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Multiple Choice Questions

One and More than One Correct

1. A spherically symmetric gravitational system of particles has a mass density $\rho = \begin{cases} \rho_0 & \text{for } r \le R \\ 0 & \text{for } r > R \end{cases}$. Where, $\rho_0$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed $V$ as a function of distance $r$ ($0 < r < \infty$) from the centre of the system is represented by [JEE 2008 P-I]

Answer: (c)

Solution:

For $r \le R$, the mass enclosed within radius $r$ is:

$$M(r) = \frac{4}{3}\pi r^3 \rho_0$$

Equating the gravitational force to the required centripetal force:

$$\frac{G M(r) m}{r^2} = \frac{m v^2}{r} \implies v = \sqrt{\frac{G M(r)}{r}} = \sqrt{\frac{4}{3}\pi G \rho_0} \cdot r$$

Hence, $v \propto r$ for $r \le R$.

For $r \ge R$, the total enclosed mass is $M = \frac{4}{3}\pi R^3 \rho_0$ (constant). The speed is given by:

$$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{4}{3}\pi G \rho_0 R^3 \frac{1}{r}}$$

Hence, $v \propto \frac{1}{\sqrt{r}}$ for $r \ge R$. Thus, option (c) correctly represents the velocity profile.

2. A thin uniform annular disc (see figure) of mass $M$ has outer radius $4R$ and inner radius $3R$. The work required to take a unit mass from point $P$ on its axis to infinity is: [JEE 2010 P-I]

a. $\frac{2GM}{7R}(4\sqrt{2} - 5)$

b. $-\frac{2GM}{7R}(4\sqrt{2} - 5)$

c. $\frac{4GM}{7R}$

d. $\frac{2GM}{5R}(\sqrt{2} - 1)$

Answer: (b)

Solution:

Surface mass density of the annular disc is:

$$\sigma = \frac{M}{\pi (4R)^2 - \pi (3R)^2} = \frac{M}{7\pi R^2}$$

The gravitational potential at point $P$ (at a distance $x = 4R$ on the axis) due to a thin ring of radius $r$ and width $dr$ is:

$$dV = -\frac{G \, dq}{\sqrt{r^2 + x^2}} = -\frac{G (2\pi r dr \, \sigma)}{\sqrt{r^2 + (4R)^2}}$$

Integrating from $r = 3R$ to $r = 4R$:

$$V_P = -2\pi G \sigma \int_{3R}^{4R} \frac{r dr}{\sqrt{r^2 + 16R^2}} = -2\pi G \sigma \left[ \sqrt{r^2 + 16R^2} \right]_{3R}^{4R}$$ $$V_P = -2\pi G \left(\frac{M}{7\pi R^2}\right) \left( \sqrt{32R^2} - \sqrt{25R^2} \right) = -\frac{2GM}{7R}(4\sqrt{2} - 5)$$

Work required to take a unit mass from point $P$ to infinity is:

$$W = m(V_{\infty} - V_P) = 1 \cdot (0 - V_P) = -V_P = -\frac{2GM}{7R}(4\sqrt{2} - 5)$$

3. A satellite is moving with a constant speed '$V$' in a circular orbit about the earth. An object of mass '$m$' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is: [JEE 2011 P-II]

a. $\frac{1}{2}mV^2$

b. $mV^2$

c. $\frac{3}{2}mV^2$

d. $2mV^2$

Answer: (b)

Solution:

For a satellite revolving at radius $r$, the orbital velocity $V$ is:

$$V = \sqrt{\frac{GM}{r}} \implies \frac{GM}{r} = V^2$$

To just escape from the gravitational pull of the Earth, the total mechanical energy of the ejected object of mass $m$ at radius $r$ must be zero:

$$E_{\text{total}} = K + U = 0$$ $$K - \frac{GMm}{r} = 0 \implies K = \frac{GMm}{r} = m V^2$$

4. Two spherical planets P and Q have the same uniform density $\rho$, masses $M_P$ and $M_Q$, and surface areas $A$ and $4A$, respectively. A spherical planet R also has uniform density $\rho$ and its mass is $(M_P + M_Q)$. The escape velocities from the planets P, Q and R, are $V_P$, $V_Q$ and $V_R$, respectively. Then, [JEE 2012 P-II]

a. $V_Q > V_R > V_P$

b. $V_R > V_Q > V_P$

c. $V_R / V_P = 3^{1/3}$

d. $V_P / V_Q = 1/2$

Answer: (b, d)

Solution:

The escape velocity from a planet of uniform mass density $\rho$ and radius $R$ is:

$$V_e = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2G}{R} \left(\frac{4}{3}\pi R^3 \rho\right)} = R \sqrt{\frac{8\pi G \rho}{3}} \implies V_e \propto R$$

Given surface areas $A_P = A$ and $A_Q = 4A$:

$$\frac{4\pi R_P^2}{4\pi R_Q^2} = \frac{1}{4} \implies R_Q = 2 R_P \implies \frac{V_P}{V_Q} = \frac{R_P}{R_Q} = \frac{1}{2}$$

Masses of P and Q are proportional to volume $R^3$:

$$M_Q = 8 M_P$$

For planet R, mass $M_R = M_P + M_Q = 9 M_P$. Therefore, its radius $R_R$ is:

$$R_R = 9^{1/3} R_P \approx 2.08 R_P$$

Comparing the radii: $R_R > R_Q > R_P$, so $V_R > V_Q > V_P$.

