CG SCERT (Chhattisgarh Board) Class 9 Science Solutions | 2026-27 Official Edition
Complete 100% textbook-grounded CG Board Class 9 Science Chapter 11 Solutions: Gravitation (गुरुत्वाकर्षण) in Bilingual Format (English & Hindi Medium).
End-of-Chapter Exercises (Solutions)
अभ्यास के सभी प्रश्नों के हल
(i) The gravitational force between two bodies does not depe…
(दो पिंडों के बीच गुरुत्वाकर्षण बल किस पर निर्भर नहीं करता:)…
(i) The gravitational force between two bodies does not depend on:
(ii) The value of G is:
(a) 7.67 × 1011 Nm2/kg2
(b) 6.67 × 1011 Nm2/kg
(c) 6.67 × 10-11 Nm2/kg2
(d) 5.67 × 1011 Nm2/kg2
(iii) The value of gravitational acceleration on the surface of earth is:
(a) 9.8 m/s2
(b) 8.8 m/s2
(c) 4.8 m/s2
(d) 8.9 m/s2
(iv) According to universal law of gravitation, the force acting between two bodies of masses m1 and m2 placed at a distance R is:
(a) F = G (m1 m2)/(R2)
(b) F = G (m1 m2)/(4R)
(c) F = G (m1 m2)/(R)
(d) F = G M/R
(v) A body is moving upwards opposing the gravitational pull of earth. What will be its velocity at the highest point reached?
(a) 0
(b) u2 / 2g
(c) h / t
(d) 2gh
(i) On earth, the weight of a body having 10 kg mass will be ……………………………
–
What will be the gravitational force between earth and an object of 1 kg placed on its surface. Here, the mass of earth is 6 × 1024 kg, and distance from the earth’s centre to the surface is 6400 km. / 1 kg ? 6 × 1024 kg 6400 km
Step-by-step Solution /
- English: A sheet of paper has a larger surface area compared to the compact paper ball. As it falls through air, it experiences much greater upward air resistance (friction), which slows down its fall.
- English:
- It explains the force that binds us to the Earth.
- It explains the motion of the Moon and artificial satellites around the Earth.
- It explains the motion of planets around the Sun.
- It explains the occurrence of ocean tides due to the gravitational pull of the Moon and Sun.
- Hindi:
- English: According to Newton’s second law (a = F/m), the acceleration produced in a body is inversely proportional to its mass. Since Earth’s mass is about 81 times greater than the Moon’s mass, the acceleration produced in the Earth by the gravitational pull of the Moon is negligible, so Earth does not visibly move toward the Moon.
A ball is thrown vertically upwards with a speed of 49 m/s. Calculate (i) Maximum height of the ball, (ii) Total time taken by the ball to reach the earth’s surface. / 49 m/s (i) , (ii)
If a body is thrown vertically upwards at a speed of 10 m/s then after how much time and with what velocity will it reach back to us? / 10 m/s ?
- English:
The gravitational force will become 4F in any of the following situations:
- If the mass of both objects is doubled (m1‘ = 2m1 and m2‘ = 2m2).
- If the distance between the two objects is reduced to half (r’ = r / 2).
- If the mass of one object is made four times (m1‘ = 4m1).
- Hindi:
- English:
- The acceleration due to gravity g = GM/R2 depends only on the Earth’s mass and radius, and is independent of the object’s mass. Thus, both objects accelerate at the same rate and reach the Earth simultaneously.
- No, the gravitational force F = m g is not equal; the heavier object experiences a larger gravitational force proportional to its mass.
- Hindi:
- English:
- Let Earth’s mass be M, radius be R, and mass of body be m.
- Gravitational force by Earth: F = G (M m)/(R2) — (Equation 1)
- By Newton’s second law of motion: F = m g — (Equation 2)
- Equating (1) and (2): m g = G (M m)/(R2) ⇒ g = (G M)/(R2).
