JEE Physics Gravitation MCQs Set 16

Download JEE MCQs for JEE Physics: Gravitation

Explore reliable objective questions for Gravitation tailored for JEE learners. Utilizing these Physics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Chapter-wise Objective Questions: Gravitation

View or download the dedicated Gravitation MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Physics evaluations.

Multiple Choice Questions

Question 1: What gravitational force does the full Moon (mass 7.34 × 10²² kg) exert on a human brain of mass 1.4 kg when the centre of the Moon is 378000 km away? (G = 6.67 × 10⁻¹¹ N m² kg⁻²)
(a) 2.4 × 10⁻⁵ N
(b) 4.8 × 10⁻⁵ N
(c) 1.8 × 10⁻⁷ N
(d) 4.8 × 10⁻³ N
Show Answer & Explanation

Answer: (b) 4.8 × 10⁻⁵ N

Explanation:
1. \( F=\frac{Gm_1m_2}{r^2}=\frac{6.67\times10^{-11}\times1.4\times7.34\times10^{22}}{(3.78\times10^{8})^2} \)
2. F ≈ 4.8 × 10⁻⁵ N.

Teacher's Note:
1. Convert the distance to metres before substituting.
2. The pull is tiny because the brain's mass is small.

Question 2: The gravitational potential in a region is V = (20x + 40y) J/kg. The magnitude of the gravitational force on a particle of mass 0.250 kg placed at the point (2, 4) is:
(a) 15 N
(b) 44.7 N
(c) 5.0 N
(d) 11.2 N
Show Answer & Explanation

Answer: (d) 11.2 N

Explanation:
1. Field \( \vec E=-\nabla V=-(20\hat i+40\hat j) \) N/kg.
2. Magnitude \( \sqrt{20^2+40^2}=44.7 \) N/kg.
3. Force = 0.25 × 44.7 = 11.2 N.

Teacher's Note:
1. Gravitational field is the negative gradient of potential.
2. Force = mass × field.

Question 3: A body, initially at rest at a height R above the surface of the earth (radius R), falls freely towards the earth. Its velocity on reaching the surface, in terms of g at the surface, is:
(a) \( \sqrt{gR} \)
(b) \( \sqrt{gR/2} \)
(c) \( \sqrt{2gR} \)
(d) \( 2\sqrt{gR} \)
Show Answer & Explanation

Answer: (a) \( \sqrt{gR} \)

Explanation:
1. Energy conservation between r = 2R and r = R: \( \tfrac12v^2=GM\left(\frac1R-\frac1{2R}\right)=\frac{GM}{2R} \).
2. So \( v^2=GM/R=gR \).

Teacher's Note:
1. Do not use v² = 2gh here, because g changes with height.
2. GM = gR².

Question 4: Two satellites move in circular orbits of radius r and 1.01 r about the same planet. By roughly what percentage is the time period of the second satellite greater than that of the first?
(a) 1%
(b) 1.5%
(c) 2%
(d) 3%
Show Answer & Explanation

Answer: (b) 1.5%

Explanation:
1. Kepler's third law: \( T\propto r^{3/2} \).
2. Ratio = (1.01)^{3/2} ≈ 1.015.
3. So the period is about 1.5% larger.

Teacher's Note:
1. For small changes, ΔT/T ≈ (3/2)(Δr/r).
2. Here (3/2) × 1% = 1.5%.

Question 5: Two identical stars of mass M orbit about their centre of mass. Each orbit is circular with radius R, and the stars are always on opposite ends of a diameter. The minimum energy required to separate the two stars to infinity is:
(a) \( \frac{GM^2}{R} \)
(b) \( \frac{GM^2}{8R} \)
(c) \( \frac{GM^2}{2R} \)
(d) \( \frac{GM^2}{4R} \)
Show Answer & Explanation

Answer: (d) \( \frac{GM^2}{4R} \)

Explanation:
1. Separation 2R; speed from \( \frac{GM^2}{4R^2}=\frac{Mv^2}{R} \) gives \( v^2=\frac{GM}{4R} \).
2. Total KE = \( 2\times\tfrac12Mv^2=\frac{GM^2}{4R} \); PE = \( -\frac{GM^2}{2R} \).
3. Total energy = −GM²/4R, so the energy needed is GM²/4R.

