JEE Physics Gravitation MCQs Set 25

Answer: (b) become one-fourth

Explanation:
1. \( F \propto \frac{1}{r^2} \), so doubling r divides F by 4.

Teacher's Note:
1. Gravitation obeys an inverse square law.
2. Force falls to one-fourth when distance doubles.

Answer: (a) acceleration

Explanation:
1. Both fall with acceleration g, independent of mass, so they reach the ground with the same speed and the same acceleration.
2. Kinetic energy, momentum and weight all depend on mass.

Teacher's Note:
1. Free-fall acceleration does not depend on mass.
2. Mass-dependent quantities differ for the two bodies.

Answer: (c) \( \frac{5}{6}mgh \)

Explanation:
1. \( \Delta U = \frac{mgh}{1 + h/R} \).
2. With \( h/R = 1/5 \): \( \Delta U = \frac{mgh}{6/5} = \frac{5}{6}mgh \).

Teacher's Note:
1. The expression mgh holds only for h much smaller than R.
2. The exact change is GMm(1/R - 1/(R+h)).

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Gravity is attractive, so work must be done against it to take a mass to infinity; hence potential is negative everywhere. Statement A is true.
2. The force does follow the inverse square law, so R is true.
3. The inverse square dependence gives the form of the potential, but the reason the potential is negative is the attractive nature of the force. R does not explain A.

Teacher's Note:
1. Check the sign of the potential from the attractive nature of the force.
2. True statements need not explain each other.

Answer: (a) \( mgR\,\frac{n}{n+1} \)

Explanation:
1. \( W = GMm\left(\frac1R - \frac{1}{R+nR}\right) = \frac{GMm}{R}\cdot\frac{n}{n+1} \).
2. Using \( GM = gR^2 \): \( W = mgR\,\frac{n}{n+1} \).

Teacher's Note:
1. Use the exact potential energy difference for large heights.
2. For n much less than 1 this reduces to mg(nR).

Answer: (b) 1 : 4

Explanation:
1. Both move about the centre of mass with the same angular velocity, so \( r \propto 1/m \) and the speed \( v \propto 1/m \).
2. \( KE = \tfrac12 mv^2 \propto 1/m \), so \( KE_{4m} : KE_m = 1 : 4 \).

Teacher's Note:
1. Heavier body is nearer to the centre of mass and moves slower.
2. Equal and opposite momenta give KE inversely proportional to mass.

Answer: (b) \( \frac{3g}{4\pi RG} \)

Explanation:
1. \( g = \frac{GM}{R^2} = \frac{4}{3}\pi G\rho R \).
2. So \( \rho = \frac{3g}{4\pi RG} \).

Teacher's Note:
1. Write M as density times volume.
2. This relation lets us weigh the earth from g.

Answer: (a) \( \frac{5g}{9} \)

Explanation:
1. \( g' = g\left(\frac{R}{1.5R}\right)^2 = \frac{4g}{9} \).
2. Decrease \( = g - \frac{4g}{9} = \frac{5g}{9} \).

Teacher's Note:
1. Be careful: the question asks for the decrease, not the new value.
2. The new value is 4g/9.

Answer: (d) \( \frac15 \)

Explanation:
1. \( g = \frac43\pi G\rho R \), so for constant g, \( \rho \propto 1/R \).
2. A five-fold radius needs a density one-fifth of the original.

Teacher's Note:
1. Use g proportional to density times radius.
2. Larger radius needs lower density to keep g fixed.

Answer: (d) the radius of its orbit becomes half

Explanation:
1. In a circular orbit \( KE = \frac{GMm}{2r} \), so \( KE \propto 1/r \).
2. Doubling KE halves r; the new orbit is lower but still circular and bound.

Teacher's Note:
1. KE of a circular orbit is inversely proportional to the radius.
2. A lower circular orbit has higher kinetic energy but lower (more negative) total energy.

Answer: (a) 2v₀

Explanation:
1. \( v_e = \sqrt{\frac{8\pi G\rho}{3}}\,R \), so for equal density \( v_e \propto R \).
2. Double the radius gives double the escape velocity.

Teacher's Note:
1. Escape speed is proportional to R times the square root of density.
2. Same density means v_e ∝ R.

Answer: (c) A body of zero total mechanical energy placed in a gravitational field will escape from the field.

