JEE Physics Gravitation MCQs Set 26

Answer: (b) \( \frac{\sqrt5-1}{2}R \)

Explanation:
1. Above: \( g\frac{R^2}{(R+h)^2} \). Below: \( g\left(1-\frac hR\right) \).
2. Equating: \( R^3 = (R-h)(R+h)^2 \) gives \( h^2 + hR - R^2 = 0 \), so \( h = \frac{\sqrt5-1}{2}R \).

Teacher's Note:
1. Use the exact inverse-square law above and the linear law below.
2. Solve the quadratic and reject the negative root.

Answer: (a) \( \frac23\sqrt{\frac{2GM}{5R}} \)

Explanation:
1. Half of the maximum height is 4R, i.e. 5R from the centre. Energy: \( \tfrac12v^2 - \frac{GM}{5R} = -\frac{GM}{9R} \).
2. \( v^2 = 2GM\left(\frac1{5R}-\frac1{9R}\right) = \frac{8GM}{45R} \), so \( v = \frac23\sqrt{\frac{2GM}{5R}} \).

Teacher's Note:
1. Use the turning point as the reference for energy.
2. Half of the height, not half of the distance from the centre.

Answer: (d) \( 1\times10^5 \) J

Explanation:
1. In a circular orbit \( E = \frac U2 \), so the new total energy is \( -1\times10^5 \) J.
2. Energy to be supplied \( = -1\times10^5 - (-2\times10^5) = 1\times10^5 \) J.

Teacher's Note:
1. For circular orbits E = U/2 = -K.
2. New orbit has the larger radius, so energy must increase.

Answer: (d) \( -\frac{2GM_em}{r} \)

Explanation:
1. The momenta were equal and opposite, so the wreckage is momentarily at rest and has no kinetic energy.
2. Its energy is purely potential: \( U = -\frac{GM_e(2m)}{r} = -\frac{2GM_em}{r} \); it then falls towards the earth.

Teacher's Note:
1. Opposite equal momenta cancel in a perfectly inelastic collision.
2. All the kinetic energy is lost.

Answer: (a) \( \frac{GMm}{2C}\left(\frac1{R_e}-\frac1r\right) \)

Explanation:
1. Total energy in a circular orbit is \( -\frac{GMm}{2r} \). Energy lost \( = \frac{GMm}{2R_e} - \frac{GMm}{2r} \).
2. Time \( = \frac{\text{energy lost}}{C} = \frac{GMm}{2C}\left(\frac1{R_e}-\frac1r\right) \).

Teacher's Note:
1. Energy lost divided by the rate of loss gives the time.
2. The slow spiral keeps the orbit nearly circular.

Answer: (c) \( \frac{2GM\rho}{L} \)

Explanation:
1. Only the components along the perpendicular survive: \( dF = \frac{GM\rho\,dx}{L^2+x^2}\cdot\frac{L}{\sqrt{L^2+x^2}} \).
2. \( F = GM\rho L\int_{-\infty}^{\infty}\frac{dx}{(L^2+x^2)^{3/2}} = \frac{2GM\rho}{L} \).

Teacher's Note:
1. Use the standard result for the field of an infinite line mass: 2Gρ/L.
2. Cancel components along the rod by symmetry.

Answer: (a) \( \frac{2\pi R^{3/2}\,(6\sqrt6)}{\sqrt{GM}\,(2\sqrt2+3\sqrt3)} \)

Explanation:
1. Orbit radii are 2R and 3R: \( \omega_P = \sqrt{\frac{GM}{8R^3}} \), \( \omega_Q = \sqrt{\frac{GM}{27R^3}} \).
2. They move oppositely, so the relative angular speed is \( \omega_P+\omega_Q = \sqrt{\frac{GM}{R^3}}\cdot\frac{2\sqrt2+3\sqrt3}{6\sqrt6} \).
3. They are aligned again after a relative turn of \( 2\pi \): \( t = \frac{2\pi}{\omega_P+\omega_Q} \), which gives the stated value.

Teacher's Note:
1. Opposite senses mean the angular speeds add.
2. They line up again after a relative angle of 2π.

Answer: (b) \( \sqrt{6.4} \) km

Explanation:
1. Jump speed: \( v^2 = g_eb \) (45° projection). On the planet \( g_p = 2g_e\frac{R_p}{R_e} \).
2. Escape needs \( v^2 = 2g_pR_p = 4g_e\frac{R_p^2}{R_e} \). Setting \( g_eb = 4g_e\frac{R_p^2}{R_e} \): \( R_p = \sqrt{\frac{bR_e}{4}} = \sqrt{6.4\times10^6}\) m \( = \sqrt{6.4} \) km (about 2.5 km).

Teacher's Note:
1. The man's takeoff speed is the same on both bodies.
2. Planet gravity scales with density times radius.

