JEE Physics Gravitation MCQs Set 22

Answer: (d) \( R/2 \)

Explanation:
1. Weight is proportional to \( g = GM/R^2 \), so \( \dfrac{W_e}{W_p} = \dfrac{M_e}{M_p}\cdot\dfrac{R_p^2}{R^2} = 9 \cdot \dfrac{R_p^2}{R^2} \).
2. Given \( W_e/W_p = 9/4 \), so \( 9R_p^2/R^2 = 9/4 \).
3. Hence \( R_p = R/2 \).

Teacher's Note:
1. Always write weight as mg with g = GM/R² when masses and radii of two bodies are compared.
2. Here the mass is given directly, so the equal-density statement is consistent and not needed in the calculation.

Answer: (b) \( 2.6\times10^{6}\text{ m} \)

Explanation:
1. At height h, \( g' = g\left(\dfrac{R}{R+h}\right)^2 \). Given \( g'/g = 1/2 \).
2. So \( R + h = \sqrt{2}\,R \) and \( h = (\sqrt2 - 1)R = 0.414 \times 6.4\times10^{6} \).
3. \( h \approx 2.65\times10^{6} \) m, closest to \( 2.6\times10^{6} \) m.

Teacher's Note:
1. Do not use the approximation g' = g(1 − 2h/R) here, because h is comparable to R.
2. Use the exact inverse-square form whenever the height is not very small.

Answer: (b) 108 N

Explanation:
1. Same density means \( g_p = g/10 = 1 \) m s⁻² at the surface; radius \( R = 6\times10^{5} \) m.
2. Inside the planet, \( g(x) = g_p\,x/R \), where x is the distance from the centre.
3. The wire extends from \( x = 4R/5 \) to \( R \): \( F = \int \lambda\,g_p\dfrac{x}{R}\,dx = \dfrac{\lambda g_p}{2R}\left(R^2 - \dfrac{16R^2}{25}\right) = \dfrac{9}{50}\lambda g_p R \).
4. \( F = 0.18 \times 10^{-3}\times 1\times 6\times10^{5} = 108 \) N.

Teacher's Note:
1. Inside a uniform sphere g falls linearly towards the centre, so the weight of a hanging wire must be found by integration.
2. Always convert the Earth data to the planet's g first.

Answer: (c) increase

Explanation:
1. \( g = GM/R^2 \) with M constant, so \( g \propto R^{-2} \).
2. A 1 percent decrease in R raises g by about 2 percent.
3. So g increases.

Teacher's Note:
1. For a small change, the percentage change in g is about twice that in R, with the opposite sign.
2. This is a quick way to answer variation questions without calculating.

Answer: (a) \( 1.24\times10^{-3} \)

Explanation:
1. Effective gravity at the equator is \( g' = g - \omega^2 R \).
2. Setting \( g' = 0 \) gives \( \omega = \sqrt{g/R} = \sqrt{9.8/(6.4\times10^6)} \).
3. \( \omega = 1.24\times10^{-3} \) rad/s (about 17 times the present rate).

Teacher's Note:
1. The centrifugal term ω²R reduces g at the equator.
2. Objects would float off the equator at this rate, which is the idea of a satellite skimming the surface.

Answer: (a) 11

Explanation:
1. Orbit radius \( r = 2.02\times10^{6} \) m; \( GM = 6.67\times10^{-11}\times 8\times10^{22} = 5.336\times10^{12} \).
2. \( T = 2\pi\sqrt{r^3/GM} = 2\pi\sqrt{8.24\times10^{18}/5.336\times10^{12}} \approx 7.8\times10^{3} \) s.
3. Revolutions in 24 h = \( 86400/7808 \approx 11.07 \), so 11 complete revolutions.

Teacher's Note:
1. Complete revolutions means we take the integer part of the ratio.
2. Use r measured from the centre, not the height above the surface.

Answer: (c) \( GM/3a^2 \)

Explanation:
1. The shell extends from r = a to r = 3a, so the point at 3a lies on its outer surface.
2. Everything is enclosed: total mass = M + 2M = 3M.
3. \( E = G(3M)/(3a)^2 = GM/3a^2 \).

Teacher's Note:
1. At a point on or outside a spherical shell, treat all enclosed mass as concentrated at the centre.
2. Check whether the point is inside, on or outside each shell first.

Answer: (b) \( E/16 \)

Explanation:
1. Minimum energy \( = GMm/R \). With equal density, \( M \propto R^3 \), so energy \( \propto R^2 \).
2. Volume ratio 64 means radius ratio \( R_E/R_M = 4 \).
3. So \( E_{Moon} = E/16 \).

Teacher's Note:
1. Escape energy per unit mass is proportional to ρR² for bodies of equal density.
2. Convert a volume ratio to a radius ratio with a cube root.

Answer: (c) \( 42\ \text{km s}^{-1} \)

Explanation:
1. Required energy per unit mass: \( \tfrac12 v_s^2 = \dfrac{GM_e}{R_e} + \dfrac{GM_s}{r} \).
2. \( \dfrac{GM_e}{R_e} = \tfrac12 v_e^2 \) and \( \dfrac{GM_s}{r} = \dfrac{3\times10^5}{2.5\times10^4}\cdot\dfrac{GM_e}{R_e} = 12\cdot\tfrac12v_e^2 \).
3. So \( v_s^2 = 13 v_e^2 \Rightarrow v_s = \sqrt{13}\times 11.2 \approx 40.4 \) km s⁻¹.
4. The closest option is 42 km s⁻¹.

Teacher's Note:
1. Escape from a two-body system needs the sum of the two potential energies to be overcome.
2. Orbital motion of the Earth is ignored here, which is why the answer is only an approximation.

