CBSE Class 11 Physics Gravitation Objective Questions

CLASS 12 PHYSICS
GRAVITATION
OBJECTIVE QUESTIONS

1. The gravitational field due to a mass distribution is E = K/x³ in the x-direction, where K is a constant. Taking the gravitational potential to be zero at infinity, its value at a distance x is
(a) K/x
(b) K/2x
(c) K/x²
(d) K/2x²

Answer: (d) K/2x²

1. Field E = -dV/dx, so V(x) = -∫ E dx taken from ∞ to x.
2. V(x) = ∫ (K/x³) dx taken from x to ∞ = K/2x².
3. Hence, the correct option is (d).

Teacher's Note:
a) Potential at a point is the work done per unit mass in bringing a body from infinity to that point.
b) Do not forget the factor 2 that appears when integrating 1/x³.

2. A planet of mass m moves around the sun of mass M in an elliptical orbit. The maximum and minimum distance of the planet from the sun are r₁ and r₂ respectively. The time-period of the planet is proportional to
(a) r₁^3/2
(b) r₂^3/2
(c) (r₁ + r₂)^3/2
(d) (r₁ - r₂)^3/2

Answer: (c) (r₁ + r₂)^3/2

1. By Kepler's third law, T² is proportional to a³, where a is the semi-major axis.
2. Semi-major axis a = (r₁ + r₂)/2.
3. So T is proportional to (r₁ + r₂)^3/2.
4. Hence, the correct option is (c).

Teacher's Note:
a) In an ellipse the semi-major axis is the average of the greatest and least distances.
b) Students often use r₁ alone, which is wrong for a non-circular orbit.

3. A particle hanging from a spring stretches it by 1 cm at earth's surface. Radius of earth is 6400 km. At a place 800 km above the earth's surface, the same particle will stretch the spring by
(a) 1 cm
(b) 8 cm
(c) 0.1 cm
(d) 0.79 cm

Answer: (d) 0.79 cm

1. Extension is proportional to the weight, which is proportional to g.
2. g at height h = g × (R/(R + h))² = g × (6400/7200)² = g × 0.79.
3. Extension = 1 cm × 0.79 = 0.79 cm.
4. Hence, the correct option is (d).

Teacher's Note:
a) Use the distance from the centre of the earth, R + h, not h alone.
b) The spring stretch is proportional to g because the mass is unchanged.

4. A satellite is to revolve around the earth in a circle of radius 8000 km. The speed at which this satellite be projected into orbit will be
(a) 8 km/s
(b) 7.08 km/sec
(c) 16 km/s
(d) 3 km/sec.

Answer: (b) 7.08 km/sec

1. Orbital speed v = √(GM/r) = R√(g/r).
2. Take g = 9.8 × 10⁻³ km/s² and R = 6400 km, so v = 6400 × √(9.8 × 10⁻³/8000) km/s.
3. v = 6400 × 1.107 × 10⁻³ km/s = 7.08 km/s.
4. Hence, the correct option is (b).

Teacher's Note:
a) GM is replaced by gR² so that only g and the earth's radius are needed.
b) Keep all lengths in the same unit, either km or m, throughout.
c) Orbital speed does not depend on the mass of the satellite.

5. If the distance between the earth and the sun were half its present value, the number of days in a year would have been
(a) 64.5
(b) 129
(c) 182.5
(d) 730

Answer: (b) 129

1. By Kepler's third law, T² is proportional to r³, so T is proportional to r^3/2.
2. T' = 365 × (1/2)^3/2 = 365/2.828 days.
3. T' = 129 days approximately.
4. Hence, the correct option is (b).

Teacher's Note:
a) Halving the distance does not halve the period; the power is 3/2.
b) Check by cubing: (1/2)³ = 1/8 and the square root of 1/8 is 0.354.

6. An artificial satellite moving in a circular orbit around the earth has a total kinetic and potential energy equal to E₀. Its potential energy is
(a) - E₀
(b) 1.5 E₀
(c) 2 E₀
(d) E₀

Answer: (c) 2 E₀

1. For a circular orbit, kinetic energy K = GMm/2r, potential energy U = -GMm/r.
2. Total energy E₀ = K + U = -GMm/2r.
3. So U = -GMm/r = 2 × (-GMm/2r) = 2E₀.
4. Hence, the correct option is (c).

