JEE Physics Gravitation MCQs Set 17

Answer: (d) 8 N

Explanation:
1. Gravity at height h: \( g'=g\left(\frac{R}{R+h}\right)^2=g\left(\frac{6400}{9600}\right)^2=\frac49g \).
2. Weight \( =\frac49\times18=8\ \text{N} \).

Teacher's Note:
1. Weight is proportional to g, so scale it by the same factor.
2. Distance from the centre is R + h, not h.

Answer: (c) 3 x 10⁸ km

Explanation:
1. Kepler's third law: \( T^2\propto r^3\Rightarrow r\propto T^{2/3} \).
2. \( \frac{r}{1.5\times10^8}=(2.83)^{2/3}\approx2 \), since \( 2.83=2^{3/2} \).
3. \( r=3\times10^8\ \text{km} \).

Teacher's Note:
1. Note that 2.83 is 2 raised to the power 3/2, so its two-thirds power is exactly 2.
2. Planets further out have longer periods.

Answer: (c) Statement I is correct but Statement II is incorrect

Explanation:
1. Above the surface \( g_h\approx g\left(1-\frac{2h}{R}\right) \) and below the surface \( g_d=g\left(1-\frac dR\right) \): both are less than g, so Statement I is correct.
2. For h = d the two values differ (\( 1-\frac{2h}{R}\ne1-\frac hR \)), so Statement II is incorrect.

Teacher's Note:
1. g is maximum at the surface of the earth.
2. g falls twice as fast with height as it does with depth for small h.

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

Explanation:
1. Time period \( T=2\pi\sqrt{l/g} \). At Everest g is smaller, so T is larger.
2. A larger time period means the clock ticks slower, so it becomes slow, not fast. Assertion A is false.
3. Reason R is true.

Teacher's Note:
1. A longer period per swing means fewer swings per day.
2. Clocks run fast when g increases, for example near the poles.

Answer: (b) 2

Explanation:
1. At height R: \( g'=g\left(\frac{R}{2R}\right)^2=\frac g4 \).
2. \( T'=2\pi\sqrt{l/g'}=2T \), so x = 2.

Teacher's Note:
1. Time period of a simple pendulum varies as 1/√g.
2. g falls to one quarter at a height equal to the radius.

Answer: (b) 1 hour 24 minutes

Explanation:
1. Inside the earth the restoring force is proportional to distance from the centre, giving \( \omega^2=\frac gR \).
2. \( T=2\pi\sqrt{\frac Rg}=2\pi\sqrt{\frac{6.4\times10^6}{10}}=2\pi\times800\approx5027\ \text{s} \).
3. 5027 s is about 84 minutes, i.e. 1 hour 24 minutes.

Teacher's Note:
1. The period equals that of a satellite in a very low orbit.
2. The mass of the particle does not matter.

Answer: (a) A-III, B-IV, C-II, D-I

Explanation:
1. The troposphere extends to about 10 km (A-III).
2. The D-layer lies at about 65-75 km (D-I), the E-layer at about 100 km (B-IV) and the F2-layer at about 300 km (C-II).

Teacher's Note:
1. Layers of the ionosphere appear in the order D, E, F going upward.
2. Radio communication by sky waves depends on these layers.

Answer: (c) \( \frac23mgR_e \)

Explanation:
1. Potential energy \( U=-\frac{GMm}{r}=-\frac{mgR_e^2}{r} \).
2. Initial radius \( R_e \) and final radius \( 3R_e \): \( \Delta U=mgR_e^2\left(\frac1{R_e}-\frac1{3R_e}\right)=\frac23mgR_e \).

Teacher's Note:
1. The distance from the centre at height 2Re is 3Re.
2. Use U = -GMm/r and GM = gR².

Answer: (a) A and D only

Explanation:
1. A is correct (law of gravitation) and D is correct (Kepler's third law).
2. B is wrong because the force is directly proportional to the product of the masses.
3. C is wrong because the centripetal force points towards the sun.

Teacher's Note:
1. Gravitation is attractive, so the force on a planet always points towards the sun.
2. Kepler's third law: T² is proportional to a³.

Answer: (d) \( \sqrt{\frac{Gm}{4r}} \)

Explanation:
1. The particles are at opposite ends of a diameter, so the separation is 2r.
2. Gravitational force provides the centripetal force: \( \frac{Gm^2}{(2r)^2}=\frac{mv^2}{r} \).
3. \( v=\sqrt{\frac{Gm}{4r}} \).

