JEE Physics Gravitation MCQs Set 06

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Multiple Choice Questions

Question 1: A particle of mass m moves in a horizontal circle of radius R under a centripetal force \( -\frac{A}{r^2} \) (A constant). Taking zero potential energy at infinity, the total energy is:
(a) \( \frac AR \)
(b) \( -\frac AR \)
(c) \( \frac{A}{2R} \)
(d) \( -\frac{A}{2R} \)
Show Answer & Explanation

Answer: (d) \( -\frac{A}{2R} \)

Explanation:
1. \( U = -\frac AR \)
2. \( \frac{mv^2}{R} = \frac{A}{R^2} \Rightarrow K = \frac{A}{2R} \)
3. \( E = K+U = -\frac{A}{2R} \)

Teacher's Note:
1. Same structure as a satellite: E equals minus kinetic energy.

Question 2: The potential energy of a 3 kg body at the surface of a planet is -54 J. The escape velocity is:
(a) 18 m/s
(b) 162 m/s
(c) 36 m/s
(d) 6 m/s
Show Answer & Explanation

Answer: (d) 6 m/s

Explanation:
1. Escape needs \( \frac12mv^2 = 54 \)
2. \( v^2 = \frac{108}{3} = 36 \)
3. \( v = 6 \) m/s

Teacher's Note:
1. Kinetic energy needed equals the magnitude of potential energy.

Question 3: The escape velocity of a 1 kg body on a planet is 100 m/s. The potential energy of the body at that planet is:
(a) -5000 J
(b) -1000 J
(c) -2400 J
(d) -10000 J
Show Answer & Explanation

Answer: (a) -5000 J

Explanation:
1. \(
2. U
3. = \frac12mv_e^2 = \frac12\times1\times100^2 = 5000 \) J
4. \( U = -5000 \) J

Teacher's Note:
1. Escape energy equals minus the potential energy.

Question 4: The ratio of the radii of two satellites is p and the ratio of their accelerations due to gravity is q. The ratio of their escape velocities is:
(a) \( \left(\frac qp\right)^{1/2} \)
(b) \( \left(\frac pq\right)^{1/2} \)
(c) \( pq \)
(d) \( \sqrt{pq} \)
Show Answer & Explanation

Answer: (d) \( \sqrt{pq} \)

Explanation:
1. \( v_e = \sqrt{2gR} \)
2. Ratio \( = \sqrt{qp} \)

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

Question 5: The escape velocity of a body from the earth is 11.2 km/s. If it is thrown at an angle of 45 degrees to the horizontal, the escape velocity will be:
(a) 11.2 km/s
(b) 22.4 km/s
(c) \( 11.2/\sqrt2 \) km/s
(d) \( 11.2\sqrt2 \) km/s
Show Answer & Explanation

Answer: (a) 11.2 km/s

Explanation:
1. Escape velocity depends only on energy, not on direction
2. It stays 11.2 km/s (ignoring air)

Teacher's Note:
1. Energy is a scalar, so direction does not matter.

Question 6: The escape velocity from the earth is 11.2 km/s. Another planet has 100 times the mass of the earth and 4 times its radius. Its escape velocity is:
(a) 56.0 km/s
(b) 280 km/s
(c) 112 km/s
(d) 11.2 km/s
Show Answer & Explanation

Answer: (a) 56.0 km/s

Explanation:
1. \( v \propto \sqrt{M/R} \)
2. \( \sqrt{100/4} = 5 \)
3. \( 5\times11.2 = 56 \) km/s

Teacher's Note:
1. Scale by the square root of the mass-to-radius ratio.

Question 7: A body is projected vertically upward from the earth's surface with a velocity equal to half the escape velocity. If R is the radius of the earth, the maximum height attained is:
(a) \( \frac R6 \)
(b) \( \frac R3 \)
(c) \( \frac{2R}{3} \)
(d) \( R \)
Show Answer & Explanation

Answer: (b) \( \frac R3 \)

Explanation:
1. \( \frac12v^2 = \frac{GM}{4R} \)
2. \( GM\left(\frac1R-\frac1{R+h}\right) = \frac{GM}{4R} \)
3. \( \frac1{R+h} = \frac{3}{4R},\ h = \frac R3 \)

Teacher's Note:
1. Energy conservation with the speed v_e/2.

Question 8: The binding energy of the moon-earth system is:
(a) \( \frac{GM_eM_m}{r_{em}} \)
(b) \( \frac{GM_eM_m}{2r_{em}} \)
(c) \( -\frac{GM_eM_m}{r_{em}} \)
(d) \( -\frac{GM_eM_m}{2r_{em}} \)
Show Answer & Explanation

Answer: (b) \( \frac{GM_eM_m}{2r_{em}} \)

Explanation:
1. Total energy \( E = -\frac{GM_eM_m}{2r_{em}} \)
2. Binding energy is its magnitude

Teacher's Note:
1. Binding energy is positive.

