JEE Physics Gravitation MCQs Set 31

Answer: (a) \( \dfrac{R}{2} \)

Explanation:
1. \( g=\dfrac{GM}{R^2} \): \( \dfrac{g_p}{g_e}=\dfrac49=\dfrac{M_p}{M_e}\cdot\dfrac{R^2}{R_p^2}=\dfrac19\cdot\dfrac{R^2}{R_p^2} \).
2. \( R_p^2=\dfrac{R^2}4\Rightarrow R_p=\dfrac R2 \).

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (b) \( 2.6\times10^6 \) m

Explanation:
1. \( g'=g\left(\dfrac R{R+h}\right)^2=\dfrac g2\Rightarrow R+h=\sqrt2R \).
2. \( h=(\sqrt2-1)(6.4\times10^6)=2.65\times10^6\approx2.6\times10^6 m \).

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (c) g increases linearly from zero at d = 0 to its maximum at d = R, and then falls as 1/d² beyond R

Explanation:
1. Inside a uniform earth \( g=\dfrac{GM}{R^3}d\propto d \) (zero at the centre).
2. At the surface it is maximum, and outside \( g=\dfrac{GM}{d^2}\propto\dfrac1{d^2} \).

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (d) 108 N

Explanation:
1. Same density: \( g_p=\dfrac{g_e}{10}=1 m/s^{2} \) at the surface. Inside, \( g(r)=g_p\dfrac rR \).
2. Wire from \( r=0.8R \) to \( R \): \( F=\lambda\int_{0.8R}^Rg_p\dfrac rR\,dr=\lambda g_p\dfrac{R(1-0.64)}2=0.18\lambda g_pR \).
3. \( R=6\times10^5 m \): \( F=0.18\times10^{-3}\times1\times6\times10^5=108 N \).

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (a) increase

Explanation:
1. \( g=\dfrac{GM}{R^2} \) with \( M \) fixed and \( R \) decreasing.
2. \( R \) falls by 1%, so \( g \) rises by about 2%.

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (b) (a) and (b) only

Explanation:
1. Inside a uniform sphere \( F=\dfrac{GMr}{R^3}\propto r \), so \( F_1/F_2=r_1/r_2 \): \( (a) \) true, \( (c) \) and \( (d) \) false.
2. Outside, \( F=\dfrac{GM}{r^2}\propto\dfrac1{r^2} \), so \( F_1/F_2=r_2^2/r_1^2 \): \( (b) \) true.

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (c) \( 1.24\times10^{-3} \)

Explanation:
1. At the equator \( g_{eff}=g-\omega^2R=0 \).
2. \( \omega=\sqrt{\dfrac gR}=\sqrt{\dfrac{9.8}{6.4\times10^6}}=1.24\times10^{-3} \) rad/s.

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (d) 11

Explanation:
1. \( r=2.02\times10^6 m \). \( T=2\pi\sqrt{\dfrac{r^3}{GM}}=2\pi\sqrt{\dfrac{8.24\times10^{18}}{5.34\times10^{12}}}\approx7.8\times10^3 s \).
2. Revolutions \( =\dfrac{86400}{7810}\approx11 \).

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (a) \( \dfrac{GM}{3a^2} \)

Explanation:
1. At \( r=3a \) (the outer surface of the shell) all the mass \( M+2M=3M \) lies inside.
2. \( g=\dfrac{G(3M)}{(3a)^2}=\dfrac{GM}{3a^2} \).

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (b) \( \dfrac{E}{16} \)

Explanation:
1. Escape energy \( =\dfrac{GMm}R=\dfrac43\pi G\rho R^2m\propto R^2 \) for equal density.
2. Volume ratio 64 gives \( R_e=4R_m \), so \( E_{moon}=\dfrac E{16} \).

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (c) \( v_s=42 \) km/s

Explanation:
1. \( \tfrac12v_s^2=\dfrac{GM_e}{R_e}+\dfrac{GM_s}{r}=\dfrac{GM_e}{R_e}\left[1+\dfrac{3\times10^5}{2.5\times10^4}\right]=13\dfrac{GM_e}{R_e} \).
2. With \( v_e^2=\dfrac{2GM_e}{R_e} \): \( v_s^2=13v_e^2\Rightarrow v_s=\sqrt{13}(11.2)\approx40\text{ km/s}\approx42 \) km/s.

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (d) \( \sqrt{gR}\,(\sqrt2-1) \)

Explanation:
1. Orbital speed \( v_o\approx\sqrt{gR} \); escape speed \( v_e=\sqrt{2gR} \).
2. Increase \( =\sqrt{gR}(\sqrt2-1) \).

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.

Answer: (a) \( \dfrac{-GM}{R} \)

Explanation:
1. Potential of the full sphere at \( r=\dfrac R2 \): \( -\dfrac{GM(3R^2-r^2)}{2R^3}=-\dfrac{11GM}{8R} \).
2. The removed sphere has mass \( \dfrac M8 \) and radius \( \dfrac R2 \); its potential at its own centre is \( -\dfrac32\cdot\dfrac{G(M/8)}{R/2}=-\dfrac{3GM}{8R} \).
3. Remaining: \( -\dfrac{11GM}{8R}+\dfrac{3GM}{8R}=-\dfrac{GM}R \).