Answer: (b) \( \sqrt{\frac{GM}{R}}\left[\sqrt{\frac23}-\sqrt{\frac{8}{15}}\right] \)

Explanation:
1. Let apogee distance be kR. Angular momentum: \( v_a = v_p/k \). Energy: \( \tfrac12\cdot\frac{6}{5} - 1 = \frac{0.6}{k^2} - \frac1k \) (in units of GM/R) gives \( 0.4k^2 - k + 0.6 = 0 \), so k = 1.5.
2. \( v_a = \sqrt{\frac{6GM}{5R}}\cdot\frac{1}{1.5} = \sqrt{\frac{8GM}{15R}} \). Circular speed at 1.5R: \( \sqrt{\frac{GM}{1.5R}} = \sqrt{\frac{2GM}{3R}} \).
3. Increase = circular speed minus apogee speed, which gives the stated value.

Teacher's Note:
1. Conserve angular momentum and energy between perigee and apogee.
2. Choose the root k = 1.5 (k = 1 is the perigee).

Answer: (a) \( 6.67\times10^{-10} \) J

Explanation:
1. \( W = \frac{GMm}{R} = \frac{6.67\times10^{-11}\times100\times0.01}{0.1} \).
2. \( W = 6.67\times10^{-10} \) J.

Teacher's Note:
1. Work against gravity to infinity equals the magnitude of the potential energy.
2. Convert grams to kilograms and centimetres to metres.

Answer: (d) d = 2h

Explanation:
1. Above: \( \Delta g \approx \frac{2h}{R}g \). Below: \( \Delta g = \frac{d}{R}g \).
2. Equal change requires d = 2h.

Teacher's Note:
1. Use first-order approximations for small h and d.
2. g decreases twice as fast with height as with depth.

Answer: (d) 1

Explanation:
1. The electronic charge is a universal constant and does not depend on gravity.
2. The experiment would give the same value of e, so the ratio is 1.

Teacher's Note:
1. Charge is quantised and the same everywhere.
2. Gravity only changes the drop's weight, not its charge.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Gauss's law for gravitation gives total flux of magnitude 4πGM through any closed surface enclosing mass M, so A is true.
2. The inverse square radial dependence is exactly what makes the flux independent of the shape and size of the surface, so R is true.
3. R is the basis of Gauss's law and hence correctly explains A.

Teacher's Note:
1. Gauss's law applies to any inverse square field.
2. R explains why the flux is 4πGM.

Answer: (c) 2R

Explanation:
1. \( \frac{R^2}{(R+h)^2} = \frac19 \) gives \( R+h = 3R \).
2. So h = 2R.

Teacher's Note:
1. The distance from the centre must triple.
2. Height is distance from the centre minus R.

Answer: (c) \( -\frac{9Gm}{r} \)

Explanation:
1. Field zero at x from m: \( \frac{m}{x^2} = \frac{4m}{(r-x)^2} \), so \( x = \frac r3 \).
2. \( V = -\frac{Gm}{r/3} - \frac{4Gm}{2r/3} = -\frac{3Gm}{r} - \frac{6Gm}{r} = -\frac{9Gm}{r} \).

Teacher's Note:
1. Potential adds as a scalar even where the field cancels.
2. Neutral point is nearer the smaller mass.

Answer: (d) \( 6.4\times10^{10} \) J

Explanation:
1. Energy needed \( = \frac{GMm}{R} = mgR \).
2. \( = 1000\times10\times6.4\times10^6 = 6.4\times10^{10} \) J.

Teacher's Note:
1. Energy to escape equals the magnitude of the surface potential energy.
2. GM = gR².

Answer: (b) \( \frac{5GmM}{6R} \)

Explanation:
1. Orbit radius is 3R, so \( E_{orbit} = -\frac{GMm}{6R} \). Initial energy \( = -\frac{GMm}{R} \).
2. Required \( = -\frac{GMm}{6R} + \frac{GMm}{R} = \frac{5GMm}{6R} \).

Teacher's Note:
1. Include both the potential and the kinetic energy of the orbit.
2. Altitude 2R means orbital radius 3R.

Answer: (c) r₁ : r₂

Explanation:
1. With the same period, \( \omega \) is equal and \( a = \omega^2r \propto r \).
2. So the ratio is r₁ : r₂, independent of mass.

Teacher's Note:
1. Same time means the same angular velocity.
2. Centripetal acceleration does not depend on mass.

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

Explanation:
1. Potential at distance R/2 inside the full sphere: \( -\frac{GM(3R^2 - R^2/4)}{2R^3} = -\frac{11GM}{8R} \).
2. The removed sphere has mass M/8, and its potential at its own centre is \( -\frac{3G(M/8)}{2(R/2)} = -\frac{3GM}{8R} \).
3. Potential of the cavity point \( = -\frac{11GM}{8R} + \frac{3GM}{8R} = -\frac{GM}{R} \).

