JEE Physics Gravitation MCQs Set 23

Answer: (b) \( R \)

Explanation:
1. Escape KE = \( GMm/R \), so the given KE is \( GMm/2R \).
2. Energy conservation: \( \dfrac{GMm}{2R} - \dfrac{GMm}{R} = -\dfrac{GMm}{r} \Rightarrow r = 2R \).
3. Height above the surface \( h = r - R = R \).

Teacher's Note:
1. Total energy at the top is purely potential, equal to −GMm/r.
2. Subtract R at the end, since h is measured from the surface.

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

Explanation:
1. Potential at the midpoint: \( V = -\dfrac{GM_1}{d/2} - \dfrac{GM_2}{d/2} = -\dfrac{2G(M_1+M_2)}{d} \).
2. For escape, \( \tfrac12 v^2 = \dfrac{2G(M_1+M_2)}{d} \).
3. \( v = 2\sqrt{\dfrac{G(M_1+M_2)}{d}} \).

Teacher's Note:
1. Potentials add as scalars, so this is simpler than adding fields.
2. The distance to each star is d/2, which doubles each potential term.

Answer: (b) 3

Explanation:
1. \( g = \tfrac43\pi G\rho R \), so \( R_p/R_e = (g_p/g_e)(\rho_e/\rho_p) = \dfrac{\sqrt6}{11}\cdot\dfrac32 \).
2. \( v_{esc} = \sqrt{2gR} \), so \( \dfrac{v_p^2}{v_e^2} = \dfrac{g_pR_p}{g_eR_e} = \dfrac{\sqrt6}{11}\cdot\dfrac{3\sqrt6}{22} = \dfrac{9}{121} \).
3. \( v_p = \dfrac{3}{11}\times 11 = 3 \) km s⁻¹.

Teacher's Note:
1. Express g and R in terms of density to compare two planets.
2. The answer is the square root of the product of the g ratio and the R ratio.

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

Explanation:
1. Both the station and the astronaut fall freely with the same acceleration towards the Earth.
2. Because there is no relative acceleration, the astronaut feels no normal reaction, which is weightlessness.
3. So both statements are true and the reason explains the assertion.

Teacher's Note:
1. Weightlessness is the absence of a normal reaction, not the absence of gravity.
2. Free fall with the same acceleration as the surroundings produces apparent weightlessness.

Answer: (b) 6400 km

Explanation:
1. Orbital speed \( v^2 = GM/r \). Given \( v = \tfrac12 v_e \), \( v^2 = \tfrac14\cdot\dfrac{2GM}{R} = \dfrac{GM}{2R} \).
2. So \( r = 2R \) and height \( h = r - R = R = 6400 \) km.

Teacher's Note:
1. Compare v² = GM/r with v_e² = 2GM/R to relate r and R.
2. Height is r − R, not r.

Answer: (a) 7.9 km/s

Explanation:
1. From the previous situation the orbit radius is \( 2R \).
2. Energy conservation from rest at \( 2R \): \( \tfrac12v^2 = GM\left(\dfrac1R - \dfrac1{2R}\right) = \dfrac{GM}{2R} \).
3. \( v = \sqrt{GM/R} = \sqrt{gR} = \sqrt{9.8\times6.4\times10^6} \approx 7.9 \) km/s.

Teacher's Note:
1. After stopping, the satellite starts from rest, so only the potential energy difference gives its kinetic energy.
2. √(gR) is the orbital speed near the surface, 7.9 km/s.

Answer: (b) It can always remain above a point on the equator only.

Explanation:
1. A satellite stays fixed over one point only if its orbital period is 24 h and its orbital plane coincides with the equator.
2. Only then does the centre of its orbit, which is the Earth's centre, lie in the plane through the point.
3. New Delhi is not on the equator, so no satellite can stay permanently above it: (a) and (c) are wrong.
4. Equatorial points can have a geostationary satellite, so (b) is correct and (d) is wrong.

