगी हैं |

प्रश्न 2.
ऋतुओं के नियंत्रण में पर्वत कैसे सहायक होते हैं?
उत्तर:
पर्वत तो अकसर ऊँचे होते हैं । पर्वत प्रकृति की संपदा है । इनसे कई लाभ हैं । हमें बहुत सहायक होते हैं, इनकी तलहटों में अनेक पेड -पौधों के होने के कारण हमेशा हरियाली और ठंडक बनी रहती हैं। खासकर समुद्र की तेज़ हवाओं से पर्वत देश की रक्षा करते हैं। तूफ़ान, आँधी आदि प्रकृति संबन्धी अनेक हालतों से ये हमें बचाते हैं । ऐसे पर्वत ऋतुओं के नियंत्रण में भी बडे सहायक होते हैं। उन्हीं के कारण मौसम समय पर आते हैं | मेघों को रोककर वर्षाएँ सही समय पर आने में इनका खास महत्व रहता है। इस तरह हम देखते हैं कि ऋतुओं के नियंत्रण में पर्वत बडे सहायकारी होते हैं।

आ) पाठ पढ़िए । अभ्यास – कार्य कीजिए।

* कविता के आधार पर उचित क्रम दीजिए।

1) सागर कहता है लहराकर। ( )
2) पर्वत कहता शीश उठाकर | ( 1 )
3) मन में गहराई लाओ। ( )
4) तुम भी ऊँचे बन जाओ । ( )
उत्तर:
1) 2
2) 4
3) 1
4) 3

स्तंभ ‘क’ को स्तंभ ‘ख’ से जोड़िए और उसका भाव बताइए ।

उत्तर:

पद्यांश पढ़िए । अब इन प्रश्नों के उत्तर दीजिए।

प्रश्न 1.
सूरज के बारे में आप क्या जानते हैं?
उत्तर:
सूरज भूमि के बहुत नज़दीक रहनेवाला एक नक्षत्र ही है । सूरज एक जलता हुआ आग का गोला है ।
पृथ्वी से सूरज कई गुना बड़ा है। पृथ्वी पर रहनेवाले समस्त प्राणि कोटि को आवश्यक जीव शक्ति सूरज से ही मिलती है। ग्रह परिवार में सूरज केंद्र स्थान में है । ग्रह, उपग्रह और लघु ग्रह सूरज के चारों ओर घूमते रहते हैं । पृथ्वी पर रहा पानी सूरज की गरमी से भाप के रूप में ऊपर उठकर मेघ बनते हैं,
ठंडी हवा लगते ही मेघ पानी बरसाते हैं ।

प्रश्न 2.
कवि ने सूरज, पंछी, हवा से क्या कहा?
उत्तर:
कवि ने सूरज से आदमियों को जगाने को पंछी (पक्षियों) से आदमियों के कानों पर चिल्लाने को और हवा से आदमी को हिलाने को कहा ।

प्रश्न 3.
वास्तव में जगने का क्या तात्पर्य है?
उत्तर:
वास्तव में जगने का तात्पर्य यह है कि जो बेखबर सपनों में खोया पडा है उसे जागरूक करना है।

अभिव्यक्ति-सृजनात्मकता

अ) प्रश्नों के उत्तर लिरिवाए।

प्रश्न 1.
पर्वत हमें सिर उठाकर जीने के लिए क्यों कह रहा होगा?
उत्तर:
पर्वत हमेशा बडे – बडे शिखरों के कारण ऊँचे रहते हैं। यदि हम अच्छे से अच्छे काम और बढ़प्पन के कार्य करें तो हम भी उन्हीं के तरह दुनिया में आदर के साथ सिर उठा करके रहेंगे। हमारा जीवन उन्नत बनेगा। हम दुनिया में आदरणीय बनेंगे। इसलिए हम अच्छे – अच्छे काम करते हुए पर्वतों के जैसे सिर खड़ा करके (उठाकर) रहने के लिए पर्वत ऐसा कह रहा होगा।

प्रश्न 2.
विपत्तियों का सामना हमें कैसे करना चाहिए?
उत्तर:
धैर्य, समयस्फूर्ति, उपाय, साहस और सोच विचार के साथ हम विपत्तियों का सामना कर सकते हैं। हम पृथ्वी से धैर्य न छोड़ने की भावना को ग्रहण कर के विपत्तियों का सामना कर सकेंगे।

प्रश्न 3.
प्रकृति के अन्य तत्व जैसे : नदियाँ, सूरज, पेड आदि हमें क्या सीख देते हैं?
उत्तर:
प्रकृति के अन्य तत्व जैसे नदियाँ, सूरज, पेड़ आदि हमें ये सीख देते हैं –

नदियाँ :
निर्मल रहने, मीठे-मीठे मृदुल उमंगों को भरने तथा परोपकार की भावना का सीख देती हैं।

सूरज :
जगने और जगाने तथा सकल जीवों का आधार बनने का सीख देता है। सूरज हमें अंधकार को दूर करके प्रकाश को फैलने का सीख भी देता है।

पेड़ :
फेड़ हमें सदा परोपकारी बने रहने का सीख देते हैं।

आ) कविता का सारांश अपने शब्दों में लिरिवए ।
(या)
“प्रकृति के विभिन्न अंगों से हम कुछ न कुछ सीखते हैं।” इस कथन की पुष्ट करते हुए प्रकृति की सीख पाठ का सार अपने शब्दों में लिखिए।
(या)
पर्वत, सागर, पृथ्वी और आसमान हमें क्या संदेश देते हैं?
(या)
प्रकृति के संदेशों से हम जीवन सफल बना सकते हैं। “प्रकृति की सीख’ कविता पाठ के आधार पर समझाइए।
(या)
प्रकृति के कण – कण में कुछ न कुछ संदेश छिपा होता है। सोहनलाल द्विवेदी जी ने प्रकृति के माध्यम से कौन – सा संदेश दिया?
उत्तर:
‘प्रकृति की सीख” कविता के कवि श्री सोहनलाल द्विवेदी जी हैं । उनका जीवनकाल 1906-1988 है।

अपनी अमूल्य रचनाओं से वे हिन्दी साहित्य में प्रसिद्ध हुए हैं। उनकी प्रमुख रचना ” सेवाग्राम’ है। भारत सरकार ने उन्हें पद्मश्री से विभूषित किया।

प्रस्तुत कविता में कवि कहते हैं कि हमारे चारों ओर की प्रकृति के कण-कण में कुछ संदेश छिपा स्हता है । उन संदेशों का आचरण करने से हम अपने जीवन को सार्थक व सफल बना सकते हैं।

पर्वत हम से कहता है कि तुम भी अपना सिर उठाकर मुझ जैसे ऊँचे (लंबे) बनजाओ । अर्थ है कि . सब में ऊँचा रहना महान गुण है। समुद्र लहराते कहता है कि मैं जैसे गहरा हूँ वैसे तुम भी अपने मन गहरे और विशाल बनाओ | माने गहराई से सोचो ।

समुद्र के चंचल तरंग उठ-उठ कर गिरते हैं। क्या तुम समझ सकते हो कि वे क्या कह रहे हैं ? वे कहना चाहते हैं कि तुम भी उनके जैसे अपने दिल में मीठी और कोमल आशाएँ भर लो । क्योंकि आशा को सफल बनाने का प्रयत्न ज़रूर करते हो।

धरती हम से कहती है कि चाहे जितनी बडी ज़िम्मेदारी तुम्हें पूरी करनी है, धैर्य के साथ उसे पूरा करो । आकाश कहता है कि मुझ जैसा पूरा फैलकर सारी दुनिया ढक लो | माने महान बनकर सब पर अपना प्रभाव दिखाओ।

इ) कविता के भाव से सूक्तियाँ लिरिवाए ।
उत्तर:

• शीश उठाकर ऊँचे बनो।
• धैर्य न छोडो।
• मन में गहराई लाओ।
• सारा संसार ढक लो।
• मन में मीठे मृदुल उमंग भर लो ।
• 

ई) नीचे दी गयी पंक्तियों के आधार पर छोटी – सी कविता लिरिवए।

हरियाली कहती ………………………. ।
महकते फूल कहते ………………….. ।
चहचहाते पक्षी कहते ……………….. ।
बहती नदियाँ कहती ………………… ।
उत्तर:
हरियाली कहती जीवन में हरा-भरा रहो ।
महकते फूल कहते हँसते-हँसते जीना सीखो।
चहचहाते पक्षी कहते आपस में मिलजुलकर रहो।
बहती नदियाँ कहती अविराम काम करते रहो।

भाषा की बात

अ) पाठ में आये पुनरुक्त शब्द रेखांकित कीजिए और वाक्य प्रयोग कीजिए ।

जैसे : मीठी-मीठी – बच्चे मीठी -मीठी बातें करते हैं |
उत्तर:
उठ – उठ = तरल तरंग उठ-उठ गिर रहे हैं।
भर लो – भर लो = भर लो – भर लो मन में मीठे उमंग।

आ) विपरीत अर्थ लिखिए और वाक्य प्रयोग कीजिए |

जैसे : न्याय x अन्याय- हमें अन्याय का विरोध करना चाहिए।
(धैर्य, हिंसा, शांति, यश, धर्म, गौरव, सत्य)
उत्तर:
धैर्य – कष्टों को देखकर अधैर्य मत होओ।
हिंसा – गाँधीजी अहिंसा का पालन करते थे।
शांति – आजकल विश्व भर में अशांति फैली हुई है।
यश – असत्य से तुम अपयश पाते हो।
धर्म – कौरव अधर्म से जीतना चाहते थे।
गौरव – बडों के प्रति अगौरव भाव मत दिखाइए।
सत्य – असत्य कभी भी मत बोलो ।

इ) इन शब्दों के बीच का अंतर समझिए और स्पष्ट कीजिए। ऐसे ही तीन शब्द लिखिए ।

उत्तर:

परियोजना कार्य

सोहनलाल द्विवेदी के बारे में जानकारी प्राप्त कीजिए । उनकी कोई एक कविता ढूँढ़कर लिखिए।
उत्तर:

सोहनलाल द्विवेदी जी का जन्म 22 फरवरी, सन् 1906 को कानपुर के पास बिंदकी नामक गाँव में हुआ था । ये राष्ट्रीय काव्य धारा के कवि थे । युगावतार, भैरवी, पूजागीत, युगाधार, विषपान, वासन्ती, चित्रा आदि उनकी काव्य कृतियाँ हैं। उनकी मृत्यु मार्च 1988 को हुई।

आपके द्वारा लिखी गयी शिशुभारती, बच्चों के बापू, बिगुल, बाँसुरी और झरना और दूध बतासा आदि रचनाएँ बच्चों को आकर्षित करती हैं | आपकी प्रसिद्ध रचना सेवाग्राम है। 1969 में भारत सरकार ने आपको पङ्मश्री उपाधि प्रदान कर सम्मानित किया था।

उने लाल अब आँखें खोलो।

उठो लाल अब आँखें खोलो
पानी लाई हूँ मुँह धो लो
बीती रात कमल दल फूले
उनके ऊपर भंवरे झूले
चिड़ियाँ चहक उठी पेड़ों पर
बहने लगी हवा अति सुंदर
नभ में न्यारी लाली छाई
धरती ने प्यारी छवि पाई
भोर हुई सूरज उग आया
जल में पड़ी सुनहरी छाया
नन्हीं नन्हीं किरणें आई
फूल खिले कलियाँ मुस्काई
इतना सुंदर समय न खोओ
मेरे प्यारे अब मत सोओ

प्रकृति की सीख Summary in English

Mountains say that everybody should hold thier head high. Sea states that everybody should be profound with pure and good thought waves, same as it (the sea) is deep with waves.

Do you understand what the rising and frothy tides of the sea tell us ? They advise us to fill our hearts with sweet, tender and joyous feelings.

The earth tells us not to desert courage how muchever burden lies upon your head. The sky tells us to spread everywhere and encircle the world.

We get message from each and every particle of the nature. Mountains, rivers, brooks and trees etc., tell us something. We should make our lives truthful with their messages. This alone is the essence of the lesson.

प्रकृति की सीख Summary in Telugu

పర్వతాలు తమ వలే అందరూ ఉన్నతంగా తల ఎత్తుకుని నిలబడమని చెబుతున్నాయి. సముద్రము తను ఏ విధంగా అలలతో కూడినదై లోతుగా ఉంటుందో అట్లే అందరూ మనస్సులో మంచి ఆలోచనా తరంగాలతో కూడి లోతును కల్గియుండమని చెబుతోంది.

ఎగిరి ఎగిరి పడుతున్న నురుగులు కక్కే సముద్ర తరంగాలు (అలలు) ఏమి చెబుతున్నవో అర్థమైనదా ? మనస్సులో తీయ తీయని సున్నితమైన సంతోష భావాలను నింపుకొమ్మని చెబుతున్నాయి.

మన తలపై ఎంత భారం ఉన్నా ధైర్యం విడువొద్దని భూమి మనతో చెబుతున్నది. ఆకాశం అంతటా తనలా వ్యాపించమని, ప్రపంచాన్ని కప్పమని చెబుతోంది.

ప్రకృతి యొక్క అణువణువు నుండి మనకు ఏదో ఒక సందేశం లభిస్తోంది. పర్వతాలు, నదులు, సెలయేర్లు, చెట్లు మొ||నవి మనతో ఏదో చెబుతాయి. వీటి సందేశంతో మనం మన జీవితాలను సఫలం చేసుకోవాలి. ఇదే ఈ పద్యభాగ సారాంశం.

अर्थग्राहयता – प्रतिक्रिया

1. निम्नलिखित पद्यांश पढ़कर दिये गये प्रश्नों के उत्तर एक वाक्य में लिखिए।

पर्वत कहता शीश उठाकर,
तुम भी ऊँचे बन जाओ।
सागर कहता है लहराकर,
मन में गहराई लाओ।
समझ रहे हो क्या कहती है,
उठ – उठ गिर गिर तरल तरंग ।
भर लो, भर लो अपने दिल में,
मीठे – मीठे मृदुल उमंग ॥

प्रश्न : 1. पर्वत क्या कहता है?
उत्तर:
पर्वत शीश उठाकर ऊँचे बन जाने को कहते हैं।

2. मन में गहराई लाने की सीख किससे मिलती है?
उत्तर:
मन में गहराई लाने की सीख सागर से मिलती है।

3. तरंग हमें कौनसी सीख देती है?
उत्तर:
तरंग हमें दिल में मीठे – मीठे मृदुल उमंग भरने

4. ‘शीश’ शब्द का अर्थ क्या है?
उत्तर:
शीश = सिर

5. इस पद्यांश के कवि कौन हैं?
उत्तर:
इस पद्यांश के कवि हैं सोहनलाल द्विवेदी।

2. निम्न लिखित पद्यांश पढ़कर नीचे दिये गये वैकल्पिक प्रश्नों के उत्तर दीजिए। सही विकल्प से संबंधित अक्षर चुनकर कोष्ठक में रखिए।

कितनी सुंदर कितनी प्यारी
त्यौहारों में सबसे ज्यारी
रंग बिरंगी होली
आई रंग बिरंगी होली।
आओ मिलकर खुशी मनाएँ
एक दूजे को रंग लगाएँ
गले मिलें, हर झगड़ा भूलें
भेदभाव हम सभी मिटाएँ।
प्रश्न :
1. त्यौहारों में सबसे न्यारी क्या है?
A) होली
B) दिवाली
C) ईद
D) बैसाखी
उत्तर:
A) होली

2. होली के दिन एक दूसरे पर क्या लगाते हैं? की सीख देती है।
A) पानी
B) फूल
C) फल
D) रंग
उत्तर:
D) रंग

3. होली के दिन क्या भूल जाते हैं?
A) झगडे करना
B) रंग लगाना
C) खुशी मनाना
D) गले मिलना
उत्तर:
A) झगडे करना

4. ‘खुशी’ – इस शब्द में प्रत्यय क्या है?
A) खु
B) खुश
C) सी
D) ई
उत्तर:
D) ई

5. होली हमारे मन से किस भावना को मिटाती
A) निस्वार्थ भाव
B) भेदभाव
C) प्यार
D) खुशी
उत्तर:
B) भेदभाव

अभिव्यक्ति – सृजनात्मकता

2. निम्नलिखित प्रश्न का उत्तर दो या तीन वाक्यों में लिखिए।

1. सोहनलाल द्विवेदी के बारे में लिखिए।
उत्तर:
★ सोहनलाल द्विवेदी जी का जन्म सन् 1906 में हुआ था।
★ आप की मृत्यु सन् 1988 में हुई।
★ ‘सेवाग्राम’ आप की प्रसिद्ध रचना है।
★ आप ‘पद्मश्री’ पुरस्कार से सम्मानित है।

अर्थग्राह्यता -प्रतिक्रिया

पठित – पद्यांश

निम्न लिखित पद्यांश पढ़कर दिये गये प्रश्नों के उत्तर एक वाक्य में दीजिए।

1. पर्वत कहता शीश उठाकर,
तुम भी ऊँचे बन जाओ।
सागर कहता है लहराकर,
मन में गहराई लाओ।
प्रश्न :
1. पर्वत क्या कहता है?
उत्तर:
1. पर्वत हमें ऊँचे बन जाने को कहते हैं।
(या)
पर्वत कहते हैं कि तुम भी हमारे जैसे ऊँचे बन जाओ।

2. सागर क्या कहता है?
उत्तर:
गहराकर मन में गहराई लाने को सागर कहता है।

3. यह उपर्युक्य पद्यांश किस पाठ से लिया गया
उत्तर:
यह उपर्युक्त पद्यांश ‘प्रकृति की सीख’ नामक पाठ से लिया गया है।

4. उपर्युक्त पद्यांश के कवि कौन है? ।
उत्तर:
उपर्युक्त पद्यांश के कवि हैं “श्री सोहनलाल द्विवेदी जी”।

5. “शीश” शब्द का पर्यायवाची शब्द क्या है?
उत्तर:
शीश शब्द का पर्यायवाची शब्द है – “सिर”

2. समझ रहे हो क्या कहती है,
उठ – उठ गिर गिर तरल तरंग ।
भर लो, भर लो अपने दिल में,
मीठे – मीठे मृदुल उमंग ॥
प्रश्न :
1. तरल तरंग क्या करती है?
उत्तर:
तरलं तरंग उठ – उठकर गिर पडती है।

2. तरल तरंग उठ-उठ गिर कर क्या कहती
उत्तर:
तरल तरंग उठ – उठ गिर कर कहती है | कि तुम अपने मन में मीठे मृदुल उमंग भर लो।

3. उपर्युक्त पद्यांश किस पाठ से लिया गया है?
उत्तर:
उपर्युक्त पद्यांश ‘प्रकृति की सीख’ नामक पद्य पाठ से लिया गया है।

4. ‘मीठे’ शब्द का विलोम क्या है?
उत्तर:
मीठे – शब्द का विलोम है – “कडुआ”

5. उपर्युक्त पद्यांश के कवि कौन हैं?
उत्तर:
उपर्युक्त पद्यांश के कवि हैं श्री सोहनलाल द्विवेदी जी।

3. पृथ्वी कहती, धैर्य न छोडो,
कितना ही हो सिर पर भार।
नभ कहता है, फैलो इतना,
ढक लो तुम सारा संसार ||
प्रश्न :
1. नभ क्या कहता है?
उत्तर:
नभ कहता है कि इतना फैलाकर सारा संसार को ढक लो।

2. पृथ्वी क्या कहती है?
उत्तर:
पृथ्वी कहती है कि “धैर्य न छोडो’ । धैर्य न छोडने को पृथ्वी कहती है।

3. पृथ्वी शब्द का पर्यायवाची शब्द लिखिए।
उत्तर:
पृथ्वी शब्द का पर्यायवाची शब्द हैं। धरती / वसुधा | वसु/ जमीन/ भूमि/ धरा आदि।

4. उपर्युक्त पद्यांश के कवि कौन है?
उत्तर:
उपर्युक्त पद्यांश के कवि हैं श्री सोहनलाल द्विवेदी जी।

5. धैर्य शब्द का विलोम क्या है?
उत्तर:
धैर्य शब्द का विलोम शब्द है डर।

अपठित – पद्यांश

निम्न लिखित पद्यांश पढ़कर दिये गये प्रश्नों के उत्तर विकल्पों में से चुनकर लिखिए।

1. कुछ तो नहीं लेते थे।
पर सब को देते छाया ये,
हमारे हित सब कुछ सहते,
उनके हेतु हम जीवित रहते।
सोखते हैं गंदी हवा ये
पर देते हैं शुद्ध हवा ये।
इनका गुण गान करें हम,
आओ पेड़ लगाये हम ॥
प्रश्न :
1. इस पद्य में किनके बारे में बताया गया है?
A) नदियों के
B) पेडों के
C) पर्वतों के
D) तारों के
उत्तर:
B) पेडों के

2. शुद्ध हवा कौन देते हैं?
A) नदी
B) पृथ्वी
C) चाँद
D) पेड
उत्तर:
D) पेड

3. हमें किनका गुण गान करना है?
A) आसमान का
B) पेडों का
C) पृथ्वी का
D) इन सबका
उत्तर:
B) पेडों का

4. हमें क्या लगाना है?
A) चूना
B) मिट्टी
C) पेड
D) पानी
उत्तर:
C) पेड

5. पेड किस प्रकार के हवा को सोखते हैं?
A) शुद्ध
B) गंदी
C) केवल शुद्ध हवा
D) इन सबको
उत्तर:
B) गंदी