- Calculation: Substituting M = 6 × 1024 kg, R = 6.4 × 106 m, G = 6.67 × 10-11:
g = (6.67 × 10-11 × 6 × 1024)/((6.4 × 106)2) = (40.02 × 1013)/(40.96 × 1012) ≈ 9.81 m/s2.
- Hindi:
- (1) (2) : m g = G (M m)/(R2) ⇒ g = (G M)/(R2)
- : M = 6 × 1024 kg, R = 6.4 × 106 m, G = 6.67 × 10-11
g = (6.67 × 10-11 × 6 × 1024)/((6.4 × 106)2) = (40.02 × 1013)/(40.96 × 1012) ≈ 9.81 m/s2
- English: Can you tell what effect will the increase or decrease of the value of G have on our daily lives? What would have happened if there was no gravitational force? Discuss with your friends in a group.
- English:
- Effect of change in G: If G increases, gravitational attraction will become much stronger, making objects significantly heavier, causing structures to collapse under their weight, and pulling atmospheric gases closer. If G decreases, weight will reduce, making walking, holding objects, and planetary orbits unstable.
- If there were no gravitational force: Objects, water, atmosphere, and living beings would float away into space. Planets would cease to revolve around the Sun, and Earth would lose its atmosphere, making life impossible.
- Hindi:
- English: Calculate the force of attraction between you and a friend sitting at a distance of 1 m from you. Do you feel this force? If not then why?
Step-by-step Solution /
- English:
- Assume mass of student m1 = 50 kg, mass of friend m2 = 50 kg, distance r = 1 m, G = 6.67 × 10-11 N m2/kg2.
- Formula: F = G (m1 m2)/(r2) = (6.67 × 10-11 × 50 × 50)/(12) = 1.6675 × 10-7 N.
- Reason for not feeling: This force (1.67 × 10-7 N) is extremely small compared to the force of friction between our bodies and the floor, so we do not feel it.
- Hindi:
- : F = G (m1 m2)/(r2) = (6.67 × 10-11 × 50 × 50)/(12) = 1.6675 × 10-7 N
- English: Is the value of ‘g’ same at all places on earth? Does the value of ‘g’ depend on the mass of the body? Is the value of ‘g’ on the equator of earth and the poles same? Where will its value be greater?
- English:
- No, the value of g is not same at all places on Earth.
- g = GM/R2 does not depend on the mass, volume, or nature of the falling body.
- The value of g is greater at the poles and smaller at the equator because Earth is flattened at the poles, making equatorial radius Re larger than polar radius Rp.
- Hindi:
| Travelled Distance s | Speed of Object A | Speed of Object B | Time taken by A | Time taken by B |
|---|---|---|---|---|
| 0 m | 0 m/s | 0 m/s | 0 s | 0 s |
| 10 m | 10√2 ≈ 14.14 m/s | 10√2 ≈ 14.14 m/s | √2 ≈ 1.41 s | √2 ≈ 1.41 s |
| 20 m | 20 m/s | 20 m/s | 2 s | 2 s |
| 100 m | 20√5 ≈ 44.72 m/s | 20√5 ≈ 44.72 m/s | 2√5 ≈ 4.47 s | 2√5 ≈ 4.47 s |
- Given: Object A (2 kg) and Object B (10 kg) dropped from 100 m height (g = 10 m/s2).
- Formulas: v = √(2gs), t = √(2s/g).
- Conclusion: Mass of an object does not affect its speed during free fall. Both objects fall at the exact same rate.
(a) The distance between both the bodies /
- Correct Option / : (c) 6.67 × 10-11 Nm2/kg2.
- Correct Option / : (a) 9.8 m/s2.
- Correct Option / : (a) F = G (m1 m2)/(R2).
- Correct Option / : (a) 0.
Answer / : 98 N (Calculation: W = mg = 10 × 9.8 = 98 N).
(ii) A body falling from rest from a height h will have a velocity of ………………………………….