Teacher's Note:
1. Binding energy is the magnitude of the (negative) total energy.
2. Total energy of a bound circular system equals minus its kinetic energy.

Question 6: Two earth satellites A and B of equal mass are in circular orbits at altitudes 6400 km and 19200 km (radius of the earth 6400 km). The ratio of the kinetic energy of A to that of B is:
(a) 4 : 1
(b) 2 : 1
(c) 1 : 2
(d) 3 : 1
Show Answer & Explanation

Answer: (b) 2 : 1

Explanation:
1. Orbit radii: r_A = 12800 km, r_B = 25600 km.
2. \( KE=\frac{GMm}{2r}\propto\frac1r \).
3. KE_A : KE_B = r_B : r_A = 2 : 1.

Teacher's Note:
1. Altitude plus the earth's radius gives the orbital radius.
2. Kinetic energy of a circular orbit varies as 1/r.

Question 7: Saturn is about six times farther from the Sun than Mars. Which statement is correct?
(a) Saturn has the greater period of revolution, and Mars has the greater orbital speed and angular speed
(b) Saturn has the greater period, orbital speed and angular speed
(c) Saturn has the greater period and orbital speed, Mars has the greater angular speed
(d) Mars has the greater period, Saturn has the greater angular speed
Show Answer & Explanation

Answer: (a) Saturn has the greater period of revolution, and Mars has the greater orbital speed and angular speed

Explanation:
1. \( T\propto r^{3/2} \), so Saturn (larger r) has the longer period (about 15 times).
2. Orbital speed \( v\propto r^{-1/2} \), so Mars is faster.
3. Angular speed is 2π/T, so Mars also has the larger angular speed.

Teacher's Note:
1. Farther planets move slower and take longer.
2. Angular speed is inversely proportional to the period.

Question 8: The acceleration due to gravity at a height R/20 above the earth's surface is 9 m/s². The approximate value at a point at the same distance R/20 below the surface is:
(a) 8.6 m/s²
(b) 9.9 m/s²
(c) 9.4 m/s²
(d) 10.4 m/s²
Show Answer & Explanation

Answer: (c) 9.4 m/s²

Explanation:
1. Surface value: \( g_0=9\times(1.05)^2=9.92 \) m/s².
2. At depth d: \( g'=g_0(1-d/R)=9.92\times0.95 \).
3. g' ≈ 9.4 m/s².

Teacher's Note:
1. Outside: g falls as 1/r². Inside: g falls linearly with depth.
2. First recover the surface value.

Question 9: A pendulum has a period of exactly 1.00 s at the equator. Assuming the earth is a sphere and including its rotation, its period at the north pole is nearly (take g at the pole = 9.8 m/s², R = 6400 km):
(a) 1.000 s
(b) 0.990 s
(c) 0.998 s
(d) 1.002 s
Show Answer & Explanation

Answer: (c) 0.998 s

Explanation:
1. At the equator \( g_{eq}=g_p-\omega^2R \) with \( \omega=7.27\times10^{-5} \) rad/s.
2. \( \omega^2R/g\approx3.5\times10^{-3} \).
3. T ∝ 1/√g, so \( T_p=T_{eq}\sqrt{g_{eq}/g_p}\approx0.998 \) s.

Teacher's Note:
1. Rotation reduces the effective gravity at the equator, so the pendulum is slower there.
2. The pole value is slightly smaller than 1.00 s.

Question 10: Four identical particles of mass m orbit on a common circle of radius r, located at the corners of a square inscribed in the circle, in the same direction and with the same speed, only due to their mutual gravitational attraction. The speed of each particle is:
(a) zero
(b) \( \sqrt{\frac{Gm}{r}(1+2\sqrt2)} \)
(c) \( \sqrt[3]{\frac{Gm}{r}} \)
(d) \( \left[\frac{Gm}{r}\left(\frac{1+2\sqrt2}{4}\right)\right]^{1/2} \)
Show Answer & Explanation

Answer: (d) \( \left[\frac{Gm}{r}\left(\frac{1+2\sqrt2}{4}\right)\right]^{1/2} \)

Explanation:
1. Side of the square = r√2; diagonal = 2r.
2. Net inward force = \( 2\cdot\frac{Gm^2}{2r^2}\cos45^\circ+\frac{Gm^2}{4r^2}=\frac{Gm^2}{r^2}\left(\frac{\sqrt2}{2}+\frac14\right) \).
3. Setting this equal to \( mv^2/r \): \( v^2=\frac{Gm}{r}\cdot\frac{1+2\sqrt2}{4} \).