Teacher's Note:
a) The printed key gives (b), which is wrong; the correct option is (c) because U is twice the total energy.
b) In a circular orbit U = 2E and K = -E, where E is the (negative) total energy.

7. A planet has twice the density of earth but the acceleration due to gravity on its surface is exactly the same as on the surface of earth. Its radius in terms of radius of earth R will be
(a) R/4
(b) R/2
(c) R/3
(d) R/8

Answer: (b) R/2

1. g = (4/3) πGρR, so g is proportional to ρR.
2. Same g and density twice that of earth gives 2ρR' = ρR.
3. R' = R/2.
4. Hence, the correct option is (b).

Teacher's Note:
a) Write g in terms of density and radius to compare two planets quickly.
b) Do not use g = GM/R² with M unchanged, since the mass here is different.

8. Two identical copper spheres each of radius R are placed in contact with each other. If the gravitational attraction between them is F, then
(a) F ∝ R²
(b) F ∝ R⁴
(c) F ∝ 1/R²
(d) F ∝ 1/R⁴

Answer: (b) F ∝ R⁴

1. Mass of each sphere m = (4/3)πR³ρ, so m is proportional to R³.
2. Distance between centres = 2R.
3. F = Gm²/(2R)², so F is proportional to R⁶/R² = R⁴.
4. Hence, the correct option is (b).

Teacher's Note:
a) Both the masses and the separation change with R, so count both.
b) Spheres in contact have their centres 2R apart, not R.

9. Suppose g_e be the acceleration due to gravity at the equator and g_p be that at the poles. Assuming earth to be a sphere of radius R_e rotating about its own axis with angular speed ω, then g_p - g_e is given by
(a) ω²/R_e
(b) R_eω²
(c) ω² R_e²
(d) ω²/R_e²

Answer: (b) R_eω²

1. At latitude λ, g' = g - Rω² cos²λ.
2. At the poles λ = 90°, so g_p = g; at the equator λ = 0°, so g_e = g - R_eω².
3. g_p - g_e = R_eω².
4. Hence, the correct option is (b).

Teacher's Note:
a) The centrifugal term Rω² is subtracted only at the equator.
b) Check the unit: radius × (angular speed)² has the unit of acceleration.

10. An infinite number of masses, each of 1 kg, are placed on the positive x-axis at 1 m, 2 m, 4 m, 8 m,.......from the origin. The magnitude of the gravitational field at the origin due to this distribution of masses is
(a) 2G
(b) \( \frac{4G}{3} \)
(c) \( \frac{3G}{4} \)
(d) ∞

Answer: (b) \( \frac{4G}{3} \)

1. All fields point along the +x direction, so the magnitudes add.
2. E = G(1/1² + 1/2² + 1/4² + 1/8² + ...) = G(1 + 1/4 + 1/16 + 1/64 + ...).
3. This is a geometric series with first term 1 and ratio 1/4, so the sum = 1/(1 - 1/4) = 4/3.
4. E = 4G/3.
5. Hence, the correct option is (b).

Teacher's Note:
a) The printed key gives (c), which is wrong; the correct option is (b) because the series sum is 4/3, not 3/4.
b) Distances are squared in the field formula, so the ratio of the series is 1/4, not 1/2.
c) Sum of an infinite geometric series is a/(1 - r).

11. A satellite is orbiting around the earth. By what percentage should we increase its velocity so as to enable it to escape away from earth ?
(a) 41.4%
(b) 50%
(c) 82.8%
(d) 100%

Answer: (a) 41.4%

1. Orbital velocity v_o = √(GM/r) and escape velocity v_e = √(2GM/r) at the same distance.
2. v_e = √2 × v_o = 1.414 v_o.
3. Increase = 0.414 v_o = 41.4%.
4. Hence, the correct option is (a).

Teacher's Note:
a) Escape velocity is √2 times the orbital velocity at the same height.
b) The percentage is based on the original orbital velocity.