–
Answer / : √(2gh) (or v2 = 2gh).
(iii) Value of universal gravitational constant is ………………………………
–
Answer / : 6.67 × 10-11 N m2/kg2.
(iv) The value of gravitational acceleration in SI units is ………………………………..
–
Answer / : m/s2 (or 9.8 m/s2).
(v) If two objects of different masses are dropped from a same height, they will reach the surface of earth at ……………………………. time.
–
Answer / : *Same.
Step-by-step Solution /
- English:
- Given: Mass of object m = 1 kg, Mass of Earth M = 6 × 1024 kg.
- Radius R = 6400 km = 6.4 × 106 m, G = 6.67 × 10-11 N m2/kg2.
- Formula: F = (G M m)/(R2) = (6.67 × 10-11 × 6 × 1024 × 1)/((6.4 × 106)2).
- F = (40.02 × 1013)/(40.96 × 1012) = (400.2)/(40.96) ≈ 9.77 N ≈ 9.8 N.
- Hindi:
- R = 6400 km = 6.4 × 106 m, G = 6.67 × 10-11 N m2/kg2
- : F = (G M m)/(R2) = (6.67 × 10-11 × 6 × 1024 × 1)/((6.4 × 106)2)
- F = (40.02 × 1013)/(40.96 × 1012) = (400.2)/(40.96) ≈ 9.77 N ≈ 9.8 N
- English:
- Original force F = G (m1 m2)/(r2).
- (i) m1‘ = 2m1 ⇒ F’ = G ((2m1)m2)/(r2) = 2F (Force gets doubled).
- (ii) r’ = 3r ⇒ F” = G (m1 m2)/((3r)2) = 1/9 F (Force becomes one-ninth).
- (iii) m1‘ = 2m1, m2‘ = 2m2 ⇒ F”’ = G ((2m1)(2m2))/(r2) = 4F (Force becomes four times).
- Hindi:
- F = G (m1 m2)/(r2)
- (ii) r’ = 3r ⇒ F” = 1/9 F
- (iii) m1‘ = 2m1, m2‘ = 2m2 ⇒ F”’ = 4F
Answer / उत्तर:
- Hindi: कागज की खुली शीट का पृष्ठ क्षेत्रफल (surface area) उसी कागज से बनी सिकुड़ी हुई गेंद की तुलना में बहुत अधिक होता है। हवा में गिरते समय शीट पर ऊपर की ओर लगने वाला वायु प्रतिरोध (घर्षण) अधिक होता है, जिससे वह धीमी गति से गिरती है।
Answer / उत्तर:
- यह हमें पृथ्वी से बाँधे रखने वाले बल की व्याख्या करता है।
- यह पृथ्वी के चारों ओर चंद्रमा और कृत्रिम उपग्रहों की गति को समझाता है।
- यह सूर्य के चारों ओर ग्रहों की गति का आधार बताता है।
- यह चंद्रमा और सूर्य के गुरुत्वाकर्षण के कारण समुद्र में ज्वार-भाटा आने के कारण को स्पष्ट करता है।
Answer / उत्तर:
- Hindi: न्यूटन के दूसरे नियमानुसार (a = F/m), किसी बल द्वारा उत्पन्न त्वरण वस्तु के द्रव्यमान के व्युत्क्रमानुपाती होता है। चूंकि पृथ्वी का द्रव्यमान चंद्रमा के द्रव्यमान से लगभग 81 गुना अधिक है, अतः चंद्रमा के आकर्षण बल के कारण पृथ्वी में उत्पन्न त्वरण नगण्य होता है और पृथ्वी चंद्रमा की ओर गति करती दिखाई नहीं देती।
Step-by-step Solution /
- English:
- Given: u = 49 m/s, v = 0 (at max height), g = 9.8 m/s2.
- (i) Maximum Height h:
- v2 = u2 – 2gh ⇒ 0 = (49)2 – 2(9.8)h ⇒ 19.6 h = 2401 ⇒ h = 122.5 m.
