CBSE Class 9 Science Gravitation Sure Shot Questions

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ASSIGNMENT QUESTIONS

GRAVITATION

MULTIPLE CHOICE QUESTIONS

1. Two objects of different masses falling freely near the surface of moon would [1 Mark]
(a) have same velocities at any instant
(b) have different accelerations
(c) experience forces of same magnitude
(d) undergo a change in their inertia

Answer: (a) have same velocities at any instant

1. Acceleration due to gravity is independent of the mass of the falling object.
2. All freely falling bodies experience the same acceleration near the surface of a given celestial body.
3. Therefore, objects of different masses acquire the same velocity at any given instant.
4. Hence, the correct option is (a).

Teacher's Note:
a) The acceleration due to gravity depends only on the mass and radius of the moon.
b) Students often confuse acceleration with gravitational force acting on the object.

2. The value of acceleration due to gravity [1 Mark]
(a) is same on equator and poles
(b) is least on poles
(c) is least on equator
(d) increases from pole to equator

Answer: (c) is least on equator

1. The Earth is flattened at the poles and bulges at the equator, making the equatorial radius greater than the polar radius.
2. Acceleration due to gravity is inversely proportional to the square of the radius (\(g = \frac{GM}{R^2}\)).
3. Due to a larger radius at the equator, the value of g is least there.
4. Hence, the correct option is (c).

Teacher's Note:
a) Centrifugal force due to Earth's rotation also reduces g at the equator.
b) Remember that polar radius is smaller, so g is maximum at the poles.

3. The gravitational force between two objects is F. If masses of both objects are halved without changing distance between them, then the gravitational force would become [1 Mark]
(a) F/4
(b) F/2
(c) F
(d) 2 F

Answer: (a) F/4

1. The gravitational force is given by \(F = \frac{G m_1 m_2}{d^2}\).
2. New masses are \(\frac{m_1}{2}\) and \(\frac{m_2}{2}\).
3. New force \(F' = \frac{G (\frac{m_1}{2})(\frac{m_2}{2})}{d^2} = \frac{1}{4} \left(\frac{G m_1 m_2}{d^2}\right) = \frac{F}{4}\).
4. Hence, the correct option is (a).

Teacher's Note:
a) Gravitational force is directly proportional to the product of masses.
b) Halving both individual masses reduces the product by a factor of 4.

4. A boy is whirling a stone tied with a string in an horizontal circular path. If the string breaks, the stone [1 Mark]
(a) will continue to move in the circular path
(b) will move along a straight line towards the centre of the circular path
(c) will move along a straight line tangential to the circular path
(d) will move along a straight line perpendicular to the circular path away from the boy

Answer: (c) will move along a straight line tangential to the circular path

1. In uniform circular motion, the velocity vector of the object is directed along the tangent to the circle at every point.
2. When the string breaks, the centripetal force ceases to act on the stone.
3. According to Newton's first law, the stone continues to move in the direction of its velocity vector due to inertia.
4. Hence, the correct option is (c).

Teacher's Note:
a) Inertia dictates that an object in motion continues along a straight line.
b) The tangent represents the instantaneous direction of motion at the point of release.

5. An object is put one by one in three liquids having different densities. The object floats with 1/9, 2/11 and 3/7 parts of their volumes outside the liquid surface in liquids of densities \(d_1\), \(d_2\) and \(d_3\) respectively. Which of the following statement is correct? [1 Mark]
(a) \(d_1 \gt d_2 \gt d_3\)
(b) \(d_1 \gt d_2 \lt d_3\)
(c) \(d_1 \lt d_2 \gt d_3\)
(d) \(d_1 \lt d_2 \lt d_3\)

Answer: (d) \(d_1 \lt d_2 \lt d_3\)

1. The fraction of volume submerged is given by \(\frac{V_{sub}}{V} = 1 - \frac{V_{outside}}{V}\).
2. For liquid 1: fraction outside = \(\frac{1}{9}\), submerged fraction = \(1 - \frac{1}{9} = \frac{8}{9} \approx 0.889\).
3. For liquid 2: fraction outside = \(\frac{2}{11}\), submerged fraction = \(1 - \frac{2}{11} = \frac{9}{11} \approx 0.818\).
4. For liquid 3: fraction outside = \(\frac{3}{7}\), submerged fraction = \(1 - \frac{3}{7} = \frac{4}{7} \approx 0.571\).
5. Since buoyant force equals weight, \(d \cdot V_{sub} = \text{constant}\), meaning density is inversely proportional to the submerged volume fraction.
6. Submerged fractions are \(0.889 (\text{liquid 1}) \gt 0.818 (\text{liquid 2}) \gt 0.571 (\text{liquid 3})\), so densities are in reverse order: \(d_1 \lt d_2 \lt d_3\).
7. Hence, the correct option is (d).