Explanation:
1. Zero total energy corresponds to parabolic motion: the body just reaches infinity with zero speed, so it escapes.
2. A concentric cavity has zero field; the actual path for sub-orbital speed is an ellipse; a satellite orbit can lie in any plane through the centre.

Teacher's Note:
1. Total energy greater than or equal to zero means an unbound orbit.
2. Only geostationary satellites need to be in the equatorial plane.

Answer: (a) \( g \propto \frac{1}{R^2} \)

Explanation:
1. \( g = \frac{GM}{R^2} \) with M fixed.
2. So g increases as the inverse square of the shrinking radius.

Teacher's Note:
1. Surface gravity depends on mass and radius only.
2. Smaller radius means much larger g.

Answer: (b) 12 hours

Explanation:
1. Satellite and earth turn in opposite senses, so the relative angular speed is \( 2\omega \).
2. The interval is 24/2 = 12 hours.

Teacher's Note:
1. Relative angular velocity adds when motions are opposite.
2. Same-direction motion would give zero relative speed (geostationary).

Answer: (b) \( \frac{r}{4^{1/3}} \)

Explanation:
1. \( \omega^2 r^3 = GM \), so \( r \propto \omega^{-2/3} \).
2. With \( \omega \to 2\omega \): \( r' = r/2^{2/3} = r/4^{1/3} \).

Teacher's Note:
1. Kepler's third law: T² ∝ r³.
2. A shorter day needs a lower orbit.

Answer: (a) \( \left[(m_1+m_2)\frac{2G}{d}\right]^{1/2} \)

Explanation:
1. Momentum conservation: \( m_1v_1 = m_2v_2 \). Energy: \( \tfrac12m_1v_1^2 + \tfrac12m_2v_2^2 = \frac{Gm_1m_2}{d} \).
2. With reduced mass \( \mu = \frac{m_1m_2}{m_1+m_2} \): \( \tfrac12\mu v_{rel}^2 = \frac{Gm_1m_2}{d} \), so \( v_{rel} = \sqrt{\frac{2G(m_1+m_2)}{d}} \).

Teacher's Note:
1. Use the reduced mass for two-body approach problems.
2. Relative speed is the sum of the two speeds.

Answer: (d) \( \frac23\frac{GMm}{R^2} \)

Explanation:
1. Just outside: force \( = \frac{GMm}{R^2} \) from the whole mass M.
2. Just inside the shell only the central mass M/3 acts: \( \frac{GMm}{3R^2} \).
3. Change \( = \frac{GMm}{R^2} - \frac{GMm}{3R^2} = \frac23\frac{GMm}{R^2} \).

Teacher's Note:
1. A uniform shell exerts no force on a body inside it.
2. Only the enclosed mass attracts a body inside.

Answer: (d) 4R

Explanation:
1. After impact: \( KE = \frac{2Gm^2}{3R} \), \( U = -\frac{8Gm^2}{3R} \), total \( E = -\frac{2Gm^2}{R} \).
2. At maximum separation d the kinetic energy is zero: \( -\frac{8Gm^2}{d} = -\frac{2Gm^2}{R} \), so \( d = 4R \).

Teacher's Note:
1. Energy conservation fixes the turning point.
2. Total energy after impact is negative, so the stars remain bound.

Answer: (a) \( R/2 \)

Explanation:
1. \( g = \frac43\pi G\rho R \), so \( \rho R \) must be the same.
2. With double density the radius is half.

Teacher's Note:
1. g depends on the product of density and radius.
2. Same g, double density, half radius.

Answer: (c) The acceleration of S is always directed towards the centre of the earth.

Explanation:
1. Gravity is a central force towards the earth's centre, so the acceleration always points there.
2. Angular momentum is conserved in both magnitude and direction, total energy is constant, and linear momentum changes.

Teacher's Note:
1. Central forces conserve angular momentum.
2. Speed varies along an ellipse, so linear momentum is not constant.

Answer: (a) gravitational potential energy

Explanation:
1. \( U = -\frac{GMm}{r} \) becomes less negative as r increases, so U increases.
2. \( \omega \propto r^{-3/2} \), \( v \propto r^{-1/2} \) and \( a \propto r^{-2} \) all decrease.

Teacher's Note:
1. Larger orbit means slower and less acceleration.
2. Total energy also increases (less negative).

