JEE Physics Gravitation MCQs Set 33

Answer: (a) \( 2\pi\sqrt{R/g} \)

Explanation:
1. Weightless means the centripetal requirement equals gravity: \( m\omega^2R=mg \), so \( \omega=\sqrt{g/R} \).
2. Time period \( T=\dfrac{2\pi}{\omega}=2\pi\sqrt{R/g} \).

Teacher's Note:
1. At weightlessness the normal reaction at the equator is zero, which is why gravity alone supplies the centripetal force.
2. This period (about 84 minutes) is also the period of a satellite skimming the earth.

Answer: (a) A

Explanation:
1. Seen from space, A moves eastwards with speed \( \omega R-v \) and B with speed \( \omega R+v \) (\( v \) is the speed relative to the ground).
2. The normal reaction is \( N=mg-\dfrac{m(\text{speed})^2}{R} \), so the train with the smaller space speed (A) has the larger \( N \).
3. By Newton's third law A presses harder on the tracks.

Teacher's Note:
1. Compare speeds in the inertial frame, not relative to the ground, when the centripetal term is needed.
2. A train moving against the earth's rotation has the smaller effective centripetal requirement and therefore a larger normal force.

Answer: (c) \( \sqrt{2H/3g} \)

Explanation:
1. The earth pulls the body with acceleration \( g \) towards the earth. The body (mass \( 2M \)) pulls the earth towards itself with acceleration \( 2g \).
2. The relative acceleration of approach is \( g+2g=3g \).
3. \( H=\tfrac12(3g)t^2 \), so \( t=\sqrt{2H/3g} \).

Teacher's Note:
1. When the falling body is as massive as the planet, the planet's own acceleration cannot be neglected.
2. Add the two accelerations to get the relative acceleration, then use \( s=\tfrac12at^2 \).

Answer: (d) \( \sqrt{4H/3g} \)

Explanation:
1. The earth accelerates towards the body with \( g/2 \), because the body has half the earth's mass.
2. Relative acceleration \( =g+\tfrac g2=\tfrac{3g}{2} \).
3. \( H=\tfrac12\cdot\tfrac{3g}{2}t^2 \), so \( t=\sqrt{4H/3g} \).

Teacher's Note:
1. The acceleration of a planet due to a body is \( Gm/r^2 \), proportional to the body's mass.
2. Always sum the two accelerations when both objects are free to move.

Answer: (b) \( 2\pi\sqrt{R/g} \)

Explanation:
1. For a pendulum of length \( l \) in the earth's field, \( T=2\pi\Big/\sqrt{g\left(\tfrac1l+\tfrac1R\right)} \).
2. As \( l\to\infty \), \( \tfrac1l\to0 \), so \( T=2\pi\sqrt{R/g} \).

Teacher's Note:
1. The simple formula \( 2\pi\sqrt{l/g} \) assumes a uniform field and fails for very long pendulums.
2. The limiting value is about 84 minutes for the earth.

Answer: (a) \( 9R \)

Explanation:
1. \( g'=g\left(\dfrac{R}{R+h}\right)^2=\dfrac{g}{100} \).
2. So \( \dfrac{R}{R+h}=\dfrac1{10} \), giving \( R+h=10R \) and \( h=9R \).

Teacher's Note:
1. Height is measured from the surface, while the distance in the formula is measured from the centre.
2. Subtract \( R \) from the centre distance to get the height.

Answer: (b) 32 km

Explanation:
1. For small \( h \): \( g'\approx g\left(1-\dfrac{2h}{R}\right) \).
2. A 1% decrease gives \( \dfrac{2h}{R}=0.01 \), so \( h=\dfrac{R}{200}=\dfrac{6400}{200}=32 \) km.

Teacher's Note:
1. For \( h\ll R \) use the binomial approximation \( (1+h/R)^{-2}\approx1-2h/R \).
2. Do not use this approximation when \( h \) is comparable to \( R \), as in the previous question type.

Answer: (c) \( 2h=d \)

Explanation:
1. Above the surface: \( g_1=g\left(1-\dfrac{2h}{R}\right) \).
2. Below the surface: \( g_2=g\left(1-\dfrac{d}{R}\right) \).
3. Setting \( g_1=g_2 \) gives \( 2h=d \).