Teacher's Note:
a) Higher density liquids exert greater upthrust, causing a smaller fraction of the object to remain submerged.
b) Compare submerged fractions rather than outside fractions directly to avoid miscalculation.

6. In the relation F = G M m / d\(^2\), the quantity G [1 Mark]
(a) depends on the value of g at the place of observation
(b) is used only when the earth is one of the two masses
(c) is greatest at the surface of the earth
(d) is universal constant of nature

Answer: (d) is universal constant of nature

1. \(G\) is the universal gravitational constant whose value is \(6.67 \times 10^{-11}\) N m\(^2\) kg\(^{-2}\).
2. It remains constant everywhere in the universe regardless of the masses or distance between them.
3. It does not depend on the medium, temperature, or the presence of Earth.
4. Hence, the correct option is (d).

Teacher's Note:
a) Do not confuse universal gravitational constant \(G\) with acceleration due to gravity \(g\).
b) Henry Cavendish experimentally determined the value of \(G\).

7. Law of gravitation gives the gravitational force between [1 Mark]
(a) the earth and a point mass only
(b) the earth and Sun only
(c) any two bodies having some mass
(d) two charged bodies only

Answer: (c) any two bodies having some mass

1. Universal law of gravitation states that every body in the universe attracts every other body with a force.
2. This force is proportional to the product of their masses and applies universally to any pair of objects with mass.
3. It is not restricted to planets or celestial bodies.
4. Hence, the correct option is (c).

Teacher's Note:
a) The word 'universal' signifies that the law applies to all objects across the universe.
b) Even everyday small objects exert gravitational attraction on each other, though it is extremely weak.

8. The value of quantity G in the law of gravitation [1 Mark]
(a) depends on mass of earth only
(b) depends on radius of earth only
(c) depends on both mass and radius of earth
(d) is independent of mass and radius of the earth

Answer: (d) is independent of mass and radius of the earth

1. The universal gravitational constant \(G\) is a fundamental physical constant.
2. Its value does not depend on the properties, mass, or dimensions of the Earth or any other celestial body.
3. It is an intrinsic constant of nature.
4. Hence, the correct option is (d).

Teacher's Note:
a) In contrast, acceleration due to gravity \(g\) depends on the mass and radius of the planet.
b) Always distinguish between constant \(G\) and variable \(g\).

9. Two particles are placed at some distance. If the mass of each of the two particles is doubled, keeping the distance between them unchanged, the value of gravitational force between them will be [1 Mark]
(A) \(\frac{1}{4}\)
(B) 4 times
(C) \(\frac{1}{2}\) times
(D) unchanged

Answer: (B) 4 times

1. According to the universal law of gravitation, F = G \(\frac{m_1 m_2}{r^2}\).
2. If both masses are doubled, the new force becomes F' = G \(\frac{(2m_1)(2m_2)}{r^2}\) = 4 G \(\frac{m_1 m_2}{r^2}\) = 4F.
3. Hence, the gravitational force becomes 4 times the original value.
4. Therefore, the correct option is (B).

Teacher's Note:
a) The gravitational force is directly proportional to the product of the masses.
b) Be careful not to confuse direct proportionality with inverse square proportionality in distance.

10. The earth attracts a body of mass 2 kg on its surface with a force of [1 Mark]
(A) 9.8 N
(B) 19.6 N
(C) 6.67 x 10⁻¹¹ N
(D) 2 x 6.67 x 10⁻¹¹ N

Answer: (B) 19.6 N

1. Force equals the weight of the object, which is given by F = mg.
2. Given mass m = 2 kg and acceleration due to gravity g = 9.8 m/s².
3. F = 2 kg × 9.8 m/s² = 19.6 N.
4. Hence, the correct option is (B).

Teacher's Note:
a) Force of attraction by the Earth on any object is simply its weight.
b) Use g = 9.8 m/s² unless specified otherwise.