12. A planet is revolving around the sun in elliptical orbit. Its closest distance from the sun is r and the farthest distance is R. If the orbital velocity of the planet closest to the sun be v, then what is the velocity at the farthest point ?
(a) v r/R
(b) v R/r
(c) v \( \left(\frac{r}{R}\right)^{1/2} \)
(d) v \( \left(\frac{R}{r}\right)^{1/2} \)

Answer: (a) v r/R

1. Angular momentum is conserved: m v r = m v' R.
2. v' = v r/R.
3. Hence, the correct option is (a).

Teacher's Note:
a) At the farthest point the planet moves slowest, so v' is less than v.
b) Conservation of angular momentum gives the ratio directly, with no square roots.

13. A body of mass m rises to a height h = R/5 from the earth's surface where R is earth's radius. If g is acceleration due to gravity at the earth's surface, the increase in potential energy is
(a) m g h
(b) (4/5) m g h
(c) (5/6) m g h
(d) (6/7) m g h

Answer: (c) (5/6) m g h

1. Increase in potential energy = GMm(1/R - 1/(R + h)) = GMm h/(R(R + h)).
2. Using GM = gR², this equals mgh × R/(R + h) = mgh/(1 + h/R).
3. With h = R/5: mgh/(1 + 1/5) = (5/6) mgh.
4. Hence, the correct option is (c).

Teacher's Note:
a) The formula mgh holds only when h is very small compared with R.
b) Rewrite GM as gR² to bring g into the answer.

14. Two particles of mass m and M are initially at rest and infinitely separated from each other. Due to mutual interaction they approach each other. Their relative velocity of approach at a distance of separation d between them is
(a) 2 G \( \left(\frac{d}{M+m}\right) \)
(b) \( \sqrt{\frac{2G(M+m)}{d}} \)
(c) \( \sqrt{\frac{2G(M-m)}{d}} \)
(d) \( \sqrt{\frac{2Gd}{M+m}} \)

Answer: (b) \( \sqrt{\frac{2G(M+m)}{d}} \)

1. Loss of potential energy = gain in kinetic energy: GMm/d = (1/2) μ v², where μ = Mm/(M + m).
2. GMm/d = (1/2) × (Mm/(M + m)) × v².
3. v² = 2G(M + m)/d.
4. So v = √(2G(M + m)/d).
5. Hence, the correct option is (b).

Teacher's Note:
a) For two moving bodies use the reduced mass and the relative velocity.
b) Check the dimensions: the quantity under the root must be (velocity)².

15. The energy needed to lift a 2 kg mass vertically upwards through 2 m distance on a planet whose size is the same and mass 4 times as that of the earth is
(a) 160 J
(b) 80 J
(c) 40 J
(d) 20 J

Answer: (a) 160 J

1. g = GM/R², so for 4 times the mass and same radius g' = 4g = 40 m/s² (taking g = 10 m/s²).
2. Energy = m g' h = 2 × 40 × 2 = 160 J.
3. Hence, the correct option is (a).

Teacher's Note:
a) The energy is the work done against gravity, W = mg'h.
b) Take g = 10 m/s² unless told otherwise, as the options are based on it.

16. The mean radius of the earth's orbit around the sun is 1.5 × 10¹¹ m and that of the orbit of mercury is 6 × 10¹⁰ m. The mercury will revolve around the sun in nearly
(a) \( \sqrt{2/5} \) years
(b) 2/5 years
(c) (2/5)² years
(d) (2/5)^3/2 years

Answer: (d) (2/5)^3/2 years

1. Kepler's law: T² is proportional to r³, so T_m/T_e = (r_m/r_e)^3/2.
2. r_m/r_e = 6 × 10¹⁰ / 1.5 × 10¹¹ = 0.4 = 2/5.
3. T_m = (2/5)^3/2 years, which is about 0.25 years.
4. Hence, the correct option is (d).

Teacher's Note:
a) The earth's period is 1 year, so the ratio itself is the answer in years.
b) Do not drop the 3/2 power.

