JEE Physics Gravitation MCQs Set 34

Answer: (c) is reduced to a quarter

Explanation:
1. Use \( F=\dfrac{Gm_1m_2}{r^2} \).
2. With \( r\to 2r \), \( F\to F/4 \).

Teacher's Note:
1. Gravitational force follows an inverse-square law.
2. Doubling the distance divides the force by four.

Answer: (d) \( 36\times10^{21} \)

Explanation:
1. The sun's pull provides the centripetal force: \( F=m\omega^2 r \).
2. \( F=6\times10^{24}\times(2\times10^{-7})^2\times1.5\times10^{11}=3.6\times10^{22} \) N, which is \( 36\times10^{21} \) N.

Teacher's Note:
1. Convert km to m before substituting.
2. The gravitational force supplies the centripetal force of the orbit.

Answer: (a) depends on the system of units only

Explanation:
1. The constant is the universal gravitational constant G.
2. It does not depend on the medium, but its numerical value changes with the system of units.

Teacher's Note:
1. G is \( 6.67\times10^{-11} \) N m\(^2\)/kg\(^2\) in SI.
2. Gravitational force is unaffected by the intervening medium.

Answer: (d) 1.62 m/s\(^2\)

Explanation:
1. \( g=\dfrac{GM}{R^2}=\dfrac{6.67\times10^{-11}\times7.34\times10^{22}}{(1.74\times10^{6})^2} \).
2. This equals \( \dfrac{4.896\times10^{12}}{3.028\times10^{12}}\approx1.62 \) m/s\(^2\).

Teacher's Note:
1. The moon's g is about one-sixth of the earth's.
2. Check the powers of ten first.

Answer: (a) \( \dfrac12 \)

Explanation:
1. \( F\propto x(1-x)M^2/r^2 \).
2. \( \dfrac{d}{dx}[x(1-x)]=1-2x=0 \) gives \( x=\dfrac12 \).

Teacher's Note:
1. The product of two numbers with a fixed sum is greatest when they are equal.

Answer: (a) \( F \)

Explanation:
1. \( F=Gm_1m_2/r^2 \) contains no term for the medium.
2. So the force stays \( F \).

Teacher's Note:
1. Unlike electrostatic force, gravity cannot be shielded by a medium.

Answer: (b) 400 gm-wt

Explanation:
1. \( g\propto M/R^2 \), so \( g'/g=\dfrac{1/7}{(1/2)^2}=\dfrac47 \).
2. Weight \( =700\times\dfrac47=400 \) gm-wt.

Teacher's Note:
1. Mass is unchanged; only weight changes.

Answer: (c) 1/800 rad/s

Explanation:
1. Weight at the equator is \( m(g-\omega^2R) \), zero when \( \omega^2R=g \).
2. \( \omega=\sqrt{10/6.4\times10^{6}}=\dfrac{1}{800} \) rad/s.

Teacher's Note:
1. Mass of the body does not matter.

Answer: (a) \( W_1<W_2>W_3 \)

Explanation:
1. Below the surface \( g \) decreases with depth, so \( W_1<W_2 \).
2. Above the surface \( g \) decreases with height, so \( W_3<W_2 \).

Teacher's Note:
1. The maximum value of g (ignoring rotation) is at the surface.

Answer: (a) 960.40 cm/s\(^2\)

Explanation:
1. Here \( h/R=0.01 \), small, so \( g'\approx g\Big(1-\dfrac{2h}{R}\Big) \).
2. \( g'=980\times0.98=960.4 \) cm/s\(^2\).

Teacher's Note:
1. Use the binomial approximation only when \( h\ll R \).

Answer: (b) \( 4g \)

Explanation:
1. \( g=GM/R^2 \) with \( R\to R/2 \).
2. So \( g'=4g \).

Teacher's Note:
1. Mass is fixed, so only the radius matters.

Answer: (a) \( \dfrac{g_1}{g_2}=\dfrac{R_1}{R_2} \)

Explanation:
1. With the same \( \rho \), \( g\propto R \).
2. So \( g_1/g_2=R_1/R_2 \).

Teacher's Note:
1. Express g in terms of density to compare planets of the same material.

