JEE Physics Gravitation MCQs Set 28

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

Explanation:
1. The moon's attraction varies across the oceans and causes tides, so the assertion is true.
2. The reason is a true fact but it does not explain the tides.

Teacher's Note:
1. Tides come from the moon's pull.

Answer: (d) None of these

Explanation:
1. \( K \) is positive, \( U \) is negative and \( E = -K \) is negative.
2. None of the three statements is correct.

Teacher's Note:
1. Bound orbit: \( E \lt 0 \).

Answer: (a) 4.22 × 10⁴ km

Explanation:
1. From \( T = 24 h, r = (GMT^{2}/4\pi^{2})^{1/3} \approx 4.22 \times 10^{7} m \).
2. That is 4.22 × 10⁴ km from the centre.

Teacher's Note:
1. About 6.6 earth radii.

Answer: (d) I, II and III

Explanation:
1. Weight = mg and \( g \) on the moon is about \( g/6 \).
2. Mass, both inertial and gravitational, is intrinsic and does not change.

Teacher's Note:
1. Mass is constant, weight varies.

Answer: (c) Kinetic energy

Explanation:
1. \( K \) = ½mv² ≥ 0.
2. The potential energy of a bound system and the total energy of a satellite are negative.

Teacher's Note:
1. Only kinetic energy is non-negative.

Answer: (b) move tangentially to the original orbit with the same velocity

Explanation:
1. With no force, the satellite moves in a straight line with its instantaneous velocity.
2. That velocity is tangent to the orbit.

Teacher's Note:
1. Newton's first law.

Answer: (d) zero

Explanation:
1. By the shell theorem, the field inside a shell is zero.
2. The potential there is constant.

Teacher's Note:
1. Zero field inside a shell.

Answer: (c) revolves round the earth at a suitable height with the same angular velocity and in the same direction as the earth does about its own axis

Explanation:
1. It must have a period of \( 24 h \) in the equatorial plane.
2. So it matches the earth's angular velocity and direction and appears fixed in the sky.

Teacher's Note:
1. Same ω as the earth.

Answer: (b) the total angular momentum remains constant

Explanation:
1. The gravitational force of the sun is central, so there is no torque.
2. Hence the angular momentum is constant, while speeds vary.

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

Answer: (a) at the equator

Explanation:
1. The earth is flattened at the poles, so the equatorial radius is larger and \( g \) is smaller there.
2. Rotation also lowers the effective \( g \) at the equator.

Teacher's Note:
1. \( g \) is minimum at the equator.

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

Explanation:
1. The escape velocity comes from energy conservation and has no direction dependence.
2. The formula √(2gR) contains only \( g \) and \( R \), which supports the assertion.

Teacher's Note:
1. Energy is a scalar.

Answer: (a) \( m^0 \)

Explanation:
1. \( v_{esc} \) = √(2GM/R) does not contain the body's mass.
2. So it varies as \( m^{0} \).

Teacher's Note:
1. Escape velocity is independent of the projectile's mass.

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

Explanation:
1. The sun's force is central, so the torque about the sun is zero.
2. Angular momentum is therefore conserved while the speeds vary.

Teacher's Note:
1. Zero torque, constant \( L \).

Answer: (b) the equator

Explanation:
1. Weight = \( mg_{eff} \), and \( g_{eff} \) is least at the equator.
2. So the body weighs least at the equator.

Teacher's Note:
1. Centrifugal effect is greatest at the equator.

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

Explanation:
1. The astronaut and the station are in free fall with the same acceleration.
2. That produces weightlessness, so the reason explains the assertion.

Teacher's Note:
1. Free fall means weightlessness.

Answer: (c) \( 6\sqrt2 \)

Explanation:
1. Orbital radii: \( 6R \) for the geostationary satellite \( (T = 24 h) \) and \( 3R \) for the other.
2. \( T ∝ r^{1.5} \), so \( T = 24 \times (1/2)^{1.5} = 24/(2\sqrt{2}) = 6\sqrt{2} h \).

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

Answer: (c) zero

Explanation:
1. \( U \) = −Gm₁m₂/r.
2. As \( r \to \infty, U \to 0 \).

Teacher's Note:
1. Potential energy is zero at \( \infty \) by convention.

Answer: (b) increase

Explanation:
1. The polar radius is smaller and there is no centrifugal reduction at the poles.
2. So \( g \) increases from the equator to the poles.

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

Answer: (a) \( \sqrt{\frac{Gm}{4R}} \)

Explanation:
1. The particles are at opposite ends of a diameter, so their separation is \( 2R \).
2. Gm²/(2R)² = mv²/R gives \( v \) = √(Gm/4R).

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

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

Explanation:
1. For small heights \( g \) is constant and the path is a parabola; for large heights \( g \) varies and the path becomes an ellipse, so the assertion is true.
2. The path depends on gravity, so the reason is false.

Teacher's Note:
1. Path depends on gravity.

Answer: (b) less than the orbital speed of the earth

Explanation:
1. \( v \) = √(GM/r) decreases with orbital radius.
2. Jupiter is farther from the sun, so it is slower.

Teacher's Note:
1. Farther planets move slower.

Answer: (d) 2R

Explanation:
1. g′ = \( g R^{2}/(R + h)^{2} = g/9 \) gives \( R + h = 3R \).
2. So \( h = 2R \).

Teacher's Note:
1. Inverse square law.

Answer: (d) \( 6.4\times10^{10}\ \text{J} \)

Explanation:
1. The energy needed equals the binding energy GMm/R = mgR.
2. = \( 1000 \times 10 \times 6.4 \times 10^{6} = 6.4 \times 10^{10} J \).

Teacher's Note:
1. \( E \) = mgR.

Answer: (b) decreases

Explanation:
1. \( V \) at the centre = −GM/R.
2. As \( R \) decreases, \( V \) becomes more negative, so it decreases.

Teacher's Note:
1. Potential at the centre depends on \( 1/R \).

Answer: (d) half the potential energy of the satellite

Explanation:
1. \( K \) = GMm/2r, \( U \) = −GMm/r and \( E \) = −GMm/2r.
2. So \( E = U/2 \).

Teacher's Note:
1. \( E = -K = U/2 \).