Teacher's Note:
1. Gravity falls twice as fast with height as with depth near the surface.
2. So the depth must be double the height for the same value of \( g \).

Answer: (c) \( g\propto R \)

Explanation:
1. \( g=\dfrac{GM}{R^2} \) and \( M=\dfrac43\pi R^3\rho \).
2. So \( g=\dfrac43\pi G\rho R \), which is proportional to \( R \) at constant density.

Teacher's Note:
1. Replace the mass by density times volume before deciding how \( g \) depends on \( R \).
2. At fixed density the radius dependence changes from \( R^{-2} \) to \( R^{1} \).

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

Explanation:
1. Inside a uniform shell the gravitational field is zero, so the potential does not change with position.
2. It equals the potential at the surface: \( V=-\dfrac{GM}{a} \).

Teacher's Note:
1. Zero field inside means constant potential, not zero potential.
2. The constant is the value at the shell's surface.

Answer: (d) the particle will remain stationary

Explanation:
1. Inside a uniform spherical shell the net gravitational force is zero at every point.
2. With no net force the particle stays at rest wherever it is placed.

Teacher's Note:
1. This is Newton's shell theorem; the pulls of the near and far parts of the shell cancel exactly.
2. The result is independent of the masses involved.

Answer: (b) \( \sqrt{\frac{2GM}{a}\left(1-\frac1{\sqrt2}\right)} \)

Explanation:
1. Potential at \( P \): \( V_P=-\dfrac{GM}{\sqrt{a^2+a^2}}=-\dfrac{GM}{\sqrt2\,a} \). Potential at \( C \): \( V_C=-\dfrac{GM}{a} \).
2. Energy conservation: \( \tfrac12v^2=V_P-V_C=\dfrac{GM}{a}\left(1-\dfrac1{\sqrt2}\right) \).
3. So \( v=\sqrt{\dfrac{2GM}{a}\left(1-\dfrac1{\sqrt2}\right)} \).

Teacher's Note:
1. The potential on the axis of a ring is \( -GM/\sqrt{a^2+x^2} \).
2. Use \( \tfrac12mv^2 \) equal to the fall in potential energy.

Answer: (b) \( \frac{v_e}{\sqrt2} \)

Explanation:
1. Potential at the surface \( =-\dfrac{GM}{R} \); at the centre \( =-\dfrac{3GM}{2R} \) for a uniform planet.
2. \( \tfrac12v^2=\dfrac{3GM}{2R}-\dfrac{GM}{R}=\dfrac{GM}{2R} \), so \( v^2=\dfrac{GM}{R} \).
3. Since \( v_e^2=\dfrac{2GM}{R} \), \( v=\dfrac{v_e}{\sqrt2} \).

Teacher's Note:
1. The centre potential of a uniform sphere is \( \tfrac32 \) times the surface potential.
2. Link the answer to \( v_e^2=2GM/R \) to express it in terms of \( v_e \).

Answer: (c) \( \frac\pi2\sqrt{R/g} \)

Explanation:
1. Inside a uniform planet the force is \( -\dfrac{mg}{R}x \), so the motion is simple harmonic with \( \omega=\sqrt{g/R} \).
2. The body starts at the amplitude and reaches the mean position (the centre) in a quarter period.
3. \( t=\dfrac T4=\dfrac{\pi}{2}\sqrt{R/g} \).

Teacher's Note:
1. Motion through a diametral tunnel is SHM with the same period as a satellite near the surface.
2. The surface to centre trip is a quarter of that period.

Answer: (c) continue to move in the same orbit as the satellite

Explanation:
1. At the moment of separation the part has the same position and velocity as the satellite.
2. Orbital motion depends only on position and velocity, not on mass, so the part follows the same orbit.

Teacher's Note:
1. Orbital speed \( \sqrt{GM/r} \) does not depend on the mass of the orbiting body.
2. Without a push or drag, nothing changes the orbit of the detached part.

Answer: (d) \( d \) will decrease and \( v \) will increase

Explanation:
1. Total energy of a satellite is \( E=-\dfrac{GMm}{2r} \). Losing energy makes \( E \) more negative, so \( r \) decreases.
2. Orbital speed \( v=\sqrt{GM/r} \) increases as \( r \) decreases.