- (ii) Total Time T:
- Time to go up t = u/g = 49/9.8 = 5 s.
- Total time T = 2 × t = 2 × 5 = 10 s.
- Hindi:
- v2 = u2 – 2gh ⇒ 0 = (49)2 – 2(9.8)h ⇒ 19.6 h = 2401 ⇒ h = 122.5 m
Step-by-step Solution /
- English:
- Given: u = 10 m/s, g = 10 m/s2.
- Time to reach maximum height t1 = u/g = 10/10 = 1 s.
- Total time to return T = 2 × t1 = 2 × 1 = 2 s.
- By conservation of energy, it reaches back with the exact same speed 10 m/s in the downward direction.
- Hindi:
- : u = 10 m/s, g = 10 m/s2
Answer / उत्तर:
गुरुत्वाकर्षण बल निम्नलिखित में से किसी भी स्थिति में 4F हो जाएगा:
- यदि दोनों पिंडों के द्रव्यमान को दोगुना कर दिया जाए (m1‘ = 2m1 तथा m2‘ = 2m2)।
- यदि दोनों पिंडों के बीच की दूरी को आधा कर दिया जाए (r’ = r / 2)।
- यदि किसी एक पिंड का द्रव्यमान चार गुना कर दिया जाए (m1‘ = 4m1)।
Answer / उत्तर:
- गुरुत्वीय त्वरण g = GM/R2 केवल पृथ्वी के द्रव्यमान और त्रिज्या पर निर्भर करता है तथा वस्तु के द्रव्यमान से स्वतंत्र होता है। इसलिए दोनों वस्तुएँ समान त्वरण से गिरती हैं और एक साथ पहुँचती हैं।
- नहीं, दोनों पर लगने वाला गुरुत्वाकर्षण बल F = m g बराबर नहीं होता; भारी वस्तु पर उसके द्रव्यमान के अनुपात में अधिक बल लगता है।
Step-by-step Derivation / चरणबद्ध व्युत्पत्ति:
- माना पृथ्वी का द्रव्यमान M, त्रिज्या R, तथा वस्तु का द्रव्यमान m है।
- पृथ्वी द्वारा आरोपित गुरुत्वाकर्षण बल: F = G (M m)/(R2) — (समीकरण 1)
- न्यूटन के द्वितीय नियम से: F = m g — (समीकरण 2)
- समीकरण (1) और (2) से: m g = G (M m)/(R2) ⇒ g = (G M)/(R2)।
- मान गणना: M = 6 × 1024 kg, R = 6.4 × 106 m, G = 6.67 × 10-11 रखने पर:
Answer / उत्तर:
- G के मान में परिवर्तन का प्रभाव: यदि G का मान बढ़ जाए, तो गुरुत्वाकर्षण आकर्षण बल अत्यधिक तीव्र हो जाएगा जिससे सभी वस्तुएँ भारी हो जाएँगी, इमारतें अपने ही भार से ढह जाएँगी और वायुमंडल दब जाएगा। यदि G का मान घट जाए, तो भार कम हो जाएगा, जिससे चलना, वस्तुओं को पकड़ना और ग्रहों की कक्षाएँ अस्थिर हो जाएँगी।
- यदि गुरुत्वाकर्षण बल न होता: पृथ्वी पर मौजूद वस्तुएँ, जल, वायुमंडल और जीव अंतरिक्ष में तैरने लगेंगे। ग्रह सूर्य के चारों ओर चक्कर लगाना बंद कर देंगे और वायुमंडल नष्ट होने से जीवन असंभव हो जाएगा।
Answer / उत्तर:
- नहीं, पृथ्वी पर सभी स्थानों पर g का मान समान नहीं होता है।
- सूत्र g = GM/R2 के अनुसार g का मान गिरते हुए पिंड के द्रव्यमान, आयतन या प्रकृति पर निर्भर नहीं करता।
- ध्रुवों (poles) पर g का मान अधिक और भूमध्य रेखा (equator) पर कम होता है क्योंकि ध्रुवीय त्रिज्या Rp, भूमध्यरेखीय त्रिज्या Re से कम होती है।
Section 113
द्रव्यमान m की एक वस्तु को पृथ्वी की…
An object of mass ‘m’ is dropped from a height ‘h’ above the surface of earth. What will be its velocity just before reaching the surface of earth?