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (b) v increases linearly with r up to r = R and then falls as \( 1/\sqrt r \)

Explanation:
1. Inside: \( g=\dfrac{GMr}{R^3} \), and \( \dfrac{v^2}r=g\Rightarrow v=\sqrt{\dfrac{GM}{R^3}}\,r\propto r \).
2. Outside: \( v=\sqrt{\dfrac{GM}r}\propto\dfrac1{\sqrt r} \).

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (c) \( \dfrac{1}{2} \) \( mgR \)

Explanation:
1. \( \Delta U=GMm\left(\dfrac1R-\dfrac1{2R}\right)=\dfrac{GMm}{2R} \).
2. With \( GM=gR^2 \): \( \Delta U=\dfrac12mgR \).

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (d) (b) and (d) only

Explanation:
1. \( v_{esc}=\sqrt{\dfrac{8\pi G\rho}3}\,R\propto R \). Areas 1 : 4 give radii \( R_P:R_Q=1:2 \), masses 1 : 8.
2. \( R \) has mass \( 9M_P \), so \( R_R=9^{1/3}R_P\approx2.08R_P \). Hence \( v_P:v_Q:v_R=1:2:2.08 \).
3. So \( v_R>v_Q>v_P (b) \) and \( v_P/v_Q=\tfrac12 (d) \) are true; \( v_R/v_P\approx2.08\neq3 \).

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (a) (a), (c) and (d) only

Explanation:
1. \( (a) \) The two cavities are symmetric about \( O \), so the fields from the full sphere (zero at the centre) and the cavities cancel: true.
2. \( (b) \) At \( B \) the full sphere gives \( \tfrac{8\pi}3G\rho \) toward \( O \), while the cavity at \( A \) gives only \( \tfrac\pi{12}G\rho \) in the opposite sense: not zero, false.
3. \( (c), (d) \) Every point on each circle (in the plane \( x = 0 \), at distance 6 or 2 from \( O \)) is equidistant from both cavity centres and from \( O \), so the potential is the same at all of them: true.

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (b) R

Explanation:
1. Escape KE \( =\dfrac{GMm}R \), so the given KE is \( \dfrac{GMm}{2R} \).
2. \( -\dfrac{GMm}R+\dfrac{GMm}{2R}=-\dfrac{GMm}{R+h}\Rightarrow R+h=2R\Rightarrow h=R \).

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (c) \( 2\sqrt{\dfrac{G(M_1+M_2)}{d}} \)

Explanation:
1. Potential at the midpoint \( =-\dfrac{GM_1}{d/2}-\dfrac{GM_2}{d/2}=-\dfrac{2G(M_1+M_2)}d \).
2. \( \tfrac12v^2=\dfrac{2G(M_1+M_2)}d\Rightarrow v=2\sqrt{\dfrac{G(M_1+M_2)}d} \).

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (d) 2

Explanation:
1. \( \dfrac g4 \) at height \( h \) means \( \left(\dfrac R{R+h}\right)^2=\dfrac14\Rightarrow h=R \).
2. Energy: \( \tfrac12v^2=GM\left(\dfrac1R-\dfrac1{2R}\right)=\dfrac{GM}{2R}\Rightarrow v^2=\dfrac{GM}R \). Then \( v_{esc}^2=\dfrac{2GM}R=2v^2 \), so \( N = 2 \).

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (a) 3

Explanation:
1. \( g=\dfrac43\pi G\rho R\Rightarrow\dfrac{R_p}{R_e}=\dfrac{g_p}{g_e}\cdot\dfrac{\rho_e}{\rho_p}=\dfrac{\sqrt6}{11}\cdot\dfrac32 \).
2. \( v_{esc}=\sqrt{2gR} \): \( \dfrac{v_p^2}{v_e^2}=\dfrac{\sqrt6}{11}\cdot\dfrac{3\sqrt6}{22}=\dfrac{9}{121} \).
3. \( v_p=11\times\dfrac3{11}=3 \) km/s.

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (b) 99.5 R

Explanation:
1. Potential inside a uniform sphere at \( r=0.99R \): \( V=-\dfrac{GM(3R^2-r^2)}{2R^3}=-1.00995\dfrac{GM}R \). KE \( =\dfrac12\cdot\dfrac{2GM}R=\dfrac{GM}R \).
2. Total energy \( E=-0.00995\dfrac{GM}R \). At the top \( -\dfrac{GM}{R+H}=E\Rightarrow R+H=100.5R \).
3. \( H\approx99.5R \).

Teacher's Note:
1. Write \( g = GM/R^2 \) and the potential or escape-energy expression first, then substitute the given ratios.
2. Check the limiting cases (inside and outside the body, or \( r \) tends to \( \infty \)) to confirm the result.

Answer: (c) \( T/R\ \text{is a constant} \)

Explanation:
1. Mass inside \( r \): \( M(r)=\int_0^r\dfrac K{r^2}4\pi r^2dr=4\pi Kr \). Field \( g=\dfrac{GM(r)}{r^2}=\dfrac{4\pi GK}r \).
2. \( \dfrac{v^2}R=\dfrac{4\pi GK}R\Rightarrow v \) is constant. \( T=\dfrac{2\pi R}v\propto R \).

Teacher's Note:
1. For circular orbits use \( GMm/r^2 = mv^2/r \), which gives \( v = \sqrt{GM/r}, T^2 \) proportional to \( r^3 \) and total energy -GMm/2r.
2. Escape needs total energy zero, so the escape speed is \( \sqrt{2} \) times the circular orbital speed at the same radius.