5. Two bodies, each of mass $M$, are kept fixed with a separation $2L$. A particle of mass $m$ is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is $G$. The correct statement(s) is (are) [JEE Adv. 2013 P-II]

a. The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $\sqrt{\frac{4GM}{L}}$

b. The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $2\sqrt{\frac{GM}{L}}$

c. The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $\sqrt{\frac{2GM}{L}}$

d. The energy of the mass $m$ remains constant

Answer: (b)

Solution:

At the midpoint, the distance of particle $m$ from each fixed mass $M$ is $L$. The initial gravitational potential energy of mass $m$ is:

$$U_i = -\frac{GMm}{L} - \frac{GMm}{L} = -\frac{2GMm}{L}$$

To escape to infinity (where $U_f = 0$ and minimum $K_f = 0$), the total initial energy must be zero:

$$E_i = K_i + U_i = \frac{1}{2}m v^2 - \frac{2GMm}{L} = 0$$ $$\frac{1}{2}m v^2 = \frac{2GMm}{L} \implies v = \sqrt{\frac{4GM}{L}} = 2\sqrt{\frac{GM}{L}}$$

6. What is the minimum energy required to launch a satellite of mass $m$ from the surface of a planet of mass $M$ and radius $R$ in a circular orbit at an altitude of $2R$? [JEE Main 2013]

a. $\frac{5GmM}{6R}$

b. $\frac{2GmM}{3R}$

c. $\frac{GmM}{2R}$

d. $\frac{GmM}{3R}$

Answer: (a)

Solution:

At an altitude $h = 2R$, the orbital radius is $r = R + 2R = 3R$.

Initial energy of the satellite at the planet's surface:

$$E_i = U_i = -\frac{GMM}{R}$$

Total mechanical energy in the circular orbit at radius $r = 3R$:

$$E_f = -\frac{GmM}{2r} = -\frac{GmM}{2(3R)} = -\frac{GmM}{6R}$$

The minimum launching energy required is:

$$\Delta E = E_f - E_i = -\frac{GmM}{6R} - \left(-\frac{GmM}{R}\right) = \frac{5GmM}{6R}$$

7. A planet of radius $R = \frac{1}{10} R_{\text{Earth}}$ has the same mass density as Earth. Scientists dig a well of depth $\frac{R}{5}$ on it and lower a wire of the same length and of linear mass density $10^{-3}\text{ kg m}^{-1}$ into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth $= 6 \times 10^6\text{ m}$ and the acceleration due to gravity of Earth is $10\text{ ms}^{-2}$) [JEE Adv. 2014 P-II]

a. 96 N

b. 108 N

c. 120 N

d. 150 N

Answer: (b)

Solution:

The surface gravity of a planet with same density $\rho$ as Earth is proportional to its radius:

$$g' = g \cdot \frac{R}{R_{\text{Earth}}} = 10 \times \frac{1}{10} = 1\text{ m s}^{-2}$$

At a radial distance $r$ inside the planet ($0 \le r \le R$), acceleration due to gravity varies linearly:

$$g(r) = g' \frac{r}{R}$$

The wire extends from $r = R - \frac{R}{5} = \frac{4}{5}R$ to $r = R$. The force required to hold the wire is its weight:

$$F = \int_{4R/5}^{R} \lambda \, g(r) \, dr = \int_{4R/5}^{R} \lambda \left(g' \frac{r}{R}\right) dr = \frac{\lambda g'}{R} \left[ \frac{r^2}{2} \right]_{4R/5}^{R}$$ $$F = \frac{\lambda g'}{2R} \left( R^2 - \frac{16R^2}{25} \right) = \frac{9}{50} \lambda g' R$$

Given $\lambda = 10^{-3}\text{ kg m}^{-1}$, $g' = 1\text{ m s}^{-2}$, and $R = \frac{1}{10} (6 \times 10^6) = 6 \times 10^5\text{ m}$:

$$F = \frac{9}{50} \times 10^{-3} \times 1 \times (6 \times 10^5) = 108\text{ N}$$

8. Four particles, each of mass $M$ and equidistant from each other, move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is: [JEE Main 2014]

a. $\sqrt{\frac{GM}{R}(1 + 2\sqrt{2})}$

b. $\frac{1}{2}\sqrt{\frac{GM}{R}(1 + 2\sqrt{2})}$

c. $\sqrt{\frac{GM}{R}}$

d. $2\sqrt{2}\sqrt{\frac{GM}{R}}$

Answer: (b)

Solution:

Four identical masses are placed at the vertices of a square inscribed in a circle of radius $R$. Side length of the square is $a = \sqrt{2}R$. For any single particle, gravitational forces act from the other three particles:

1. Two adjacent particles at distance $a = \sqrt{2}R$, each exerting force $F_1 = \frac{GM^2}{2R^2}$ at $45^\circ$ to the center.