- Initial velocity u = 0, acceleration a = +g, distance s = h.
- Using third equation of motion: v2 = u2 + 2gh ⇒ v2 = 0 + 2gh ⇒ v = √(2gh).
Section 113
एक गेंद को 5 ms की चाल से…
Throw a ball vertically upwards with a speed of 5 m/s. How much height will the ball reach and after how long will it return back to your hands? In which position of the ball, its speed will be minimum and what will be its value?
- Given: u = 5 m/s, final velocity at max height v = 0, g = 10 m/s2.
- Maximum height h: v2 = u2 – 2gh ⇒ 0 = 52 – 2(10)h ⇒ 20h = 25 ⇒ h = 1.25 m.
- Time to reach top t1: v = u – gt ⇒ 0 = 5 – 10t1 ⇒ t1 = 0.5 s.
- Total time to return T = 2 × t1 = 2 × 0.5 = 1 s.
- Speed is minimum (0 m/s) at the highest point (1.25 m).
Section 114
चंद्रमा का द्रव्यमान Msubmoonsub 7 10sup22sup kg है…
Mass of the moon is Mmoon = 7 × 1022 kg and its radius is Rmoon = 1700 km. Find out the value of acceleration on the surface of moon.
- M = 7 × 1022 kg, R = 1700 km = 1.7 × 106 m, G = 6.67 × 10-11 N m2/kg2.
- Formula: gmoon = (G M)/(R2) = (6.67 × 10-11 × 7 × 1022)/((1.7 × 106)2) = (46.69 × 1011)/(2.89 × 1012) ≈ 1.616 m/s2 ≈ 1.62 m/s2.
Section 114
पृथ्वी और चंद्रमा की सतह पर द्रव्यमान m…
Compare the weight of an object of mass ‘m’ on the surface of earth and the moon.
- Weight on Earth We = m ge = m × 9.8 N.
- Weight on Moon Wm = m gm = m × 1.62 N.
- Ratio: Wm/We = (1.62)/(9.8) ≈ 1/6 ⇒ Wm = 1/6 We.
- Weight of an object on the Moon is one-sixth of its weight on Earth.
Section 114
पृथ्वी और चंद्रमा के केंद्रों के बीच की…
The distance between the centres of earth and the moon is 3.84 × 105 km. How much force will they exert on each other?
- Me = 6 × 1024 kg, Mm = 7 × 1022 kg (given in text), r = 3.84 × 108 m, G = 6.67 × 10-11.
- F = (G Me Mm)/(r2) = (6.67 × 10-11 × 6 × 1024 × 7 × 1022)/((3.84 × 108)2) = (280.14 × 1035)/(14.7456 × 1016) ≈ 1.90 × 1020 N.
- (Note: If standard moon mass 7.36 × 1022 kg is used, F ≈ 2.01 × 1020 N).
Where will be the centre of gravity of…
Where will be the centre of gravity of…
For a uniform sphere, it is at its geometric centre. For a triangular table, it is at the centroid (intersection of medians).
Can a body have more than one centre…
Can a body have more than one centre…
No, a body can have only one unique centre of gravity.
Why does the leaning tower of pisa not…
Why does the leaning tower of pisa not…
Because the vertical line passing through its centre of gravity falls well within its base area.
Why do you bend forward when you lift…
Why do you bend forward when you lift…
To bring the combined centre of gravity of the body and the load vertically above our feet (base), maintaining stability.