Teacher's Note:
1. Use superposition: full sphere minus the removed sphere.
2. Potential inside a solid sphere is -GM(3R² - r²)/2R³.

Answer: (b) g increases linearly with d from zero at the centre up to R, then decreases as 1/d² beyond R

Explanation:
1. Inside a uniform earth \( g = \frac{GMd}{R^3} \propto d \), with maximum value at the surface.
2. Outside \( g = \frac{GM}{d^2} \propto 1/d^2 \).

Teacher's Note:
1. The graph rises linearly then falls like an inverse square curve.
2. The peak is at the surface.

Answer: (d) \( T_A = T_B \)

Explanation:
1. Both stars revolve about their common centre of mass and stay on a line through it, so they complete a revolution in the same time.
2. Hence the periods are equal.

Teacher's Note:
1. Binary partners share the same angular velocity.
2. Kepler's law relates period to the separation of the pair, not to individual radii.

Answer: (c) v increases linearly with r up to R, then decreases as \( 1/\sqrt r \)

Explanation:
1. For \( r \le R \): \( \frac{v^2}{r} = \frac{GM(r)}{r^2} = \frac43\pi G\rho_0r \), so \( v \propto r \).
2. For \( r > R \): \( v = \sqrt{GM/r} \propto r^{-1/2} \).

Teacher's Note:
1. Use the enclosed mass for each region.
2. The speed peaks at the surface of the distribution.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Gravity supplies the entire centripetal force, so the astronaut and the station fall freely together; no normal force acts, giving apparent weightlessness. A is true.
2. R describes this free-fall state and is the reason for A.

Teacher's Note:
1. Weightlessness is due to free fall, not absence of gravity.
2. Orbital motion is continuous free fall.

Answer: (a) 6

Explanation:
1. Each star has \( L = mr^2\omega \) with \( r \propto 1/m \), so \( L \propto 1/m \).
2. \( L_A : L_B = 11 : 2.2 = 5 : 1 \). Total / \( L_B \) \( = \frac{5+1}{1} = 6 \).

Teacher's Note:
1. Angular momentum of each star is inversely proportional to its mass.
2. Add the two angular momenta.

Answer: (d) 3

Explanation:
1. \( g \propto \rho R \): \( \frac{R_p}{R_e} = \frac{\sqrt6/11}{2/3} = \frac{3\sqrt6}{22} \).
2. \( v_e \propto R\sqrt\rho \): \( \frac{v_p}{v_e} = \frac{3\sqrt6}{22}\sqrt{\frac23} = \frac{3}{11} \), so \( v_p = 3 \) km/s.

Teacher's Note:
1. First find the radius ratio from g.
2. Escape speed is proportional to R times √ρ.

Answer: (b) 2 and 4 only

Explanation:
1. Area ratio 1 : 4 gives \( R_Q = 2R_P \), so \( M_Q = 8M_P \). Planet R has \( M_R = 9M_P \) and \( R_R = 9^{1/3}R_P \approx 2.08R_P \).
2. With equal density \( V \propto R \): \( V_P : V_Q : V_R = 1 : 2 : 2.08 \). So V_R > V_Q > V_P and V_P/V_Q = 1/2, while V_R/V_P is about 2.08, not 3.

Teacher's Note:
1. Equal density makes escape speed proportional to radius.
2. Mass is proportional to R³.

Answer: (c) 2 and 4 only

Explanation:
1. Potential at the midpoint \( = -\frac{GM}{L}-\frac{GM}{L} = -\frac{2GM}{L} \). Escape: \( \tfrac12v^2 = \frac{2GM}{L} \), so \( v = 2\sqrt{GM/L} \).
2. The two large masses are fixed and the force is conservative, so the energy of m stays constant.

Teacher's Note:
1. Add the potentials of the two masses.
2. Conservative forces conserve mechanical energy.

Answer: (d) 7

Explanation:
1. Tension zero means both point masses have equal acceleration. Near mass: \( \frac{GM}{(3l)^2} - \frac{Gm}{l^2} \). Far mass (at 4l): \( \frac{GM}{(4l)^2} + \frac{Gm}{l^2} \).
2. Equating: \( GM\left(\frac19-\frac1{16}\right) = \frac{2Gm}{1} \) per \( l^{-2} \), so \( m = \frac{7M}{288} \) and k = 7.

Teacher's Note:
1. Equal accelerations mean the rod is slack.
2. The far mass attracts the near one outward.

Answer: (b) 2

Explanation:
1. Acceleration is g/4 at distance 2R from the centre, so the highest point is at 2R.
2. Energy: \( \tfrac12v^2 - \frac{GM}{R} = -\frac{GM}{2R} \), so \( v^2 = \frac{GM}{R} \). Then \( v_{esc}^2 = \frac{2GM}{R} = 2v^2 \), so N = 2.

Teacher's Note:
1. Locate the maximum height from the inverse square law.
2. Compare v² with 2GM/R.