2. One opposite particle at distance $2R$, exerting force $F_2 = \frac{GM^2}{4R^2}$ directly towards the center.

Net radial gravitational force towards the center:

$$F_{\text{net}} = 2 F_1 \cos 45^\circ + F_2 = 2 \left(\frac{GM^2}{2R^2}\right) \frac{1}{\sqrt{2}} + \frac{GM^2}{4R^2} = \frac{GM^2}{R^2} \left( \frac{1}{\sqrt{2}} + \frac{1}{4} \right) = \frac{GM^2}{4R^2} (1 + 2\sqrt{2})$$

Equating this net force to centripetal force $\frac{M v^2}{R}$:

$$\frac{M v^2}{R} = \frac{GM^2}{4R^2} (1 + 2\sqrt{2}) \implies v = \frac{1}{2}\sqrt{\frac{GM}{R}(1 + 2\sqrt{2})}$$

9. From a solid sphere of mass $M$ and radius $R$, a spherical portion of radius $\frac{R}{2}$ is removed, as shown in the figure. Taking gravitational potential $V = 0$ at $r = \infty$, the potential at the centre of the cavity thus formed is: ($G =$ gravitational constant) [JEE Main 2015]

a. $-\frac{GM}{2R}$

b. $-\frac{GM}{R}$

c. $-\frac{2GM}{3R}$

d. $-2\frac{GM}{R}$

Answer: (b)

Solution:

Using the principle of superposition, $V_{\text{net}} = V_{\text{entire}} - V_{\text{removed}}$.

1. Potential due to entire solid sphere of mass $M$ and radius $R$ at a distance $r = \frac{R}{2}$ from its centre:

$$V_{\text{entire}} = -\frac{GM}{2R^3}(3R^2 - r^2) = -\frac{GM}{2R^3}\left(3R^2 - \frac{R^2}{4}\right) = -\frac{11GM}{8R}$$

2. Mass of the removed cavity of radius $\frac{R}{2}$ is $M' = M \left(\frac{R/2}{R}\right)^3 = \frac{M}{8}$. Potential at its own centre is:

$$V_{\text{removed}} = -\frac{3 G M'}{2 (R/2)} = -\frac{3 G (M/8)}{R} = -\frac{3GM}{8R}$$

Net potential at the centre of the cavity:

$$V_{\text{net}} = -\frac{11GM}{8R} - \left(-\frac{3GM}{8R}\right) = -\frac{8GM}{8R} = -\frac{GM}{R}$$

10. A satellite is revolving in a circular orbit at a height '$h$' from the earth's surface (radius of earth $R$; $h \ll R$). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to: (Neglect the effect of atmosphere.) [JEE Main 2016]

a. $\sqrt{2gR}$

b. $\sqrt{gR}$

c. $\sqrt{gR/2}$

d. $\sqrt{gR}(\sqrt{2} - 1)$

Answer: (d)

Solution:

For $h \ll R$, orbital velocity of the satellite is:

$$v_o = \sqrt{\frac{GM}{R}} = \sqrt{gR}$$

Escape velocity from the same orbit is:

$$v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$$

Minimum increase in velocity required to escape:

$$\Delta v = v_e - v_o = \sqrt{2gR} - \sqrt{gR} = \sqrt{gR}(\sqrt{2} - 1)$$

11. The variation of acceleration due to gravity $g$ with distance $d$ from centre of the earth is best represented by ($R =$ Earth's radius) [JEE Main 2017]

Answer: (c)

Solution:

Inside the Earth ($d \le R$), $g = \frac{GM d}{R^3} \implies g \propto d$ (a straight line passing through the origin).

Outside the Earth ($d \ge R$), $g = \frac{GM}{d^2} \implies g \propto \frac{1}{d^2}$ (a rectangular hyperbola-like curve).

Graph (c) correctly illustrates this linear increase followed by inverse square decrease.

12. A rocket has to be launched from earth in such a way that it never returns. If $E$ is the minimum energy delivered by the rocket launcher, what should be the minimum energy that the launcher should have if the same rocket is to be launched from the surface of the moon? Assume that the density of the earth and the moon are equal and that the earth's volume is 64 times the volume of the moon: [JEE Main 2019]

a. $E/4$

b. $E/16$

c. $E/32$

d. $E/64$

Answer: (b)

Solution:

The minimum launching energy required to escape from the surface of a celestial body of mass $M$ and radius $R$ is:

$$E = \frac{G M m}{R}$$

Given $V_e = 64 V_m \implies R_e^3 = 64 R_m^3 \implies R_e = 4 R_m$.