Teacher's Note:
1. Resolve the forces from the two neighbours along the radius and add the pull from the opposite particle.
2. The tangential components of the neighbours cancel.

Question 11: Three identical stars of mass M are at the vertices of an equilateral triangle of side L. They revolve in a circular orbit circumscribing the triangle, keeping the shape. The speed of each star is:
(a) not possible
(b) \( \sqrt{\frac{GM}{L}} \)
(c) \( \sqrt{\frac{2GM}{L}} \)
(d) \( 2\sqrt{\frac{GM}{L}} \)
Show Answer & Explanation

Answer: (b) \( \sqrt{\frac{GM}{L}} \)

Explanation:
1. Circumradius = L/√3.
2. Net force on a star = \( 2\frac{GM^2}{L^2}\cos30^\circ=\frac{\sqrt3GM^2}{L^2} \).
3. \( \frac{Mv^2}{L/\sqrt3}=\frac{\sqrt3GM^2}{L^2}\Rightarrow v=\sqrt{GM/L} \).

Teacher's Note:
1. The net force is directed towards the centroid and provides the centripetal force.
2. Circumradius of an equilateral triangle is side/√3.

Question 12: The gravitational field in a region is E = −k/r (k constant). If the reference point is at distance d_i, where the potential is V_i, the potential at distance r is:
(a) \( V=k\ln\frac1{d_i} \)
(b) \( V=\ln\frac r{d_i}+\frac{V_i}{k} \)
(c) \( V=k\ln\frac r{d_i}+V_i \)
(d) \( V=\ln\frac r{d_i}+kV_i \)
Show Answer & Explanation

Answer: (c) \( V=k\ln\frac r{d_i}+V_i \)

Explanation:
1. \( V-V_i=-\int_{d_i}^{r}E\,dr=\int_{d_i}^{r}\frac kr\,dr=k\ln\frac r{d_i} \).
2. So \( V=k\ln(r/d_i)+V_i \).

Teacher's Note:
1. Potential difference is minus the integral of field along the path.
2. The integral of 1/r gives a natural logarithm.

Question 13: A very large number of particles, each of mass m, are placed at distances 1 m, 2 m, 4 m, 8 m and so on from the origin along a line. The total gravitational potential at the origin is:
(a) −2Gm
(b) −8Gm
(c) −4Gm
(d) −3Gm
Show Answer & Explanation

Answer: (a) −2Gm

Explanation:
1. \( V=-Gm\left(1+\tfrac12+\tfrac14+\tfrac18+\dots\right) \).
2. The geometric series sums to 2, so V = −2Gm.

Teacher's Note:
1. Potential is a scalar and simply adds.
2. Infinite geometric series with ratio 1/2 sums to 1/(1 − 1/2).

Question 14: A body starts from rest at a point at distance R₀ from the centre of the earth (mass M, radius R). The velocity acquired when it reaches the surface of the earth is:
(a) \( \sqrt{2GM\left(\frac1R-\frac1{R_0}\right)} \)
(b) \( 2GM\sqrt{\frac1R-\frac1{R_0}} \)
(c) \( GM\left(\frac1R-\frac1{R_0}\right) \)
(d) \( 2GM\left(\frac1R-\frac1{R_0}\right) \)
Show Answer & Explanation

Answer: (a) \( \sqrt{2GM\left(\frac1R-\frac1{R_0}\right)} \)

Explanation:
1. \( \tfrac12v^2=GM\left(\frac1R-\frac1{R_0}\right) \) from energy conservation.
2. So \( v=\sqrt{2GM(1/R-1/R_0)} \).