Question 9: Two artificial satellites A and B are at heights r_A and r_B above the earth's surface (earth's radius R). The ratio of their speeds v_A : v_B is:
(a) \( \left(\frac{r_B+R}{r_A+R}\right)^{1/2} \)
(b) \( \left(\frac{r_B+R}{r_A+R}\right)^{2} \)
(c) \( \left(\frac{r_B}{r_A}\right)^{2} \)
(d) \( \left(\frac{r_B}{r_A}\right)^{1/2} \)
Show Answer & Explanation

Answer: (a) \( \left(\frac{r_B+R}{r_A+R}\right)^{1/2} \)

Explanation:
1. \( v = \sqrt{GM/(R+h)} \)
2. \( \frac{v_A}{v_B} = \sqrt{\frac{R+r_B}{R+r_A}} \)

Teacher's Note:
1. Orbital radius is measured from the centre.

Question 10: The average radii of the orbits of Mercury and the earth around the sun are \( 6\times10^7 \) km and \( 1.5\times10^8 \) km. The ratio of their orbital speeds is:
(a) \( \sqrt5:\sqrt2 \)
(b) \( \sqrt2:\sqrt5 \)
(c) 2.5 : 1
(d) 1 : 25
Show Answer & Explanation

Answer: (a) \( \sqrt5:\sqrt2 \)

Explanation:
1. \( v \propto r^{-1/2} \)
2. \( \frac{v_M}{v_E} = \sqrt{\frac{1.5\times10^8}{6\times10^7}} = \sqrt{2.5} = \sqrt5:\sqrt2 \)

Teacher's Note:
1. Nearer planets move faster.

Question 11: A body is dropped from a satellite in its geostationary orbit:
(a) it will burn on entering the atmosphere
(b) it will remain in the same place with respect to the earth
(c) it will reach the earth in 24 hours
(d) it will perform uncertain motion
Show Answer & Explanation

Answer: (b) it will remain in the same place with respect to the earth

Explanation:
1. The body has the same orbital velocity as the satellite
2. It keeps orbiting with it, staying in the same place relative to the earth

Teacher's Note:
1. Release gives no relative velocity.

Question 12: Two ordinary satellites revolve round the earth in the same elliptical orbit. Which quantity is conserved?
(a) velocity
(b) angular velocity
(c) angular momentum
(d) none of the above
Show Answer & Explanation

Answer: (c) angular momentum

Explanation:
1. Gravity is a central force
2. So torque about the earth is zero and angular momentum is constant

Teacher's Note:
1. Speed and angular velocity change along an ellipse.

Question 13: Kepler's second law is a consequence of:
(a) conservation of kinetic energy
(b) conservation of linear momentum
(c) conservation of angular momentum
(d) conservation of speed
Show Answer & Explanation

Answer: (c) conservation of angular momentum

Explanation:
1. Equal areas in equal times: \( \frac{dA}{dt} = \frac{L}{2m} \)
2. Constant L gives constant areal velocity

Teacher's Note:
1. The gravitational force is central.

Question 14: A projectile starts moving round the earth in a circular path of radius equal to nine times the earth's radius R. Its time period will be:
(a) \( 2\pi\sqrt{\frac Rg} \)
(b) \( 27\times2\pi\sqrt{\frac Rg} \)
(c) \( \pi\sqrt{\frac Rg} \)
(d) \( 0.8\times3\pi\sqrt{\frac Rg} \)
Show Answer & Explanation

Answer: (b) \( 27\times2\pi\sqrt{\frac Rg} \)

Explanation:
1. \( T = 2\pi\sqrt{\frac{r^3}{GM}} = 2\pi\sqrt{\frac{(9R)^3}{gR^2}} \)
2. \( = 2\pi\cdot27\sqrt{\frac Rg} \)

Teacher's Note:
1. Replace GM by gR squared.

Question 15: A satellite of the earth of mass m is taken from orbital radius 2R to 3R. The minimum work done is:
(a) \( \frac{GMm}{6R} \)
(b) \( \frac{GMm}{12R} \)
(c) \( \frac{GMm}{24R} \)
(d) \( \frac{GMm}{3R} \)
Show Answer & Explanation

Answer: (b) \( \frac{GMm}{12R} \)

Explanation:
1. \( E = -\frac{GMm}{2r} \)
2. \( W = \Delta E = \frac{GMm}{2}\left(\frac1{2R}-\frac1{3R}\right) = \frac{GMm}{12R} \)

Teacher's Note:
1. Work equals change in total orbital energy.