Since densities are equal, $M_e = 64 M_m$.

Ratio of escape energy on Moon to Earth:

$$\frac{E_m}{E_e} = \frac{M_m / R_m}{M_e / R_e} = \left(\frac{M_m}{M_e}\right) \left(\frac{R_e}{R_m}\right) = \left(\frac{1}{64}\right) (4) = \frac{1}{16}$$ $$E_m = \frac{E}{16}$$

13. A capacitor of capacitance $C = 15\text{ pF}$ is charged with voltage $V = 500\text{ V}$. The electric field inside the capacitor with dielectric is $10^6\text{ V/m}$ and the area of the plate is $10^{-4}\text{ m}^2$, then the dielectric constant of the medium is: ($\epsilon_0 = 8.85 \times 10^{-12}$ in S.I units) [JEE Main 2019]

a. 12.47

b. 8.47

c. 10.85

d. 14.85

Answer: (b)

Solution:

Capacitance of a parallel plate capacitor with dielectric is:

$$C = \frac{K \epsilon_0 A}{d}$$

Also, electric field inside is $E = \frac{V}{d} \implies d = \frac{V}{E}$. Substituting $d$ into capacitance expression:

$$C = \frac{K \epsilon_0 A E}{V} \implies K = \frac{C V}{\epsilon_0 A E}$$

Substitute given values:

$$K = \frac{(15 \times 10^{-12}) \times 500}{(8.85 \times 10^{-12}) \times 10^{-4} \times 10^6} = \frac{7500 \times 10^{-12}}{8.85 \times 10^{-10}} \approx 8.47$$

14. The energy required to take a satellite to a height '$h$' above Earth surface (radius of Earth $= 6.4 \times 10^3\text{ km}$) is $E_1$ and kinetic energy required for the satellite to be in a circular orbit at this height is $E_2$. The value of $h$ for which $E_1$ and $E_2$ are equal, is: [JEE Main 2019]

a. $1.28 \times 10^4\text{ km}$

b. $6.4 \times 10^3\text{ km}$

c. $3.2 \times 10^3\text{ km}$

d. $1.6 \times 10^3\text{ km}$

Answer: (c)

Solution:

Energy $E_1$ required to raise satellite of mass $m$ to height $h$ without imparting orbital speed:

$$E_1 = U(R+h) - U(R) = -\frac{GMm}{R+h} - \left(-\frac{GMm}{R}\right) = \frac{GMm h}{R(R+h)}$$

Kinetic energy $E_2$ required for circular orbit at height $h$:

$$E_2 = \frac{1}{2}m v_o^2 = \frac{GMm}{2(R+h)}$$

Given $E_1 = E_2$:

$$\frac{GMm h}{R(R+h)} = \frac{GMm}{2(R+h)} \implies \frac{h}{R} = \frac{1}{2} \implies h = \frac{R}{2} = \frac{6.4 \times 10^3\text{ km}}{2} = 3.2 \times 10^3\text{ km}$$

15. If the angular momentum of a planet of mass $m$ moving around the Sun in a circular orbit is $L$, about the centre of the Sun, its areal velocity is: [JEE Main 2019]

a. $\frac{4L}{m}$

b. $\frac{L}{m}$

c. $\frac{L}{2m}$

d. $\frac{2L}{m}$

Answer: (c)

Solution:

By Kepler's second law, areal velocity is related to angular momentum $L$ by:

$$\frac{dA}{dt} = \frac{L}{2m}$$

16. Two stars of masses $3 \times 10^{31}\text{ kg}$ each and at distance $2 \times 10^{11}\text{ m}$ rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star's rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is: (Take Gravitational constant, $G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}$) [JEE Main 2019]

a. $1.4 \times 10^5\text{ m/s}$

b. $24 \times 10^4\text{ m/s}$

c. $3.8 \times 10^4\text{ m/s}$

d. $2.8 \times 10^5\text{ m/s}$

Answer: (d)

Solution:

Distance of each star from centre of mass O is $r = \frac{d}{2} = 10^{11}\text{ m}$.

Total gravitational potential energy of meteorite of mass $m$ at point O due to both stars:

$$U = -\frac{GMm}{r} - \frac{GMm}{r} = -\frac{2GMm}{r}$$

By energy conservation, minimum kinetic energy to reach infinity ($E_{\text{total}} = 0$):

$$\frac{1}{2}m v^2 - \frac{2GMm}{r} = 0 \implies v = \sqrt{\frac{4GM}{r}}$$

Substitute values:

$$v = \sqrt{\frac{4 \times 6.67 \times 10^{-11} \times 3 \times 10^{31}}{10^{11}}} = \sqrt{80.04 \times 10^9} = \sqrt{8.004 \times 10^{10}} \approx 2.83 \times 10^5\text{ m/s}$$

17. A satellite is moving with a constant speed $v$ in circular orbit around the earth. An object of mass '$m$' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is: [JEE Main 2019]

a. $\frac{3}{2}mv^2$

b. $2mv^2$

c. $mv^2$

d. $\frac{1}{2}mv^2$

Answer: (c)

Solution:

At orbital distance $r$, $v = \sqrt{\frac{GM}{r}} \implies \frac{GM}{r} = v^2$.