Teacher's Note:
1. Loss of potential energy appears as kinetic energy.
2. Check the dimensions: GM/R is velocity squared.

Question 15: Three equal masses m are at the corners of an equilateral triangle of side a. A fourth particle of mass m is placed at the mid-point of one side. The net gravitational force on it is:
(a) \( \frac{Gm^2}{a^2} \)
(b) zero
(c) \( \frac{3Gm^2}{a^2} \)
(d) \( \frac{4Gm^2}{3a^2} \)
Show Answer & Explanation

Answer: (d) \( \frac{4Gm^2}{3a^2} \)

Explanation:
1. The two masses at the ends of that side pull in opposite directions with equal force and cancel.
2. The third mass is at distance \( \frac{\sqrt3}2a \), so force \( \frac{Gm^2}{3a^2/4}=\frac{4Gm^2}{3a^2} \).

Teacher's Note:
1. Use symmetry to cancel the equal and opposite pulls.
2. Distance to the opposite vertex is the altitude of the triangle.

Question 16: Three equal masses m are at the corners of an equilateral triangle of side a. The work done by an external agent in changing the side to 2a is:
(a) \( \frac{Gm^2}{a} \)
(b) \( \frac{3Gm^2}{a} \)
(c) \( \frac{4Gm^2}{3a} \)
(d) \( \frac{3Gm^2}{2a} \)
Show Answer & Explanation

Answer: (d) \( \frac{3Gm^2}{2a} \)

Explanation:
1. Initial energy \( U_i=-\frac{3Gm^2}{a} \); final \( U_f=-\frac{3Gm^2}{2a} \).
2. Work = U_f − U_i = 3Gm²/(2a).

Teacher's Note:
1. Work done by the external agent equals the increase in potential energy.
2. Three pairs, each at the same separation.

Question 17: Two masses m are held fixed at a distance a apart, and a third mass m is at a distance a from both (equilateral arrangement). The third mass is released. Its speed when it reaches the mid-point of the line joining the other two is:
(a) \( 2\sqrt{\frac{Gm}{a}} \)
(b) \( \sqrt{\frac{Gm}{2a}} \)
(c) \( \sqrt{\frac{2Gm}{a}} \)
(d) \( \sqrt{\frac{Gm}{a}} \)
Show Answer & Explanation

Answer: (a) \( 2\sqrt{\frac{Gm}{a}} \)

Explanation:
1. Potential energy of the moving mass at start: \( -\frac{2Gm^2}{a} \); at the mid-point: \( -2\cdot\frac{Gm^2}{a/2}=-\frac{4Gm^2}{a} \).
2. Kinetic energy gained = 2Gm²/a, so \( \tfrac12mv^2=\frac{2Gm^2}{a} \).
3. v = 2√(Gm/a).

Teacher's Note:
1. At the mid-point it is a/2 from each fixed mass.
2. Use energy conservation rather than forces.

Question 18: A satellite orbits very close to the surface of a planet of uniform density ρ. Its period of revolution is:
(a) proportional to 1/ρ
(b) independent of ρ
(c) proportional to ρ
(d) \( \text{proportional to }1/\sqrt\rho \)
Show Answer & Explanation

Answer: (d) \( \text{proportional to }1/\sqrt\rho \)

Explanation:
1. \( T=2\pi\sqrt{R^3/GM} \) and \( M=\tfrac43\pi R^3\rho \).
2. So \( T=\sqrt{3\pi/(G\rho)} \), i.e. T ∝ 1/√ρ.

Teacher's Note:
1. For a near-surface orbit the radius equals the planet's radius, so it cancels out.
2. Only the density of the planet determines the period.

Question 19: An artificial satellite of the earth releases a package. If air resistance is neglected, the package will:
(a) never reach the earth
(b) hit the earth exactly below the point of release
(c) hit the earth ahead of the point of release
(d) hit the earth behind the point of release
Show Answer & Explanation

Answer: (a) never reach the earth

Explanation:
1. On release the package has the same velocity as the satellite, so it is at the same height with the orbital speed.
2. It therefore stays in the same orbit and continues to move along with the satellite.