2. घिर आये हैं बादल काले
गरज रहे हैं बन मतवाले|
खुश हो होकर मोर कूकते।
झूम – झूम के नाच दिखाते।
रिमझिम – रिमझिम पानी पडता
जन – जन का मन खूब फडकता।
प्रश्न:
1. कैसे बादल घिर आये हैं?
A) लाल
B) काले
C) हरे
D) सफ़ेद
उत्तर:
B) काले

2. बादल क्या कर रहे हैं?
A) गरज
B) खेल
C) दौड
D) रो
उत्तर:
A) गरज

3. खुशी से कौन नाच दिखाते हैं?
A) बादल
B) नदियाँ
C) कागज़
D) मोर
उत्तर:
D) मोर

4. पानी कैसे पडता है?
A) टर – टर
B) टन – टन
C) रिम-झिम
D) झम – झम
उत्तर:
C) रिम-झिम

5. किनका मन खूब फडकता है?
A) पानी
B) जन
C) बादल
उत्तर:
B) जन

3. क्षुधार्थ रंतिदेव ने दिया करस्थ थाल भी,
तथा दधीचि ने दिया परार्थ अस्थिजाल भी।
उशीनर क्षितीश ने स्वमांस दान भी किया,
सहर्ष वीर कर्ण ने शरीर – चर्म भी दिया ।
अनित्य देह के लिए अनादि जीव क्या डरे ?
वही मनुष्य है कि जो मनुष्य के लिए मरे ॥
प्रश्न :
1. मनुष्य कौन है?
A) जो पक्षियों के लिए मरता है।
B) जो मनुष्य के लिए मरता है।
C) जो पशुओं के लिए मरता है।
D) जो अपने लिए मरता है।
उत्तर:
B) जो मनुष्य के लिए मरता है।

2. उशीनर क्षितीश ने किसे दान किया?
A) पक्षीमांस
B) स्वमांस
C) पत्नी का मांस
D) बकरी का मांस
उत्तर:
B) स्वमांस

3. परार्थ अस्थिजाल किसने दिया?
A) रंतिदेव
B) उशीनर
C) दधीचि
D) वीर कर्ण
उत्तर:
C) दधीचि

4. रंतिदेव ने क्या दिया ?
A) करस्थ थाल
B) शरीर – कवच
C) अस्थिजाल
D) स्वमांस
उत्तर:
A) करस्थ थाल

5. यह देह कैसा है?
A) नित्य
B) सत्य
C) असत्य
D) अनित्य
उत्तर:
D) अनित्य

4. हम बादल कहते सरिता – सर,
कण – कण कहता हम उसका घर,
धरती कहती रोम रोम में
समा गया है वह चिर उज्ज्वल !
करुणामय चिरंजीवी बादल !
कहाँ गया वह श्यामल बादल?
प्रश्न :
1. करुणामय और चिरंजीवी कौन है?
A) नभ
B) पृथ्वी
C) वर्षा
D) बादल
उत्तर:
D) बादल

2. बादल शब्द का अर्थ क्या है?
A) मेघ
B) सरिता सर
C) पर्वत
D) झील
उत्तर:
A) मेघ

3. कण – कण क्या कहता है?
A) हम उसका घर
B) हम किसी का घर नहीं
C) हम अपने घर
D) ये सब ठीक
उत्तर:
A) हम उसका घर

4. इस पद्य में इसके बारे में बताया गया है।
A) श्यामल बादल
B) नील गगन
C) नीली नदी
D) इन सबके
उत्तर:
A) श्यामल बादल

5. बादल इस रंग के होते है।
A) लाल
B) पीला
C) सफ़ेद
D) श्यामल
उत्तर:
D) श्यामल

5. सच पूछो तो बड़ा आदमी,
होना सब से आसान काम ।
शील व गुणों से होता,
सदा बड़े लोगों में नाम ॥
जो तुम होना बड़ा चाहते,
तो उसका है एक उपाय।
दुर्बल, दीन, अनाथ जनों का,
तन – मन – धन से करो सहाय ॥
प्रश्न :
1. सबसे आसान काम क्या है?
A) सच पूछना
B) बड़ा आदमी होना
C) शील व गुणों से होना
D) साहस करना
उत्तर:
B) बड़ा आदमी होना

2. सदा बडे लोगों में नाम कैसा होता है?
A) सच पूछना
B) शील व गुणों से होना
C) दुर्बल रहना
D) दीन रहना
उत्तर:
B) शील व गुणों से होना

3. इस पद्य में कवि क्या बनने उपाय बताते हैं?
A) पक्षी बनने
B) पर्वत बनने
C) बडा बनने
D) साधु बनने
उत्तर:
C) बडा बनने

4. बड़ां होने का एक उपाय पहचानिए।
A) सहायता करना
B) नाश करना
C) पूजा करना
D) शाप देना
उत्तर:
A) सहायता करना

5. दुर्बल, दीन, अनाथ जनों को सहाय कैसा करना है?
A) तन से
B) मन से
C) तन – मन – धन से
D) किसी से नहीं
उत्तर:
C) तन – मन – धन से

AP State Syllabus AP Board 9th Class Physical Science Important Questions Chapter 8 Gravitation

AP State Syllabus 9th Class Physical Science Important Questions 8th Lesson Gravitation

9th Class Physical Science 8th Lesson Gravitation 1 Mark Important Questions and Answers

Question 1.
Write the precautions to be taken while doing an activity to locate centre of gravity of a regular object?
Answer:

1. Tie the rope at geometrical centre of an regular object.
2. Draw the line along the rope after object come to rest.

Question 2.
What path will moon take when the moon stops rotating round the earth?
Answer:
Moon take the path of tangent to the orbit, in which it rotates around the earth.

9th Class Physical Science 8th Lesson Gravitation 2 Marks Important Questions and Answers

Question 1.
If a person of mass 60 kg went to the Moon, then
a) Is there any change in his Mass and Weight?
Answer:
Mass does not change and remains 60 kgs. But, weight changes,

b) What will be his weight on the Earth?
Answer:

Question 2.
Consider the gravitational force formula \(\mathbf{F}=\mathbf{G} \frac{\mathbf{M}_{1} \mathbf{M}_{2}}{\mathbf{R}^{2}}\) and answer the following
questions.
i) What is the value of G?
Answer:
G = 6.67 × 10^-11 Nm² kg^-2

ii) What does the term ‘R’ indicate here?
Answer:
R indicates “distance between two objects.”

Question 3.
To know about the center of gravity of the atmosphere of the earth, Ravi asked some questions to his teacher. Guess which question he asked and the answer given by the teacher.
Answer:

1. Where would be the centre of gravity of atmosphere of earth?
2. What is centre of gravity?

Answer from the teacher :

1. The centre of gravity of the earth’s atmosphere is in the centre of the earth.
2. The point where total weight appears to act is called centre of gravity.

Question 4.
Consider the following equation of motion and rewrite them for a free fall body v = u + at
s = ut + ½ at²
Answer:
For a free fall body u = 0; a = g and s
i) v = u + at ⇒ v = 0 + gt ⇒ v = gt
ii) s = ut + ½at² ⇒ h = 0 + ½gt² ⇒ h = ½gt²

9th Class Physical Science 8th Lesson Gravitation 4 Marks Important Questions and Answers

Question 1.
Two spherical balls of mass 20 kg each one placed with their centres 20 cm apart. Find the gravitational force of attraction between them.
(G = 6.67 × 10^-11 Nm². Kg^-2)
Answer:
Masses of balls M₁ and M₂ = 20 kg each
distance (d) = 20 cm = 0.2 m.
Gravitational force of attraction between M₁ and M₂

Question 2.
a) Draw the centre of gravity of the following uniform objects.
i) equilateral triangle
ii) square
iii) circle
Answer:

b) What is the speciality of centre of gravity of a material?
Answer:

1. Stability depends upon the centre of gravity of a material.
2. If we draw a line straight down from the centre of gravity of an object of any
   shape and it falls inside the base of the object, then the object will be stable.

9th Class Physical Science 8th Lesson Gravitation Important Questions and Answers

9th Class Physical Science 8th Lesson Gravitation 1 Mark Important Questions and Answers

Question 1.
Define centripetal force.
Answer:
The net force which can change only the direction of the velocity of a body is called “centripetal force”.

Question 2.
Define centripetal acceleration.
Answer:
The acceleration which can change only the direction of velocity of a body is called “centripetal acceleration”.

Question 3.
State Newton’s universal law of Gravitation.
Answer:
The universal law of gravitation states that every body in the universe attracts other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Question 4.
What do you mean by acceleration due to gravity?
Answer:
The acceleration produced due to gravitational force of the earth near the surface is called free-fall acceleration or acceleration due to gravity.

Question 5.
What is universal gravitational constant?
Answer:
The value of ‘G’ i.e., the universal gravitational constant is equal to the magnitude of force between a pair of 1 kg – masses that are lm apart.
G = 6.67 × 10^-11 Nm² kg^-2.

Question 6.
When do you say that a body is freely falling?
Answer:
A body is said to be free-fall body when only one gravitational force acts on that body.

Question 7.
Define weight.
Answer:
Weight of a body is the force of attraction on the body due to earth.
W = mg

Question 8.
What is centre of gravity?
Answer:
The point where total weight appears to act is called centre of gravity.

Question 9.
When will a body be stable?
Answer:
If we draw a line straight down from the centre of gravity of an object of any shape and it falls inside the base of the object, then the object will be stable.

Question 10.
What is time period?
Answer:
The time taken by a body executing uniform circular motion, to complete one revolution is called time period.

Question 11.
The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?
Answer:
Moon attracts with same gravitational forces as earth attracts moon. They form action reaction pair.

9th Class Physical Science 8th Lesson Gravitation 2 Marks Important Questions and Answers

Question 1.
Draw a diagram showing the balancing of fork and spoon.
Answer:

Question 2.
When could we say that an object is stable?
Answer:
If we draw a line straight down from the centre of gravity of an object of any shape and it falls inside the base of the object, then the object will stable.

If the line through the centre of gravity falls outside the base then the object will be unstable.

Question 3.
The mass of the earth is 6 × 10²⁴ kg and that of the moon is 7.4 × 10²² kg. If the distance between the earth and the moon is 3.84 × 10⁵ km. Calculate the force exerted by the earth on the moon.
Answer:

Thus the force exerted by the earth on the moon is 2.01 × 10²⁰ N.

Question 4.
A car falls off an edge and drops to the ground in 0.5 s.
i) What is its speed on striking the ground?
ii) What is the average speed during the 0.5 s?
iii) What is the height of the edge from the ground?
Answer:

Question 5.
An object is thrown vertically upwards and rises to a height of 10 m. Calculate,
i) the velocity with which the object was thrown upwards.
ii) the time taken by the object to reach the highest point.
Answer:
i) s = 10 m, v = 0 m/s, a = – g = – 9.8 m/s2.
v² – u² = 2as
0 – u² = 2 × – 9.8 × 10 ⇒ u = \(\sqrt{196}\) ⇒ u = 14 m/s.

ii) v = u + at
0 = 14 – 9.8 × t
t = 1.43 s

Question 6.
Mass of an object is 10 kg. What is its weight on the earth?
Answer:
m = 10 kg, g = 9.8 m/s².
W = mg = 10 × 9.8 = 98 N.

Question 7.
Why is the weight of an object on moon 1/6th its weight on the earth?
Answer:
The acceleration due to gravity on moon is 1/6th of its value on earth.
We know, weight = mass × acceleration due to gravity.
∴ The weight on moon is 1 /6th its weight on earth.

Question 8.
How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
The force between two objects is inversely proportional to square of distance.
\(\mathrm{F} \propto \frac{1}{\mathrm{~d}^{2}}\)
If the distance between the objects reduced to half then force increases by four times.

Question 9.
Gravitational force that acts on all objects is proportional to their masses. It is when a heavy object does not fall faster than a light object.
Answer:
The acceleration due to gravity attained by an object due to gravitational force does not depend on mass of the object. So heavy object and lighter object falls at a time on earth.

Question 10.
If the moon attracts the earth, why the earth does not move towards the moon?
Answer:
Moon attracts with same force as earth attracts moon. But we know acceleration is inversely proportional to mass. As the mass of earth is greater than moon. So earth does not move towards the moon.

Question 11.
What happens to force between two objects, if
i) the mass of one object is doubled?
ii) the distance between objects boubled and tripled?
iii) the masses of both objects are doubled?
Answer:
i) If mass of one object is doubled the force also would be doubled. Since F ∝ m.
ii) If the distance between the objects boubled and tripled the force is reduced by 4 times and 9 times respectively.
Scince, \(\mathrm{F} \propto \frac{1}{\mathrm{~d}^{2}}\)
iii) If masses of both the objects are doubled, then force is increased by 4 times.
Scince, F ∝ m₁m₂.

Question 12.
A ball is thrown vertically upward with a velocity of 49 m/s. Calculate,
i) the maximum height to which it rises.
ii) the total time to return to the surface of the earth.
Answer:
i) u = 49 m/s, g = 9.8 m/s².

Question 13.
A stone is released from the top of tower of height 19.6 m. Calculate the final velocity just before touching the ground.
Answer:
h = 19.6
u = 0
a = + g = 9.8 m/s²
v² = 2gh

Question 14.
What is universal gravitational constant?
Answer:
The value of ‘G’ i.e., the universal gravitational constant is equal to the magnitude of force between a pair of 1 kg – masses that are lm apart.
G = 6.67 × 10^-11 Nm² kg^-2.

Question 15.
When does an object get stability?
(OR)
Write a short notes on “stability”.
Answer:

• The location of the center of gravity is important for stability.
• If we draw a line straight down from the center of gravity of an object of any shape and it falls inside the base of the object, then the object will be stable.
• If the line through the center of gravity falls outside the base, then the object will be unstable.

Question 16.
State the universal law of gravitation and explain it.
Answer:
The universal law of gravitation :
The universal law of’gravitation states that every body in the universe attracts other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Explanation :
Let two bodies of masses M₁ and M₂ be separated by a distance of ’d’. Then the force of gravitation between them

G is a proportionality constant, called universal gravitational constant.
The value of G = 6.67 × 10^-11 Nm² kg^-2

9th Class Physical Science 8th Lesson Gravitation 4 Marks Important Questions and Answers

Question 1.
Define centripetal force and derive an expression for it.
Answer:
Centripetal force :
The net force which can change only the direction of the velocity of a body is called “centripetal force.”

Derivation :
Newton’s second law says that the net force on a moving body produces an acceleration in it, which is directed along the net force.

So in uniform circular motion, a net force acts towards the centre, called as centripetal force.

According to Newton’s second law of motion,

Here R is the radius of the circle.

Question 2.
Calculate the speed of moon.
Answer:
1) We know that the motion of the moon around the earth is approximately uniform circular motion.

2) We can calculate the speed of the moon using the equation.
\(\mathrm{v}=\frac{2 \pi \mathrm{R}}{\mathrm{T}}\) ……… (1)
R = distance of the moon from the centre of the earth
T = Time period of the moon

3) Thus the acceleration of the moon towards the centre of the earth

Question 3.
What is the ‘universal law of gravitation’? Derive a formula to it.
(OR)
Derive \(\mathbf{F}_{\text {grav }}=\frac{\mathbf{G M}_{1} \mathbf{M}_{2}}{\mathbf{d}^{2}}\)
Answer:
1) The universal law of gravitation states that every body in the universe attracts other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

2) The direction of the force of attraction is along the line joining the centers of the two bodies.

3) Let two bodies of masses M₁ and M₂ be separated by a distance of ’d’. Then the force of gravitation between them.

6) G is a proportionality constant called universal gravitational constant and found by Henry cavendish to be G = 6.67 × 10_-11 Nm² Kg^-2

Question 4.
Prove that acceleration due to gravity is independent of masses.
Answer:

Let us drop a body of mass’m’ near the earth’s surface.
Let ‘M’ be the mass of the earth and ‘R’ be the radius of the earth.
Now the force of attraction on the mass is given by
\(\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{r}^{2}} \Rightarrow \frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
From Newton’s second law, F/m is equal to acceleration.
Here this acceleration is denoted by ’g’.
Hence, \(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
From above equation, we can conclude that g’ is independent of the mass of the body.

Question 5.
Find the acceleration of a body which is in uniform circular motion with speed ‘v’ and radius ‘R’.
(OR)
Derive \(\mathrm{a}_{\mathrm{c}}=\frac{\mathrm{V}^{2}}{\mathrm{R}}\)
1) Let a body move with a constant speed ‘v’ in a circular path of radius ‘R’.

2) DV will be represented by the base of the isosceles triangle as shown in the figure.

3) The exact value of the sum of the magnitudes of the changes in velocity of the body during the course of revolution will be equal to the circumference ‘2πu” of the circle.

4) We know that the magnitude of the acceleration is equal to the ratio of magnitude of change in velocity for one revolution and the time period.

5) Let ac be magnitude of acceleration of the body in uniform circular motion.

6) That is a_c = 2π v/T.

7) Where ‘T’ is time required to complete one revolution.

8) We know that T = \(\frac{2 \pi \mathrm{R}}{\mathrm{v}}\)

9) Substituting this expression we get \(a_{c} \frac{v^{2}}{R}\)

Question 6.
How can you find the center of gravity of an irregular body?
Answer:

1. Take an irregular shaped object.
2. Suspend a plum in one position and draw a line along the rope of the plum.
3. Change the position of the object and do the same.
4. In this way draw lines in all possible positions.
5. The center of gravity lies where the lines intersect.
6. In this way we can find the center of gravity of an irregular shaped object.

AP State Syllabus AP Board 9th Class Hindi Textbook Solutions Chapter 10 अमर वाणी Textbook Questions and Answers.

AP State Syllabus 9th Class Hindi Solutions Chapter 10 अमर वाणी

9th Class Hindi Chapter 10 अमर वाणी Textbook Questions and Answers

InText Questions (Textbook Page No. 44)

प्रश्न 1.
चित्र में क्या दिखायी दे रहा है?
उत्तर:
चित्र में दो हाथ हैं। अपनी मुट्ठियों से एक रस्सी को पकडे हुए हैं।

प्रश्न 2.
यहाँ हाथों से क्या किया जा रहा है?
उत्तर:
यहाँ हाथों से एक रस्सी को तोडा जा रहा है।

प्रश्न 3.
जीवन में दोस्ती का क्या महत्व है?
उत्तर:
दोस्ती के बिना जीवन अधूरा है | हमारे कष्टों के समय दोस्तों के सहारे हम सहायता पाते हैं | एक बात हमें याद रखना है कि अच्छी दोस्ती ही करनी चाहिए |

अर्थवाह्यता-प्रतिक्रिया

अ) इन प्रश्नों के उत्तर सोचकर दीजिए।

प्रश्न 1.
कुछ प्रमुख हिन्दी कवियों के नाम बताइए ।
उत्तर:
हिन्दी साहित्य महान और विशिष्ट है। इसमें अनेक कवि हैं।, उनमें. कुछ हैं – चंदबरदाई, जगनिक, तुलसीदास, मीराबाई, रसखान, कबीरदास, रहीम, नंददास, वृन्द, मैथिलीशरणगुप्त, जयशंकर प्रसाद, महादेवीवर्मा, सुमित्रानंदन पंत, सुभद्रा कुमारी चौहान, सियारामशरण गुप्त, दिनकर, बिहारीलाल।

प्रश्न 2.
कबीर के बारे में बताइए |
उत्तर:
कबीर के जन्म और मृत्यु के संबंध में मतभेद हैं । फिर भी कुछ विद्वानों के अनुसार कबीरदास का जन्म सन् 1398 में काशी में एक हिन्दू परिवार में हुआ था । किन्तु नीरु – नीमा नामक एक जुलाहा दंपति ने इनका पालन-पोषण किया था । कबीरदास रामानंद के शिष्य थे | कबीर पढे-लिखे. नहीं थे । कबीर ज्ञानमार्गी शाखा के भक्तिकाल के कवि थे । उनकी समस्त कृतियाँ तीन भागों में निहित हैं – साखी, रमैनी, और पदावली । उनका काव्य संग्रह बीजक है । उनकी मृत्यु सन् 1528 में हुई ।

प्रश्न 3.
अभ्यास का क्या महत्व है ?
(या)
कवि वृंद ने जडमति को सुजान बनने का क्या उपाय बताया?
उत्तर:
मानव जीवन में अभ्यास का महत्वपूर्ण स्थान है। अभ्यास करते रहने से असंभव काम भी संभव ज़रूर होता है। मूर्ख व्यक्ति भी अभ्यास करने से महा पंडित बन सकता है। महाकवि वाल्मीकि ही इसका सच्चा उदाहरण है। कविवर वृंद भी इसी विषय का समर्थन करते हुए कहते हैं – शिला पर निशान पडना बहुत मुश्किल है। लेकिन रस्सी की रगड से उस पर भी निशान पड़ जाता है। इसी तरह अभ्यास करते – करते मूर्ख (जडमति) भी ज्ञानी (सुजान) बन सकता है। इसलिए अभ्यास के द्वारा मानव सब कुछ प्राप्त कर सकता है।

आ) पाठ पढकर उत्तर दीजिए।

• भाव से संबंधित दोहा लिखिए |

प्रश्न 1.
सुख में ईश्वर का ध्यान रखनेवाले को दुःख नहीं होता |
उत्तर:
दुःख में सुमिरन सब करे, सुख में करे न कोय |
जो सुख में सुमिरन करे, दुःख काहे को होय |