Teacher's Note:
1. Losing energy makes a satellite speed up because kinetic energy is \( -E \) in a circular orbit.
2. Atmospheric drag therefore causes a spiral inwards with increasing speed.

Answer: (b) \( 1:8 \)

Explanation:
1. The orbit radii (from the earth's centre) are \( 2R \) and \( 8R \).
2. \( T\propto r^{3/2} \), so \( \dfrac{T_1}{T_2}=\left(\dfrac{2R}{8R}\right)^{3/2}=\dfrac18 \).

Teacher's Note:
1. Add the radius of the earth to each height before applying Kepler's third law.
2. \( \left(\tfrac14\right)^{3/2}=\tfrac18 \).

Answer: (c) The water and the glass have the same acceleration, equal to \( g \), towards the centre of the earth, hence there is no relative motion between them

Explanation:
1. Both the glass and the water are in free fall with the same acceleration towards the earth.
2. With no relative acceleration, the water needs no support from the glass and does not fall out.

Teacher's Note:
1. Apparent weightlessness is due to free fall, not the absence of gravity.
2. Gravity at that height is almost equal to \( g \), so the earth's pull is certainly present.

Answer: (c) i, iii and iv only

Explanation:
1. A beam balance compares masses, so it reads the same at both places: \( W_B=W_B' \), and at the pole it agrees with the spring balance, \( W_B=W_S \).
2. At the equator the spring balance reads \( mg-m\omega^2R \), which is less than its pole reading, so \( W_S'<W_S \).
3. Hence \( W_B'=W_S' \) is false because \( W_B'=mg>W_S' \).

Teacher's Note:
1. A beam balance measures mass (it compares weights at the same place); a spring balance measures effective weight.
2. Rotation reduces only the spring balance reading at the equator.

Answer: (b) \( \frac{\omega^2R^2}{2g} \)

Explanation:
1. Equator reading: \( g_{eff}=g-\omega^2R \).
2. Above the pole: \( g\left(1-\dfrac{2h}{R}\right) \).
3. Equate: \( \dfrac{2gh}{R}=\omega^2R \), so \( h=\dfrac{\omega^2R^2}{2g} \).

Teacher's Note:
1. Use the small-height approximation \( 1-2h/R \) for gravity above the surface.
2. The equatorial reduction is \( m\omega^2R \), the centripetal term.

Answer: (a) \( \frac{\omega^2R^2}{g} \)

Explanation:
1. Equator reading: \( g-\omega^2R \).
2. Below the surface: \( g\left(1-\dfrac dR\right) \).
3. Equate: \( \dfrac{gd}{R}=\omega^2R \), so \( d=\dfrac{\omega^2R^2}{g} \).

Teacher's Note:
1. Gravity decreases at half the rate with depth that it does with height near the surface.
2. So the depth is twice the height for the same reading.

Answer: (b) i and iv only

Explanation:
1. For the star of mass \( m \) at distance \( r_1 \) from the centre of mass: \( \dfrac{G(2m)m}{l^2}=m\omega^2r_1 \), with \( r_1=\dfrac{2l}{3} \).
2. This gives \( \omega^2=\dfrac{3Gm}{l^3} \), so \( T=2\pi\sqrt{\dfrac{l^3}{3Gm}} \).
3. Hence \( T\propto l^{3/2} \) and \( T\propto m^{-1/2} \).

Teacher's Note:
1. The centripetal force on each star is the mutual gravitational force, with each star circling the centre of mass.
2. Kepler's third law holds for binary systems with total mass in the denominator.

Answer: (b) i and iv only

Explanation:
1. The distance of each mass from the centre is \( r=\dfrac{a}{\sqrt3} \), and the net gravitational force on each mass is \( \dfrac{\sqrt3Gm^2}{a^2} \) towards the centre.
2. \( m\omega^2\dfrac{a}{\sqrt3}=\dfrac{\sqrt3Gm^2}{a^2} \) gives \( \omega^2=\dfrac{3Gm}{a^3} \).
3. So \( T=2\pi\sqrt{\dfrac{a^3}{3Gm}} \), proportional to \( a^{3/2} \) and \( m^{-1/2} \).

Teacher's Note:
1. Resolve the two equal pulls along the line to the centre: \( 2\cos30^\circ \) times \( Gm^2/a^2 \).
2. Equal masses on an equilateral triangle rotate rigidly about the centroid.