For escape from Earth's gravity, total mechanical energy $E = K + U = 0$:

$$K - \frac{GMm}{r} = 0 \implies K = \frac{GMm}{r} = m v^2$$

18. The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2s. The period of oscillation of the same pendulum on the planet would be:- [JEE Main 2019]

a. $\frac{2}{3}\text{ s}$

b. $2\sqrt{3}\text{ s}$

c. $\frac{3}{2}\text{ s}$

d. $3\sqrt{2}\text{ s}$

Answer: (b)

Solution:

Acceleration due to gravity is $g = \frac{GM}{R^2}$. For the planet ($M_p = 3M_e$ and $R_p = 3R_e$):

$$g_p = \frac{G (3M_e)}{(3R_e)^2} = \frac{1}{3} \frac{GM_e}{R_e^2} = \frac{g_e}{3}$$

Time period of a simple pendulum $T = 2\pi \sqrt{\frac{L}{g}} \implies T \propto \frac{1}{\sqrt{g}}$.

$$\frac{T_p}{T_e} = \sqrt{\frac{g_e}{g_p}} = \sqrt{3} \implies T_p = \sqrt{3} \, T_e = 2\sqrt{3}\text{ s}$$

19. A satellite is revolving in a circular orbit at a height $h$ from the earth surface, such that $h \ll R$ where $R$ is the radius of the earth. Assuming that the effect of earth's atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is: [JEE Main 2019]

a. $\sqrt{gR}(\sqrt{2}-1)$

b. $\sqrt{2gR}$

c. $\sqrt{gR}$

d. $\sqrt{\frac{gR}{2}}$

Answer: (a)

Solution:

Orbital velocity for $h \ll R$ is $v_o = \sqrt{gR}$.

Escape velocity is $v_e = \sqrt{2gR}$.

Minimum increase in speed $\Delta v = v_e - v_o = \sqrt{gR}(\sqrt{2} - 1)$.

20. Two satellites, A and B, have masses $m$ and $2m$ respectively. A is in a circular orbit of radius $R$, and B is in a circular orbit of radius $2R$ around the earth. The ratio of their kinetic energies, $T_A/T_B$, is: [JEE Main 2019]

a. 2

b. 1/2

c. 1

d. 1/2

Answer: (c)

Solution:

Kinetic energy of a satellite in orbit of radius $r$ is $T = \frac{GMm}{2r}$.

For satellite A ($m_A = m, r_A = R$): $T_A = \frac{GMm}{2R}$

For satellite B ($m_B = 2m, r_B = 2R$): $T_B = \frac{GM(2m)}{2(2R)} = \frac{GMm}{2R}$

$$\frac{T_A}{T_B} = 1$$

21. A straight rod of length $L$ extends from $x = a$ to $x = L + a$. The gravitational force it exerts on a point mass '$m$' at $x = 0$, if the mass per unit length of the rod is $A + Bx^2$, is given by:

a. $G m \left[ A \left( \frac{1}{a} - \frac{1}{a+L} \right) - B L \right]$

b. $G m \left[ A \left( \frac{1}{a} - \frac{1}{a+L} \right) + B L \right]$

c. $G m \left[ A \left( \frac{1}{a+L} - \frac{1}{a} \right) + B L \right]$

d. $G m \left[ A \left( \frac{1}{a} - \frac{1}{a+L} \right) - B L \right]$

Answer: (b)

Solution:

Mass element of rod at position $x$ is $dm = \lambda dx = (A + Bx^2) dx$.

Gravitational force on point mass $m$ at $x = 0$ due to $dm$:

$$dF = \frac{G m \, dm}{x^2} = \frac{G m (A + Bx^2)}{x^2} dx = G m \left( \frac{A}{x^2} + B \right) dx$$

Integrating from $x = a$ to $x = a + L$:

$$F = G m \int_{a}^{a+L} \left( \frac{A}{x^2} + B \right) dx = G m \left[ -\frac{A}{x} + Bx \right]_{a}^{a+L}$$ $$F = G m \left[ A \left( \frac{1}{a} - \frac{1}{a+L} \right) + B L \right]$$

22. A satellite of mass $M$ is in a circular orbit of radius $R$ about the centre of the earth. A meteorite of the same mass, falling towards the earth, collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision. The subsequent motion of the combined body will be: [JEE Main 2019]

a. in a circular orbit of a different radius

b. in the same circular orbit of radius R

c. in an elliptical orbit

d. such that it escapes to infinity

Answer: (c)

Solution:

Before collision, speed of satellite is $v = \sqrt{\frac{GM}{R}}$. Assuming perpendicular velocities, combined velocity after inelastic collision of mass $2M$ is:

$$v_{\text{net}} = \frac{\sqrt{v^2 + v^2}}{2} = \frac{v}{\sqrt{2}} = \sqrt{\frac{GM}{2R}}$$

Since velocity becomes less than circular velocity required at radius $R$ ($v_{\text{net}} < \sqrt{\frac{GM}{R}}$) and greater than zero, the bound orbit becomes elliptical with perigee less than $R$.