Teacher's Note:
1. Release does not change the velocity, and orbital motion is just free fall around the earth.
2. The package becomes a tiny satellite itself.

Question 20: Which of the following would make an orbiting satellite escape from the earth? (A) Its speed is increased by (√2 − 1) × 100%. (B) Its speed is made √1.5 times its initial value. (C) Its kinetic energy is doubled. (D) It stops moving in the orbit. Which statements are correct?
(a) A, B and C only
(b) A and C only
(c) B and D only
(d) A, B, C and D
Show Answer & Explanation

Answer: (b) A and C only

Explanation:
1. Escape speed = √2 × orbital speed, so the speed must become √2 v, an increase of (√2 − 1)100%: (A) correct.
2. (B) √1.5 < √2, so it only goes to a bigger orbit: wrong.
3. (C) Doubling KE multiplies v by √2: correct.
4. (D) Stopping makes it fall to the earth: wrong.

Teacher's Note:
1. Escape needs total energy zero, i.e. KE equal to the magnitude of PE.
2. Orbital KE is half of |PE|, so doubling KE reaches zero total energy.

Question 21: If the radius of the orbit of a satellite is decreased, which statements are correct? (A) Its kinetic energy decreases. (B) Its potential energy decreases. (C) Its mechanical energy decreases. (D) Its speed decreases.
(a) C and D only
(b) A and B only
(c) B and C only
(d) A and D only
Show Answer & Explanation

Answer: (c) B and C only

Explanation:
1. \( KE=\frac{GMm}{2r},\ PE=-\frac{GMm}{r},\ E=-\frac{GMm}{2r} \).
2. As r decreases, KE and speed increase, PE becomes more negative (decreases), and E decreases (more negative).
3. So B and C are correct.

Teacher's Note:
1. Total energy of a bound satellite is negative and becomes more negative in a lower orbit.
2. A satellite in a lower orbit moves faster.

Question 22: A planet of mass m revolves around the Sun of mass M in an elliptical orbit with minimum and maximum distances r₁ and r₂. If its minimum speed is \( \sqrt{\frac{2GMr_1}{(r_1+r_2)r_2}} \), its maximum speed is:
(a) \( \sqrt{\frac{2GM}{r_1+r_2}} \)
(b) \( \sqrt{\frac{2GMr_2}{(r_1+r_2)r_1}} \)
(c) \( \sqrt{\frac{2GMr_1}{(r_1+r_2)r_2}} \)
(d) \( \sqrt{\frac{2Gmr_2}{(r_1+r_2)r_1}} \)
Show Answer & Explanation

Answer: (b) \( \sqrt{\frac{2GMr_2}{(r_1+r_2)r_1}} \)

Explanation:
1. Angular momentum is conserved: \( v_{max}r_1=v_{min}r_2 \).
2. \( v_{max}=v_{min}\frac{r_2}{r_1}=\sqrt{\frac{2GMr_1}{(r_1+r_2)r_2}}\cdot\frac{r_2}{r_1}=\sqrt{\frac{2GMr_2}{(r_1+r_2)r_1}} \).

Teacher's Note:
1. The planet is fastest at its closest approach (perihelion).
2. Conservation of angular momentum links the two extreme points.

Question 23: For the earth, which statements are correct? (A) The gravitational field is zero both at the centre and at infinity. (B) The gravitational potential is zero both at the centre and at infinity. (C) The potential is the same at the centre and at infinity but not zero. (D) The gravitational potential is minimum at the centre.
(a) A and B only
(b) B and C only
(c) A and D only
(d) C and D only
Show Answer & Explanation

Answer: (c) A and D only

Explanation:
1. Field inside a uniform sphere is proportional to r, so zero at the centre; it is also zero at infinity: (A) correct.
2. Potential at the centre = −3GM/2R and at infinity = 0: (B) and (C) wrong.
3. The potential is lowest (most negative) at the centre: (D) correct.

Teacher's Note:
1. Potential at the centre of a uniform solid sphere is 1.5 times its surface value.
2. Field zero does not mean potential zero.