17. If a planet has a mass twice that of the earth and its radius is three times that of the earth, then a 1 kg mass on its surface would weigh about
(a) 1.1 N
(b) 2.2 N
(c) 3.3 N
(d) 4.4 N

Answer: (b) 2.2 N

1. g' = g × (M'/M) × (R/R')² = 9.8 × 2 × (1/9) = 2.18 m/s².
2. Weight = 1 × 2.18 = 2.2 N approximately.
3. Hence, the correct option is (b).

Teacher's Note:
a) Radius enters as the square in the denominator, so tripling it divides g by 9.
b) Weight is mass times local g, so mass stays 1 kg but weight changes.

18. The radius of the earth is 6400 km and g = 10 m/s². In order that a body of 5 kg weighs zero at the equator, the angular speed of the earth should be
(a) \( \frac{1}{180} \) rad/s
(b) \( \frac{1}{400} \) rad/s
(c) \( \frac{1}{800} \) rad/s
(d) \( \frac{1}{1600} \) rad/s

Answer: (c) \( \frac{1}{800} \) rad/s

1. Apparent weight at the equator is zero when g = Rω².
2. ω = √(g/R) = √(10/6.4 × 10⁶) = √(1.5625 × 10⁻⁶).
3. ω = 1.25 × 10⁻³ rad/s = 1/800 rad/s.
4. Hence, the correct option is (c).

Teacher's Note:
a) The mass of the body (5 kg) is not needed; it is a distractor.
b) Convert the radius to metres before using g in m/s².

19. At what depth below the surface of the earth is the value of g same as that at a height of 5 km ?
(a) 10 km
(b) 7.5 km
(c) 5 km
(d) 2.5 km

Answer: (a) 10 km

1. At a small height h, g' = g(1 - 2h/R).
2. At a depth d, g' = g(1 - d/R).
3. Equating, d/R = 2h/R, so d = 2h = 2 × 5 = 10 km.
4. Hence, the correct option is (a).

Teacher's Note:
a) Both formulas are approximations valid when h and d are much smaller than R.
b) Gravity falls twice as fast with height as with depth.

20. If R is the radius of the earth and g is the acceleration due to gravity on the surface of the earth, the mean density of the earth is
(a) \( \frac{3πR}{4gG} \)
(b) \( \frac{4πG}{3gR} \)
(c) \( \frac{4RG}{3πg} \)
(d) \( \frac{3g}{4πRG} \)

Answer: (d) \( \frac{3g}{4πRG} \)

1. g = GM/R² and M = (4/3)πR³ρ.
2. g = (4/3)πGρR.
3. ρ = 3g/(4πRG).
4. Hence, the correct option is (d).

Teacher's Note:
a) Express the mass through density and volume to link g and ρ.
b) Check the dimensions of the final expression to avoid swapping R and g.

21. Acceleration due to gravity on the surface of the moon is (1/6)th of that on the surface of the earth and the diameter of the moon is (1/4)th of that of the earth. If the escape velocity from the earth is 11 km/s then the escape velocity from the moon is
(a) \( \frac{11}{24} \) km/s
(b) 11√24 km/s
(c) \( \frac{11}{\sqrt{24}} \) km/s
(d) 11 × 24 km/s

Answer: (c) \( \frac{11}{\sqrt{24}} \) km/s

1. Escape velocity v = √(2gR).
2. g_m = g/6 and R_m = R/4 (the radius ratio equals the diameter ratio).
3. v_m/v_e = √((1/6) × (1/4)) = 1/√24.
4. v_m = 11/√24 km/s, which is about 2.2 km/s.
5. Hence, the correct option is (c).

Teacher's Note:
a) Diameter ratio and radius ratio are the same, so no extra factor of 2 is needed.
b) Both g and R enter under the root.

22. A particle is projected vertically upwards with a velocity \( \sqrt{gR} \), where R denotes the radius of the earth and g the acceleration due to gravity on the surface of the earth. Then the maximum height ascended by the particle is
(a) R/2
(b) R
(c) 2R
(d) 5R/4

Answer: (b) R

1. Energy conservation: (1/2)mv² - GMm/R = -GMm/(R + h).
2. With v² = gR and GM = gR²: gR/2 - gR = -gR²/(R + h).
3. -gR/2 = -gR²/(R + h), so R + h = 2R.
4. h = R.
5. Hence, the correct option is (b).