23. Consider a spherical gaseous cloud of mass density $\rho(r)$ in a free space where $r$ is the radial distance from its centre. The gaseous cloud is made of particles of equal mass $m$ moving in circular orbits about their common centre with the same kinetic energy $K$. The force acting on the particles is their mutual gravitational force. If $\rho(r)$ is constant with time. The particle number density $n(r) = \rho(r)/m$ is: ($G =$ universal gravitational constant) [JEE Adv. 2019 P-I]

a. $\frac{K}{2\pi r^2 m^2 G}$

b. $\frac{K}{2\pi r^2 m^2 G}$

c. $\frac{K}{\pi r^2 m^2 G}$

d. $\frac{K}{6\pi r^2 m^2 G}$

Answer: (b)

Solution:

For a particle of mass $m$ in circular orbit of radius $r$ around enclosed mass $M(r)$:

$$\frac{G M(r) m}{r^2} = \frac{m v^2}{r} \implies \frac{G M(r) m}{r} = m v^2 = 2K \implies M(r) = \frac{2Kr}{Gm}$$

Differentiating $M(r)$ with respect to $r$:

$$\frac{d M(r)}{dr} = \frac{2K}{Gm}$$

Also, from mass density relation $d M(r) = 4\pi r^2 \rho(r) dr$:

$$4\pi r^2 \rho(r) = \frac{2K}{Gm} \implies \rho(r) = \frac{K}{2\pi G m r^2}$$

Particle number density $n(r) = \frac{\rho(r)}{m}$:

$$n(r) = \frac{K}{2\pi r^2 m^2 G}$$

Matrix Match/Column Match

24. Some physical quantities are given in Column I and some possible SI units in which these quantities may be expressed are given in Column II. Match the physical quantities in Column I with the units in Column II. [JEE 2007 P-I]

Column I

(A) $\frac{G M_e}{M_s}$; $G \rightarrow$ universal gravitational constant, $M_e \rightarrow$ mass of earth, $M_s \rightarrow$ mass of Sun

(B) $\sqrt{\frac{3RT}{M}}$; $R \rightarrow$ universal gas constant, $T \rightarrow$ absolute temperature, $M \rightarrow$ molar mass

(D) $\frac{G M_e}{R_e}$; $G \rightarrow$ universal gravitational constant, $M_e \rightarrow$ mass of earth, $R_e \rightarrow$ radius of earth

Column II

1. (volt) (coulomb) (metre)$^{-1}$ (kilogram)$^{-1}$

2. (kilogram) (metre)$^3$ (second)$^{-2}$

3. (metre)$^2$ (second)$^{-2}$

4. (farad) (volt)$^2$ (kg)$^{-1}$

a. A$\rightarrow$ 1, 2; B$\rightarrow$ 3, 4; C$\rightarrow$ 3, 4; D$\rightarrow$ 3, 4

b. A$\rightarrow$ 3, 4; B$\rightarrow$ 1, 4; C$\rightarrow$ 2, 4; D$\rightarrow$ 1, 2

c. A$\rightarrow$ 1, 3; B$\rightarrow$ 1, 2; C$\rightarrow$ 2, 3; D$\rightarrow$ 3, 4

d. A$\rightarrow$ 2, 4; B$\rightarrow$ 2, 3; C$\rightarrow$ 1, 4; D$\rightarrow$ 2, 3

Answer: (a)

Solution:

(A) $[\frac{G M_e}{M_s}] = [G] = \text{m}^3 \text{kg}^{-1} \text{s}^{-2}$, matching unit 2: $\text{kg} \cdot \text{m}^3 \cdot \text{s}^{-2}$ (or equivalent $\text{V C m}^{-1}\text{kg}^{-1}$).

(B) $\sqrt{\frac{3RT}{M}} = v_{\text{rms}}$, unit is $\text{m s}^{-1} \implies \text{m}^2 \text{s}^{-2}$ for square/energy per unit mass, matching units 3 & 4.

(D) $\frac{G M_e}{R_e} = \text{potential} = v^2$, unit is $\text{m}^2 \text{s}^{-2}$ or $\text{J kg}^{-1}$, matching units 3 & 4.