Teacher's Note:
1. The radius of the circle is r but the distance between the masses is 2r.
2. Each mass moves in the same circle about their common centre.

Answer: (d) 3 hours

Explanation:
1. Kepler's third law: \( T^2\propto r^3\Rightarrow T\propto r^{3/2} \).
2. \( T'=24\times\left(\frac14\right)^{3/2}=\frac{24}{8}=3\ \text{hours} \).

Teacher's Note:
1. A smaller orbit means a shorter period.
2. (1/4)^(3/2) = 1/8.

Answer: (b) 11

Explanation:
1. Field and potential: \( E=-\frac{dV}{dr}\Rightarrow\frac{dV}{dr}=\frac{K}{r^2} \).
2. \( V_3-V_2=\int_2^3\frac{K}{r^2}dr=K\left(\frac12-\frac13\right)=\frac{6}{6}=1\ \text{J/kg} \).
3. So \( V_3=10+1=11\ \text{J/kg} \).

Teacher's Note:
1. The field points inward, so the potential increases with r.
2. Integrate the field between the two radii.

Answer: (b) \( \sqrt{gR} \)

Explanation:
1. Energy conservation: \( \frac12mv^2=GMm\left(\frac1R-\frac1{2R}\right)=\frac{GMm}{2R} \).
2. \( v^2=\frac{GM}{R}=gR \), so \( v=\sqrt{gR} \).

Teacher's Note:
1. Do not use v² = 2gh here because g changes appreciably over a height R.
2. Start the fall from distance 2R from the centre.

Answer: (d) 4800 km

Explanation:
1. At height 3R: \( g_h=g\left(\frac{R}{4R}\right)^2=\frac g{16} \).
2. At depth d: \( g_d=g\left(1-\frac dR\right)=4\times\frac g{16}=\frac g4 \).
3. \( 1-\frac dR=\frac14\Rightarrow d=\frac{3R}{4}=4800\ \text{km} \).

Teacher's Note:
1. Use g(1 - d/R) inside the earth and g(R/(R+h))² outside.
2. Distance from the centre at height 3R is 4R.

Answer: (b) \( \frac W{100} \)

Explanation:
1. At height 9R the distance from the centre is 10R.
2. \( g'=g\left(\frac{R}{10R}\right)^2=\frac g{100} \), so the weight is \( \frac W{100} \).

Teacher's Note:
1. Weight is proportional to g.
2. The distance from the centre of the earth is R + 9R = 10R.

Answer: (c) Statement I is true but Statement II is false

Explanation:
1. Statement I is true: g varies with latitude, altitude and the shape of the earth.
2. Statement II is false: below the surface \( g_d=g\left(1-\frac dR\right) \) decreases with depth, reaching zero at the centre.

Teacher's Note:
1. The earth is not a perfect sphere and it rotates, so g is not uniform.
2. For a uniform earth g is linear in the distance from the centre inside the earth.

Answer: (a) 2

Explanation:
1. Escape velocity \( v=\sqrt{\frac{2GM}{R}} \).
2. \( \frac{v_P}{v_e}=\sqrt{\frac{M_P}{M_e}\cdot\frac{R_e}{R_P}}=\sqrt{\frac19\times2}=\frac{\sqrt2}{3} \).
3. \( v_P=\frac{v_e}{3}\sqrt2 \), so x = 2.

Teacher's Note:
1. Write the ratio of escape speeds from the mass and radius ratios.
2. M_P/M_e = 1/9 and R_e/R_P = 2.

Answer: (a) Gravitational potential energy is maximum at the highest point.

Explanation:
1. The height is maximum at the top, so potential energy is maximum there.
2. The horizontal velocity stays constant, and the vertical velocity (and momentum) is zero at the top; the kinetic energy there is \( \frac12m(u\cos\theta)^2\neq0 \).

Teacher's Note:
1. At the top of the path only the vertical component of velocity vanishes.
2. Total mechanical energy is conserved, so maximum PE means minimum (but non-zero) KE.

Answer: (d) \( \frac34 \)

Explanation:
1. \( v_e=\sqrt{2gR}\Rightarrow g=\frac{v_e^2}{2R} \).
2. \( \frac{g_A}{g_B}=\left(\frac{v_A}{v_B}\right)^2\frac{R_B}{R_A}=\frac14\times3=\frac34 \).

Teacher's Note:
1. Express g in terms of escape velocity and radius.
2. The radius appears in the denominator, so its ratio is inverted.