Question 24: The gravitational field in a region is \( \vec E=(3\hat i-4\hat j) \) N/kg. The work done (in joule) in displacing a particle of mass 1 kg through 1 m along the line 4y = 3x + 9 is:
(a) 7
(b) 5
(c) −5
(d) 0
Show Answer & Explanation

Answer: (d) 0

Explanation:
1. A unit vector along the line (slope 3/4) is \( (4\hat i+3\hat j)/5 \).
2. Work = \( m\vec E\cdot\vec d=(3\cdot4-4\cdot3)/5=0 \) J.

Teacher's Note:
1. Work by a field is the dot product of force and displacement.
2. The field is perpendicular to this line.

Question 25: The two stars of a binary system, of masses M₁ and M₂, move in circles about their centre of mass (which does not accelerate), with orbital radii R₁ and R₂. Two such stars have orbital radii 1.00 × 10⁹ km and 5.00 × 10⁸ km and orbital period 44.5 years. The total mass of the system is approximately (G = 6.67 × 10⁻¹¹ SI, 1 yr = 3.16 × 10⁷ s):
(a) 2.0 × 10³⁰ kg
(b) 1.0 × 10³⁰ kg
(c) 5.0 × 10²⁹ kg
(d) 1.0 × 10³² kg
Show Answer & Explanation

Answer: (b) 1.0 × 10³⁰ kg

Explanation:
1. Separation = R₁ + R₂ = 1.5 × 10¹² m; T = 1.405 × 10⁹ s.
2. \( M_1+M_2=\frac{4\pi^2(R_1+R_2)^3}{GT^2} \).
3. = 4π² (3.375 × 10³⁶) / (6.67 × 10⁻¹¹ × 1.97 × 10¹⁸) ≈ 1.0 × 10³⁰ kg.

Teacher's Note:
1. Both stars have the same period and the same angular speed.
2. Use Kepler's law with the separation, not an individual radius.

Question 26: The escape speed from the Sun's gravitational field for an object at the distance of the earth's orbit (radius R) is how many times the earth's orbital speed?
(a) 1
(b) 2
(c) \( 1/\sqrt2 \)
(d) \( \sqrt2 \)
Show Answer & Explanation

Answer: (d) \( \sqrt2 \)

Explanation:
1. Orbital speed \( v_o=\sqrt{GM_s/R} \).
2. Escape speed \( v_e=\sqrt{2GM_s/R}=\sqrt2\,v_o \).

Teacher's Note:
1. The escape speed is always √2 times the circular orbital speed at the same distance.
2. This is true for any central body.

Question 27: An object far from the Sun has speed v₀ and aiming parameter ℓ (the perpendicular distance between its line of motion and the Sun's centre). The minimum distance it gets to the Sun (mass M) is:
(a) \( \frac{GM+\sqrt{G^2M^2+v_0^4\ell^2}}{v_0^2} \)
(b) \( \frac{\ell v_0}{\sqrt{2GM}} \)
(c) \( \frac{2GM}{v_0^2} \)
(d) \( \frac{-GM+\sqrt{G^2M^2+v_0^4\ell^2}}{v_0^2} \)
Show Answer & Explanation

Answer: (d) \( \frac{-GM+\sqrt{G^2M^2+v_0^4\ell^2}}{v_0^2} \)

Explanation:
1. At closest approach, angular momentum: \( v_0\ell=vr \); energy: \( \tfrac12v_0^2=\tfrac12v^2-\frac{GM}r \).
2. Eliminating v: \( v_0^2r^2+2GMr-v_0^2\ell^2=0 \).
3. The positive root is \( r=\frac{-GM+\sqrt{G^2M^2+v_0^4\ell^2}}{v_0^2} \).

Teacher's Note:
1. Use two conservation laws (angular momentum and energy) at closest approach.
2. Choose the positive root of the quadratic.