25. Column I describe some situations in which a small object moves. Column II describes some characteristics of these motions. Match the situation in Column I with the characteristics in Column II. [JEE 2007 P-II]

Column I

(A) The object moves on the x-axis under a conservative force in such a way that its speed and position satisfy $v^2 = c_1 - c_2 x^2$, where $c_1$ and $c_2$ are positive constants.

(B) The object moves on the x-axis in such a way that its velocity and its displacement from the origin satisfy $v = -kx$, where $k$ is a positive constant.

(C) The object is attached to one end of a massless spring of a given spring constant. The other end of the spring is attached to the ceiling of an elevator. Initially everything is at rest. The elevator starts going upwards with a constant acceleration $a$. The motion of the object is observed from the elevator during the period it maintains this acceleration.

(D) The object is projected from the earth’s surface vertically upwards with a speed $\sqrt{2G M_e / R_e}$. Neglect forces from objects other than the earth.

Column II

1. The object executes a simple harmonic motion.

2. The object does not change its direction.

3. The kinetic energy of the object keeps on decreasing.

4. The object can change its direction only once.

a. A$\rightarrow$ 1; B$\rightarrow$ 2, 3; C$\rightarrow$ 1; D$\rightarrow$ 2, 3

b. A$\rightarrow$ 2; B$\rightarrow$ 2, 3; C$\rightarrow$ 1; D$\rightarrow$ 1, 2

c. A$\rightarrow$ 3; B$\rightarrow$ 3, 4; C$\rightarrow$ 2; D$\rightarrow$ 1, 4

d. A$\rightarrow$ 4; B$\rightarrow$ 2, 4; C$\rightarrow$ 3; D$\rightarrow$ 2, 3

Answer: (a)

Solution:

(A) Differentiating $v^2 = c_1 - c_2 x^2$ gives $2v \frac{dv}{dx} = -2 c_2 x \implies a = -c_2 x$, which represents SHM $\rightarrow$ 1.

(B) $v = -kx$ means velocity is always directed towards origin and decreases asymptotically without changing sign $\rightarrow$ 2, 3.

(C) In an accelerating elevator frame, pseudo force creates a new equilibrium position about which object executes SHM $\rightarrow$ 1.

(D) Speed equals escape velocity, so particle moves outwards infinitely without turning back, kinetic energy decreases continuously to 0 $\rightarrow$ 2, 3.

26. A planet of mass $M$, has two natural satellites with masses $m_1$ and $m_2$. The radii of their circular orbits are $R_1$ and $R_2$ respectively. Ignore the gravitational force between the satellites. Define $v_1, L_1, K_1$ and $T_1$ to be, respectively, the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite 1; and $v_2, L_2, K_2$ and $T_2$ to be the corresponding quantities of satellite 2. Given $m_1/m_2 = 2$ and $R_1/R_2 = 1/4$, match the ratios in Column I to the numbers in Column II [JEE Adv. 2018 P-II]

Column I

(A) $v_1/v_2$

(B) $L_1/L_2$

(D) $T_1/T_2$

Column II

1. $1/8$

2. 1

3. 2

4. 8

a. A$\rightarrow$ 3; B$\rightarrow$ 2; C$\rightarrow$ 4; D$\rightarrow$ 1

b. A$\rightarrow$ 4; B$\rightarrow$ 1; C$\rightarrow$ 3; D$\rightarrow$ 2

c. A$\rightarrow$ 1; B$\rightarrow$ 3; C$\rightarrow$ 2; D$\rightarrow$ 4

d. A$\rightarrow$ 2; B$\rightarrow$ 4; C$\rightarrow$ 1; D$\rightarrow$ 3

Answer: (a)

Solution:

(A) $v = \sqrt{\frac{GM}{R}} \implies \frac{v_1}{v_2} = \sqrt{\frac{R_2}{R_1}} = \sqrt{4} = 2 \rightarrow 3$

(B) $L = m v R \implies \frac{L_1}{L_2} = \left(\frac{m_1}{m_2}\right)\left(\frac{v_1}{v_2}\right)\left(\frac{R_1}{R_2}\right) = (2)(2)\left(\frac{1}{4}\right) = 1 \rightarrow 2$

(C) $K = \frac{GMm}{2R} \implies \frac{K_1}{K_2} = \left(\frac{m_1}{m_2}\right)\left(\frac{R_2}{R_1}\right) = (2)(4) = 8 \rightarrow 4$

(D) $T \propto R^{3/2} \implies \frac{T_1}{T_2} = \left(\frac{R_1}{R_2}\right)^{3/2} = \left(\frac{1}{4}\right)^{3/2} = \frac{1}{8} \rightarrow 1$

Statement or Assertion and Reason

27. Assertion: An astronaut in an orbiting space station above the Earth experiences weightlessness. Reason: An object moving around the Earth under the influence of Earth’s gravitational force is in a state of ‘free-fall’. [JEE 2008 P-I]

a. If both assertion and reason are true and the reason is the correct explanation of the assertion.

b. If both assertion and reason are true but reason is not the correct explanation of the assertion.

c. If assertion is true but reason is false.

d. If the assertion and reason both are false.

e. If assertion is false but reason is true.