प्रश्न 2.
सज्जन के साथ रहने से सुख मिलता है ।
उत्तर:
उत्तम जन के संग में, सहजे ही सुखभासि |
जैसे नृप लावै इतर, लेत सभा जनवासि ||

इ) दोहा पढ़िए । भाव अपने शब्दों में बताइए।

प्रश्न 1.
धीरे-धीरे रे मना, धीरे सब कुछ होय ।
माली सींचै सौ घड़ा, ऋतु आये फल होय ॥
उत्तर:
कवि कबीरदास इस दोहे में धैर्यक्के महत्व को समझाते हैं । कवि कहते हैं कि हे मन ! धीरज रखो | जल्द बाज़ी मत करो। तुम्हारा काम धीरे-धीरे पूरा होगा | ऋतु के आने पर ही पेड़ को फल लगेंगे | वनमाली सौ घडे पानी डालकर पेडों को सींचने पर भी कोई फायदा नहीं | समय आने पर ही कार्य की सिद्धि होती है।

अभिव्यक्ति – सृजनात्मकता
अ) प्रश्नोत्तर

प्रश्न 1.
कबीर की दृष्टि में – “किसी काम को धीरे – धीरे करना” इसका क्या तात्पर्य है?
उत्तर:
कबीरदास जी कहते हैं कि सारे काम धीरे – धीरे निश्चित समय पर ही होते हैं। शीघ्रता दिखाने मात्र से कोई काम नहीं होता | फैल जल्दी पाने की अभिलाषा से माली सौ घडे पानी से पेड सींचता है तो भी असमय में पेड से कोई फल नहीं मिलता। मौसम के आने पर ही पेड से फल निकलते हैं इसी तरह जल्दीबाज़ी करने मात्र से कोई काम नहीं होता।

प्रश्न 2.
हमें किन प्रयत्नों से सफलता मिल सकती है?उत्तर:
उत्तर:
मानव जीवन में कोई भी काम असंभव नहीं है। मन लगाकर प्रयत्न करने से हर काम संभव होता है। मन में श्रद्धा रखकर, काम की सफलता पर विश्वास रखते, भगवान पर भरोसा रखकर, संदेह या संशय छोडकर एकाग्र चित्त होकर काम करते रहने से ज़रूर सफलता मिल सकती है। इस तरह हमें श्रद्धा, विश्वास, कडी मेहनत, शक्ति से युक्त प्रयत्नों से सफलता ज़रूर मिल सकती है।

प्रश्न 3.
भगवान का स्मरण कब करना चाहिए? क्यों?
उत्तर:
सब लोग भगवान (ईश्वर) का स्मरण दुख में ही करते हैं। अर्थात् भगवान दुख में ही याद आता है। सुखों में हम भगवान का स्मरण नहीं करते। अर्थात् सुखों में हमें भगवान याद नहीं आता। सुख में ईश्वर का ध्यान
रखनेवाले को दुख ही नहीं होता। इसलिए हमें सुख में भी भगवान का स्मरण करना चाहिए।

आ) तुलना कीजिए।

कबीर व वृन्द के दोहों में क्या अंतर स्पष्ट होता है?
उत्तर:
कबीर और वृन्द दोनों ने दोहे लिखे थे । इन दोनों के दोहों में निम्न अंतर हैं।

इ) कुछ नीति संबंधी नारें लिखिए।
उत्तर:

• सत्यमेव जयते।
• बड़ों की सेवा करो।
• सदा सच ही बोलिए।
• स्त्रियों का आदर करो।
• धर्मो रक्षति रक्षितः
• सभी जीव जंतुओं से प्यार करो।
• वृक्षो रक्षति रक्षित :
• ईमानदारी जीवन बिताइए।
• विद्यावान सर्वत्र पूज्यते।

ई) कबीर और वृंद के दोहे हमारे जीवन को किस प्रकार प्रभावित करते हैं? बताइए।
(या)
‘अमर वाणी’ कविता पाठ में किन नैतिक मूल्यों के बारे में बताया गया? लिखिए|
उत्तर:
कबीर और वृंद के दोहे हमारे जीवन को इस प्रकार प्रभावित करते हैं –

कबीर

• कबीर के दोहे हमें भगवान को सुख समय में भी स्मरण करने का आदत दिलाते हैं।
• कबीर के दोहों से हम जीवन में किसी भी काम को जल्दबाजी से न करने का आदत को सीख लेते हैं। हर एक काम बहुत सोच विचार के साथ धीरे – धीरे करने का आदत हम सीख लेते हैं।
• कबीर के दोहों के द्वारा हम भगवान से हमारे लिये जितना आवश्यक है उतना ही देने की प्रार्थना कर सकेंगे।

वंद

• वृंद के दोहों के द्वारा हम उत्तम जन से स्नेह करने का सीख सीख लेंगे। हम यह जान सकेंगे कि सज्जनों से ही दोस्ती करना चाहिए।
• वृंद के दोहों के द्वारा हम जीवन में हर हमेशा अच्छे से अच्छे काम ही करने का सीख सीख लेंगे। क्योंकि हमें अच्छे कामों से ही सुख और पुण्य मिलता है। बुरे कामों से पाप मिलेगा। हम दूसरों को जो करेंगे हमें भगवान वही करेगा।
• वृंद के दोहों के द्वारा हमें निरंतर अभ्यास करने का आदत मिलता है। निरंतर अभ्यास से जडमति छात्र भी बुद्धिमान बनते हैं।
• इस प्रकार कबीर और वृंद के दोहे हमारे जीवन को प्रभावित करते हैं।

भाषा की बात

‘जड़मति’ शब्द जड़ और मति शब्दों के जोड़ने पर बना है। ऐसे ही कुछ और शब्द बनाइए।
उत्तर:
जड़प्रकृति, जड़वाद, जड़हीन, जड़ाव, जड़ाऊ, जड़हन, जड़ताई, जड़काला आदि।

परियोजना कार्य

कुछ दोहे संकलित कीजिए।

★ जाति न पूछो साधु की, पूछ लीजिए ज्ञान ।
मोल करो तरवार का, पड़ा रहन दो म्यान ।।

★ वृच्छ कबहुँ नहिं फल भखै, नदी न संचै नीर |
परमारथ के कारने, साधु न धरा शरीर ||

★ विद्या धन उद्यम बिना, कहौ जु पावै कौन |
बिना डुलाए ना मिले, ज्यौं पंखा का पौन ||

★ जो जल बाढ़े नाव में, घर में बाढ़े दाम |
दोऊ हाथ उलीचिये, यह सज्जन कौ काम ||

★ मधुर वचन तै जात मिट, उत्तम जन अभिमान |
तनिक सीत जल सों मिटै, जैसे दूध उफ़ान ||

★ अपनी पहुँच विचारि कै, करतब करियौ दौर |
तेते पांव पसारिये, जेती लॉबी सौर ||

अमर वाणी Summary in English

Kabirdas :
One remembers God in sorrow. No one remembers him in pleasure.
Where will be the sorrow when we remember God in pleasure.

+ O mind slowness is essential. Everything happens slowly.
Even the gardener waters the plants with hundred pots of water, do they give fruit if it is not season?

O God, provide my family with only that much it requires, I shouldn’t be hungry. My guests too shouldn’t be hungry i.e., Give me the power to attend to my guests too.

Vrind (Vrindavandas) :

Everybody gets pleasure usually in the company of virtuous people.
Everybody in the court feel a pleasant smell if the king comes using a perfume?

How can we have pleasure by doing wrong deeds?
If the seeds of calatropis are planted, how are the mangoes produced?

By constant practice, even a dunce can be made wise. If the rope is drawn repeatedly, its marks are formed on the rock. Likewise, if one practises well, one will become proficient.

अमर वाणी Summary in Telugu

కబీర్ దాస్

దు:ఖంలో భగవంతుణ్ణి అందరూ స్మరిస్తారు. సుఖంలో ఎవ్వరూ స్మరించరు.
సుఖంలో కూడా భగవంతుణ్ణి స్మరిస్తే ఇంకా దు:ఖమెక్కడిది?

ఓ మనస్సా , నిదానమే ప్రధానం. నిదానంగా అన్నీ జరుగుతాయి.
తోటమాలి వంద కడవలు నీరు పోసినా ఋతువు రానిదే కాయలు కాయునా?

ఓ భగవంతుడా నా కుటుంబానికి ఎంత కావాలో (అవసరమో) అంతే ఇవ్వు (ప్రసాదించు). నేను ఆకలి (నిరాహారంతో)గా ఉండకూడదు. నా ఇంటికి వచ్చిన వారెవరైనా ఆకలితో ఉండకూడదు. అనగా నా ఇంటికి వచ్చిన వారికి కూడా తిండి పెట్టగలిగే శక్తిని ఇవ్వు.

వృంద్ (వృందావన్ దాస్)

మంచివాళ్ళ సాంగత్యంలో సహజంగానే అందరికీ సుఖం లభిస్తుంది అదెట్లనగా
‘రాజులు అత్తరు పూసుకుని వస్తే సభలోనున్న అందరూ సువాసన గ్రహిస్తారు కదా !

చెడ్డ పనులు చేసి సుఖాన్ని కోరుకుంటే ఎలా పొందుతాము? జిల్లేడు విత్తనాలను నాటితే మామిడికాయలు ఎక్కడి నుండి వస్తాయి? (మామిడిచెట్టు ఎక్కడి నుండి మొలుస్తుంది?

అభ్యాసం చేయగా చేయగా అనగా నిరంతర అభ్యాసంతో మందమతి (మూర్ఖుడు) కూడా బుద్ధిమంతుడు . (సుగుణవంతుడు) అవుతాడు. తాడును పదేపదే లాగగా రాతిపై దాని గుర్తులేర్పడతాయి. అదే విధంగా అభ్యాసం చేయగా చేయగా బుద్ధి వికసించి పండితులగుతారు.

अर्थग्राहयता – प्रतिक्रिया

निम्नलिखित प्रश्न का उत्तर दो या तीन पंक्तियों में लिखिए।

1. मानव जीवन में अच्छी संगति का क्या प्रभाव होता है?
उत्तर:
मानव जीवन में अच्छी संगति का प्रभाव बहुत अच्छा होता है। अच्छी संगति से हम भी जग में अच्छे बन जाएँगे। उस अच्छाई का फल सभी भोगते हैं। नृप अत्तर लगाकर आए तो सभी सभाजन वासी उस इतर को ग्रहण करेंगे।

अर्थग्राह्यता – प्रतिक्रिया

पठित – पद्यांश

निम्न लिखित पद्यांश पढ़कर दिये गये प्रश्नों के उत्तर एक वाक्य में दीजिए।

1. दुःख में सुमिरन सब करे,सुख में करे न कोय।
जो सुख में सुमिरन करे, दुःख काहे को होय॥
धीरे – धीरे रे मना, धीरे सब कुछ होय। |
माली सींचे सौ घड़ा, ऋतु आये फल होय।।
प्रश्न :
1. भगवान का स्मरण कब नहीं करते हैं?
उत्तर:
सुख में भगवान का स्मरण नहीं करते हैं।

2. भगवान का स्मरण सब कब करते है?
उत्तर:
सब दुख में भगवान का स्मरण करते हैं।

3. फल कब फलते हैं?
उत्तर:
ऋतु आने से फल – फलते हैं।

4. ये दोहे किस पाठ से लिये गये हैं?
उत्तर:
ये दोहे अमर वाणी पाठ से लिये गये हैं।

5. माली कितने घडे सींचता है?
उत्तर:
माली सौ घडे सीचंता है।

2. साई इतना दीजिए, जामें कुटुम्ब समाय
मैं भी भूखा ना रहूँ, साधु न भूखा जाय॥
उत्तम जन के संग में, सहजे ही सुखभासि।
जैसे नृप लावै इतर, लेत सभा जनवासि॥
प्रश्न:
1. कबीर भगवान से कितना धन देने की प्रार्थना करते हैं?
उत्तर:
कबीर भगवान से जितना धन अपना परिवार के लिए, आवश्यकता है उतना धन ही को देने की प्रार्थना करते हैं।

2. इतर कौन लाते हैं?
उत्तर:
इतर नृप लाते हैं।

3. उत्तम जन के संग से हमें क्या मिलता है?
उत्तर:
उत्तम जन के संग से हमें सुख मिलता है।

4. उपर्युक्त इन दोहों में से दूसरे दोहे के कवि कौन है?
उत्तर:
उपर्युक्त इन दोहों में से दूसरे दोहे के कवि है श्री बंद। (वृंदावन दास)

3. करै बुराई सुख चहै, कैसे पावै कोई।
रोपै बिरवा आक को, आम कहाँ तें होइ॥
करत – करत अभ्यास ते, जड़मति होत सुजान।
रसरी आवत जात तें, सिल पर परत निसान॥
प्रश्न :
1. जडमति सुजान कैसा बनता है?
उत्तर:
अभ्यास करते – करते जडमति सुजान बनता है।

2. रस्सी रगडने से सिल पर क्या पडता है?
उत्तर:
रस्सी रगडने से सिल पर निशान पडता है।

3. लोग बुराई करने से किसे नहीं पाते हैं?
उत्तर:
लोग बुराई करने से सुख नहीं पाते हैं।

4. इन दोहों के कवि कौन है?
उत्तर:
इन दोहों के कवि हैं श्री वृंद (वृंदावन दास)

5. इन दोहों को किस पाठ से लिया गया है?
उत्तर:
इन दोहों को ‘अमर वाणी’ नामक पाठ से लिया गया है।

अपठित – पद्यांश

निम्न लिखित पद्यांश पढ़कर दिये गये प्रश्नों के उत्तर विकल्पों में से चुनकर लिखिए।

1. पुष्कर सोता है निज सर में,
भ्रमर सो रहा है पुष्कर में,
गुंजन सोया कभी भ्रमर में,
सो, मेरे गृह – गुंजन, सो !
सो, मेरे अंचल – धन, सो !
प्रश्न :
1. पुष्कर यहाँ सोता है
A) निज सर में
B) सागर में
C) नाल में
D) झील में
उत्तर:
A) निज सर में

2. कभी भ्रमर में कौन सोया है?
A) भ्रमर
B) पुष्कर
C) गुंजन
D) सर
उत्तर:
C) गुंजन

3. भ्रमर कहाँ सो रहा है?
A) पुष्कर में
B) निज सर में
C) गृह में
D) गुंजन में
उत्तर:
A) पुष्कर में

4. सो, मेरे …… सो। रिक्त स्थान की पूर्ति करो।।
A) भ्रमर
B) पुष्कर
C) गुंजन
D) अंचल धन
उत्तर:
D) अंचल धन

5. गृह शब्द का अर्थ पहचानिए।
A) घर
B) वन
C) कमल
D) नयन
उत्तर:
A) घर

2. नहीं बजती उसके हाथों में कोई वीणा,
नहीं होता कोई अनुराग – राग – आलाप,
नूपुरों में भी रुनझुन -रुनझुन नहीं,
सिर्फ एक अव्यक्त शब्द – सा ‘चुप, चुप, चुप’,
है गूंज रहा सब कहीं।
प्रश्न :
1. उसके हाथों में क्या नहीं बजती?
A) कोई वीणा
B) कोई राग
C) नूपुर
D) रुनझुन
उत्तर:
A) कोई वीणा

2. इनमें भी रुनझुन – रुनझुन नहीं
A) वीणा में
B) अनुराग में
C) नूपुरों में
D) हाथों में
उत्तर:
C) नूपुरों में

3. सब कहीं क्या गूंज रहा है?
A) वीणा
B) चुप, चुप, चुप
C) नूपुर
D) राग
उत्तर:
B) चुप, चुप, चुप

4. क्या – क्या नहीं होता है?
A) अनुराग
B) राग
C) आलाप
D) ये सब
उत्तर:
D) ये सब

5. हाथ शब्द का पर्यायवाची शब्द पहचानिए।
A) पैर
B) पग
C) कर
D) त्रिभुज
उत्तर:
C) कर

3. झर – झर, झर – झर झरता झरना।
आलस कभी न करता झरना।
थक कर कभी न सोता झरना।
प्यास सभी की हरता झरना ॥
प्रश्न :
1. प्यास सभी की कौन हरता है?
A) झरना
B) सागर
C) कुआ
D) नल
उत्तर:
A) झरना

2. यह थक कर कभी नहीं सोता है
A) कौआ
B) मोर
C) झरना
D) हिरण
उत्तर:
C) झरना

3. झरना कभी – भी यह नहीं करता
A) गृह कार्य
B) आलस
C) दुख
D) शब्द
उत्तर:
B) आलस

4. झरना ऐसा झरता है –
A) टर – टर
B) धन – धन
C) चम – चम
D) झर – झर
उत्तर:
D) झर – झर

5. इस पद्य का उचित शीर्षक पहचानिए।
A) सागर
B) पर्वत
C) झरना
D) नदी
उत्तर:
C) झरना

4. बचो अर्चना से, फूल माला से,
अंधी अनुशंसा की हाला से,
बचो वंदना की वंचना से, आत्म रति से,
चलो आत्म पोषण से, आत्म की क्षति से।
प्रश्न :
1. हमें किससे बचना है?
A) साँप से
B) सिहं से
C) बाघ से
D) अर्चना से
उत्तर:
D) अर्चना से

2. हमें इसकी वंचना से बचना है –
A) हाला की
B) वंदना की
C) अंधी की
D) फूलमाला की
उत्तर:
B) वंदना की

3. हमें किस पोषण से चलना है?
A) आत्म
B) शरीर
C) हृदय
D) मन
उत्तर:
A) आत्म

4. अंधी अनुशंसा की हाला से हमें क्या करना चाहिए?
A) बचना
B) भागना
C) फ़सना
D) फैलना
उत्तर:
A) बचना

5. हमें इससे भी बचना चाहिए
A) शिक्षा से
B) दंड से
C) आत्मरति से
D) इन सबसे
उत्तर:
C) आत्मरति से

5. निश्चल आत्मा है अक्षय,
निश्चल मृण्मय तन नश्वर,
यह जीवन चक्र चिरंतन .
तू हँस – हँस जी, हँस – हँस मर ॥
प्रश्न :
1. निश्चल आत्मा कैसा है?
A) अक्षय
B) क्षय
C) व्यय
D) माया
उत्तर:
A) अक्षय

2. किस प्रकार का तन नश्वर है?
A) निश्चल आत्मा
B) निश्चल मृण्मय तन
C) जीवन चक्र
D) चिरंतन
उत्तर:
B) निश्चल मृण्मय तन

3. यह जीवन चक्र कैसा है?
A) नश्वर
B) शास्वत
C) शुभप्रद
D) चिरंतन
उत्तर:
D) चिरंतन

4. मरना शब्द का विलोम पहचानिए।
A) जीत
B) जीना
C) जलना
D) जागना
उत्तर:
B) जीना

5. इस पद्य में आये पुनरुक्ति शब्द पहचानिए।
A) हँस – हँस
B) आत्मा
C) मुण्मय
D) ये सब
उत्तर:
A) हँस – हँस

AP State Syllabus AP Board 9th Class Physical Science Important Questions Chapter 4 Atoms and Molecules

AP State Syllabus 9th Class Physical Science Important Questions 4th Lesson Atoms and Molecules

9th Class Physical Science 4th Lesson Atoms and Molecules 1 Mark Important Questions and Answers

Question 1.
What is the difference between 2N and N₂?
Answer:
2N means two nitrogen atoms.
N₂ means one nitrogen molecule.

Question 2.
Mohan said, “O₂ differs from O,”. Do you agree? Justify.
Answer:
1) Yes, I agree with Mohan’s statement.
2) ‘O’ means single oxygen atom.
‘O₂‘ means single oxygen molecule.

Question 3.
The atomic number (Z) of an element is 6. Name the element.
Answer:
Carbon.

Question 4.
Write any one precaution in doing the experiment chromotography.
Answer:
Make sure that the ink line or mark does not touch the water.

Question 5.
In a class, a teacher asked students to write the molecular formula of Oxygen, Namitha wrote the formula as “O₂“, Raju as “O”. Which one is correct?
Answer:
Namitha is correct.
Molecular formula of oxygen is O₂.

9th Class Physical Science 4th Lesson Atoms and Molecules 2 Marks Important Questions and Answers

Question 1.
Fill the following table.

Answer:

Question 2.
Compare the subatomic particles electron, proton and neutron.
Answer:

Question 3.
What are the materials used in “Conservation of mass” experiment?
Answer:
Material required for Conservation of mass:
Sodium sulphate, Barium chloride, distilled water, conical flask, spring balance, small test tube, rubber cork, thread, retort stand.

9th Class Physical Science 4th Lesson Atoms and Molecules 4 Marks Important Questions and Answers

Question 1.
Fill in the blanks in the table using the given information.

Answer:

Question 2.
Read the following information and answer.
Molecular mass of a molecule is the sum of the atomic masses of individual atoms in it.

i) Find the molecular mass of Na₂CO₃.
Answer:
Molecular mass of Na₂CO₃ = (23 × 2) + 12 + (16 × 3) = 46 + 12 + 48 = 106

ii) If the molecular weight of a compound is 44. Which is made with carbon and oxygen. What its molecular formula?
Answer:
Molecular weight of a compound = 44
At. mass of Carbon = 12 Remaining at. mass = 44-12 = 32 = 16 × 2
Hence, 16 × 2 ⇒ (Oxygen atomic mass × 2) ⇒ O₂
Molecular formula = CO₂

iii) What is the unit of atomic mass?
Answer:
Unit of atomic mass = amu
atomic mass unit (amu) is defined as precisely \(\frac{1}{12}\) the mass of an atom of carbon -12.

iv) On the basis of molecular weights of NaOH and H₂O, which is heavier?
Answer:
Molecular weight of NaOH = 23+16 + 1= 40
Molecular weight of H₂O = (1 × 2)+ 16 = 2 + 16=18
Hence, NaOH is heavier than H₂O.