Teacher's Note:
a) The height is measured from the surface, not from the centre of the earth.
b) The formula h = v²R/(2gR - v²) gives the same value.

23. Two bodies of masses 10³ kg and 10⁵ kg are separated by a distance of 2 m. At what distance from the smaller body will the intensity of the gravitational field be zero ?
(a) \( \frac{1}{11} \) m
(b) \( \frac{2}{11} \) m
(c) \( \frac{1}{9} \) m
(d) \( \frac{2}{9} \) m

Answer: (b) \( \frac{2}{11} \) m

1. The null point lies between the masses at a distance x from the smaller mass.
2. G × 10³/x² = G × 10⁵/(2 - x)², so (2 - x)/x = √100 = 10.
3. 2 - x = 10x, so x = 2/11 m.
4. Hence, the correct option is (b).

Teacher's Note:
a) The null point is always nearer to the smaller mass.
b) Equate the two field magnitudes, then take the square root of the ratio of masses.

24. The figure shows the motion of a planet around the sun in an elliptic orbit with the sun at one focus. The shaded areas A and B can be assumed to be equal. If t₁ and t₂ represent the times taken by the planet to move from a to b and from c to d respectively, then
(a) t₁ < t₂
(b) t₁ > t₂
(c) t₁ = t₂
(d) from the given information the relation between t₁ and t₂ cannot be determined.

[Figure: An ellipse with the sun S at a focus. Near the top, points b and a on the orbit; the thin shaded triangular region A is between S and the arc ba. Near the bottom, points c and d on the orbit; the shaded region B is between S and the arc cd. Arrows on the orbit show the direction of motion.]

Answer: (c) t₁ = t₂

1. By Kepler's second law, the line joining the sun and the planet sweeps equal areas in equal times.
2. Areas A and B are equal, so the times taken are equal.
3. Hence, t₁ = t₂ and the correct option is (c).

Teacher's Note:
a) The areal velocity is constant, so equal swept areas mean equal times even though the arc lengths differ.
b) Do not compare the arc lengths ab and cd.

25. If g is the acceleration due to gravity at the surface of the earth, then the energy required to launch a satellite of mass m from the surface of the earth into a circular orbit at an altitude of 2R, R being the radius of the earth, is
(a) \( \frac{mgR}{6} \)
(b) \( \frac{mgR}{3} \)
(c) \( \frac{2mgR}{3} \)
(d) \( \frac{5mgR}{6} \)

Answer: (d) \( \frac{5mgR}{6} \)

1. Orbit radius = R + 2R = 3R.
2. Energy in orbit = -GMm/(2 × 3R) = -GMm/6R and energy on the surface = -GMm/R.
3. Energy required = -GMm/6R + GMm/R = (5/6)GMm/R.
4. Using GM = gR², energy required = 5mgR/6.
5. Hence, the correct option is (d).

Teacher's Note:
a) The satellite needs both potential energy and orbital kinetic energy, so use the total energy in orbit.
b) Altitude 2R means the orbit radius is 3R, a common trap.

26. Two satellites S₁ and S₂ revolve round a planet in coplanar circular orbits in the same sense. Their periods of revolution are 1 hour and 8 hours, respectively. The radius of the orbit of S₁ is 10⁴ km. The speed of S₂ relative to S₁, when they are closest, in km/hr, is
(a) 10⁴ π
(b) 2 × 10⁴ π
(c) 10⁴ π /2
(d) 4 × 10⁴ π

Answer: (a) 10⁴ π

1. T² is proportional to r³, so r₂/r₁ = (8/1)^2/3 = 4 and r₂ = 4 × 10⁴ km.
2. v₁ = 2πr₁/T₁ = 2π × 10⁴ km/hr.
3. v₂ = 2π × 4 × 10⁴/8 = π × 10⁴ km/hr.
4. When closest, both move in the same direction, so relative speed = 2π × 10⁴ - π × 10⁴ = 10⁴ π km/hr.
5. Hence, the correct option is (a).

Teacher's Note:
a) Satellites moving in the same sense are in the same direction at the closest point, so subtract the speeds.
b) Use Kepler's law first to find the unknown orbit radius.
c) Speed = circumference divided by period.