Teacher's Note:
1. Every orbit lies in a plane through the Earth's centre.
2. A point not on the equator cannot stay in such a plane as the Earth rotates.

Answer: (c) \( L/2m \)

Explanation:
1. Area swept in time dt is \( dA = \tfrac12 r\,(v\,dt) \).
2. Areal velocity \( dA/dt = \tfrac12 rv = \dfrac{L}{2m} \) since \( L = mvr \).

Teacher's Note:
1. Areal velocity is constant for a central force (Kepler's second law).
2. It equals half the specific angular momentum.

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

Explanation:
1. The two stars of a binary system revolve about their common centre of mass and are always on opposite sides of it.
2. In every revolution each star traces its orbit once in the same time, so the periods are equal.
3. Hence \( T_A = T_B \).

Teacher's Note:
1. The line joining the stars rotates as a rigid body about the centre of mass.
2. Kepler's law with R as the star separation does not relate the individual orbit radii.

Answer: (b) 129

Explanation:
1. \( T^2 \propto r^3 \Rightarrow T' = 365\left(\tfrac12\right)^{3/2} \).
2. \( T' = 365/2.828 \approx 129 \) days.

Teacher's Note:
1. Use Kepler's third law in ratio form.
2. A closer planet has a shorter year.

Answer: (b) \( T^2\propto R^{7/2} \)

Explanation:
1. \( \dfrac{mv^2}{R}\propto R^{-5/2} \Rightarrow v^2\propto R^{-3/2} \).
2. \( T^2 = \dfrac{4\pi^2R^2}{v^2}\propto R^{2+3/2} = R^{7/2} \).

Teacher's Note:
1. Substitute v² from the force law into T = 2πR/v.
2. For F ∝ R⁻² this method gives T² ∝ R³.

Answer: (b) angular momentum

Explanation:
1. Gravity is a central force, so its torque about the Sun is zero.
2. Hence the angular momentum of the planet about the Sun is conserved.
3. Constant angular momentum means constant areal velocity.

Teacher's Note:
1. Kepler's second law is equivalent to conservation of angular momentum.
2. It holds for any central force, not just gravity.

Answer: (c) \( 2\sqrt3\text{ s} \)

Explanation:
1. \( g_p = \dfrac{G(3M)}{(3R)^2} = \dfrac{g}{3} \).
2. \( T = 2\pi\sqrt{l/g}\propto g^{-1/2} \), so \( T_p = 2\sqrt3 \) s.

Teacher's Note:
1. Diameter ratio is the same as the radius ratio.
2. A smaller g makes the pendulum slower.

Answer: (a) \( 2.8\times10^{5}\text{ m/s} \)

Explanation:
1. Each star is \( 10^{11} \) m from O, so \( V = -\dfrac{2GM}{10^{11}} \).
2. \( \tfrac12v^2 = \dfrac{2\times6.67\times10^{-11}\times3\times10^{31}}{10^{11}} = 4.0\times10^{10} \).
3. \( v = \sqrt{8.0\times10^{10}} \approx 2.8\times10^{5} \) m/s.

Teacher's Note:
1. The gravitational potential at O is the sum of the two star potentials.
2. Escape means kinetic energy plus potential energy equals zero.

Answer: (d) \( \dfrac12\sqrt{\dfrac{GM}{R}(1+2\sqrt2)} \)

Explanation:
1. The particles sit at the corners of a square with diagonal 2R, so side \( a = \sqrt2 R \).
2. Net force towards the centre: \( 2\cdot\dfrac{GM^2}{2R^2}\cos45^\circ + \dfrac{GM^2}{4R^2} = \dfrac{GM^2}{R^2}\left(\dfrac{\sqrt2}{2}+\dfrac14\right) \).
3. \( \dfrac{Mv^2}{R} = \dfrac{GM^2}{4R^2}(2\sqrt2+1) \), so \( v = \dfrac12\sqrt{\dfrac{GM}{R}(1+2\sqrt2)} \).