Question 3.
Complete the table.

Answer:

9th Class Physical Science 4th Lesson Atoms and Molecules Important Questions and Answers

9th Class Physical Science 4th Lesson Atoms and Molecules 1 Mark Important Questions and Answers

Question 1.
How did the element Helium get its name?
Answer:
Place of discovery of element can also play a role in its naming. The gas which was first discovered in the sun was named Helium. Because, Greek name for sun is helios’.

Question 2.
What is an atom?
Answer:
An atom is the smallest particle of an element that can participate in chemical reaction and retain all its properties.

Question 3.
Is ‘Na’ an element or compound? Why?
Answer:
‘Na’ is an element, because it was formed by the same atoms of ‘Na’.

Question 4.
Is O₂ an element or compound? Why?
Answer:
O₂ is a compound, because O₂ is formed by combining two oxygen atoms.

Question 5.
What is an Avogadro number? What is its value?
Answer:
The number of particles present in one mole of any substance is the Avogadro number (N_A).
This is equal to 6.022 × 10²³.

Question 6.
Which instrument is used to calculate the atomic mass exactly?
Answer:
Mass spectrometer.

Question 7.
What is atomic mass?
Answer:
The number of times one atom of given element is heavier than 1/12th part of atomic mass of carbon -12 is called its atomic mass.

Question 8.
How molecules are formed?
Answer:
A molecule is formed by the combination of different kinds of atoms that are chemically bonded together by attractive forces.

Question 9.
Why is it not possible to see an atom with naked eye?
Answer:
The size of an atom is so small i.e., less than 1 nm (1 nano metre). So we are unable to see an atom.

Question 10.
What is a chemical formula?
Answer:
The chemical formula of a compound is a symbolic representation of its composition.

Question 11.
How many atoms are present in.
i) H₂S molecule and
ii) PO₄^3- ion?
Answer:
i) The number of atoms in H₂S molecule = 2 + 1=3
ii) The number of atoms in PO₄^3- ion =1 + 4 = 5

9th Class Physical Science 4th Lesson Atoms and Molecules 2 Marks Important Questions and Answers

Question 1.
Who is called as the father of modern chemistry? What are his main contributions?
Answer:

1. Antoine Lavoisier a French scientist is called as “Father of modern chemistry”.
2. He made many important contributions to chemistry.
3. One of his contributions is law of conservation of mass.

Question 2.
State the following.
a) Law of conservation of mass
b) Law of constant proportions
Answer:
a) Law of conservation of mass :
Matter is neither created nor destroyed during a chemical reaction.
(or)
The mass of the reactants is equal to the mass of the products of chemical reaction.

b) Law of constant proportions :
A given chemical substance always contains the same elements combined in a fixed proportions by weight.

Question 3.
What are the Dalton’s proposals about the nature of matter?
Answer:
John Dalton proposed the basic theory about the nature of matter. His proposals are :

1. If mass was to be conserved, then all elements must be made up of extremely small particles called atoms.
2. If law of constant proportions is to be followed, the particles of same substance couldn’t be dissimilar.

Question 4.
What is the proposal of Indian sage Kanada, about atom?
Answer:

1. About 2600 years ago, an Indian sage called Kanada, postulated atoms in his Vaishesika sutra”.
2. He proposed that all forms of matter are composed of very small particles known as “Anu”.
3. Each “Anu” may be made up of still smaller particles called “Paramanu”.

Question 5.
What is the use of symbols for elements?
Answer:

1. We know that chemistry involves a lot of reactions.
2. It will be a waste of time to write the full name of the elements and compounds every time to describe a reaction.
3. To avoid this we use symbols for naming the elements and formulae to represent compounds.

Question 6.
What are the characteristics of a symbol?
Answer:
1) A symbol can have either one or two letters of English.
Ex : H, He, N, Ne, etc.

2) The first letter of the symbol is always upper case and the second letter is always lower case.
Ex :
1) Al, Cr, Cl, etc. is the right method to represent elements.
2) CL, bE, he, etc. is the wrong method to represent elements.

Question 7.
In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water, and 8.2 g of sodium ethanoate. Show that these observations are in agreement with law of conservation of mass.
Answer:
Sodium carbonate + Ethanoic acid → Sodium ethanoate + Carbon dioxide + Water
The total mass of reactants = 5.3 + 6 = 11.3 g
The total mass of products = 2.2 + 0.9 + 8.2 = 11.3 g
∴ Total mass of reactants = Total mass of products
So these observations are in agreement with law of conservation of mass.

Question 8.
Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
Answer:
According to data 1 g of hydrogen is reacting with 8 g of oxygen.
So amount of oxygen would be required to react completely with 3 g of hydrogen
= 3 x 8 = 24 g

Question 9.
Calculate formula unit weight of ZnO, Na₂O, K₂CO₃ (Given atomic weight of Zn = 654)
Answer:
Formula unit weight of ZnO = 65 + 16 = 81 u
Formula unit weight of Na₂O = 2 × 23 +16 = 62u
Formula unit weight of K₂CO₃ = 2 × 39 + 12 + 3 × 16 = 138 u

Question 10.
Calculate the number of moles for the following.
i) 52 g of He (finding mole from mass).
ii) 12.044 × 10₂₃ number of He atoms (finding mole from number of particles).
Answer:

Question 11.
If one mole of carbon atom weighs 12 g. What is the mass of 1 atom of carbon?
Answer:

Question 12.
When 3 g of carbon is burnt in 8 g of oxygen 11 g of carbon dioxide is produced. What mass of carbon dioxide is formed when 3 g of carbon is burnt in 50 g of oxygen.
Answer:
3 g of carbon react with 8 g of oxygen to form 11 g of carbon dioxide.
Ratio of carbon and oxygen = 3 : 8
The amount oxygen react with 3g of carbon = 3 × 8 = 24.
So the amount of carbon dioxide formed = 3 + 24 = 27 g.

Question 13.
Given the names of the elements present in the following compounds.
a) Quick lime
b) Hydrogen bromide
c) Baking soda
d) Potassium sulphate
Answer:

Question 14.
Calculate number of sulphur (Sg) present in 16 g of solid sulphur.
Answer:
Molecular weight of sulphur = 8 × 32 = 256 g.
256 g of sulphur contains 6.022 × 10₂₃ molecules.
Number of molecules present in 16 g of sulphur = \(\frac{16}{256}\) × 6.022 × 10²³
= 3.77 × 10²² molecules

Question 15.
Anitha wrote the formula for oxygen molecule as 20. Is it correct or not? Why?
Answer:

1. It is not correct.
2. The formula of oxygen molecule is O₂.
3. 2O shows two separate atoms of oxygen.
4. O₂ shows molecule of oxygen.

9th Class Physical Science 4th Lesson Atoms and Molecules 4 Marks Important Questions and Answers

Question 1.
How elements got their names? Explain with examples.
Answer:
I: Sometimes elements are named based on their property.
Ex :
1) The Latin name for water is “hydro”.
So the element that combined with oxygen to give water was named as ‘hydrogen’.

2) The Latin word for acid is ‘oxy’.
Hence the gas that forms acid is ‘oxygen’.

II. Place of discovery of element can also play a role in its naming.
Ex : The gas that was first discovered in the sun (Greek name for the sun is “helios”) was named as ‘helium’.

III. Sometimes the elements were named to honour the scientists.
Ex : Einsteinium, Fermium, Rutherfordium and Mendelevium, etc.

Question 2.
Define the following terms.
a) Atomicity
b) Valency
c) Ions
Answer:
a) Atomicity :
The number of atoms constituting a molecule is known as its atomicity.

Ex :
1) Na is monoatomic
2) O₂ is diatomic
3) O₃ is triatomic

b) Valency :
Atoms of the elements have power to combine with atoms of other elements. This is known as valency.
Ex :
1) Valency of carbon is 4.
2) Valency of oxygen is 2.
3) Valency of hydrogen is 1.

c) Ions:
1) Ions may be a single charged atoms or a group of atoms that have a net charge on them.
2) A negatively charged ion is called anion and the positive charge ion is cation.
Ex : Na⁺, Cu⁺², S^2-, N^3-, etc.

Question 3.
Explain the method of writing a formula to a compound using criss – cross method, with the help of an example.
Answer:
The following steps should be taken while attempting to write a chemical formula using criss – cross method.
Ex : Take sodium carbonate as an example.
1) Write the symbols of atoms or group of atoms side by side, usually the cation first.
NaCO₃

2) Write the valency of each atom or group of atoms on the top of its symbol.
Na¹ (CO₃)²

3) Divide the valency number by their highest common factor if any to get the simple ratio.
Na¹ (CO₃)²

4) Interchange the valency and write the numbers to the lower right of the constituents as subscript.
Na₂(CO₃)₁

5) If any constituent receives the number 1, ignore it while writing the formula.
Na₂CO₃

6) If group of atoms received the number more than 1, enclose it within brackets. Hence the formula for the sodium carbonate is Na₂CO₃.

Question 4.
Define the terms :
a) molecular mass,
b) formula unit mass,
c) mole and
d) molar mass.
Answer:
a) Molecular mass :
The molecular mass of a substance is the sum of the atomic masses of all the atoms in a molecule of a substance.

b) Formula unit mass :
The formula unit mass of a substance is a sum of the atomic masses of all atoms in a formula unit of a compound.

c) Mole :
1) One mole of a substance is the amount of the substance which contains as many particles or entities that are equal to the atoms present in exactly 12 grams of C¹² isotope.
2) The number of particles present in one mole of any substance has a fixed value of 6.022 × 10²³. This number is called Avogadro’s constant (N_A).

d) Molar mass :
The mass of 1 mole of a substance which is expressed in grams is called its molar mass.

Question 5.
What are the main postulates of Dalton’s atomic theory?
Answer:
The main postulates of Dalton’s atomic theory are :

1. Matter consists of indivisible particles called atoms.
2. Atoms are neither created nor destroyed in a chemical reaction, but atoms will reorganize.
3. All the atoms of a given element have identical mass and chemical properties.
4. Atoms of different elements have different masses and chemical properties.
5. Compounds are formed when atoms of different elements combine in simple whole number ratios i.e., chemical change is the union or separation of atoms as a whole number.

Question 6.
Draw the pie diagram to understand the atomic mass and explain it.
Answer:

Explanation :

1. Let us assume the circle in the diagram represents mass of one carbon – 12.
2. It is divided into 12 equal parts as shown in the figure.
3. Each part represents 1/12th of atomic mass of one carbon -12.

Atomic mass unit :
One atomic mass unit is defined as the mass exactly one-twelfth the atomic mass of carbon – 12 isotope.

Question 7.
Illustrate the concept of mole through a diagram.
Answer:
Mole :

1. One mole of a substance is the amount of the substance which contains as many particles (atoms, molecules etc.) or entities that are equal to the atoms present in exactly 12 grams of ¹²C isotope.
2. The number of particles present in one mole of any substance has a fixed value of 6.022 × 10²³. This number is called Avogadro’s constant (N_A).

Question 8.
How can you appreciate John Dalton for proposing his atomic theory?
Answer:

• We know, development of science and technology is a combined effort of Roman scientists.
• The zeal to know or discover something new, leads the scientists for their discoveries.
• One of such is the atomic theory proposed by Dalton, based on Lavoisier’s law of conservation of mass and Proust’s law of constant proportions.
• Dalton said atom is indivisible.
• This proposal lead to discover many new things by various scientists and to unveil the complete structure of atom.
• Hence Dalton’s contribution is highly appreciable.

Question 9.
How do you feel after studying the symbols and formulae of different elements and compounds?
Answer:

• In my earlier class I came to know that chemistry is nowhere but in our kitchen.
• I used to call so many chemicals with their names.
• Now I am able to call them with their symbols and formulae.
• For example 1 can call the common salt as NaCl.
• This thrills me alot, and I am curious to know the formulae of so many common household substances.
• This is very helpful for my higher studies.

Question 10.
Describe the experiment conducted by Joseph L. Proust, which lead him to propose law of constant proportions.
Answer:

• Proust took two samples of copper carbonate, one from nature and another prepared in the lab.
• These two samples are chemically decomposed to find percentage of copper, carbon and oxygen.
• The results obtained are given in the table.

• From the above table we observe that the percentage of copper, carbon and oxygen atoms in two samples are same.
• Based on this observation Proust proposed the law of constant proportions as “A given chemical substance always contains the same elements combined in a fixed proportions by weight”.

Question 11.
Write down the formulae of these compounds, using criss – cross method?
i) Sodium oxide
ii) Aluminium chloride
iii) Sodium sulphate
iv) Magnesium hydroxide
Answer:

Question 12.
Write down the names of compounds represented by the following formulae and also write the anion, cation.
i) Al₂(SO₄)₃
ii) CaCl₂
iii) K₂SO₄
iv) KNO₃
v) CaCO₃
Answer:

Question 13.
What are the rules you have to follow while writing a chemical formula?
Answer:

1. The valencies or charges on the ion must balance.
2. When a compound consists of metal and non-metal, the name of the symbol of metal is written first, eg : Calcium oxide (CaO)
3. In compounds formed with poly atomic ions, the ion is enclosed in a bracket before writing the number to indicate the ratio.

Question 14.
Calculate the molecular weight H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.
Answer:
Molecular weight of H₂ = 2 × 1 = 2 u
Molecular weight of O₂ = 2 × 16 = 32 u
Molecular weight of Cl₂ = 2 × 35.5 = 71 U
Molecular weight of CO₂ = 12 + 2 × 16 = 44 u
Molecular weight of CH₄ = 12 + 4 × 1 = 16 u
Molecular weight of C₂H₆ = 2 × 12 + 6 × 1= 30 u
Molecular weight of C₂H₄ = 2 × 12 + 4 × 1= 28 u
Molecular weight of NH₃ = 14 + 3 × 1 = 17 u
Molecular weight of CH₃OH = 12 + 4 × 1 + 16 = 32 u

Question 15.
a) What is atomicity?
b) Give examples.
c) Why do elements have different atomicities?
Answer:
a) Atomicity:
The number of atoms constituting a molecule is known as its ‘atomicity’.

b) Examples :
A molecule of hydrogen consists of two atoms of hydrogen. Here the atomocity is two. Hence it is known as a diatomic molecule. Helium (He), Argon (Ar) exist as single atom. Hence they are known as monoatomic.
Ozone (O₃) has tetratomic
Sulphur (S₈) has octatomic

c) Every element has a definite combining capacity, that determines the atomicity of its molecules. Every element reacts with other element according to its combining capacity, which we call as its valency.

AP State Syllabus AP Board 9th Class Physical Science Important Questions Chapter 5 What is inside the Atom?

AP State Syllabus 9th Class Physical Science Important Questions 5th Lesson What is inside the Atom?

9th Class Physical Science 5th Lesson What is inside the Atom? 1 Mark Important Questions and Answers

Question 1.
Write any two limitations of Rutherford’s atomic model.
Answer:

• The revolving electron would lose energy continuously and get directed towards positively charged nucleus and eventually crash into the nucleus.
• If this is true, the atoms would become highly unstable and the matter would not exist, but matter exists.

Question 2.
In Rutherford’s experiment, when the alpha particles hit the foil, Rutherford expected that all the alpha particles would be deflected by small angles. Why did Rutherford expect the above observation?
Answer:
Rutherford thought that positive charge is distributed throughout the atom.

Question 3.
Write an application of isotopes in the field of medicine.
Answer:

• The isotope of iodine is used in the treatment of goitre.
• The istope of cobalt is used in the treatment of cancer.

9th Class Physical Science 5th Lesson What is inside the Atom? 2 Marks Important Questions and Answers

Question 1.
What is valency? Write valency of hydrogen.
Answer:

• The number of electrons present in outer most orbit of an atom is called its valency.
• Valency of hydrogen is 1.

Question 2.
Observe the table given below.

Answer the following questions.
i) Which Shell has highest number of electrons?
ii) Write the general form of the formula to find maximum number of electrons in each shell?
Answer:
i) 0 Shell ‘N’.
ii) 2n² (n = 1, 2, 3, ……..)

Question 3.
Convert 36g of water into mole?
Answer:
Water molecular formula = H₂O
At.wt. of water = (2 × 1) + 16 = 18
1 mole, water = 18 gr.
36g. of water = \(\frac{36}{18}\) = 2 moles

9th Class Physical Science 5th Lesson What is inside the Atom? 4 Marks Important Questions and Answers

Question 1.
Fill the blanks in the table using the given information. (Isotopes are not included).

Answer:

Question 2.
Fill in the missing information in the table given below.

Answer:

Question 3.
Draw the figures showing arrangement of electrons for the given elements.
1. Helium, Oxygen, Argon.
Answer:
1) Helium
2) Oxygen
3) Argon

2. How many neutrons are present in the nucleus of Sodium?
Answer:
Sodium = ²³₁₁Na
Neutrons = 23-11 = 12

Question 4.
a) Draw neat diagrams indicating the nucleus and arragement of electrons in different shell for the following elements?
i) Helium
ii) Carbon
iii) Argon
Answer:

b) Which of the above element is unstable? Why?
Answer:
Carbon is unstable. The nudes of carbon-14 atoms are unstable because they have too many neutrons relative to protons, so they gradually decay.

9th Class Physical Science 5th Lesson What is inside the Atom? Important Questions and Answers

9th Class Physical Science 5th Lesson What is inside the Atom? 1 Mark Important Questions and Answers

Question 1.
Write the names of isotopes of Hydrogen.
Answer:
Hydrogen, deuterium, tritium.

Question 2.
Which element has maximum number of isotopes?
Answer:
Cesium and Helium are the elements having maximum number of isotopes.

Question 3.
What is maximum number of electrons present in M-shell?
Answer:
Maximum number of electrons present in M-shell is 2 × 32 = 2 × 9 = 18.

Question 4.
Which atom doesn’t contain neutron in its nuclear?
Answer:
Hydrogen.

Question 5.
Show the arrangement of electrons in phosphorus through a diagram.
Answer:
Atomic number of phosphorus is 15.
Distribution of electrons : 2, 8, 5.

Question 6.
Who proposed rules for electron distribution in an atom?
Answer:
Bohr and Bury.

Question 7.
What is the first rule of Bohr and Bury for electron distribution in an atom?
Answer:
The maximum number of electrons present in a shell is given by the formula 2n², where ‘n’ is the shell number, which takes values 1, 2, 3, 4,

Question 8.
Write the second principle of Bohr – Bury.
Answer:
Each energy level or electron shell is further divided into sub-shells. The maximum number of electrons that can be accommodated in each shell is 8.

Question 9.
Write the third law of Bohr – Bury.
Answer:
Electrons cannot be filled in a given shell unless the inner shells are completely filled i.e., shells are filled in step-wise manner.

Question 10.
What is an octet?
Answer:
An outermost shell which has 8 electrons is said to possess an octet.

Question 11.
Define atomic number.
Answer:
Atomic number :
Atomic number is the number of protons in the nucleus of an atom, denoted by ‘Z’.

Question 12.
Define atomic mass number.
Answer:
Atomic mass number:
Atomic mass number is the number of protons plus the number of neutrons, denoted by A.
∵ A = Z + N

Question 13.
On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.
Answer:
The negative and positive charges are equal in magnitude are present in atom. So, the atom as a whole is electrically neutral as whole according to Thomson model.

Question 14.
On the basis of Rutherford model of an atom, which sub-atomic particle is present in the nuclear of an atom?
Answer:
α – particles are repelled by the nucleus. So it contains positive charged particle that is proton.

9th Class Physical Science 5th Lesson What is inside the Atom? 2 Marks Important Questions and Answers

Question 1.
What are the postulates of Thomson’s model of the atom?
Answer:
J.J. Thomson proposed a model of atom in 1898. The main postulates are

1. An atom is considered to be a sphere of uniform positive charge and electrons are embedded into it.
2. The total mass of the atom is considered to be uniformly distributed throughout the atom.
3. The negative and the positive charges are supposed to cancel out the atom as a whole is electrically neutral.

This model is commonly known as plum pudding model or watermelon model.

Question 2.
What are the features of Rutherford’s model of atom?
Answer:
In 1909, Rutherford proposed a model of atom based on the alpha particle scattering experiment. The main features are

1. All the positively charged material in an atom formed a small dense centre, called the nucleus of the atom. The electrons were not a part of nucleus.
2. The negatively charged electrons revolve around the nucleus in well-defined orbits.
3. The size of the nucleus is very small as compared to the size of the atom.

This model is known as planetary model because the electrons revolve around the nucleus as planets revolve around the sun.

This model failed to explain the stability of atom.

Question 3.
Write the rules proposed by Bohr – Bury for electron distribution.
Answer:
Bohr-Bury proposed the following rules for electron distribution.
Rule – 1 :
The maximum number of electrons that can be accommodated in each shell is given by a formula 2n2. Where n is the shell number, which takes the values 1, 2, 3,….
Ex : For K shell, n = 1

∴ Maximum number of electrons in K shell = 2(1)2 = 2 × 1=2

Rule – 2 : Each energy level or electron shell is further divided into subshell. The maximum number of electrons that can be accommodated in each subshell is 8.