Answer: (b) 1 and 4 only

Explanation:
1. Accelerations are \( \frac{Gm_2}{r^2} \) and \( \frac{Gm_1}{r^2} \), so the lighter mass accelerates more (1 true, 2 false).
2. The centre of mass is at rest only in the frame in which it started at rest, so 3 is false.
3. Only internal conservative forces act, so total energy is constant (4 true).

Teacher's Note:
1. The lighter body has the larger acceleration.
2. Rest is frame-dependent.

Answer: (b) 2, 3 and 4

Explanation:
1. Potential inside is constant, equal to \( -\frac{GM}{R} \), not zero.
2. Field is zero everywhere inside, which is the same value everywhere (so statements 2, 3 and 4 hold; 1 is false).

Teacher's Note:
1. Zero field does not mean zero potential.
2. A constant potential has zero gradient.

Answer: (a) 2 and 3 only

Explanation:
1. Radius is the circumradius \( a/\sqrt3 \). Net force on each \( = \sqrt3\frac{Gm^2}{a^2} = \frac{mv^2}{a/\sqrt3} \), so \( v = \sqrt{Gm/a} \).
2. \( U = -\frac{3Gm^2}{a} \), \( K = \frac{3}{2}\cdot\frac{Gm^2}{a} \), \( E = -\frac{1.5Gm^2}{a} \), so binding energy is 1.5Gm²/a.
3. \( T = \frac{2\pi (a/\sqrt3)}{v} = 2\pi\sqrt{\frac{a^3}{3Gm}} \), so 4 is false.

Teacher's Note:
1. Radius of the path is the centroid distance, a/√3.
2. Binding energy is the magnitude of total energy.

Answer: (d) may be equal to 10 N/kg or greater than 10 N/kg

Explanation:
1. The x-component of the field is \( -\frac{dV}{dx} = \frac{20}{2} = 10 \) N/kg.
2. Other components along y or z may exist, so the magnitude is at least 10 N/kg.

Teacher's Note:
1. Only the component along the x-axis is fixed by the data.
2. Magnitude is not less than any one of its components.

Answer: (d) 1 and 3 only

Explanation:
1. Drag removes energy; the orbit shrinks and total energy decreases but speed increases, so kinetic energy rises.
2. Drag exerts a retarding torque, so angular momentum falls, and the period decreases.

Teacher's Note:
1. Satellites speed up as they spiral inwards.
2. Angular momentum is not conserved with a dissipative torque.

Answer: (a) 2 and 4 only

Explanation:
1. To stay above one point it must rotate with the earth, west to east, with a 24 hour period.
2. This is possible only in the equatorial plane, directly above points on the equator.

Teacher's Note:
1. Geostationary means fixed relative to the ground.
2. Only the equatorial plane keeps a satellite overhead.

Answer: (c) \( -\frac{3Gm^2}{a} \)

Explanation:
1. There are 6 pairs, so \( U_i = -\frac{6Gm^2}{a} \) and \( U_f = -\frac{6Gm^2}{2a} = -\frac{3Gm^2}{a} \).
2. Work by the system \( = U_i - U_f = -\frac{3Gm^2}{a} \) (negative, because the masses are pulled apart against attraction).

Teacher's Note:
1. Work done by a conservative force equals the decrease in potential energy.
2. Separating attracting bodies makes this work negative.

Answer: (a) \( \frac{Gm^2}{3L^2} \)

Explanation:
1. Mass per unit length \( = \frac{m}{2L} \). \( F = Gm\int_L^{3L}\frac{(m/2L)\,dx}{x^2} \).
2. \( = \frac{Gm^2}{2L}\left(\frac1L - \frac1{3L}\right) = \frac{Gm^2}{3L^2} \).

Teacher's Note:
1. Integrate the force from small elements of the rod.
2. Distances run from L to 3L from the small mass.

Answer: (b) \( -\frac{4Gm^2}{L}\left(3+\frac{3}{\sqrt2}+\frac1{\sqrt3}\right) \)

Explanation:
1. There are 28 pairs: 12 edges (distance L), 12 face diagonals (\( \sqrt2L \)) and 4 body diagonals (\( \sqrt3L \)).
2. \( U = -\frac{Gm^2}{L}\left(12 + \frac{12}{\sqrt2} + \frac{4}{\sqrt3}\right) = -\frac{4Gm^2}{L}\left(3+\frac3{\sqrt2}+\frac1{\sqrt3}\right) \).

Teacher's Note:
1. Count every pair once: 8 × 7 / 2 = 28.
2. Group pairs by separation.