Answer: (d) \( \sqrt{gR}\,(\sqrt2-1) \)

Explanation:
1. Orbital speed for h << R: \( v_o = \sqrt{gR} \).
2. Escape speed: \( v_e = \sqrt{2gR} = \sqrt2\,v_o \).
3. Minimum increase = \( v_e - v_o = \sqrt{gR}(\sqrt2-1) \).

Teacher's Note:
1. The escape speed is √2 times the circular orbital speed at the same radius.
2. This relation is used again and again in satellite problems.

Answer: (a) (1) and (3) only

Explanation:
1. Areas A and 4A give radii \( r \) and \( 2r \), so \( M_Q = 8M_P \).
2. For equal density \( v_{esc} = \sqrt{2GM/R} \propto R \).
3. Radius of R: \( \tfrac43\pi R_R^3 = 9\cdot\tfrac43\pi r^3 \Rightarrow R_R = 9^{1/3} r \approx 2.08r \).
4. So \( v_P : v_Q : v_R = 1 : 2 : 2.08 \). Thus (1) and (3) hold; \( v_R/v_P = 2.08 \), not 3.

Teacher's Note:
1. For bodies of equal density, escape speed is proportional to the radius.
2. Add masses (volumes), not radii, when two planets are merged.

Answer: (b) 2

Explanation:
1. \( g' = g/4 \Rightarrow r = 2R \) at the highest point.
2. Energy: \( \tfrac12v^2 - \dfrac{GM}{R} = -\dfrac{GM}{2R} \Rightarrow v^2 = \dfrac{GM}{R} \).
3. \( v_{esc}^2 = 2GM/R = 2v^2 \), so \( N = 2 \).

Teacher's Note:
1. Find the maximum distance from g ∝ 1/r², then use conservation of energy.
2. Integer-type answers can be checked by comparing v² with GM/R.

Answer: (d) \( T/R \text{ is a constant} \)

Explanation:
1. Mass inside radius r: \( M(r) = \int_0^r 4\pi r^2\dfrac{K}{r^2}dr = 4\pi K r \).
2. Gravitational force gives \( \dfrac{mv^2}{r} = \dfrac{GM(r)m}{r^2} = \dfrac{4\pi GKm}{r} \), so \( v^2 = 4\pi GK \).
3. Speed is independent of r, so \( T = 2\pi R/v \propto R \) and T/R is constant.

Teacher's Note:
1. Find the enclosed mass by integration when the density varies with r.
2. A constant orbital speed is a typical signature of a 1/r² density profile.

Answer: (d) 1

Explanation:
1. Orbital kinetic energy \( K = \dfrac{GMm}{2r} \).
2. \( \dfrac{K_A}{K_B} = \dfrac{m/R}{2m/2R} = 1 \).

Teacher's Note:
1. Kinetic energy of a circular orbit equals half the magnitude of the potential energy.
2. Both mass and radius change here, and the changes cancel.

Answer: (a) \( 3.2\times10^{3}\text{ km} \)

Explanation:
1. \( E_1 = \dfrac{GMm}{R} - \dfrac{GMm}{R+h} = \dfrac{GMm\,h}{R(R+h)} \).
2. \( E_2 = \tfrac12mv^2 = \dfrac{GMm}{2(R+h)} \).
3. Equating: \( h/R = 1/2 \), so \( h = R/2 = 3.2\times10^{3} \) km.

Teacher's Note:
1. E1 is the gain in potential energy; E2 is the orbital kinetic energy.
2. Cancel the common factors GMm/(R+h) before solving.

Answer: (a) \( \dfrac{5GmM}{6R} \)

Explanation:
1. Orbit radius \( r = 3R \). Total energy in orbit \( = -\dfrac{GMm}{2r} = -\dfrac{GMm}{6R} \).
2. Initial energy on the surface \( = -\dfrac{GMm}{R} \).
3. Energy required \( = -\dfrac{GMm}{6R} + \dfrac{GMm}{R} = \dfrac{5GMm}{6R} \).

Teacher's Note:
1. Altitude 2R means the orbital radius is 3R.
2. Energy required equals final total energy minus initial total energy.

Answer: (c) 2 h

Explanation:
1. By Kepler's third law \( T^2 \propto r^3 \). Take the spy satellite radius as about 6400 km.
2. \( T = 24\left(\dfrac{6400}{36000}\right)^{3/2} = 24 \times 0.075 = 1.8 \) h.
3. This is approximately 2 h.

Teacher's Note:
1. A geostationary satellite has a period of 24 h.
2. Use the ratio form of Kepler's law to avoid computing GM.

Answer: (a) The acceleration of S is always directed towards the centre of the Earth.

Explanation:
1. Gravity is a central force, so acceleration always points towards the centre of the Earth: (a) is correct.
2. Central force means torque is zero, so angular momentum is constant in both magnitude and direction: (b) is wrong.
3. Gravity is conservative, so total mechanical energy is constant: (c) is wrong.
4. Speed changes along an ellipse (faster near perigee), so linear momentum changes in magnitude: (d) is wrong.

Teacher's Note:
1. Central force implies constant angular momentum vector.
2. In an ellipse the speed is not constant, even though energy is.

Answer: (b) \( mv^2 \)

Explanation:
1. At orbital radius r, \( v^2 = GM/r \), so \( U = -GMm/r = -mv^2 \).
2. For just escaping, total energy is zero: \( K - mv^2 = 0 \).
3. So \( K = mv^2 \).

Teacher's Note:
1. Just escaping means total energy equals zero.
2. Use v² = GM/r to write the potential energy in terms of the orbital speed.