JEE Physics Gravitation MCQs Set 29

Answer: (a) 100 km

Explanation:
1. At height \( h \), g′ ≈ \( g(1 - 2h/R) \), and at depth \( d \), g′ = \( g(1 - d/R) \).
2. Equating gives \( d = 2h = 100 \) km.

Teacher's Note:
1. \( d = 2h \) for small values.

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

Explanation:
1. \( G = 6.67 \times 10^{-11} N m^{2} \) kg⁻² \( (A-3) \).
2. \( g_h = g_{0}(1 - 2h/R) (B-4) \).
3. \( T^{2} ∝ R^{3} \) is the law of periods \( (C-1) \).
4. A geostationary satellite has \( T = 24 h (D-2) \).

Teacher's Note:
1. Match each item with its value or law.

Answer: (c) zero

Explanation:
1. The three forces have equal magnitude and are 120° apart.
2. Their vector sum is zero.

Teacher's Note:
1. Symmetry gives zero net force.

Answer: (a) \( T^2=\frac{4\pi^2R^3}{GM_m} \)

Explanation:
1. \( GM_m m/R^{2} = m\omega^{2}R \) with \( \omega = 2\pi/T \).
2. \( T^{2} = 4\pi^{2}R^{3}/(GM_m) \).

Teacher's Note:
1. Kepler's third law with the central mass.

Answer: (d) the semi-major axis

Explanation:
1. Kepler's third law uses the semi-major axis for an elliptical orbit.
2. For a circle this equals the radius.

Teacher's Note:
1. \( R \) is the semi-major axis.

Answer: (b) infinite

Explanation:
1. Inside the satellite the effective \( g \) is zero (free fall).
2. \( T = 2\pi\sqrt{l/g} \to \infty \).

Teacher's Note:
1. No restoring force in free fall.

Answer: (c) directly proportional to the fourth power of the radius

Explanation:
1. \( m \) = (4/3)πr³ρ and the centre distance is \( 2r \).
2. \( F \) = Gm²/(2r)² \( ∝ r^{6}/r^{2} = r^{4} \).

Teacher's Note:
1. Mass scales as \( r^{3} \), distance as \( r \).

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

Explanation:
1. \( T^{2} ∝ R^{3} \), so a smaller \( R \) gives a smaller \( T \).
2. The reason is the law behind the assertion.

Teacher's Note:
1. Kepler's third law.

Answer: (c) \( 110\ \text{km s}^{-1} \)

Explanation:
1. \( v ∝ \sqrt{M/R} \).
2. The factor is \( \sqrt (10/(1/10)) = \sqrt{100} = 10 \), so \( v = 110 \) km/s.

Teacher's Note:
1. \( v_{esc} ∝ \sqrt{M/R} \).

Answer: (b) \( t_1=2t_2 \)

Explanation:
1. Areal velocity is constant, so the time is proportional to the area swept.
2. Area SCD = 2 × area SAB gives \( t_{1} = 2t_{2} \).

Teacher's Note:
1. Equal areas in equal times.

Answer: (b) negative

Explanation:
1. Total energy \( E \) = ½mv² − GMm/R, and \( v \lt v_{esc} \) gives \( E \lt 0 \).
2. The missile is bound to the earth.

Teacher's Note:
1. Bound means negative energy.

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

Explanation:
1. Gravitational force is always attractive \( (A-2) \).
2. \( g \) is maximum at the earth's surface \( (B-1) \).
3. \( v_{esc} = \sqrt{2} \) √(gR) \( (C-4) \) and \( v_{orb} \) = √(gR) \( (D-3) \).

Teacher's Note:
1. Recall \( v_{esc} = \sqrt{2} v_{orb} \).

Answer: (a) The time taken in travelling D to A to B is less than that for B to C to D

Explanation:
1. By Kepler's second law, the planet moves faster near the sun.
2. The half of the orbit through the nearest point \( A \) is covered more quickly than the half through the farthest point \( C \).

Teacher's Note:
1. Faster near perihelion.

Answer: (a) Linear momentum

Explanation:
1. The gravitational force is central and conservative, so angular momentum and total energy are conserved.
2. The force changes the object's velocity, so its linear momentum is not conserved.

Teacher's Note:
1. Central force: angular momentum conserved.

Answer: (c) the radiation emitted by the sun exerts a radiation pressure on the comet, throwing its tail away from the sun

Explanation:
1. Radiation pressure and the solar wind push the light particles away from the sun.
2. So the tail always points away from the sun.

Teacher's Note:
1. Radiation pressure on dust.

Answer: (b) angular momentum

Explanation:
1. Areal velocity = \( L/2m \).
2. It is constant because angular momentum is conserved under a central force.

Teacher's Note:
1. dA/dt = \( L/2m \).

Answer: (d) 129

Explanation:
1. \( T^{2} ∝ R^{3} \), so \( T ∝ R^{3/2} \).
2. \( T = 365 \times (1/2)^{1.5} = 365/2.83 \approx 129 \) days.

Teacher's Note:
1. Kepler's third law.

Answer: (a) variable (proportional to distance from the centre)

Explanation:
1. Inside a uniform solid sphere \( E \) = GMr/R³.
2. So \( E ∝ r \).

Teacher's Note:
1. Field grows linearly from the centre.

Answer: (b) \( v_1 \lt v_2 \)

Explanation:
1. \( v \) = √(GM/r) is independent of the satellite's mass.
2. Since \( r_{1} \gt r_{2}, v_{1} \lt v_{2} \).

Teacher's Note:
1. Larger orbit means slower speed.

Answer: (d) \( g'=3g \)

Explanation:
1. \( g ∝ \) ρR for the same density.
2. The radius is 3 times, so g′ = \( 3g \).

Teacher's Note:
1. \( g ∝ R \) at fixed density.

Answer: (c) \( 10\sqrt{10} \)

Explanation:
1. \( T^{2} ∝ r^{3} \), so \( T_N/T_S = (10)^{3/2} \).
2. = 10√10.

Teacher's Note:
1. Use \( T ∝ r^{1.5} \).

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

Explanation:
1. \( \DeltaU \) = GMm/R − GMm/2R = GMm/2R = ½mgR.
2. The reason uses the same result through energy conservation.

Teacher's Note:
1. \( \DeltaU \) = ½mgR for \( h = R \).

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

Explanation:
1. The weights are not tied for that purpose (the traveller's mass is the same), so the assertion is false.
2. The reason is a true statement.

Teacher's Note:
1. The assertion is false.

Answer: (c) the normal force is zero

Explanation:
1. The person and the satellite fall with the same acceleration.
2. So the normal force between them is zero, which is what weightlessness means.

Teacher's Note:
1. Apparent weight equals the normal force.

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

Explanation:
1. \( g \) = GM/R² \( (A-3) \).
2. \( v_{esc} = \sqrt{2gR_E} (B-1) \).
3. \( E \) = −Gm₁m₂/2(R \( + h) (C-4 \)).
4. \( U \) = −Gm₁m₂/r \( (D-2) \).

Teacher's Note:
1. Standard formulas.