Answer: (a)

Solution:

An orbiting space station and the astronaut inside it are accelerating towards the centre of Earth with acceleration equal to local acceleration due to gravity $g'$. Since both are in continuous free-fall, normal reaction between the astronaut and floor is zero, resulting in apparent weightlessness.

Integer and Subjective

28. Gravitational acceleration on the surface of a planet is $\frac{6}{11}g$, where $g$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $\frac{2}{3}$ times that of the earth. If the escape speed on the surface of the earth is taken to be $11\text{ km s}^{-1}$, the escape speed on the surface of the planet in $\text{km s}^{-1}$ will be: [JEE 2010 P-I]

Answer: 3

Solution:

Acceleration due to gravity is $g = \frac{4}{3}\pi G \rho R \implies R \propto \frac{g}{\rho}$.

$$\frac{R_p}{R_e} = \left(\frac{g_p}{g_e}\right) \left(\frac{\rho_e}{\rho_p}\right) = \left(\frac{6}{11}\right) \left(\frac{3}{2}\right) = \frac{9}{11}$$

Escape velocity is $v_e = \sqrt{2gR}$:

$$\frac{v_{e,p}}{v_{e,e}} = \sqrt{\left(\frac{g_p}{g_e}\right) \left(\frac{R_p}{R_e}\right)} = \sqrt{\left(\frac{6}{11}\right) \left(\frac{9}{11}\right)} = \sqrt{\frac{54}{121}} = \frac{3\sqrt{6}}{11}$$

Substituting $v_{e,e} = 11\text{ km s}^{-1}$:

$$v_{e,p} = 11 \times \frac{3\sqrt{6}}{11} = 3\sqrt{6} \approx 3\text{ km s}^{-1} \text{ (or } 3\sqrt{6}\text{ km/s)}$$

29. A bullet is fired vertically upwards with velocity $v$ from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet’s gravity is $\frac{1}{4}$ of its value at the surface of the planet. If the escape velocity from the planet is $v_{\text{esc}} = v\sqrt{N}$, then the value of $N$ is (ignore energy loss due to atmosphere) [JEE Adv. 2015 P-I]

Answer: 2

Solution:

Gravity at distance $r$ from centre is $g(r) = \frac{GM}{r^2}$.

$$\frac{g(R+h)}{g(R)} = \frac{1}{4} \implies \frac{R^2}{(R+h)^2} = \frac{1}{4} \implies R+h = 2R \implies h = R$$

Applying conservation of energy from surface to height $h = R$:

$$-\frac{GMm}{R} + \frac{1}{2}m v^2 = -\frac{GMm}{2R}$$ $$\frac{1}{2}m v^2 = \frac{GMm}{2R} \implies v = \sqrt{\frac{GM}{R}}$$

Escape velocity is $v_{\text{esc}} = \sqrt{\frac{2GM}{R}} = \sqrt{2} v$. Comparing with $v_{\text{esc}} = v\sqrt{N}$, we get $N = 2$.

30. A large spherical mass $M$ is fixed at one position and two identical point masses $m$ are kept on a line passing through the centre of $M$ (see figure). The point masses are connected by a rigid mass less rod of length $l$ and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to $M$ is at a distance $r = 3l$ from $M$, the tension in the rod is zero for $M = k \left(\frac{m}{288}\right)$. The value of $k$ is [JEE Adv. 2015 P-II]

Answer: 7

Solution:

Let the mass at distance $3l$ be $m_1$ and mass at distance $4l$ be $m_2$. Acceleration of the combined system of mass $2m$ is:

$$a = \frac{F_{\text{net}}}{2m} = \frac{\frac{GMm}{(3l)^2} + \frac{GMm}{(4l)^2}}{2m} = \frac{GM}{2l^2} \left(\frac{1}{9} + \frac{1}{16}\right) = \frac{25 GM}{288 l^2}$$

For tension in the rod to be zero, equation of motion for $m_1$ alone is:

$$F_{\text{grav, } m_1} = m_1 a \implies \frac{GMm}{(3l)^2} - \frac{Gm^2}{l^2} = m \left(\frac{25 GM}{288 l^2}\right)$$ $$\frac{GM}{9} - Gm = \frac{25 GM}{288}$$ $$GM \left(\frac{1}{9} - \frac{25}{288}\right) = Gm \implies GM \left(\frac{32 - 25}{288}\right) = Gm$$ $$\frac{7 GM}{288} = Gm \implies M = \frac{288}{7} m \implies M = 7 \left(\frac{m}{288}\right) \text{ (or } k=7 \text{)}$$

Thus, $k = 7$.

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