Rule – 3: Electron cannot be filled in a given shell unless the inner shells are completely filled.

Question 4.
Explain the distribution of electrons in oxygen atom using Bohr-Bury principle.
Answer:

1. Atomic number of oxygen is 8.
2. Hence it has 8 protons and 8 electrons.

Step -1 :
The K-shell can accommodate maximum 2 electrons, so the first 2 electrons fill the shell of n = 1.

Step – 2 :
The other 6 electrons will fill the higher shell n = 2 or the L-shell.
Step – 3 :
Then the electronic structure for oxygen atom is 2, 6.

Question 5.
Fluorine atom contains 7 electrons in the outermost shell. But its valency is ‘1’. Explain.
Answer:

• Valency is the number of electrons present in the outer most orbit of an atom.
• The distribution of electrons in fluorine (atomic number 9) is 2, 7.
• Hence the valency of fluorine could be 7.
• But it is easier to fluorine to gain one electron for becoming octet.
• Hence its valency is determined by subtracting seven electrons from 8 and which gives you a valency ‘1’ for fluorine.

Question 6.
What is the importance of valency?
Answer:
1) An atom with 8 electrons or an octet in their outer most shell is chemically stable or does not combine other atoms.
Ex : Ne, Ar, etc.

2) An atom with duplet or 2 electrons in its outer most shell is also more stable when there is only one shell present in it.
Ex : He

3) Atoms of an element thus react with other atoms. So as to achieve an octet in their shell.

4) When elements reacts to form compounds, their atoms must be combining in such a way that they can attain the stable electron distribution of noble gases or inert gases.

Question 7.
How can an atom achieve octet?
Answer:
An atom can achieve an octet by two ways.

1. One by transfer of electrons.
2. Other by sharing of electrons.
3. Both the processes results in the formation of bonds between atoms.

Question 8.
Explain the notation : \({ }_{9}^{19} \mathrm{F}\)
Answer:

• F is the symbol of element (Fluorine).
• 9 is the atomic number.
• 19 is the mass number.
• Hence fluorine has 9 protons and (19 – 9 = 10) 10 neutrons in its nucleus.
• 9 electrons are revolving around the nucleus.

Question 9.
Define isotope and give two examples.
Answer:
Isotope :
The atoms of the same element which have the same number of protons but have different number of neutrons are called isotopes.

Question 10.
“Sulphur shows multiple valency.” Explain this.
Answer:
a) Atomic number of sulphur is 16.
b) The distribution of electrons is 2, 8, 6.
c) Hence the valency should be ’6′.
d) But it is easier to sulphur to gain 2 electrons than loosing 6 electrons to become octet. Hence its valency would become 2.
e) So, sulphur shows multiple valency, i.e., ‘2’ or ‘6’.

Question 11.
Show the arrangement of electrons in first 18 elements schematically.
Answer:

Question 12.
What are the applications of isotopes in our daily life?
Answer:
Applications of isotopes :

1. Some isotopes are used for solving chemical and medical mysteries.
2. Isotopes are also commonly used in the laboratory to investigate the steps of a chemical reaction.
3. The isotope of uranium is used as a fuel in nuclear reactors.
4. The isotope of iodine is used in the treatment of goitre.
5. The isotope of cobalt is used in the treatment of cancer.

Question 13.
If an atom contains one electron and one proton, will it carry any charge or not?
Answer:
It will not carry any charge because proton is positively charged particle and electron is negatively charged particle. So they would neutralise each other.

Question 14.
What do you think would be the observation if the a – particle scattering experiment is carried out using a foil of metal other than gold?
Answer:
Gold has highest malleability and ductility. Extremely very thin foil can be prepared by using Gold, which is not possible with other metals. So, we will not get same type of results using other metals.

Question 15.
Helium atom has an atomic mass of 4u and two protons in its nucleus. How many neutrons does it have?
Answer:
Atomic mass due to protons = 2 × 1 = 2 u
Neutrons have almost same mass as protons.
Mass of neutrons = 4 – 2 = 2u
∴ Number of neutrons = 2

Question 16.
Write the distribution of electrons in carbon and sodium atoms.
Answer:
Distribution of electrons in carbon is 2, 4.
Distribution of electrons in sodium is 2, 8, 1.

Question 17.
If K and L shells of an atom are full, then what would be the total number of electrons in the atom?
Answer:
Number of electrons present in K shell = 2(1)² = 2 × 1=2
Number of electrons present in L shell = 2(2)² = 2 × 4 = 8
Total number of electrons in the atom = 2 + 8 = 10

Question 18.
If number of electrons in an atom is 8 and number of protons is also 8, then
i) what is the atomic number of atom?
ii) what is the charge on the atom?
Answer:
i) The atomic number of atom is 8.
ii) The electronic configuration of atom is 2, 6. By gaining two electrons it will get nearest inert gas configuration Neon. So, the charge on the atom is – 2.

Question 19.
Chlorine occurs in two isotopic forms that is \({ }_{17}^{35} \mathrm{Cl}\) and \({ }_{17}^{37} \mathrm{Cl}\). The percentage of these forms are 75% and 25% respectively. Find atomic weight of chlorine atom?
Answer:
The atomic mass of an element is taken as an average mass of all the naturally occurring atoms of the sample element.
The average atomic mass of chlorine atom on the bases of above data

Question 20.
For the symbol H, D and T tabulate three sub atomic particles found in each of them.
Answer:

Question 21.
Na⁺ has completely filled K and L shells. Explain.
Electron configuration of Na+ is 2, 8.
So, sodium has completed filled K and L shells because the maximum number of electrons filled in K and L shells are 2 and 8 electrons respectively.

Question 22.
If Bromine atom is available in the form of say, two isotopes \({ }_{35}^{79} \mathrm{Br}\)(49.7%) and \({ }_{35}^{81} \mathrm{Br}\) (50.3%), calculate the average atomic mass of Bromine atom.
Answer:

Question 23.
The average atomic mass of sample of an element X is 16.2 u. What are the percentages of isotopes \({ }_{8}^{16} \mathrm{X}\) and \({ }_{8}^{18} \mathrm{X}\)?
Answer:

∴ The percentage of first isotope is 90 and percentage of second isotope is 10.

Question 24.
If Z = 3, what would be the valency of the elements? Also, name the element?
Answer:
Electronic configuration of element is 2, 1 and the element is Lithium.
The element will get nearest inert gas Helium configuration by loosing one electron. So its valency is 1.

Question 25.
Composition of nuclei of two atomic species X and Y are given as under.

Give the mass number of X and Y. What is the relation between the two species?
Answer:
Mass number of X = 6 + 6 = 12
Mass number of Y = 6 + 8 = 14
So these two species have same atomic number (same number of protons) and different mass numbers. Therefore these two species are isotopes.

9th Class Physical Science 5th Lesson What is inside the Atom? 4 Marks Important Questions and Answers

Question 1.
How did the idea of sub-atomic particles evolve?
Answer:

• According to Dalton, atom is indivisible.
• But Michael Faraday’s experiments on electrolysis proved that atoms were acquiring negative charge during electrolysis.
• This is contradiction to Dalton’s theory.
• This lead to an idea that there must exist some tiny particles in atom which are responsible for atom to behave sometimes as charged particles.
• As atom is considered as electrically neutral, it probably had equal number of positive and negative constituents to maintain electrical neutrality.
• This gave scope to think about sub-atomic particles.

Question 2.
How do we determine the atomic mass of an element with its isotopes?
Answer:

• In nature, most elements occur as a mixture of two or more isotopes, each isotope has a certain percentage of natural occurrence.
• The atomic mass of an element is taken as an average mass of all the naturally occuring atoms of the sample element.
  Ex : Let us calculate the atomic mass of chlorine.
• Isotope of Cl occurs in nature, in two isotopic forms, with masses 35 units and 37 units.
• The isotope with mass 35 is present in 75% in nature.
• The isotope with mass 37 is present in 25% in nature.
• The average mass of chlorine is
  

Question 3.
Describe the Rutherford’s alpha particle scattering experiment. What are the conclu¬sions of this experiment.
Answer:

Rutherford conducted alpha particle scattering experiment in 1909 to study the atom.

1. The stream of alpha particles from a source having considerable energy is directed towards a very thin gold foil.
2. The gold foil was placed inside a detector in such a way that the detector would show a flash of light when an alpha particle struck it.
3. The entire arrangement was kept in a vacuum chamber.
4. Rutherford did not expect to see large deflections of alpha particles based on Thomson’s model.
5. But Rutherford observed the deflection of alpha particles through very large angles and a few alpha particles were reflected right back.
6. From this observation, Rutherford concluded as follows.

Conclusions :

1. Most of the space inside the atom is empty.
2. All the positive charge must be concentrated in a very small space within the atom and he named it as ‘nucleus’.

Question 4.
What information do you know from \({ }_{17}^{35} \mathrm{X}\)?
Answer:
Given that \({ }_{17}^{35} \mathrm{X}\)
a) The atomic number of element is 17.
b) Hence the element is chlorine, symbol is ‘Cl’.
c) Number of protons = 17.
d) Number of electrons = 17.
e) Mass number = 35.
f) Number of neutrons = 35-17 = 18
g) Distribution of electrons in shells

h) Valency is ‘l’.
i) It gains one electron to become octet.

AP State Syllabus AP Board 9th Class Physical Science Important Questions Chapter 6 Chemical Reactions and Equations

AP State Syllabus 9th Class Physical Science Important Questions 6th Lesson Chemical Reactions and Equations

9th Class Physical Science6th Lesson Chemical Reactions and Equations 1 Mark Important Questions and Answers

Question 1.
Fe₂O₃ + 2 Al → Al₂O₃ + 2Fe
Name the compound which is oxidized in the above reaction.
Answer:
In the reaction Fe₂O₃ + 2Al → Al₂O₃ + 2Fe,
Aluminium (Al) is oxidised and formed as Al₂O₃.

Question 2.
Give an example for displacement reaction.
Answer:
1) In displacement reaction one element displaces another element from its compound and takes its place there in.

2) Eg : Zinc pieces react with dilute hydrochloric acid and liberate Hydrogen gas.
Zn_(s)+ 2HCl_(aq) → ZnCl_2(aq) + H_2(g)

Question 3.
A shiny brown coloured element ‘X’ on heating in air becomes black in colour. Assume and write the name of the element ‘X’ and also predict the substance formed black in colour.
Answer:
Element X – Copper.
Black substance – Copper Oxide

Question 4.
Iron gets rust but Gold doesn’t, why?
Answer:
Gold does not oxidized.
(OR)
Gold is least reactive metal.

Question 5.
What happens if iron articles are exposed to moist air? Write the chemical equation to represent that reaction.
Answer:
Rusting takes place on iron articles when they are exposed to moist air.
2Fe_(s) + O_2(g) + 4H⁺_(aq) → 2Fe²⁺_(aq) + 2H₂O_(l)

Question 6.
On adding dilute hydrochloric acid to copper oxide powder, the solution formed is blue green. Write the new compound formed.
Answer:
Copper oxide reacts with hydrochloric acid and forms copper chloride and water. Copper chloride is in blue green colour.
CuO + 2 HCl → CuCl₂ + H₂O

Question 7.
Write the equation for the chemical decomposition reaction of silver chloride in the presence of sunlight.
Answer:

Question 8.
Balance the following chemical equation.
Na + H₂O → NaOH + H₂
Answer:
2Na + 2H₂O → 2NaOH + H₂

Question 9.
If you keep an iron piece in solid state CuSO₄ crystals, does it get any reaction? Guess the reason.
Answer:
Reaction will not takes place if an iron piece is placed in solid state CuSO₄ crystals because there will not exist separate Cu⁺², SO₄^-2 ions in CuSO₄ Crystals. In aqueous solution, they exists. So iron can not displace copper.

Question 10.
What is a chemical equation?
Answer:
Chemical Equation :
Describing a chemical reaction using least possible words or symbols is called a chemical equation.
Ex : CaO + H₂O → Ca(OH)₂

Question 11.
What are “Reactants” and “Products”?
Answer:
1) Reactants :
The substances which undergo chemical change in the reaction are called ‘Reactants’.

2) Products:
The new substances formed in a chemical reaction are called ‘Products’.
Ex : Zn + 2HCl → ZnCl₂ + H₂ ↑
3) In above reaction Zn and 2HCl are called reactants, ZnCl₂ and H₂ are called products.

Question 12.
What is a balanced chemical equation?
Answer:
Balanced Chemical Equation :
A chemical reaction in which the number of atoms of different elements on the reactants side (left side) are same as those on product side (right side) is called a balanced chemical equation.
Ex : Mg + O₂ → MgO (Unbalanced chemical equation)
2 Mg + O₂ → MgO (Balanced chemical equation)

Question 13.
What is chemical change?
Answer:
Chemical Change :
The process in which the change is affecting the identity of a molecule by a change in chemical composition is called a “chemical change”.

Question 14.
What is physical change?
Answer:
Physical Change: The process where a change is occurring only in physical properties, without affecting the identity of the molecules is called a “physical change”.

Question 15.
What is the meaning of “Exo”, “Endo” and “Thermo”?
Answer:
The term ‘Exo’ means outside, ‘Endo’ means inside and ‘Thermo’ means heat.

Question 16.
What is oxidation? Give examples.
Answer:
Oxidation is a reaction that involves the addition of oxygen or loss of hydrogen or electrons.

Question 17.
What is a reduction? Give examples.
Answer:
The process in which a substance loses oxygen or gains hydrogen or electrons is known as reduction.

Question 18.
What is corrosion?
Answer:
Corrosion :
When some metals are exposed to moisture, acids etc. they tarnish due to the formation of respective metal oxide on their surface. This process is called “corrosion”.

Question 19.
What is “galvanizing”?
Answer:
It is a method of protecting iron from rusting by coating them a thin layer of zinc.

Question 20.
What is combustion?
Answer:
The process of burning a substance in the presence of oxygen is called “combustion”.

Question 21.
What do you mean by “rancidity”?
Answer:
Food materials containing fat / oil are exposed to air, for a long time they react with atmospheric oxygen and it is responsible for spoiling of food. This process is called “rancidity”.

Question 22.
How do you know that respiration is an exothermic reaction?
Answer:
During the respiration reaction energy releases, so it is an exothermic reaction.

Question 23.
What will happen in chemical reaction?
Answer:
New substances are formed in chemical reaction.

Question 24.
What is precipitate?
Answer:
A precipitate is a solid product’ which separates out from the solution during a chemical reaction.

Question 25.
How does precipitation reaction occur? Explain with an example.
Answer:

1. Prepare lead nitrate and potassium iodide solutions in separate test tubes.
2. Mix the two solutions.
3. A yellow colour substance which is insoluble in water, is formed. This insoluble substance is known as precipitate. The precipitate here in this reaction is lead iodide.
   Pb(NO₃)₂ + 2KI → Pbl₂ + 2KNO₃

Question 26.
What is a balanced chemical equation?
Answer:
A balanced chemical equation has an equal number of atoms of different elements in the reactants and products.

Question 27.
What is an unbalanced equation?
Answer:
An unbalanced chemical equation has an unequal number of atoms of one or more elements in the reactants and products.

Question 28.
Why do the smell and taste of food items change?
Answer:
When fats and oils are oxidized they become rancid. So their smell and taste change.

Question 29.
“Freshly cut apple turning brown, the iron articles shiny when new, but gradually become reddish brown when left for sometime ………”. How do these changes occur?
Answer:
Oxygen molecules interact with different substance from metal to living tissue which may come into contact with it. The above changes occur. These are all the examples of the process of oxidation.

Question 30.
What are antioxidants?
Answer:
Antioxidants :
The spoilage of food can be prevented by adding preservatives which prevent oxidation. The substances which prevent oxidation are called antioxidants.
(OR)
The substances which prevent oxidation added to food containing fats and oil are called antioxidants.

Question 31.
Complete the following reaction.
Pb(NO₃)₂ + 2 KI → …………….. + …………………
Answer:
Pb(NO₃)₂ + 2 KI → Pbl₂ + 2KNO₃

Question 32.
If iron nail is dipped in copper sulphate solution, after sometime copper will be formed. Write the chemical equation for this reaction.
Answer:
Fe + CuSO₄ → FeSO₄ + Cu

Question 33.
Which metal is used in the manufacture of Diwali crackers?
Answer:
The metal used in manufacture of Diwali crackers is Magnesium.

Question 34.
What are new substances formed due to decomposition of lead nitrate?
Answer:
The new substances formed are lead oxide, nitrogen dioxide and oxygen.

Question 35.
Balance the following chemical equation. C₂H₆ + O₂ → CO₂ + H₂O
Answer:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Question 36.
NH₄Cl → NH₃ + HCl. Which type of reaction is this?
Answer:
This is decomposition reaction.

Question 37.
Which chemical reaction is involved in the corrosion of iron?
Answer:
The chemical reaction involved in corrosion of iron is oxidation reaction.

Question 38.
Which metal is used for wrapping food material? Why?
Answer:
Aluminium is used to wrap food material because it reacts with oxygen and forms a protective layer of aluminium oxide which prevents further oxidation.

Question 39.
Write states of the reactants and products and also write conditions required for the completion of reaction.
Answer:
2H₂O → 2H₂ + O₂
H₂O is in liquid state. H₂ and 0₂ are in gaseous state.
The reaction is carried out in the presence of electricity.

Question 40.
Give an example for chemical combination reaction where only elements take part.
Answer:
2 H₂ + O₂ → 2H₂O
Here hydrogen and oxygen both are elements.

Question 41.
Give an example for chemical combination where element and compound take part in the reaction.
Answer:
2 NaN02 + O₂ → 2 NaNO₃
Here O₂ is element and NaNO₂ is compound.

Question 42.
Give an example where two compounds combine together to form a compound.
Answer:
CaO + H₂O → Ca(OH)₂
Here both calcium oxide, water are compounds which combine together to form calcium hydroxide.

Question 43.
Before burning in air, why is Mg ribbon cleaned by rubbing with a sand paper?
Answer:
To remove the protection layer of basic magnesium carbonate from the surface of magnesium

Question 44.
X Pb(NO₃)₂ → Y PbO + Z NO₂ + W O₂.
If the equation is a balanced equation, what are values of coefficients X, Y, Z and W?
Answer:
The balanced equation is 2 Pb(NO₃)₂ → 2 PbO + 4 NO₂ + O₂.
∴ The values are X = 2,Y = 2, Z = 4 and W = 1.

Question 45.
Why does a layer of zinc prevent rusting of iron?
Answer:
Zinc reacts with oxygen and forms layer of zinc oxide which prevents further oxidation. So a layer of zinc protects iron from rusting.

Question 46.
Why does jewellery made of gold not rust?
Answer:
Gold does not react with air because it has least reactivity. So gold jewellery does not rust.

Question 47.

What type of reaction is this? Which element between A and C is more reactive? Why?
Answer:
This reaction is chemical displacement. ‘C’ is more reactive than ‘A’. So C displaces ‘A’ from AB and occupies its place.

Question 48.
Given AB + CD → AD + CB. What type of chemical reaction is this? What is the condition which makes the reaction possible?
Answer:
This is double displacement reaction. This reaction can be possible if ionisation of both the salt solutions are possible.
i.e., AB → A⁺ + B^–
CD → C⁺ + D^–

Question 49.
An iron nail is completely immersed in a test tube containing oil. Do you observe any rust on the iron nail? If not, why?
Answer:
No, I do not observe any rust because both air and moisture are required for rusting of iron.

Question 50.
How can we prevent rusting of iron?
Answer:
Rusting of iron can be prevented or at least minimised by shielding the metal surface from oxygen and moisture. It can be prevented by painting, oiling, greasing, galvanizing, chrome plating or making alloys.

Question 51.
Suggest few methods to avoid corrosion.
Answer:
Some methods to avoid corrosion

1. Painting, oiling, greasing on the surfaces of the metals.
2. Galvanizing the metal with thin layer of zinc.
3. Making alloy metals prevents corrosion,
   eg : brass, bronze and steel.

Question 52.
Some metals react with oxygen to form their oxides. It is serious problem. Give some examples for oxidation of metals and write balanced equations.
Answer:
1) Iron reacts with oxygen and forms iron oxide.
4Fe + 3O₂ → 2Fe₂O₃.

2) Copper reacts with oxygen and forms copper oxide.
2Cu + O₂ → 2CuO

3) Silver tarnishes when reacts with hydrogen sulphide and oxygen
4Ag + 2H₂S + O₂ → 2Ag₂S + 2H₂O

Question 53.
Which pipes are suggestable/suitable for water supply? Justify your answer.
Answer:
Steel, PVC and CPVC pipes are suitable/suggestable for water supply.
Reasons:

1. They are not oxidised in moisture.
2. They do not form oxides.
3. They are not corroded.
4. So, they are durable and safe to use in supply of water.

Question 54.
Which pipes are used by you for water supply to your house?
Answer:

1. PVC, CPVC and steel pipes are used for water supply in my house.
2. Because they do not corrode.

Question 55.
List of metals are given below. Classify them into corroded and non-corroded metals. Aluminium, Silver, Iroh, Copper, Gold, Tin, Tungsten, Platinum.
Answer:
Corroded metals :
Aluminium, Silver, Iron, Copper

Non-corroded metals :
Gold, Tin, Tungsten, Platinum.

