JEE Physics Gravitation MCQs Set 27

Answer: (a) \( \frac12 \)

Explanation:
1. \( F_{1} \) = GMm/r² and \( F_{2} \) = GM(2m)/(2r)² = GMm/2r².
2. The minimum \( (F_{2}) \) to maximum \( (F_{1}) \) ratio is 1/2.

Teacher's Note:
1. Compute each force separately.

Answer: (d) the escape velocity of gas molecules is lesser than their root mean square velocity

Explanation:
1. The moon's escape velocity is only about 2.4 km/s.
2. Gas molecules have a higher rms speed and escape.

Teacher's Note:
1. Escape speed below molecular speed.

Answer: (c) both I and II

Explanation:
1. Angular momentum is constant in direction and magnitude under a central force, so the motion stays in a plane.
2. The constant magnitude gives equal areas in equal times.

Teacher's Note:
1. Central force: planar motion and areal velocity.

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

Explanation:
1. Because the polar radius is smaller, \( g \) is larger at the poles.
2. This variation in radius (along with rotation) explains the difference in \( g \).

Teacher's Note:
1. \( g \) depends on the radius.

Answer: (c) the mass of the satellite

Explanation:
1. \( T \) = 2π√(R³/GM) does not involve the satellite's mass.
2. It does depend on the planet's mass and radius.

Teacher's Note:
1. Orbital motion is independent of the satellite's mass.

Answer: (b) A-2, B-4, C-3, D-1

Explanation:
1. \( F \) = GMm/R² \( (A-2) \).
2. \( g \) = GM/R² \( (B-4) \).
3. \( v_{esc} \) = √(2gR) \( (C-3) \).
4. \( g_h = g(1 - 2h/R) (D-1) \).

Teacher's Note:
1. Standard gravitation formulas.

Answer: (d) is directly proportional to g

Explanation:
1. \( g \) = (4/3)πGρR, so ρ = 3g/4πGR.
2. For fixed \( R \), ρ \( ∝ g \).

Teacher's Note:
1. Density follows \( g \).

Answer: (b) the poles

Explanation:
1. A spring balance reads mg.
2. \( g \) is greatest at the poles, so the reading is largest there.

Teacher's Note:
1. \( g \) is maximum at the poles.

Answer: (d) period is equal to the period of rotation of the earth about its axis

Explanation:
1. A geostationary satellite has a period of \( 24 h \).
2. It stays above the same point on the earth.

Teacher's Note:
1. \( T = 24 h \).

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

Explanation:
1. The minimum speed is the escape speed.
2. \( \frac{1}{2}mu^{2} \) = GMm/R gives \( u \) = √(2GM/R).

Teacher's Note:
1. Escape velocity.

Answer: (a) \( -\frac{3GM}{a} \)

Explanation:
1. Shell contributes −GM/a at every point inside it.
2. The central particle contributes −GM/(a/2) = −2GM/a.
3. Total = −3GM/a.

Teacher's Note:
1. Add the potentials of the shell and the particle.

Answer: (b) escape velocity

Explanation:
1. This is the definition of escape velocity.
2. Its value for the earth is 11.2 km/s.

Teacher's Note:
1. Definition.

Answer: (a) five times smaller

Explanation:
1. \( v ∝ \) √(gR).
2. For the moon \( g \) is about 1/6 and \( R \) about 1/3.7 of the earth's, so \( v_{moon} \approx v_{earth}/\sqrt{22} \approx v_{earth}/4.7 \), about five times smaller.

Teacher's Note:
1. Use \( v \) = √(2gR).

Answer: (c) Both (a) and (b)

Explanation:
1. An object can reach \( \infty \) if \( E \ge 0 \).
2. So it escapes when \( E \) is positive or zero.

Teacher's Note:
1. Escape needs non-negative total energy.

Answer: (c) \( 1-\frac dR \)

Explanation:
1. g′ = \( g(1 - d/R) \).
2. So \( g \) decreases by the factor \( (1 - d/R) \).

Teacher's Note:
1. \( g \) falls linearly with depth.

Answer: (b) will decrease at the equator but remain unchanged at the poles

Explanation:
1. Effective \( g = g - \omega^{2}R \cos^{2}\lambda \).
2. At the equator the reduction grows with ω, while at the poles \( \cos \lambda = 0 \) and nothing changes.

Teacher's Note:
1. The rotation term vanishes at the poles.

Answer: (c) A is true, but R is false.

Explanation:
1. The person feels weightless because the satellite and the person are in free fall.
2. Gravity does act on the satellite (it provides the centripetal force), so the reason is false.

Teacher's Note:
1. Weightless does not mean gravity-free.

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

Explanation:
1. \( g \) falls to zero at the centre, so the weight is zero.
2. The reason correctly explains the assertion.

Teacher's Note:
1. \( W \) = mg.

Answer: (c) A is true, but R is false.

Explanation:
1. Jupiter's escape velocity (about 60 km/s) is much greater than the earth's, so it holds light gases.
2. The reason is false.

Teacher's Note:
1. Large escape velocity retains light gases.

Answer: (c) the centrifugal force being more at the equator than at the poles

Explanation:
1. The centrifugal force \( m\omega^{2}r \cos \lambda \) is greatest at the equator.
2. It pushes matter outward there, producing the bulge.

Teacher's Note:
1. Centrifugal effect shapes the earth.

Answer: (c) \( -4\sqrt2\frac{Gm}{a} \)

Explanation:
1. Each corner is at distance \( a/\sqrt{2} \) from the centre.
2. \( V = 4 \) × (−Gm)/(a/√2) = −4√2 Gm/a.

Teacher's Note:
1. Potentials add as scalars.

Answer: (d) produces the necessary acceleration of its motion in a curved path

Explanation:
1. The gravitational force supplies the centripetal force.
2. The satellite is continually falling around the earth.

Teacher's Note:
1. Gravity provides the centripetal force.

Answer: (b) 6V

Explanation:
1. \( v ∝ 1/\sqrt{r} \).
2. \( B's \) radius is one-fourth, so its speed is twice, \( 6V \).

Teacher's Note:
1. Orbital speed varies as \( r^{-1/2} \).

Answer: (b) II only

Explanation:
1. Gravity is always attractive, so I is false.
2. It is a long-range force, so III is false.
3. The force between two particles is unaffected by other particles, so II is true.

Teacher's Note:
1. Gravity is attractive and long range.

Answer: (a) \( \frac12R \)

Explanation:
1. \( g \) = (4/3)πGρR.
2. Equal \( g \) with twice the density requires half the radius.

Teacher's Note:
1. \( g ∝ \) ρR.