Answer: (c) \( 4GM_0/D_0^2 \)

Explanation:
1. The radius is \( D_0/2 \).
2. \( a=\dfrac{GM_0}{(D_0/2)^2}=\dfrac{4GM_0}{D_0^2} \).

Teacher's Note:
1. Acceleration is independent of the falling particle's mass.

Answer: (d) infinite

Explanation:
1. Inside a satellite everything is in free fall, so effective \( g=0 \).
2. \( T=2\pi\sqrt{l/g_{eff}} \) tends to infinity, so the pendulum does not oscillate.

Teacher's Note:
1. A pendulum clock is useless in orbit.

Answer: (b) will go on decreasing continuously

Explanation:
1. Spring balance measures weight \( mg \).
2. \( g \) falls continuously with height, so the reading falls continuously.

Teacher's Note:
1. The reading would become zero only at infinity.

Answer: (b) decreases at the equator but remains unchanged at the poles

Explanation:
1. At the poles \( \lambda=90^\circ \), so the rotational term vanishes.
2. At the equator \( g_e=g-\omega^2R \) falls as \( \omega \) rises.

Teacher's Note:
1. The axis of rotation passes through the poles.

Answer: (b) \( R\Big(\dfrac{n-1}{n}\Big) \)

Explanation:
1. Below the surface \( g'=g\Big(1-\dfrac dR\Big) \).
2. Setting it equal to \( g/n \): \( d=R\Big(1-\dfrac1n\Big)=R\dfrac{n-1}{n} \).

Teacher's Note:
1. g varies linearly with depth.

Answer: (a) 32 km

Explanation:
1. For small \( h \), \( \dfrac{\Delta g}{g}=\dfrac{2h}{R} \).
2. \( h=\dfrac{0.01\times6400}{2}=32 \) km.

Teacher's Note:
1. The approximation is valid because \( h\ll R \).

Answer: (c) 2 : 1

Explanation:
1. \( g=\frac43\pi G\rho R\propto\rho R \).
2. Ratio \( =\dfrac{\rho_1R_1}{\rho_2R_2}=\dfrac12\times4=2:1 \).

Teacher's Note:
1. Diameter ratio equals radius ratio.

Answer: (b) \( R \)

Explanation:
1. \( g'=g\Big(\dfrac{R}{R+h}\Big)^2=\dfrac g4 \).
2. So \( \dfrac{R}{R+h}=\dfrac12 \) and \( h=R \).

Teacher's Note:
1. Altitude means height above the surface.

Answer: (a) \( -5000 \) J

Explanation:
1. Potential energy at the surface is \( -\dfrac{GMm}{R}=-\tfrac12mv_e^2 \).
2. \( U=-\tfrac12\times1\times100^2=-5000 \) J.

Teacher's Note:
1. Escape velocity is the speed that makes total energy zero.
2. Potential energy at the surface is the negative of the kinetic energy needed.

Answer: (b) \( \dfrac34mgR \)

Explanation:
1. \( \Delta U=GMm\Big(\dfrac1R-\dfrac1{4R}\Big)=\dfrac{3GMm}{4R} \).
2. Using \( GM=gR^2 \), \( \Delta U=\dfrac34mgR \).

Teacher's Note:
1. For large heights \( mgh \) is not valid.
2. Use \( \Delta U=\dfrac{mgh}{1+h/R} \).

Answer: (b) \( \sqrt{2gR} \)

Explanation:
1. Energy conservation: \( \tfrac12mv^2=\dfrac{GMm}{R}=mgR \).
2. So \( v=\sqrt{2gR} \).

Teacher's Note:
1. Falling from infinity is the reverse of escaping.

Answer: (b) must be constant

Explanation:
1. The field is the negative gradient of the potential.
2. A zero field means no change in potential, so it is constant (not necessarily zero).

Teacher's Note:
1. Potential is defined only up to an additive constant.

Answer: (b) \( 2v_e \)

Explanation:
1. \( v_e=\sqrt{\dfrac{2GM}{R}}=R\sqrt{\dfrac{8\pi G\rho}{3}} \).
2. Same \( \rho \), double \( R \) gives double velocity.

Teacher's Note:
1. Escape velocity grows with radius at constant density.