9th Class Physical Science6th Lesson Chemical Reactions and Equations 2 Marks Important Questions and Answers

Question 1.
A light yellow coloured compound ‘X’ is exposed to sunlight for some time. It is turned into gray coloured material. What is the name of ‘X’? Predict the type of chemical reaction occured in it.
Answer:
1) The name of the compound ‘X’ is Silver Bromide (AgBr).

This is a decomposition reaction and also it occurs in the presence of sunlight. Hence, it is a photo chemical reaction.

Question 2.
Oil and fat containing food item packets are flushed with Nitrogen gas. Why?
Answer:
By stopping the oxidation it prevents the rancidity of food material.

Question 3.
N_2(g) + O_2(g) + heat → 2NO_(g)
What information do you get from the above equation? Comment.
Answer:

• Nitrogen gas reacts with oxygen gas in the presence of heat energy and forms nitric oxide gas.
• It is an endothermic reaction.
• This is an example for chemical combination reaction.
• In this reaction, the reactants are gases and products are also gases.

Question 4.
Write an activity about how you conduct an experiment to show that more reactive metals replace less reactive metals from their compounds.
Answer:

• Take two iron nails and clean then by rubbing with sand paper.
• Take two test tubes and mark them ‘A’ and ‘B’.
• Pour copper sulphate solution in the test tube ‘A’ and Zinc Chloride solution in the test tube ‘B’.
• Dip iron nails in both test tubes.
• Keep them without disturbing for 20 min.

Observation :

1. The nail which is dipped in Copper Sulphate solution becomes brownish colour.
2. The nail which is dipped in Zinc Chloride solution doesn’t change.

Reaction :
In test tube ‘A’: CuSO₄ + Fe → FeSO₄ + Cu
In test tube ‘B’: ZnCl₂ + Fe → No reaction.

Conclusion :

1. Iron is more reactive than copper. So it displaces copper from Copper Sulphate solution.
2. Iron is less reactive than zinc. So, it doesn’t displace zinc from ZnClr

Question 5.
i) CaCO_3(s) → CaO_(s) + CO_3(g)
ii) 2Ag Br_(s) → 2Ag_(s) + Br_2(g)
Mention the types of reactions to which the above equations belong. Also mention which of them is a photochemical reaction.
Answer:
1) i) CaCO_3(s) → CaO_(s) + CO_2(g). It is a chemical decomposition reaction.
ii) 2Ag Br_(s) → 2Ag_(s) + Br_2(g). It is also a chemical decomposition reaction.

2) 2Ag Br_(s) → 2Ag_(s) + Br_2(g). It is a photochemical reaction. Because, this reaction
takes place in the presence of sunlight only.

Question 6.
Write the products of given reactions, if any. Give reason.
FeCl₂ + Zn →
ZnCl₂ + Fe →
Answer:
FeCl₂ + Zn → ZnCl₂ + Fe (Displacement reaction)
ZnCl₂ + Fe → No reaction. (Low reactive metals cannot displace high reactive metals)

Question 7.
Balance the following chemical equations:
i) Na + H₂O → NaOH + H₂
ii) K₂CO₃ + HCl → KCl + H₂O + CO₂
Answer:
i) 2Na + 2H₂O → 2NaOH + H₂
ii) K₂CO₃ + 2HCl → 2KCl + H₂O + CO₂

Question 8.
Observe the following balanced chemical equation and answer the given questions.
C₃H_8(g) + 5O_2(g) → 3CO_2(g) + 4H₂O_(g)
i) How many molecules of Oxygen are involved in this chemical reaction?
ii) How many moles of Propane are required to get 20 moles of Water?
Answer:
i) In this chemical reaction five molecules of oxygen are involved.
ii) Five moles of propane are required to get 20 moles of water.

Question 9.
What do you do to prevent rusting of copper and silver articles?
Answer:
I can follow some rules given below to prevent rusting of copper and silver articles.

1. Shielding the metal surface from oxygen and moisture.
2. By painting.
3. By oiling, greasing.
4. By galvanizing, chrome plating.
5. By making alloys.

Question 10.
What are the important characteristics of chemical reactions?
Answer:
The important characteristics of chemical reactions are

1. Evolution of a gas
2. Formation of a precipitate
3. Change in colour
4. Change in temperature
5. Change in state

Question 11.
What symbols do we use to indicate the physical state of reactants and products in an equation?
Answer:

1. Solid state is indicated by the symbol (s)
2. Liquid state is indicated by the symbol (l)
3. Gaseous state is indicated by the symbol (g)
4. Aqueous solution is indicated by the symbol (aq)

Question 12.
What can we do to make a chemical equation more informative?
(OR)
How can chemical equation be made more informative by knowing?
Answer:

1. Physical state,
2. Heat changes (exothermic and endothermic reactions),
3. Gas evolved,
4. Precipitate formed.

Question 13.
Commemt on “C_(s) + O_2(g) → CO_2(g) + Heat” equation.
Answer:

• The burning of carbon in oxygen is an exothermic reaction because heat is evolved in this reaction.
• An exothermic reaction is indicated by writing + Heat or + Heat energy or just + Energy on the products side of an equation.

Question 14.
Comment on “N_2(g) + O_2(g) + Heat → 2 NO_(g)” equation.
Answer:

• The reaction between nitrogen and oxygen to form nitric oxide is an endothermic reaction because heat is absorbed in this reaction.
• An endotheumic reaction is usually indicated by writing + Heat or + Heat energy or just”+ Energy” on the reactants side of an equation.

Question 15.
Balance the following equations.
1) Na + O₂ → Na₂O
2) H₂O₂ → H₂O + O₂
3) Mg(OH)₂ + HCl → MgCl₂ + H₂O
4) Fe + O₂ → Fe₂O₂
Answer:

1. 4 Na + O₂ → 2 Na₂O
2. 2 H₂O₂ → 2 H₂O + O₂
3. Mg(OH)₂ + 2 HCl → MgCl₂ + 2 H₂O
4. 4 Fe + 3 O₂ → 2 Fe₂O₃

Question 16.
2 Cu + O₂ → 2 CuO
What information do you get from above equation?
Answer:
The above equation tells us that,

1. Copper reacts with oxygen to form copper oxide.
2. The formula of copper oxide is CuO and that of oxygen is O₂.
3. 2 moles of copper atoms react with 1 mole of oxygen molecules (O₂) to produce 2 moles of copper oxide (CuO).

Question 17.
Write examples for oxidation reaction.
Answer:
Oxidation :
It is a reaction involving addition of oxygen or removal of hydrogen from a substance.
1) Magnesium reacts with oxygen to form magnesium oxide.
2 Mg + O₂ → 2 MgO

2) Copper reacts with oxygen to form copper oxide.
2 Cu + O₂ → 2 CuO

3) Iron reacts with oxygen to form ferric oxide.
4 Fe + 3 O₂ → 2Fe₂O₃

Question 18.
Write examples for reduction reaction.
Answer:
Reduction :
It is a reaction involving addition of hydrogen or removal of oxygen from a substance.
1) Nitrogen gas reacts with hydrogen gas to produce Ammonia.
N₂ + 3H₂ → 2NH₃

2) Oil reacts with hydrogen to form fat.
Oil + H₂ → Fat

Question 19.
Write the examples for corrosion reaction.
Answer:
1) The black coatings on silver.
4 Ag + 2H₂S + O₂ → 2 Ag₂S + 2H₂O

2) Green coating on copper.
2 Cu + O₂ → 2 CuO

Question 20.
Name the reactants and products in the following chemical equations.
Na₂SO₄ + BaCl₂ → BaSO₄ + NaCl
Answer:
Reactants are Sodium sulphate and Barium chloride. Products are Barium sulphate and Sodium chloride.

Question 21.
Balance the following chemical equation and follow the steps involved in balancing a chemical equation.
Cu₂S + O₂ → Cu₂O + SO₂
Answer:
Step – 1 : Write the unbalanced equation using correct chemical formula for all substances.
Cu₂S + O₂ → Cu₂O + SO₂

Step – 2 : Compare number of atoms of each element on both sides.

Balancing Cu, S, O atoms both sides
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
The equation is balanced.

Step – 3 : Write the coefficient of smallest ratio.
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂

Step – 4 : Verify above equation for balancing of atoms of each element on both sides.
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂

Question 22.
Write the difference between oxidation and reduction. Give example.
Answer:

Question 23.
Give two examples for chemical reactions in which precipitate is formed.
Answer:
The reactions in which a substance insoluble in water is formed are called precipitation reactions.
e.g.:
1) When lead nitrate solution reacts with potassium iodide solution it forms a yellow precipitate of lead iodide.
Pb(NO₃)_2(aq) + 2KI_(aq) → PbI_2(s) + 2KNO_3(aq)

2) When sodium sulphate solution reacts with barium chloride solution it forms a white precipitate of barium sulphate.
Na₂SO4_(aq) + BaCl_2(aq) → BaSO_4(s) + 2NaCl_(aq)

Question 24.
What are exothermic and endothermic reactions?
Answer:
Exothermic reaction :
A chemical reaction in which heat is released is called exothermic reaction.
Ex :
C + O₂ → CO₂ + Heat,
CaO + H₂O → Ca(OH)₂ + Heat

Endothermic reaction :
A chemical reaction in which heat is absorbed is called endothermic reaction.
Ex :
2NaHCO₃ + Heat → Na₂CO₃ + H₂O + CO₂

Question 25.
What is an alloy? Give one example.
Answer:
A metallic substance made by mixing and fusing two or more metals or a metal and a non-metal, to obtain desirable qualities such as hardness, lightness and strength is called alloy.
Ex : Brass, bronze, steel.

Question 26.
What is photochemical reaction? Give example.
Answer:
The decomposition reaction occurs in the presence of sunlight is called photochemical.. reaction.
Ex : 2AgBr_(s) → 2Ag_(s) + Br_2(g)

Question 27.
Why is power supply to our home from the electrical pole interrupted?
Answer:

• Sometimes during rainy season the power supply to our home from the electric pole will be interrupted due to the formation of the metal oxide layer on the electric wire.
• This metal oxide is an electrical insulator.

Question 28.
Name the reactions involved in the following reactions with reasons.

Answer:
a) It is a decomposition reaction carried out in the presence of heat. So it is thermal decomposition reaction.
b) It is a decomposition reaction carried out in the presence of light. So it is photochemical reaction.
c) It is a decomposition taking place in the presence of electricity. So it is electrolysis reaction.
d) Lead is more reactive than copper. So lead displaces copper from salt solution. So the reaction is chemical displacement reaction.

Question 29.
Ramu told Ravi that all material made of iron and its alloys are rusted when exposed to air. How do you correct him by asking suitable questions?
Answer:

1. What are the material we will generally use in our cooking utensils?
2.  What are the material used for surgical equipments?
3. What happens when stainless steel vessels are exposed to air?

Question 30.
What are the gases released when lead nitrate is heated and how do you identify those gases?
Answer:
When lead nitrate is heated it decomposes into lead oxide, nitrogen dioxide and oxygen.

Nitrogen diox-ide is identified by its reddish brown co|our. Oxygen makes the burning splinter to burn brightly when it is placed in test tube containing oxygen.

Question 31.
Write some chemical reactions occurring in our daily life.
Answer:

1. Souring of milk
2. Formation of curd from milk
3. Cooking of food
4. Digestion of food in our body
5. Fermentation of grapes
6. Rusting of iron
7. Burning of fuels
8. Burning of candle wax
9. Ripening of fruits.

Question 32.
If 40 gm of methane is burnt, then how much amount of CO₂ is released ?
Answer:

When 16 g of methane is burnt it releases 44 g of CO₂. The amount of CO₂ released when 40 g methane burnt 44
= \(\frac{44}{16}\) × 40=110g

Question 33.
Calculate the amount of calcium oxide formed when 2 kg of calcium carbonate is decomposed. (The atomic masses of Ca = 40 U, C = 12 U, O = 16 U).
Answer:
The balanced equation is
CaCO_3(s) → CaO_(s) + CO_2(g)
(40 + 12 + 3 x 16) U → (40 + 16) U + (12 + 2 x 16) U
100 U → 56 U + 44 U
100 g → 56g + 44g
100 g calcium carbonate on decomposition produces 56 g of calcium oxide.
Amount of calcium oxide formed due to decomposition of 2 kg of calcium carbonate

Question 34.
A solution of common salt when added to silver nitrate solution yields a precipitate of silver chloride (0.28 g). Find the mass of sodium chloride in the solution and also the mass of sodium nitrate formed.
Answer:

Question 35.
0.29 g of hydrocarbon when burnt completely in oxygen produces 448 ml of carbon dioxide at STP. From the information, calculate the
i) mass of carbon dioxide formed
ii) mass of element carbon in carbon dioxide
iii) mass of hydrogen in hydrocarbon.
Answer:
i) Gram molecular weight of carbon dioxide = 12 + 2 × 16g = 44 g.
22.4 litres of CO₂ at STP weighs 44 g.

The weight of 448 ml of CO₂ at STP = \(\frac{44}{22400}\) × 448 = 0.88 g

ii) 44g of CO₂ contains 12 g of carbon.
The amount of carbon present in 0.88 g of CO₂ = \(\frac{12}{44}\) × 0.88 = 0.24 g

iii) Weight of hydrogen = Weight of hydrocarbon – Weight of carbon
= 0.29 – 0.24 = 0.05 g

Question 36.
Calculate the weights of carbon dioxide and water that will be obtained by completely burning 0.25 g of an organic compound having molecular formula C₄H₄O₄.
Answer:

Question 37.
What weight of sulphuric acid will be required to completely dissolve 3g of magnesium carbonate? Calculate the volume of carbon dioxide evolved at STP.
Answer:

Question 38.
If 16.4 g of calcium nitrate is heated : (a) Calculate the volume of Nitrogen dioxide obtained at STP and (b) The mass of calcium oxide obtained.
Answer:

Question 39.
Give some examples for corroded and non-corroded metals and give the reasons for non-corrosion of metals.
Answer:
Examples to corroded metals :

1. Iron
2. Copper
3. Silver

Reactions:
4Fe + 3O₂ → 2Fe₂O₃
2Cu + O₂ → 2CuO
4Ag + 2H₂S + O₂ → 2Ag₂S + 2H₂O

Exmples to non-corroded metals :

1. Gold
2. Platinum
3. Brass
4. Bronze
5. Steel

Reasons:

1. Gold and platinum metals do not react with oxygen and have resistance to corrosion.
2. Brass, bronze and steel are alloys. So they have a quality of resistance to corrosion.

Question 40.
Iron is a corroded metal. Through alloying we can prevent corrosion. Justify.
Answer:

• Alloying is a very good method of improving properties of metal.
• Generally pure form of iron is very soft and stretches easily when hot.
• When iron is mixed with carbon, nickel and chromium, an alloy stainless steel is obtained.
• The stainless steel is hard and does not rust.
• So, through alloying iron we can prevent corrosion.

Question 41.
“Through alloying corrosion can be prevented.” For the justification pose some questions.
Answer:

1. What is alloying?
2. How are alloys prepared?
3. What are the qualities of alloys?
4. Which quantity prevents the corrosion in alloys?
5. Can we prevent corrosion by alloying a metal ?

9th Class Physical Science6th Lesson Chemical Reactions and Equations 4 Marks Important Questions and Answers

Question 1.
Write the balanced chemical reaction for the following and identify the type of reaction in each case.
A) Magnesium_(s) + Iodine_(g) → Magnesium iodide_(s)
B) Zinc_(s) + Hydrochloric acid_(aq) → Zinc chloride_(aq) + Hydrogen_(g)
Answer:
A) Magnesium_(s) + Iodine_(g) → Magnesium iodide_(s)
Mg + I₂ → Mgl₂. This reaction is chemical combination.

B) Zinc_(s) + Hydrochloric acid_(aq) → Zinc chloride_(aq) + Hydrogen_(g)
Zn + 2HCl → ZnCl₂ + H₂ ↑
This reaction is chemical displacement.

Question 2.
Write an activity to each of the following chemical reaction.
A) Photo chemical reaction
B) Chemical displacement reaction.
Answer:
A) Photo chemical reaction :
1) Take a pinch of Silver Bromide in a watch glass and it in the presence of sunlight.
2) Silver Bromide decomposes to silver and Bromine in sunlight.

B) Chemical displacement:

1. Take a small quantity of zinc dust in a conical flask and add some drops of dilute hydrochloric acid slowly.
2. Immediately we can observe the gas bubbles coming out from the solution.
3. This is because zinc reacts with dilute hydrochloric acid and liberates hydrogen gas.
   Zn + 2HCl → ZnCl₂ + H₂ ↑
4. In this reaction the element zinc has displaced hydrogen from hydrochloric acid.

Question 3.
Why should we balance a chemical equation ? Take any one chemical equation and explain the procedure of balancing it.
Answer:
Chemical reactions obey law of conservation of mass. So the total number of atoms of each elements in the reactants must be equal to the total number of atoms of each element in the

products. So we should have to balance chemical equation.
Eg : H₂ + O₂ → H₂O

Step -1 : Unbalanced equation = H₂ + O₂ → H₂O
Step – 2 : Compare no.of atoms of each element on both sides.

No.of 0 atoms balancing – H₂ + O₂ → 2H₂O
No.of H atoms balancing – 2H₂ + O₂ → 2H₂O

Step – 3 : The above equation is balanced and write the coefficients in the smallest ratio. 2H₂ + O₂ → 2H₂O

Step – 4 : Verify above equation for balancing of atoms each element on both sides. Hence the equation is balanced.
∴ 2H₂ + O₂ → 2H₂O

Question 4.
Balance the following chemical equations.
i) Zn_(s) + Ag NO_3(aq) → Zn (NO₃)_2(aq) + Ag_(s)
ii) Fe₂O_3(s) + C_(s) → Fe_(s) + CO_2(g)
iii) Ag_(s) + H₂S_(g) → Ag₂S_(s) + H₂O_(l)
iv) Cu_(s) + O_2(g) → CuO_(g)
Answer:
i) Zn_(s) + Ag NO_3(aq) → Zn (NO₃)_2(aq) + Ag_(s)
Balanced equation : Zn_(s) + 2 AgNO_3(aq) → Zn (NO₃)_2(aq) + 2Ag_(s)

ii) Fe₂O_3(s) + C_(s) → Fe_(s) + CO_2(g)
Balanced equation : 2Fe₂O_3(s) + 3C_(s) → 4Fe_(s) + 3CO_2(g)

iii) Ag_(s) + H₂S_(g) → Ag₂S_(s) + H₂O_(l)
Balanced equation :
We cannot balance the equation because atom ‘O’ does not exist in the reactants.

iv) Cu_(s) + O_2(g) → CuO_(s)
Balanced equation :
2Cu_(s) + O_2(g) → 2CuO_(s)

Question 5.
Write the equation for the reaction of zinc with hydrochloric acid and balance the equation. Find out the number of molecules of hydrogen gas produced in this reaction, when 1 mole of HCl completely reacts at S.T.P.
[Gram molar volume is 22.4 liters at S.T.P., Avogadro’s number is 6.023 × 10²³]
Answer:
Zn + 2HCl → ZnCl₂ + H₂

• In the reaction 2 moles of HC/ produce 1 mole of H₂.
• If 1 mole of HCl participate in reaction, Vi mole of hydrogen will be produced.
• 1 mole of H₂ gas contains 6.023 × 10²³ molecules at STP.
  Number of molecules in ½ mole of H₂ gas = 6.023 × 10²³ × ½ = 3.011 × 10²³

Question 6.
Write the examples for chemical combination.
Answer:
1) Magnesium burns in oxygen to form magnesium oxide.
2 Mg + O₂ → 2 MgO

2) When coal is burnt in oxygen, carbon dioxide is produced.
C + O₂ → CO₂ + Q (cheat energy)

3) Slaked lime is prepared by adding water to quick lime.
CaO + H₂O → Ca(OH)₂ + Q (heat energy)

4) When hydrogen reacts with oxygen, it gives water.
H₂ + O₂ → 2 H₂O

5) Hydrochloric acid is obtained by adding hydrogen to chlorine.
H₂ + Cl₂ → 2 HCl

6) Magnesium reacts with iodine to magnesium iodide.
Mg + l₂ → Mgl₂

7) Sodium reacts with chlorine to form sodium chloride.
2 Na +Cl₂ → 2 NaCl

8) Iron reacts with oxygen to form haematite.
Fe + 3O₂ → 2 Fe₂O₃

Question 7.
Write examples for chemical decomposition reaction.
Answer:
1) Calcium carbonate on heating decomposes to calcium oxide and carbon dioxide.

2) On electrolysis, water decomposes to water and hydrogen.
2 H₂O → 2 H₂ + O₂

3) Silver bromide decomposes to silver and bromine in sunlight.
2 AgBr → 2 Ag + Br₂

4) Silver chloride decomposes to silver and chlorine.
2 AgCl → 2 Ag + Cl₂

5) Glucose decomposes to ethanol and carbon dioxide.
C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂

6) Sodium bicarbonate decomposes to sodium carbonate, water and carbon dioxide.
2 NaHCO₃ + Heat → Na₂CO₃ + H₂O + CO₂

7) On heating lead nitrate decomposes to lead oxide, oxygen and nitrogen dioxide.