11. A stone dropped from a building takes 4 s to reach the ground. The height of the building is [1 Mark]
(A) 19.6 m
(B) 80.4 m
(C) 78.4 m
(D) 156.8 m

Answer: (C) 78.4 m

1. Initial velocity u = 0, time t = 4 s, and acceleration a = g = 9.8 m/s².
2. Using the equation of motion s = ut + \(\frac{1}{2}\) gt².
3. Height s = 0 + \(\frac{1}{2}\) × 9.8 × (4)² = 4.9 × 16 = 78.4 m.
4. Hence, the correct option is (C).

Teacher's Note:
a) For freely falling bodies, initial velocity is taken as zero.
b) Watch out for calculation errors when squaring the time value.

12. If ge is acceleration due to gravity on earth and gm is acceleration due to gravity on moon, then [1 Mark]
(A) ge = gm
(B) ge < gm
(C) ge = \(\frac{1}{6}\) gm
(D) gm = \(\frac{1}{6}\) ge

Answer: (D) gm = \(\frac{1}{6}\) ge

1. The acceleration due to gravity on the moon is known to be one-sixth of that on the earth.
2. Mathematically, this relation is expressed as gm = \(\frac{1}{6}\) ge.
3. Hence, the correct option is (D).

Teacher's Note:
a) Gravity depends on the mass and radius of the celestial body.
b) Moon's mass and radius are smaller than Earth's, resulting in lower gravity.

13. The mass of a body on the surface of earth is 12 kg. If acceleration due to gravity on moon is \(\frac{1}{6}\) of acceleration due to gravity on earth, then its mass on moon will be [1 Mark]
(A) 2 kgf
(B) 72 kg
(C) 12 kg
(D) zero

Answer: (C) 12 kg

1. The mass of an object is a measure of its inertia and remains constant everywhere.
2. Gravity affects weight, not mass.
3. Therefore, the mass on the moon remains the same as on the earth, which is 12 kg.
4. Hence, the correct option is (C).

Teacher's Note:
a) Mass is an intrinsic property and never changes with location.
b) Students often mistakenly divide mass by 6 instead of weight.

14. The atmosphere is held to the earth by [1 Mark]
(A) gravity
(B) wind
(C) clouds
(D) earth's magnetic field

Answer: (A) gravity

1. The gravitational pull of the earth attracts all matter, including atmospheric gases, towards its center.
2. This gravitational force prevents the gases from escaping into space.
3. Hence, the correct option is (A).

Teacher's Note:
a) Gravity acts on gaseous molecules just like any other matter.
b) Magnetic fields deflect charged particles but do not bind the neutral atmospheric envelope.

15. The force of attraction between two unit point masses separated by a unit distance is called [1 Mark]
(A) gravitational potential
(B) acceleration due to gravity
(C) gravitational field
(D) universal gravitational constant

Answer: (D) universal gravitational constant

1. From F = G \(\frac{m_1 m_2}{r^2}\), if m₁ = 1, m₂ = 1, and r = 1, then F = G.
2. Here G is the universal gravitational constant.
3. Hence, the correct option is (D).

Teacher's Note:
a) G has a constant value of 6.673 × 10⁻¹¹ N m² / kg² across the universe.
b) Distinguish carefully between gravitational force, field, and constant definitions.

16. The weight of an object at the centre of the earth of radius R is [1 Mark]
(A) zero
(B) infinite
(C) R times the weight at the surface of the earth
(D) 1/R² times the weight at surface of the earth

Answer: (A) zero

1. At the center of the earth, the gravitational pull from all directions cancels out.
2. Consequently, the acceleration due to gravity g becomes zero at the earth's center.
3. Since weight W = mg, the weight of the object becomes zero.
4. Hence, the correct option is (A).

Teacher's Note:
a) Gravitational force inside a uniform spherical shell is zero.
b) Mass remains non-zero while weight becomes zero at the center.

17. What is the source of centripetal force that a planet requires to revolve around the Sun? On what factors does that force depend? [2 Marks]

Answer:
1. The gravitational force exerted by the Sun on the planet provides the required centripetal force.
2. This force depends on the mass of the Sun, the mass of the planet, and the distance between them.