Teacher's Note:
1. Here R is the circle radius, not the side of the square.
2. The diagonal particle is at distance 2R, giving force GM²/4R².

Answer: (a) \( \dfrac{2GM}{7R}(4\sqrt2-5) \)

Explanation:
1. Potential on the axis at distance h: \( V = -\dfrac{2GM}{R_2^2-R_1^2}\left(\sqrt{R_2^2+h^2}-\sqrt{R_1^2+h^2}\right) \).
2. With \( R_2 = 4R,\ R_1 = 3R,\ h = 4R \): \( V = -\dfrac{2GM}{7R^2}(4\sqrt2R - 5R) = -\dfrac{2GM}{7R}(4\sqrt2-5) \).
3. Work needed to take unit mass to infinity \( = 0 - V = \dfrac{2GM}{7R}(4\sqrt2-5) \).

Teacher's Note:
1. Integrate the potential of thin rings, with surface density M/π(R₂²−R₁²).
2. Work against gravity to reach infinity is positive and equals −V.

Answer: (b) (2) and (3) only

Explanation:
1. Hydrostatic equilibrium: \( P(r) = \tfrac23\pi G\rho^2\,(R^2 - r^2) \).
2. So P at the centre is maximum, not zero: (1) is wrong.
3. \( \dfrac{1-9/16}{1-4/9} = \dfrac{7/16}{5/9} = \dfrac{63}{80} \): (2) is correct.
4. \( \dfrac{1-9/25}{1-4/25} = \dfrac{16}{21} \): (3) is correct.
5. \( \dfrac{3/4}{8/9} = \dfrac{27}{32} \ne \dfrac{20}{27} \): (4) is wrong.

Teacher's Note:
1. Pressure is zero at the surface and maximum at the centre.
2. Find P(r) by integrating dP = −ρ g(r) dr with g(r) = (4/3)πGρr.

Answer: (a) \( \dfrac{3\sqrt5}{2}\sqrt{\dfrac{GM}{a}} \)

Explanation:
1. Find the neutral point where the fields cancel: \( \dfrac{16M}{x^2} = \dfrac{M}{(10a-x)^2} \Rightarrow x = 8a \) from the larger centre.
2. Potential at the start (2a from 16M, 8a from M): \( V_1 = -\dfrac{16GM}{2a}-\dfrac{GM}{8a} = -\dfrac{65GM}{8a} \).
3. Potential at the neutral point (8a from 16M, 2a from M): \( V_2 = -\dfrac{16GM}{8a}-\dfrac{GM}{2a} = -\dfrac{5GM}{2a} \).
4. \( \tfrac12v^2 = V_2 - V_1 = \dfrac{45GM}{8a} \Rightarrow v = \dfrac{3\sqrt5}{2}\sqrt{GM/a} \).

Teacher's Note:
1. Past the neutral point the smaller star's pull takes over, so the body only has to reach that point.
2. Use energy conservation between the surface and the neutral point.

Answer: (c) 7

Explanation:
1. With zero tension both masses must have the same acceleration (towards M).
2. Nearer: \( a_1 = \dfrac{GM}{9l^2} - \dfrac{Gm}{l^2} \). Farther: \( a_2 = \dfrac{GM}{16l^2} + \dfrac{Gm}{l^2} \) (their mutual attraction pulls them towards each other).
3. Equating: \( GM\left(\dfrac19-\dfrac1{16}\right) = \dfrac{2Gm}{1}\cdot\dfrac1{1} \Rightarrow \dfrac{7M}{144} = 2m \Rightarrow m = \dfrac{7M}{288} \).
4. So k = 7.

Teacher's Note:
1. With zero tension in a rigid rod, both masses must have equal acceleration.
2. The mutual force between the two masses pulls the near mass away from M.