Question 8.
Write the examples for chemical displacement reaction.
Answer:
1) Zinc pieces react with dilute hydrochloric acid and liberate hydrogen gas
Zn + 2 HCl → ZnCl₂ + H₂

2) Iron reacts copper sulphate to form iron sulphate and copper.
Fe + CuSO₄ → FeSO₄ + Cu

3) Zinc is mixed with silver nitrate to form zinc nitrate and silver.
Zn + 2 AgNO₃ → Zn(NO₃)₂ + 2Ag

4) Lead reacts with copper chloride to form lead chloride and copper.
Pb + CuCl₂ → PbCl₂ + Cu

5) Sodium reacts with water to form sodium hydroxide and hydrogen.
2 Na + 2H₂O → 2 NaOH + H₂

6) Aluminium reacts with copper chloride to form aluminium chloride and copper.
2 Al + 3 CuCl₂ → 2 AlCl₃ + 3 Cu

7) Zinc reacts with sulphuric acid to form zinc sulphate and hydrogen.
Zn + H₂SO₄ → ZnSO₄ + H₂

Question 9.
Write examples for chemical double displacement reaction.
Answer:
1) Sodium sulphate solution on mixing with barium chloride solution forms a white precipitate of barium sulphate and soluble sodium chloride.
Na₂SO₄ + BaCl₂ → BaSO₄ + 2 NaCl

2) Sodium hydroxide reacts with hydrochloric acid to form sodium chloride and water.
NaOH + HCl → NaCl + H₂O

3) Sodium chloride spontaneously combines with silver nitrate in solution giving silver chloride precipitate.
NaCl + AgNO₃ → AgCl + NaNO₃

4) Mix lead nitrate solution and potassium iodide solution to form a yellow precipitate of lead iodide and potassium nitrate.
Pb(NO₃)₂ + 2Kl → PbI₂ + 2KNO₃

5) Calcium hydroxide reacts with nitric acid to form water and calcium nitrate.
Ca(OH)₂ + 2 HNO₃ → 2H₂O + Ca(NO₃)₂

6) Magnesium chloride reacts with potassium hydroxide to form magnesium hydroxide and potassium chloride.
MgCl₂ + 2 KOH → Mg(OH)₂ + 2 KCl

Question 10.
Balance the following equations.
1) Al(OH)₃ → Al₂O₃ + H₂O
2) NH₃ + CuO → Cu + N₂ + H₂O
3) Al₂(SO₄)₃ + NaOH → Al(OH)₃ + Na₂SO₄
4) HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + H₂O
5) NaOH + H₂SO₄ → Na₂SO₄ + H₂O
6) BaCl₂ + H₂SO₄ → BaSO₄ + HCl
Answer:

1. 2 Al(OH)₃ → Al₂O₃ + 3 H₂O
2. 2 NH₃ + 3 CuO → 3 Cu + N₂ + 3 H₂O
3. Al₂(SO)₃ + 6 NaOH → 2 Al(OH)₃ + 3 Na₂SO₄
4. 2 HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2 H₂O
5. 2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O
6. BaCl₂ + H₂SO₄ → BaSO₄ + 2 HCl

Question 11.
How many types of chemical reactions are there? Explain with examples.
Answer:
There are four types of chemical reactions. They are :
i) Chemical combination :
A chemical reaction in which two or more substances combine together to form a new substance is called chemical combination.
e.g.: 2H₂ + O₂ → 2H₂O
CaO + H₂O → Ca(OH)₂

ii) Chemical decomposition :
The reaction in which a compound breaks up into two or more simpler substances are known as decomposition reaction. These reactions are generally carried out by means of heat, light, electricity or catalysts.
e.g.: CaCO₃ → CaO + CO₂
2Pb(NO₃)₂ → 2 PbO + 4NO₂ + O₂

iii) Chemical displacement:
The chemical reaction in which one element takes the place of another element in a compound is called displacement reaction.
In these reactions, an atom or group of atoms in a molecule is replaced by another atom or a group of atoms.
e.g.: Zn + 2HCl → ZnCl₂ + H₂ ↑
Zn + CuSO₄ → ZnSO₄ + Cu

iv) Chemical double displacement: The reaction in which two compounds react to form two other compounds by mutual exchange of their ions is called double displacement reaction.
e.g.: NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)
NaNO_3(aq) + AgCl_(aq) → AgNO_3(s) + NaCl_(aq)

Question 12.
Balance the following chemical equations.
a) Na₂SO₄ + BaCl₂ → BaSO₄ + NaCl
b) Al₄C₃ + H₂O → CH₄ + Al(OH)₃
c) Pb(NO₃)₂ → PbO + NO₂ + O₂
d) Fe₂O₃ + Al → Al₂O₃ + Fe
Answer:
a) Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl
b) Al₄C₃ + 12H₂O → 3CH₄ + 4Al(OH)₃
c) 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂
d) Fe₂O₃ + 2Al → Al₂O₃ + 2Fe

Question 13.
How can we make a chemical equation information?
Chemical equations can be made more informative by expressing following characteristics of the reactants and products.
1. Expressing the physical state :
The different states, i.e. gaseous, liquid and solid states are represented by the notations (g), (l) and (s) respectively. If the substance is present as a solution in water the word aqueous is written.

2. Expressing the heat changes :
Q is heat energy which is shown with plus ’+’ sign on product side for exothermic reactions and minus sign on product side for endothermic reactions.
e.g.: a) C_(s) + O_2(g) → CO_2(g) + Q (exothermic reaction)
N_2(g) + O_2(g) → 2NO_(g) ” Q (endothermic reaction)

3. Expressing the gas evolved :
If a gas evolved in a reaction, it is denoted by an upward arrow ↑ or (g).
Zn_(s) + H₂SO_4(aq) → ZnSO_4(aq) + H_2(g)↑

4. Expressing precipitate formed: If a precipitate is formed in the reaction, it is denoted by downward arrow ↓.
AgNO_3(aq) + NaCl_(aq) → AgCl_(s) ↓ + NaNO_3(aq)

Question 14.
Give daily life examples of oxidation.
Answer:
Daily life examples :

1. Combustion of fuels.
2. Corrosion of metals.
3. Change of colour of fruits like apples, bananas, when they are cut.
4. Burning of crackers.
5. Rancidity of food material.
6. During rainy season the power supply to our home from the electric pole will be interrupted due to formation of metal oxide layer on the electric wire.
7. Rising of dough with yeast depends on oxidation of sugars to carbon dioxide and water.
8. Bleaching of coloured objects using moist chlorine.
9. Respiration.

Question 15.
What is the information giyen by balanced chemical equation?
Answer:

• A chemical equation gives information about the reactants and products through their symbols and formulae.
• It gives the ratio of molecules of reactants and products.
• As molecular masses are expressed in unified masses, the relative masses of reactants and products are known from the equation.
• If the masses are expressed in grams, then the equation also gives the molar ratios of reactants and products.
• If gases are involved, we can equate the masses to their volumes.
• Using molar mass and Avagadro’s number we can calculate the number of molecules and atoms of different substances from the equation.

Question 16.
Write the balanced equation and identify the type of reaction.
1) Magnesium Hydroxide_(aq) + Nitric Acid_(aq) → Magnesium Nitrate_(aq) + Water_(l)
2) Magnesium_(s) + Carbon Monoxide_(g) → Magnesium Oxide_(s) + Carbon_(g)
3) Barium Chloride_(aq) + Sodium Sulphate_(aq) → Barium Sulphate_(s) + Sodium Chloride_(aq)
4) Sodium Nitrate_(s) → Sodium Nitrite_(s) + Oxygen_(g)
Answer:
1) Mg(OH)_2(aq) + 2 HNO_3(aq) → Mg(NO₃)_2(aq) + 2 H₂O_(l)
It is both double displacement and neutralisation reaction.

2) Mg_(s) + CO_(g) → MgO_(s) + C_(s)
It is a redox reaction in which magnesium is oxidised and carbon monoxide is reduced.

3) BaCl_2(aq) + Na₂SO_4(aq) → BaSO_4(s) ↓ + 2 NaCl_(aq)
It is both precipitation and double displacement reaction.

4) 2 NaNO_3(s) → 2 NaNO_2(s) + O_2(g)
It is both endothermic and decomposition reaction.

Question 17.
We write symbol of water as H₂O. State why we should not write it as HO₂.
Answer:
1) The hydrogen atom has only T electron in its outermost shell, so it needs 1 more electron to achieve the stability. ‘2’ electrons are required to get inert gas electronic configuration.

2) The oxygen atom has ‘6’ electrons in its outermost shell and it needs ‘2’ more electrons to compare the stability. ‘8’ electron arrangement of inert gas is neon.

In the water molecule, central oxygen atom has two pairs of unshared electrons which have not been utilised in the formation of bonds.

So, we write symbol of water as H20 and we should not write it as HO₂.

Question 18.
Latha took some quantity of powder of a substance in a test tube. Heated it with spirit lamp. A gas was liberated. She sent the gas into another test tube. The colour of solution in the second test tube turned into milk white.
Answer the following questions :
a) Which substance was heated?
b) Which gas was liberated?
c) What was the solution taken in second test tube?
d) Which type of chemical reactions involved the experiment?
Answer:
a) The substance is calcium carbonate.
b) The gas liberated is carbon dioxide.
c) The solution taken in the second test tube was solution of slaked lime.
d) Two types of reactions took place in this experiment, i.e. decomposition and double displacement.
CaCO_3(s) → CaO_(s) . CO₂ ↑
Ca(OH)_2(aq) + CO_2(g) → CaCO_3(s) + H₂O

Question 19.
A light yellow colour substance (some quantity) on a watch glass is put in the sunlight. It changes into grey colour substance.
a) What is the light yellow colour substance?
b) What is the grey colour substance?
c) Which type of chemical reaction it is?
d) Write the chemical equation for the reaction.
Answer:
a) The light yellow colour substance is silver bromide.
b) The grey colour substance is silver.
c) The type of chemical reaction is photochemical reaction.

Question 20.
Heat is liberated in the reactions where water is added to calcium oxide and hydrochloric acid added to zinc pieces.
Rarnu says that they are same type of chemical reactions. Eswar’s opinion is that the reactions are not similar. What is the basis of Eswar’s opinion ? Write equations for the above reaction.
Answer:
Eswar’s thinking is correct. Although both are exothermic reactions, the type of chemical reactions is different.
Case (i) : When water is added to calcium oxide it forms calcium hydroxide. It is an example for combination reaction.
CaO_(s) + H₂O → Ca(OH)_2(aq)

Case (ii) : When hydrochloric acid is added to zinc pieces it would liberate hydrogen gas which is an example for displacement reaction.
Zn_(s) + HCl_(aq) → ZnCl_2(aq) + H_2(g) ↑

Question 21.
We see many combustion and oxidation reactions in our daily life. Among them every combustion reaction is an oxidation reaction. But not all the oxidation reactions are combustion reactions. Do you agree or disagree with the statement? Explain with proper reasons.
Answer:
1) Yes, I agree with the statement because combustion reaction is nothing but burning of a substance in the presence of oxygen so it is an oxidation reaction. So every combustion reaction is an oxidation reaction.

2) Whereas some reactions which do not require burning still they are oxidation reactions.
Eg :

1. Corrosion of metals.
2. Change of colour of fruits like apples, bananas when they are cut.
3. Rancidity of food materials.
4. Respiration.
5. Bleaching of coloured objects using moist chlorine.
   So all oxidation reactions are not combustion reactions.

Question 22.
Explain the following :
1) What happens when iron filings are added to zinc sulphate solution?
2) What happens when solid silver nitrate is added to solid sodium chloride?
Answer:
1) No reaction takes place because zinc is more reactive than iron. So iron cannot displace zinc from its salt solution.

2) No reaction takes place because in solid state silver nitrate as well as sodium chloride is unable to dissociate into constituent ions. So exchange of ions is not possible.

Question 23.
Why is steel not used for surgical equipment and what is the material used for surgical equipment? Why is that material used for surgical equipment?
Answer:
1) Steel undergoes rusting when It exposed to air. So it is not useful for preparation of surgical equipment. If we use it in surgical equipment it may cause septic of wound.

2) The material used for surgical equipment is stainless steel which is an alloy of iron, carbon, nickel and chromium. Chromium does not easily react with oxygen. So, addition of chromium makes the stainless steel free from rusting. Therefore it is used in surgical equipment.

Question 24.
Give reasons for the following.
1) Why do we add salt to water during electrolysis of water?
2) Why does hydrogen gas put off the burning splinter and it burns with blue flame?
3) Why does carbon dioxide turn lime water milky?
Answer:
1) Pure water is a bad conductor of electricity. By adding a small amount of salt the solution becomes conductor of electricity.

2) Hydrogen does not support combustion. So it puts off the burning splinter and also it is combustible. So burns with blue flame.

3) When we pass carbon dioxide through lime water it turns into milky because lime water (calcium hydroxide) reacts with carbon dioxide and forms a white milky substance, i.e. calcium carbonate.

Question 25.
A student was given the following substances and was asked to show types of chemical reactions through experiment. Write how he would have done that. Copper sulphate solution, barium chloride solution, ferrous sulphate crystals, iron nails, calcium oxide, water.
Answer:
Given chemicals are CuSO₄ solution, BaSO₄ solution, iron nails (Fe), Ferrous sulphate (FeSO₄), Calcium oxide (CaO) and water (H₂O).

i) Chemical combination :
Chemicals chosen : CaO, H₂O
CaO + H₂O → Ca(OH)₂

When we add water to calcium oxide it produces Calcium hydroxide. This is an example for chemical combination.

ii) Chemical decomposition:
Chemicals chosen : Ferrous sulphate

When we heat FeSO₄ it dissociates into Ferric oxide, Sulphur dioxide and Sulphur trioxide respectively.

iii) Chemical displacement :
Chemicals chosen : Iron nail – CuSO₄
Fe + CuSO₄ → FeSO₄ + Cu ↓

When iron nail is placed in CuSO₄ solution, the solution turns into light green due to formation of FeSO₄ and reddish brown deposit of Copper.

iv) Double displacement:
Chemicals chosen : CuSO₄ solution and BaCl2 solution.
CuSO_4(aq) + BaCl_2(aq) → BaSO_4(s) ↓ + CuCl_2(s)

When CuSO₄ is mixed with BaCl₂ solution it forms white precipitate of BaSO₄ and aqueous solution of Copper (II) Chloride.

This is an example for double displacement reaction.

Question 26.
Take two beakers and prepare lead nitrate aqueous solution and potassium iodide aqueous solutions. What are the colours of the solutions? Now mix them in another beaker. What happens? What type of chemical reaction it is? What are products?
Answer:
The colours of the solutions are white or colourless. When lead nitrate solution is mixed with potassium iodide solution we will get yellow precipitate of lead iodide. This is double displacement reaction. The products are lead iodide and potassium nitrate.
Pb(NO₃)_2(aq) + 2Kl_(aq) → Pbl_2(s) + 2KNO_3(aq)

Question 27.
Observe the following equation which shows the action of heat on Calcium Nitrate
2 Ca(NO₃)₂ → 2 CaO + 4 NO₂ + O₂
a) How many moles of NO₂ are formed when a mole of 2 Ca(NO₃)₂ is decomposed?
b) What is the volume of NO₂ produced when 164 gm of Ca(NO₃)₂ is heated at constant temperature and pressure?
c) Calculate the mass of Calcium Oxide formed when 82 gm of Ca(NO₃)₂ is heated.
d) What is the quantity of Ca(NO₃)₂ required to produce 5 moles of gaseous products?
Answer:
Given balanced equation is
a) From the balanced equation 2 moles of Ca(NO₃)₂ releases 4 moles of NO₂.
b) Molecular weight of Ca(NO₃)₂ and CaO respectively are 164 and 56.
From the equation at STP 2 × 164 g. of Ca(NO₃)₂ releases 4 × 22.4 litres of NO₂.
At similar conditions 164 g. of Ca(NO₃)₂ releasing NO₂ in litres is
= \(\frac{164}{2 \times 164}\) × 22.4 × 4 = 2 × 22.4 = 44.8 liters.

c) From the balanced equation 164 g. of Ca(NO₃)₂ decomposes and forms 112 g. of CaO. Similarly 82 g. of Ca(NO₃)₂ decomposes and forms
\(\frac{82}{164}\) × 112 = 56 g. of CaO.

d) From the above balanced equation 2 moles of Ca(NO₃)₂ releases 5 moles of gaseous products.
∴ The mass of Ca(NO₃)₂ required is 2 × 164 = 328 g.

Question 28.
Zn + HCl → ZnCl₂ + H₂
Calculate amount of zinc required to release 500g of hydrogen. (Zn = 65 U, H = 1 U, Cl = 35.5 U are the atomic masses).
Answer:
The balanced chemical equation is
Zn + 2 HCl → ZnCl₂ + H₂
65 U + (2 × 36.5) U → (65 + 2 × 35.5) U + 2 U
65 g + 73 g → 136 g + 2g
As per the balanced equation
65 g Zinc is reacting with hydrochloric acid to produce 2g of Hydrogen.

Question 29.
Calculate the volume, mass and number of molecules of carbon dioxide when 104 g of acetylene (C₂H₂) burnt in air. (Atomic masses of C = 12 U, H = 1 U, O = 16 U).
Answer:
The balanced chemical equation is
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
2 × (2 × 12U + 2 × lU) + 5 × (2 × 16U) → 4 × (12U + 2 × 16U) + 2(2 × 1U + 16U)
52 g + 160g → 176 g + 36 g
52 g of acetylene when burnt in air produces 176 g carbon dioxide.
The amount of carbon dioxide produced when 104 g acetylene burnt

At S.T.P. 1 gram molar mass of any gas occupies 22.4 litres.
So 44g carbon dioxide occupies 22.4 litres volume.
The volume occupied by 352 g carbon dioxide = \(\frac{352}{44}\) × 22.4 = 8 × 22.4 = 179.2 litres.
44 g of carbon dioxide i.e., 1 mole of CO₂ contains 6.02 × 10²³ molecules. So the number of molecules present in 352 g of carbon dioxide
352
= \(\frac{352}{44}\) × 6.02 × 10²³ = 8 × 6.02 × 10²³
= 4.816 × 10²⁴ molecules.

AP State Syllabus AP Board 9th Class Physical Science Important Questions Chapter 1 Motion.

AP State Syllabus 9th Class Physical Science Important Questions 1st Lesson Motion

9th Class Physical Science 1st Lesson Motion 1 Mark Important Questions and Answers

Question 1.

From the above data, what can you say about the motion of the object?
Answer:
The object is in the uniform motion and its speed is constant.

Question 2.
Frame any one question to understand Newton’s third law of motion.
Answer:
How does rocket engine work?

9th Class Physical Science 1st Lesson Motion 2 Marks Important Questions and Answers

Question 1.
Distinguish between speed and velocity. jawjMB
Answer:

Question 2.
What happens to the speed and the direction of motion of a ball rolling down an inclined plane?
Answer:

1. Speed of the rolling ball changes.
2. But the direction of motion remains constant.

Question 3.
A motor cyclist drive from A to B with uniform speed of 30 km/hour and returns back with a speed of 20 km/hour. Find the average speed.
Answer:
Speed of the motor cyclist from A to B = 30 km/hr.
Speed of the motor cyclist from B to A = 20 km/hr.

Question 4.
A train of length 100 m. is moving with a constant speed of 10 m/s. Calculate the- time taken by the train to cross the electric pole.
Answer:
Length of the train = 100 m; Speed of the train = 10 m/s
Time taken to cross the elctrical pole (t) = s/v = \(\frac{100}{10}\) =10 sec.
[∵ Distance travelled while crossing an electrical pole = Length of the train]

Question 5.
Calculate the average speed of “Ussahi Bolt” who sprints 100m in 9.81 sec. during 2016 Rio Olympics to win Gold medal.
Answer:
Distance covered by Ussain Bolt = 100 m
time = 9.81s.

Question 6.
Explain the terms in the formula v = u + at.
Answer:
v = u + at
u = initial velocity
v = final velocity
a = acceleration
t = time

9th Class Physical Science 1st Lesson Motion Important Questions and Answers

9th Class Physical Science 1st Lesson Motion 1 Mark Important Questions and Answers

Question 1.
Define uniform acceleration.
Answer:
Acceleration is uniform when in equal intervals of time, equal changes in velocity occur.

Question 2.
Write an equation of motion to find the distance travelled when initial velocity, time, and acceleration are given.
Answer:
s = ut + \(\frac{1}{2}\) at²

Question 3.
Draw a displacement vector from Visakhapatnam to Hyderabad to the following diagram.

Answer:

Question 4.
Difine acceleration.
Answer:
The rate of change of velocity in an object is known as its acceleration.

Question 5.
What is the key difference between distance and displacement?
Answer:
Distance is the length of the path traversed by an object in a given time interval and displacement is the shortest distance covered by the object in a specific direction.

Question 6.
The distance travelled by a particle in time’t’ is given by s = (2.5 m/s²) t². Find the average speed of the particle during the time ‘O’ to ‘5’ sec.
Answer:
The distance travelled by the particle from 0 to 5 sec is s = 2.5 m/s² × 5² = 62.5 m

Question 7.
A table clock has its minutes hand 4 cm long. Find the average velocity of the tip of the minute hand between 6.00 AM to 6.30 AM.
Answer:
The minute hand of a clock comes into a straight line from 6.00 AM to 6.30 AM.
Hence the displacement of the minute hand S = 2 x 4 = 8 cm
Time = 30 minutes = 1800 sec.

Question 8.
A ball is thrown up with an initial speed of 4m/sec. Find the maximum height reached by the body.
Answer:

Question 9.
The adjacent distance – time graph indicates
A) A particle travels constantly along X-axis.
B) Particle is at rest.
C) The velocity of the particle increases up to a time t0 and then remains constant.
D) The particle travels up to time t(( with constant velocity and then stops.