Teacher's Note:
a) Gravitation acts as the central pulling force in circular orbits.
b) Mention both the attracting bodies and the separation distance.

18. On the earth, a stone is thrown from a height in a direction parallel to the earth's surface while another stone is simultaneously dropped from the same height. Which stone would reach the ground first and why? [2 Marks]

Answer:
1. Both stones will reach the ground at the same time.
2. Vertical motion is independent of horizontal motion, so both experience the same downward acceleration due to gravity.

Teacher's Note:
a) Horizontal velocity does not affect the vertical free fall time.
b) Time depends only on initial vertical velocity and height.

19. Suppose gravity of earth suddenly becomes zero, then in which direction will the moon begin to move if no other celestial body affects it? [1 Mark]

Answer:
1. The moon will begin to move in a straight line along the tangent to its circular orbit at that point.

Teacher's Note:
a) Inertia keeps an object moving in the direction of its instantaneous velocity when centripetal force vanishes.

20. Identical packets are dropped from two aeroplanes, one above the equator and the other above the north pole, both at height h. Assuming all conditions are identical, will those packets take same time to reach the surface of earth. Justify your answer. [2 Marks]

Answer:
1. No, the packet dropped above the north pole will reach the ground slightly earlier.
2. The acceleration due to gravity is greater at the poles than at the equator due to the earth's shape and rotation.

Teacher's Note:
a) Earth is flattened at the poles, making the polar radius smaller.
b) Higher value of g at poles causes faster downward acceleration.

21. The weight of any person on the moon is about 1/6 times that on the earth. He can lift a mass of 15 kg on the earth. What will be the maximum mass, which can be lifted by the same force applied by the person on the moon? [2 Marks]

Answer:
1. Force exerted by the person depends on muscle strength, which remains constant, but weight is less on the moon.
2. Maximum mass = 15 kg × 6 = 90 kg.

Teacher's Note:
a) Muscle force available is the same in both locations.
b) Since gravity is one-sixth, six times more mass can be balanced by the same force.

22. Calculate the average density of the earth in terms of g, G and R. [3 Marks]

Answer:
1. Acceleration due to gravity is given by \(g = \frac{GM}{R^2}\), where M is mass and R is radius of the earth.
2. Mass M can be written as volume multiplied by density (\(\rho\)), so \(M = \frac{4}{3} \pi R^3 \rho\).
3. Substituting M, \(g = \frac{G (\frac{4}{3} \pi R^3 \rho)}{R^2} = \frac{4}{3} \pi G R \rho\), which gives density \(\rho = \frac{3g}{4 \pi G R}\).

Teacher's Note:
a) Assume earth is a uniform sphere of radius R.
b) Substitute volume formula for a sphere correctly.
c) Final expression must feature g, G, R, and pi.

23. The earth is acted upon by gravitation of Sun, even though it does not fall into the Sun. Why? [2 Marks]

Answer:
1. The gravitational force of the Sun provides the necessary centripetal force for the earth's orbital motion.
2. This results in revolution around the Sun rather than a direct inward fall.

Teacher's Note:
a) Orbital velocity balances the gravitational pull perpendicular to motion.
b) Similar to a stone whirled in a string.

24. Calculate the density of Earth from Newton's law of gravitation. [3 Marks]

Answer:
1. Universal law of gravitation gives surface gravity as \(g = \frac{GM}{R^2}\).
2. Expressing mass as \(M = \frac{4}{3} \pi R^3 \rho\), substitute into the gravity equation.
3. Solving for density \(\rho\) yields \(\rho = \frac{3g}{4 \pi G R}\).

Teacher's Note:
a) Relate surface gravity to mass and radius.
b) Express mass in terms of density and spherical volume.
c) State standard values or final formula clearly.

25. A body weighs more at poles than at the equator of earth. Why? [2 Marks]

Answer:
1. The radius of the earth is smaller at the poles than at the equator.
2. Since \(g\) is inversely proportional to the square of the radius, gravity and weight are greater at the poles.

Teacher's Note:
a) Earth is an oblate spheroid, not a perfect sphere.
b) Equatorial bulge reduces effective gravity there.