Question 28: A rod of length ℓ extends from x = α to x = α + ℓ on the x-axis. Its mass per unit length is a + bx². The gravitational force it exerts on a point mass m at the origin is:
(a) \( Gm\left[a\left(\frac1{\alpha+\ell}-\frac1\alpha\right)+b\ell\right] \)
(b) \( Gm\left[a\left(\frac1\alpha-\frac1{\alpha+\ell}\right)+b\ell\right] \)
(c) \( \frac{Gm(a+bx^2)}{\ell^2} \)
(d) \( Gm\left[\alpha\left(\frac1a-\frac1{a+\ell}\right)+b\ell\right] \)
Show Answer & Explanation

Answer: (b) \( Gm\left[a\left(\frac1\alpha-\frac1{\alpha+\ell}\right)+b\ell\right] \)

Explanation:
1. \( dF=\frac{Gm(a+bx^2)dx}{x^2} \).
2. \( F=Gm\int_\alpha^{\alpha+\ell}\left(\frac a{x^2}+b\right)dx=Gm\left[a\left(\frac1\alpha-\frac1{\alpha+\ell}\right)+b\ell\right] \).

Teacher's Note:
1. Integrate the force from each small element, all of which act in the same direction.
2. ∫dx/x² = −1/x.

Question 29: A uniform ring of mass M and radius R lies in a plane with its axis through the centre of a uniform sphere of mass m and radius R, the centre of the ring being at distance √3 R from the centre of the sphere. The gravitational force exerted by the ring on the sphere is:
(a) \( \frac{GMm}{8R^2} \)
(b) \( \frac{GMm}{3R^2} \)
(c) \( \sqrt3\,\frac{GMm}{8R^2} \)
(d) \( \sqrt3\,\frac{GMm}{R^2} \)
Show Answer & Explanation

Answer: (c) \( \sqrt3\,\frac{GMm}{8R^2} \)

Explanation:
1. For a point on the axis of a ring: \( F=\frac{GMmx}{(R^2+x^2)^{3/2}} \).
2. The sphere acts as a point mass at its centre, x = √3 R:
3. \( F=\frac{GMm\sqrt3R}{(4R^2)^{3/2}}=\frac{\sqrt3GMm}{8R^2} \).

Teacher's Note:
1. A uniform sphere attracts outside bodies as if its mass were at its centre.
2. The ring's axial formula is a standard result.

Question 30: The gravitational potentials at the centres of two homogeneous spherical shells A and B of the same surface mass density are in the ratio 3 : 4 (they are far apart). If the two shells coalesce into a single shell with the same surface mass density, the ratio of the potential at an internal point of the new shell to that of shell A is:
(a) 5 : 3
(b) 4 : 3
(c) 3 : 5
(d) 3 : 2
Show Answer & Explanation

Answer: (a) 5 : 3

Explanation:
1. Potential inside a shell: \( V=-\frac{GM}R=-4\pi G\sigma R \), so V ∝ R.
2. R_A : R_B = 3 : 4. The new shell's area is the sum: \( R^2=3^2+4^2=25 \), R = 5.
3. Ratio new : A = 5 : 3.

Teacher's Note:
1. With constant surface density, mass is proportional to area.
2. Add areas, not radii.

Free study material for Physics

Multiple Choice Questions (MCQs) for JEE Physics Gravitation

Chapter MCQs with Answers for JEE Physics

Review structured objective questions for JEE Physics Gravitation. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.

Expert Practice Material for JEE Physics

Built using the official NCERT book for JEE, these Physics objective sets provide reliable academic guidance. Pair your practice with our recommended NCERT solutions to master optimal problem-solving approaches.

Complete Your Chapter Revision

Explore our broader library of printable assignments, chapter notes, and mock tests designed to support continuous revision and secure higher marks in JEE assessments.

FAQs

Where can I access latest JEE Physics Gravitation MCQs Set 16?

You can get most exhaustive JEE Physics Gravitation MCQs Set 16 for free on StudiesToday.com. These MCQs for JEE Physics are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Physics JEE material?

Yes, our JEE Physics Gravitation MCQs Set 16 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Physics MCQs help in scoring full marks in JEE exams?

By solving our JEE Physics Gravitation MCQs Set 16, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Physics.

Do you provide answers and explanations for JEE Physics Gravitation MCQs Set 16?

Yes, Physics MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Physics JEE MCQs online?

Yes, you can also access online interactive tests for JEE Physics Gravitation MCQs Set 16 on StudiesToday.com as they provide instant answers and score to help you track your progress in Physics.