Question 16: If a graph is plotted between T² and r³ for a planet, its slope will be:
(a) \( \frac{4\pi^2}{GM} \)
(b) \( \frac{GM}{4\pi^2} \)
(c) \( 4\pi GM \)
(d) zero
Show Answer & Explanation

Answer: (a) \( \frac{4\pi^2}{GM} \)

Explanation:
1. \( T^2 = \frac{4\pi^2}{GM}r^3 \)
2. Slope \( = \frac{4\pi^2}{GM} \)

Teacher's Note:
1. Kepler's third law in graphical form.

Question 17: A planet revolves around the sun. Its distances at apogee and perigee are r_A and r_P, speeds v_A and v_P, and the masses of planet and sun are m and M. Which of the following are true? (a) T² = (π²/2Gm)(r_A+r_P)² (b) T² = (π²/2GM)(r_A+r_P)³ (c) v_A r_A = v_P r_P (d) v_A < v_P and r_A > r_P
(a) a, b, c
(b) a, b, d
(c) b, c, d
(d) all
Show Answer & Explanation

Answer: (c) b, c, d

Explanation:
1. Semi-major axis \( a = \frac{r_A+r_P}{2} \)
2. \( T^2 = \frac{4\pi^2a^3}{GM} = \frac{\pi^2(r_A+r_P)^3}{2GM} \), so (b) is true and (a) is false
3. Angular momentum conservation gives (c)
4. Farther point has smaller speed, (d)

Teacher's Note:
1. Use the semi-major axis in Kepler's third law.

Question 18: A satellite launching station should be:
(a) near the equatorial region
(b) near the polar region
(c) on the polar axis
(d) all locations are equally good
Show Answer & Explanation

Answer: (a) near the equatorial region

Explanation:
1. The earth's rotation adds speed towards the east
2. This is greatest at the equator, saving fuel

Teacher's Note:
1. Launch eastwards from low latitude.

Question 19: A space shuttle is launched in a circular orbit near the earth's surface. The additional velocity to be given to the shuttle to free it from the earth's gravitational influence is:
(a) 1.52 km/s
(b) 2.75 km/s
(c) 3.28 km/s
(d) 5.18 km/s
Show Answer & Explanation

Answer: (c) 3.28 km/s

Explanation:
1. Orbital speed is about 7.9 km/s and escape speed about 11.2 km/s
2. The extra speed is 11.2 - 7.9 = 3.3 km/s

Teacher's Note:
1. Escape speed is root 2 times orbital speed.

Question 20: A satellite is moving in a circular orbit around the earth with speed v. If its mass is m, its total energy is:
(a) \( \frac34mv^2 \)
(b) \( mv^2 \)
(c) \( \frac12mv^2 \)
(d) \( -\frac12mv^2 \)
Show Answer & Explanation

Answer: (d) \( -\frac12mv^2 \)

Explanation:
1. KE \( = \frac12mv^2 \), PE \( = -mv^2 \)
2. Total \( = -\frac12mv^2 \)

Teacher's Note:
1. Total energy is minus the kinetic energy.

Question 21: If the length of the day is T, the height of a TV satellite above the earth's surface that always appears stationary is:
(a) \( \left[\frac{4\pi^2GM}{T^2}\right]^{1/3} \)
(b) \( \left[\frac{4\pi^2GM}{T^2}\right]^{1/2}-R \)
(c) \( \left[\frac{GMT^2}{4\pi^2}\right]^{1/3}-R \)
(d) \( \left[\frac{GMT^2}{4\pi^2}\right]^{1/3}+R \)
Show Answer & Explanation

Answer: (c) \( \left[\frac{GMT^2}{4\pi^2}\right]^{1/3}-R \)

Explanation:
1. \( T^2 = \frac{4\pi^2r^3}{GM} \)
2. \( r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3} \)
3. Height \( = r - R \)

Teacher's Note:
1. Subtract the earth's radius to get height.

Question 22: Two bodies of mass M and m revolve about their common centre of mass in circular orbits of radii R and r due to mutual interaction. Which formula is applicable?
(a) \( \frac{GMm}{(R+r)^2} = m\omega^2r \)
(b) \( \frac{GMm}{R^2} = m\omega^2r \)
(c) \( \frac{GMm}{r^2} = m\omega^2R \)
(d) \( \frac{GMm}{R^2+r^2} = m\omega^2r \)
Show Answer & Explanation

Answer: (a) \( \frac{GMm}{(R+r)^2} = m\omega^2r \)

Explanation:
1. The separation of the bodies is R + r
2. Gravity supplies the centripetal force for m at radius r

Teacher's Note:
1. Use the separation, not an orbit radius, in the force law.

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