Answer:
C

Question 10.
A man used his car. The initial and final odometer readings are 4849 and 5549 respectively. The journey time is 70h. What is average speed of the journey?
Answer:
Distance covered = 5549 – 4849 = 700 km. ; Time = 70 h.

9th Class Physical Science 1st Lesson Motion 2 Marks Important Questions and Answers

Question 1.
An object is moving in a circular path of radius 7 m. What is the distance and displacement of an object after one revolution ?
Answer:
Radius (r) = 7 m
22
Distance covered by object = 2πr = 2 × \(\frac{22}{7}\) × 7 = 44 m
The object come back to original position displacement of the object is 0.

Question 2.
An object reaches other end of a diameter in a circular path of radius 7 m in 7s. Find speed and velocity of the object?
Answer:

Question 3.
A person moves 3 km towards east and turned toward north and travelled a distance of 4 km. Find total distance and total displacement.
Answer:

Question 4.
An object completes 1/4th revolution of a circular path of radius r. Then find the ratio of distance and displacement.
Answer:

Question 5.
a) What is a vector? Give example.
b) What is a scalar? Give example.
Answer:
a) Vector :
The physical quantity which is specified with magnitude and direction is called a vector.
Eg : Displacement, velocity are vectors.

b) Scalar :
The physical quantity which does not require any direction for its specification is called ‘scalar’.
Eg : Distance, time are scalars.

Question 6.
How does a vector represents? Explain.
Answer:

1. A vector can be represented as a directed line segment.
2. It’s length indicates magnitude and arrow indicates it’s direction.
3. Vector is represented by an arrow as shown in the figure.
   
   Here point ‘A’ is called tail and point ‘B’ is called head.

Question 7.
Bhavan did not recognise the difference between 40 meters displacement and distance. Then Anitha ex- plained the difference by asking some questions to Bhavan. What would be those questions?

Note : She used this diagram.
Answer:
By using the diagram Anitha asked the following questions :

1. What is the distance between A and D?
2. Which is the shortest distance from A to D?
3. What is the direction of an object if it travelled in the route A → B → C → D? Can you determine its direction?
4. On which direction the distance is short from A to D? Measure it.
5. Which path has specified direction?
   Either A → B → C → D or A → D?
6. Which path has specified magnitude?
   Either A → B → C → DorA → Dor Both?
7. Which is the shortest, distance among the two paths?
8. If the displacement has both magnitude and specified direction, which path is considered as a displacement?
9. What is the difference between distance and displacement?

Question 8.
What happens to the direction of velocity and acceleration when the
i) speed of an object increases?
ii) speed of an object decreases?
Answer:
i) If the speed of an object increases, the direction of velocity and acceleration are one and the same.
ii) If the speed of an object decreases, the direction of velocity and acceleration are in opposite directions. Here, at a certain instant the speed becomes zero.

Question 9.
Draw a displacement vector and a velocity vector to the given path of motion of a body.

Answer:

Question 10.
Find the acceleration of a bus if its speed increases from 0 m/s to – 600 m/s in 1 minute?
Answer:

The acceleration of the bus is 10 m/sec².

9th Class Physical Science 1st Lesson Motion 4 Marks Important Questions and Answers

Question 1.
Distinguish between distance and displacement.
Answer:

Question 2.
Explain the terms uniform and non-uniform motions.
Answer:
Uniform motion :

1. If a body covers equal distances in equal intervals of time, assuming the direction of motion is constant, then the motion is said to be uniform motion.
2. The motion of a body is said to be uniform when its velocity is constant.
3. The s -1 graph for a uniform motion is a straight line.
4. Ex : Motion of hands in a clock.

Non-uniform motion :

1. If a body covers unequal distances in equal intervals of time, assuming the direction of motion is constant, then the motion is said to be non-uniform motion.
2. A body is said to be in non-uniform motion if its velocity changes with time.
3. The s -1 graph for non-uniform motion is not a straight line.
4. Ex : Motion of a car/bus between two stations on a road.

Question 3.
Derive equations of uniform accelerated motion.
Answer:
1) Let ‘u’ be the velocity at the time t = 0 and V be the velocity at the time’t’ and let ‘s’ be the displacement covered by the body during time’t’ as shown in figure.
2) From the definition of uniform acceleration,

Question 4.
Describe the graphical method to calculate the instantaneous speed at a given point.
Answer:

1. Consider a car moving along a straight road with varying speed.
2. Take time elapsed on X – axis and distance cov¬ered on Y – axis and plot a graph for its motion at regular intervals of time.
3. A general case of motion with varying speed is shown in the figure.
4. The average speed during the time interval from
   
5. Then we calculate average speed for a very short time interval encompassing the time at an instant t3 which is so short interval, that value of average speed would not change materially if it was made even shorter.
6. The instantaneous speed is represented by the slope of the curve at a given instant of time.
7. The slope can be found by drawing a tangent to the curve at that point.
8. The slope of the curve gives speed of the car at that instant.

Question 5.
What is difference between speed and velocity?
Answer:

Question 6.
Derive s = ut + \(\frac{1}{2}\) at²
Derive one equation for displacement of a body which is in the uniform acceleration.
Answer:
1) Let u be the initial velocity,
v be the velocity at time ‘t’.
a be the acceleration,
s be the displacement

Question 7.
A car travelled from A to E station in 3 minutes. Viewing path given below in the figure. find
i) distance
ii) displacement
iii) speed
iv) velocity

Answer:
i) Distance = AB + BC + CD + DE = 540 + 460 + 540 + 260 = 1800 m

ii) Displacement = Distance between A and E

Question 8.
Give an example to each situation in daily life.
i) Speed changes when direction remains constant.
ii) Direction of motion changes when speed remains constant.
iii) Speed and direction simultaneously change.
Answer:
i) The palk taken by a ball which was released from the top of the inclined plane. Here the direction of motion remains constant but velocity gradually increases.

ii) Whirl a stone which is tied to the end of the string continuously.
Here it’s direction of motion changes but speed remains constant.

iii) The path taken by a stone, which was thrown into the air.
Here speed and direction of the motion both changes simultaneously.

AP State Syllabus AP Board 9th Class Physical Science Important Questions Chapter 2 Laws of Motion

AP State Syllabus 9th Class Physical Science Important Questions 2nd Lesson Laws of Motion

9th Class Physical Science 2nd Lesson Laws of Motion 1 Mark Important Questions and Answers

Question 1.
Calculate the momentum of a fast moving ball with a velocity 2.2 m/s of mass 6 kg.
Answer:
mass (m) = 6 kg
velocity (v) = 2.2 m/s
momentum = mv = 6 × 2.2 = 13.2 kg.m.s^-1
The momentum of a fast moving ball is 13.2 kg ms^-1.

Question 2.
What is Newton’s first law of motion?
Answer:
Every object will remains at rest or in a state of uniform motion, unless compelled to change its state by the action of a net force.

Question 3.
From a fixed height, two eggs are dropped such that one falls on a concrete floor and the other on a cushioned pillow. The egg falling on the concrete floor breaks and the one falling on the pillow does not break. Explain why in terms of momentum.
Answer:

• The change of momentum of an egg takes place in shorter time in case of concrete surface.
• The change of momentum of the egg takes place in longer time in case of cushioned pillow.
• Hence, egg does not break on cushioned pillow.

9th Class Physical Science 2nd Lesson Laws of Motion 2 Marks Important Questions and Answers

Question 1.
A vehicle of mass 300 kg travels at a velocity of 90 km/h. Find its momentum.
Answer:
Mass of the vehicle (m) = 300 kg
Velocity of the vehicle (v) = 90 km/h = 90 × \(\frac{5}{18}\) m/s = 25 m/s
Momentum = mv = 300 × 25 = 7500 kg.m.s^-1

Question 2.
“With our day to day experience, we must exert some force on an object to keep it moving”.
i) By which law, you can support the above lines?
ii) Which act on the body ? Either force or net force?
Answer:
i) Newtons first law of motion.
ii) Net force. The position of object is changed due to net force only.

9th Class Physical Science 2nd Lesson Laws of Motion 4 Marks Important Questions and Answers

Question 1.
Fill the following table.

Answer:

Question 2.
What is impulse? Derive the formula for impulse.
Answer:
Impulse :
The change of momentum of an object is called impulse.
Formulae for impulse ⇒ ∆p = F_net∆t.

Derivation :
We can express the second law of motion as

From the above equation we know that the product of net force and interaction time is called impulse of net force (∆p).

Question 3.
State and prove the law of conservation of linear momentum.
Answer:
Law of conservation of momentum : Law of conservation of momentum states, in the absence of a net external force on the system, the momentum of the system remains unchanged.
Explanation :

• Let two marbles with masses nij and m2 travel with different velocities ut and u2 in the same direction along a straight line.
• If u₁ > u₂, they collide each other and the collision lasts for time’t’.
• During collision, each marble exerts force on other marble. [F₁₂ and F₂₁]
• Let v₁ and v₂ be the velocities of the marbles after collision.

Now look at the table.

• The total momentum before collision is equal to total momentum after collision.
• Hence, the total momentum remains unchanged before and after collision.

Question 4.
a) Name the ‘machine’ use to prove Newton’s laws of motion. Draw a neat diagram of it.
(OR)
The masses ‘m ‘ and ‘m2’ are attached to a string and roll over a pully, which is attached to a rigid support. Identify the machine and draw the neat diagram of it.
Answer:
a) Atwood machine.
Answer:

b) Draw FBD of both, masses on it.
Answer:

9th Class Physical Science 2nd Lesson Laws of Motion 1 Mark Important Questions and Answers

Question 1.
Newton’s first law of motion is also called
Answer:
Law of inertia.

Question 2.
State Newton’s second law of motion.
Answer:
The rate of change of momentum of a body directly proportional to the net force acting on it and it takes place in the direction of net force.

Question 3.
What is inertia?
Answer:
The natural tendency of objects to resist a change in their state of rest or of uniform motion is called inertia.

Question 4.
What is momentum?
Answer:
Linear momentum of a body is the product of its mass and velocity.
Momentum (p) = mass (m) × velocity (v)

Question 5.
State Newton’s third law of motion.
Answer:
When an object exerts a force on the other object, the second object exerts a force on the first one with equal magnitude and in opposite direction.

Question 6.
Write the law of conservation of momentum.
ANswer:
The law of conservation of momentum states that in the absence of net external force on the system, the momentum of the system remains unchanged.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Question 7.
What is an impulse?
Answer:
The product of net force and interaction time of a system is called impulse of net force.
Impulse ∆p = F_net. ∆t.

Question 8.
What is the Aristotle’s conclusion about the natural state of an object?
Answer:
Aristotle concluded that the natural state of an earthly object is to be at rest. Hence the object at rest requires no explanation as any moving object naturally comes to rest.

Question 9.
What is the Galileo’s conclusion on the motion of an object?
Answer:
Galileo gave birth to modern science by stating that an object in motion will remain in same motion as long as no external force is applied on it.

Question 10.
What is the effect of mass on inertia?
Answer:
As mass increases, inertia increases.
Mass is a property of an object that specifies how much inertia the object has.

Question 11.
Who proposed laws of motion?
Answer:
Sir Isaac Newton proposed the laws of motion.

Question 12.
Give two daily life examples where you experience inertia.
Answer:

• When the bus which is at rest begins to move suddenly, the person standing in the bus falls backward.
• When you are travelling in a bus the sudden stop of the bus makes you fall forward.

Question 13.
If action is always equal to the reaction, explain how a horse can pull a cart.
Answer:
Horse and cart constitutes a single system. So they cannot form action-reaction pair. Therefore, horse can pull the cart.

9th Class Physical Science 2nd Lesson Laws of Motion 2 Marks Important Questions and Answers

Question 1.
What is momentum? What are its units in S.I. system?
Answer:
Momentum (p) of a body is defined as the product of its mass (m) and velocity (v).
Momentum (p) = mass (m) × velocity (v)
p = mv
S.I. units of momentum are kg. m /sec (or) N – sec.

Question 2.
A mass of 0.5 kg has been suspended to a roof as shown in the figure. What is the force exerted by the rope on the object?

Answer:
The forces acting on the object of mass 0.5 kg are
i) gravitational force (downward) = 0.5 x 9.8 = 4.9 N
ii) Tension in the string T (upward).
The object is at rest. Hence the two forces are equal.
∴ The force exerted by the string on the object is 4.9 N (upwards).

Question 3.
Write few incidents if there were no friction.
(OR)
What happens when there were no friction in this world?
Answer:
If there were no friction,

1. we could not be able to walk on ground.
2. we would not be able to stop a fast moving car.
3. any object could not be able to stay on the ground.

Question 4.
What are the factors influencing acceleration? Explain.
Answer:
Factors influencing acceleration are force and mass.
Force :
As the net force increases, the acceleration increases when mass remains constant.

Mass :
When force is constant, as the mass increases, the acceleration decreases.

Question 5.
Why does a fielder catch a fast moving cricket ball by pulling back his arms while catching it?
(OR)
Why does a cricketer move his hands backwards while catching a fast moving cricket ball?
Answer:

• When he pulls back his arms he experiences a smaller force for a longer time.
• The ball stops only when your arms stop.
• This is to avoid a large impulse.
• If he doesn’t pull his arms back, the ball will hurt him.
• This is due to a larger force for a smaller time.

Question 6.
What is the state of an object when no net force is acting on an object?
Answer:
If the net force acting on an object is zero, the object which is at rest remains at rest or if the object is already moving with a certain velocity it continue to move with same velocity.

Question 7.
Which of the following has more inertia?
a) A rubber ball and a stone of the same size.
b) A bicycle and a train.
c) A five rupees coin and a one-rupee coin.
Answer:
As the mass of the object increases the inertia of the object increases.
a) Stone has more mass than rubber ball. So stone has more inertia.
b) Train has more inertia.
c) 5 rupee coin has more inertia.

Question 8.
Explain why some of the leaves may get detached from a tree, if we vigorously shake its branch?
Answer:
The leaves are at the state of rest. When the tree is vigorously shaken the state of rest is disturbed. So the leaves are detached from the tree due to inertia of rest.

Question 9.
Which would require a greater force – accelerating a 2 kg mass at 5 ms^-2 or a 4 kg mass at 2 ms^-2?
Answer:
Force F = ma
Here we have m₁ = 2kg, a₁ = 5 ms^-2 and m₂ = 4 kg, a₂ = 2ms^-2.
Thus F₁ = m₁a₁
Thus accelerating a 2 kg mass at 5 ms^-2 would require a greater force.

Question 10.
A motor car is moving with a velocity of 108 km/h and it takes 4s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motor car if its mass along with the passengers is 1000 Kg.
Answer:

Question 11.
Explain why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
AnswerL
It is difficult for a fireman to hold a hose which ejects large amounts of water at a high velocity because due to high velocity of water the hose moves backward directiondue to Newton’s third law.

Question 12.
From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 ms^-1. Calculate the initial recoil velocity of the rifle.
Answer:
Mass of rifle m₁ = 4 kg
Mass of bullet m₂ = 50 g = \(\frac{50}{1000}\) = 0.05 kg
Initial velocity of rifle u₁ = 0
Initial velocity of bullet u₂ = 0
Final velocity of rifle (recoil velocity) v₁ = v
Final velocity of bullet v₂ = 35 ms^-1
According to law of conservation of momentum
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
4(0) + 0.05 (0) = 4v + 0.05 × 35
4v = – 1.75
v = – 0.44 m/s

Question 13.
An automobile vehicle has a mass of 1500 kg. What must the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms^-2?
Answer:
m = 1500 kg ;
a = – 1.7 ms^-2
F = ma = 1500 ×- 1.7 = -2550 N
Negative sign indicates that it is a retarding force.

Question 14.
What happens to a person standing in a bus when the bus which is at rest begins to move suddenly?
(OR)
Explain static inertia with an example.
Answer:

• The property of an object at rest will try to remain at rest is called “static inertia”.
• For example, when the bus which is at rest begins to move suddenly, the person standing in the bus falls backward because of static inertia of the body.
• Bus pulls the legs forward while body is in the rest position (static inertia).

Question 15.
Observe the given figure.
a) What happens to the coin, when we pull the paper with a jerk of force?
b) Why?
Answer:

1. The coin falls inside the glass.
2. This is due to static inertia of the coin.
3. This means coin in the state of ‘rest’ before and after pulling the paper. It falls down due to gravity only.

9th Class Physical Science 2nd Lesson Laws of Motion 4 Marks Important Questions and Answers

Question 1.
Give some examples for Newton’s third law of motion from your day-to-day observations.
Answer:
1) Flying of birds :
a) When birds fly, they push the air downward with their wings.
b) The air pushes back the bird in opposite upward direction.
c) These two forces are equal in magnitude and opposite in direction. Hence the bird can fly.

2) Swimming of fish :
a) When a fish swims in water, it pushes back the water.
b) The water pushes the fish with equal but in opposite direction.
c) Hence the fish moves forward.

3) Launching of a rocket:
a) A rocket accelerates by expelling gas at high velocity.
b) The reaction force of the gas on the rocket accelerates the rocket in a direction opposite to the expelled gases.

Question 2.
Derive F = ma.
Answer:
1) Newton’s second law of motion states that the rate of change of momentum of an object is proportional to the net force applied on the object in the direction of net force.

Question 3.
How did Galileo differ the statement of Aristotle ‘The natural state of an earthy object is to be at rest?
(OR)
Write the great experiments of Galileo on inclined planes.
(OR)
How do you appreciate Galileo, the father of modern science?
Answer:

• Galileo came up with two ingenious thought experiments.
• He did his experiments on inclined planes with smooth surfaces.
• He observed that the smoother the surface the farther the ball travelled.
• He extended this ajgument and concluded that if the surface was perfectly smooth, the ball will travel indefinitely until encountered by another object.

• As shown in the figure 1 (a) he observed that when a marble rolls down a slope it picks up speed due to the force of gravity of the earth.
• As shown in the figure 1(b) when the object rolls up an inclined plane, its speed decreases.
• Now let us assume that a marble is moving on a level surface as shown in the figure 1(C), it has no reason to speed up or slow down.
• So, it will continue to move with constant velocity.
• By this experiment, Galileo came to a conclusion which was in contrast to Aristotle’s belief that the state of an earthy object is to be at rest.

Question 4.
Explain Newton’s third Law of motion with an activity.
(OR)
Take two spring balances of equal calibrations. Connect the two spring balances and pull them in opposite direction. Does they show different readings? Why?
Answer:
Newton’s third law of motion : When an object exerts a force on an other object, the second object also exerts a force on the first one which is equal in magnitude but opposite in direction.

Activity:

1. Let us take two spring balances of equal calibrations.
2. Connect the two springs balances as shown in the figure.
3. Pull the balances in opposite direction.
4. Observation : Spring balances shows same readings.
5. Here, one spring balance shows reading of ‘action’ force, and other spring balance shows reading of ‘reaction’ force.
6. Forces are equal, direction are opposite.
7. This shows Newton’s third law of motion.

Question 5.
Take two marbles of m₁, m₂ mass respectively. Make a collision in one second with marbles using u₁, u₂ initial velocities respectively. Take the final velocities as v₁, v₂ You observed the following in the experiment.
m₁ = 10 kg u₁ = 5 m/s v₁ = 15 m/s
m₂ = 100 kg u₂ = 2 m/s -v₂ = 1 m/s
t = 1 sec
Fill the table with suitable answers and write the conclusion from it.

Answer:

Conclusion :

1. The table momentum is unchanged before and after collision.
2. We can say that the momentum is conserved.

Question 6.
How do you appreciate ‘Laws of motion’.
(OR)
How do you appreciate Newton, the great scientist.
Answer:

1. Built upon ideas primarily developed by Aristotle and Galileo, Sir Isaac Newton proposed his three fundamental Laws.
2. These Laws explain the connection between force and a change in motion.
3. These Laws are popularly known as Newton’s Law of motion.
4. Newton’s first law of motion : A body continuous its state of rest or of uniform motion unless a net force acts on it.
5. This law is very useful to study about inertia, mass, weight of an object.
6. This law is used to prepare cranes, machines, vehicles, etc.
7. Newton’s second Law of motion : The rate of change of memontum of a body is directly proportional to the net force acting on it and it takes place in the direction of net force.
8. This law is useful to find the attraction between two objects.
9. It is very useful to study the behaviour of substances like velocity, acceleration, force, momentum, etc.
10. Newton third law of motion : If an object exerts a force on an other object, the second object also exerts a force on the first one which is equal magnitude but opposite in direction.
11. This law is used to prepare rockets.
12. In this way Newtons Law’s are very useful in the fields of automobiles space centres and various industries.
13. In this way Newton laws changed the life of human beings.
14. So, I am appreciating the laws of motion and Newton for his great service to us.

Question 7.
Write suitable law and concept to the given situation or vice versa in the given table.

Answer:
A = Newton’s first law of motion
B = Static inertia
C = Newton’s third law of motion
D = Force of action = force of reaction
E = Newton’s second law of motion
F = Less impulse of net force
G = Newton’s third law of motion
H = Force of action = force of reaction
I = Fish swims in water
J = Force of action = force of reaction
K = Dropping egg on a cushion
L = Object thrown in universe (No external forces acts)
M = Newton’s first Law of motion
N = Bullet hurts
O = Newton’s second law of motion
P = Less impulse of net force
Q = Newton’s Second law of motion
R = Less momentum