26. Two particles of equal mass(m) move in a circle of radius (r) under the action of their mutual gravitational attraction. Find the speed of each particle. [3 Marks]

Answer:
1. The distance between the two particles moving in a circle of radius r is 2r.
2. Gravitational force between them = \(\frac{G m \cdot m}{(2r)^2} = \frac{G m^2}{4r^2}\).
3. This force provides the centripetal force \(\frac{m v^2}{r}\), so \(\frac{m v^2}{r} = \frac{G m^2}{4r^2}\), which gives speed \(v = \sqrt{\frac{Gm}{4r}}\).

Teacher's Note:
a) Note that the diameter separation between particles is 2r.
b) Equate gravitational attraction directly to centripetal force.
c) Simplify carefully to isolate velocity v.

27. How does the weight of an object vary with respect to mass and radius of the earth. In a hypothetical case, if the diameter of the earth becomes half of its present value and its mass becomes four times of its present value, then how would the weight of any object on the surface of the earth be affected? [4 Marks]

Answer:
1. Weight \(W = mg = \frac{G M m}{R^2}\), showing weight is directly proportional to earth mass M and inversely proportional to square of radius R.
2. Let new mass \(M' = 4M\) and new radius \(R' = \frac{R}{2}\).
3. New weight \(W' = \frac{G (4M) m}{(R/2)^2} = \frac{4 G M m}{\frac{R^2}{4}} = 16 \left(\frac{G M m}{R^2}\right) = 16W\).
4. Hence, the weight of the object becomes 16 times its present value.

Teacher's Note:
a) Use the universal gravitation formula for weight.
b) Substitute modified mass and radius terms.
c) Watch out for squaring the halved radius term.
d) Conclude with the final factor increase.

28. How does the force of attraction between the two bodies depend upon their masses and distance between them? A student thought that two bricks tied together would fall faster than a single one under the action of gravity. Do you agree with his hypothesis or not? Comment. [4 Marks]

Answer:
1. Gravitational force is directly proportional to the product of masses and inversely proportional to the square of the distance between them.
2. No, we do not agree with the student's hypothesis.
3. Acceleration due to gravity is independent of the mass of the falling object, as seen in Galileo's free fall principle.
4. Two tied bricks have greater mass but also greater weight proportionately, maintaining the same acceleration as a single brick in a vacuum.

Teacher's Note:
a) State Newton's law of gravitation clearly.
b) Address the student misconception directly.
c) Use acceleration formula \(g = \frac{GM}{R^2}\) to explain independence from object mass.
d) Mention negligible air resistance assumptions.

29. Two objects of masses m1 and m2 having the same size are dropped simultaneously from heights h1 and h2 respectively. Find out the ratio of time they would take in reaching the ground. Will this ratio remain the same if (i) one of the objects is hollow and the other one is solid and (ii) both of them are hollow, size remaining the same in each case. Give reason. [4 Marks]

Answer:
1. Using second equation of motion with initial velocity zero, \(h = \frac{1}{2} g t^2\), so time \(t = \sqrt{\frac{2h}{g}}\).
2. Ratio of times \(\frac{t_1}{t_2} = \sqrt{\frac{h_1}{h_2}}\).
3. (i) If one is hollow and the other is solid of same size, their masses differ, but acceleration due to gravity is independent of mass, so the ratio remains the same.
4. (ii) If both are hollow and of same size, masses may still be equal or unequal, but free fall time ratio depends only on heights, so the ratio remains unchanged.

Teacher's Note:
a) Apply kinematic equations for vertical motion.
b) Derive time ratio purely in terms of heights h1 and h2.
c) Explain why mass differences do not affect free fall time.
d) Conclude both sub-cases with proper justification.

30. Distinguish between mass and weight. Show that mass of a body numerically equal to weight of the body except at the centre of earth. [4 Marks]

Answer:
1. Mass is the quantity of matter contained in a body and is constant, whereas weight is the gravitational force acting on the body and varies with location.
2. Mass is a scalar quantity measured in kg, while weight is a vector quantity measured in Newtons.
3. Weight is given by \(W = mg\). Numerically, if \(g = 1\) unit, then \(W = m \times 1 = m\).
4. At the centre of the earth, gravity \(g = 0\), making weight zero, so mass is not numerically equal to weight there.

Teacher's Note:
a) Provide clear differences between mass and weight.
b) Set up the numerical relation using \(W = mg\).
c) Explain conditions where numerical values match.
d) Highlight the exception at the earth's center where g is zero.

(c) ge = 1/6gm

(d) gm = 1/6ge