JEE Gravitation Questions

Let the incident (heavier) particle have mass $$m_1 = 2m$$ and initial speed $$u$$ along the +x-axis. The target particle has mass $$m_2 = m$$ and is initially at rest in the laboratory (lab) frame.

Step 1: Velocity of the centre of mass (COM)
$$V = \frac{m_1 u + m_2 \times 0}{m_1 + m_2} = \frac{2m\,u}{2m+m} = \frac{2u}{3}$$

Step 2: Velocities in the COM frame before collision
Velocity of $$m_1$$ in COM frame: $$u_{1c} = u - V = u - \frac{2u}{3} = \frac{u}{3}$$ (along +x)
Magnitude of this velocity: $$|u_{1c}| = \frac{u}{3}$$

Step 3: Velocities in the COM frame after an elastic collision
In a perfectly elastic, two-body collision the magnitudes of the velocities in the COM frame remain unchanged; only their directions change. Hence the heavier particle still has speed $$\dfrac{u}{3}$$ in the COM frame, but its direction makes some angle $$\phi$$ with the +x-axis.

Write the post-collision COM-frame velocity of $$m_1$$ as $$\vec{u}_{1c}' = \frac{u}{3}\bigl(\cos\phi\,\hat{i} + \sin\phi\,\hat{j}\bigr)$$

Step 4: Transform back to the lab frame
Lab-frame velocity of $$m_1$$ after collision: $$\vec{v}_{1}' = \vec{V} + \vec{u}_{1c}' = \frac{2u}{3}\,\hat{i} + \frac{u}{3}\bigl(\cos\phi\,\hat{i} + \sin\phi\,\hat{j}\bigr)$$
Components:
$$v_{x} = \frac{2u}{3} + \frac{u}{3}\cos\phi$$ $$v_{y} = \frac{u}{3}\sin\phi$$

Step 5: Angular deviation of the heavier particle
Let $$\theta$$ be the angle between $$\vec{v}_{1}'$$ and the initial direction (+x). Then $$\tan\theta = \frac{v_{y}}{v_{x}} = \frac{\dfrac{u}{3}\sin\phi}{\dfrac{2u}{3} + \dfrac{u}{3}\cos\phi} = \frac{\sin\phi}{2 + \cos\phi} \quad -(1)$$

Step 6: Maximising $$\theta$$
Because $$0 \lt \theta \lt \dfrac{\pi}{2}$$, maximising $$\theta$$ is equivalent to maximising $$\tan\theta$$. Define $$f(\phi) = \dfrac{\sin\phi}{2 + \cos\phi}$$.

Differentiating and setting $$\dfrac{df}{d\phi}=0$$ gives $$\frac{df}{d\phi} = \frac{\cos\phi(2+\cos\phi) + \sin^{2}\phi}{(2+\cos\phi)^{2}} = \frac{2\cos\phi + \cos^{2}\phi + \sin^{2}\phi}{(2+\cos\phi)^{2}} = \frac{2\cos\phi + 1}{(2+\cos\phi)^{2}}$$

Setting the numerator to zero:
$$2\cos\phi + 1 = 0 \;\;\Longrightarrow\;\; \cos\phi = -\frac{1}{2} \;\;\Longrightarrow\;\; \phi = \frac{2\pi}{3}$$ (Only the value within $$0 \le \phi \le \pi$$ is relevant.) For $$\phi \lt \dfrac{2\pi}{3}$$, the derivative is positive and for $$\phi \gt \dfrac{2\pi}{3}$$ it is negative, so $$\phi = \dfrac{2\pi}{3}$$ indeed gives the maximum.

Step 7: Maximum value of $$\theta$$
Insert $$\phi = \dfrac{2\pi}{3}$$ into equation $$(1)$$:
$$\tan\theta_{\max} = \frac{\sin(2\pi/3)}{2 + \cos(2\pi/3)} = \frac{\dfrac{\sqrt{3}}{2}}{2 - \dfrac{1}{2}} = \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{3}{2}} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$$

Therefore $$\theta_{\max} = \tan^{-1}\!\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$$

Option D which is: $$\dfrac{\pi}{6}$$

The geostationary satellite revolves in the same sense as the earth’s rotation, so its period in an inertial frame is the sidereal day:

$$T_1 = 24 \text{ h}$$ (to the accuracy required in the problem)

According to Kepler’s third law, the time period is proportional to $$r^{3/2}$$:

$$\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2}$$

Given $$r_1 = 1.21\,r_2$$, we obtain

$$T_2 = T_1 \left(\frac{1}{1.21}\right)^{3/2}$$

Compute the numerical factor:
$$1.21^{3/2} = (1.21)^{1.5} \approx 1.331$$
$$\left(\frac{1}{1.21}\right)^{3/2} \approx \frac{1}{1.331} \approx 0.752$$

Hence

$$T_2 = 24 \times 0.752 \text{ h} \approx 18.0 \text{ h}$$

The second satellite moves in the opposite (retrograde) direction, so its angular velocity is opposite in sign to that of the geostationary satellite. In magnitude:

$$\omega_1 = \frac{2\pi}{T_1}, \quad \omega_2 = -\,\frac{2\pi}{T_2}$$

Their relative angular speed is

$$\omega_{\text{rel}} = \omega_1 - \omega_2 = \frac{2\pi}{T_1} + \frac{2\pi}{T_2} = 2\pi\left(\frac{1}{T_1} + \frac{1}{T_2}\right)$$

The time period with which the second satellite is seen to pass the geostationary satellite (as measured from the geostationary frame) is therefore

$$T_{\text{rel}} = \frac{2\pi}{\omega_{\text{rel}}} = \frac{1}{\dfrac{1}{T_1} + \dfrac{1}{T_2}} = \frac{T_1\,T_2}{T_1 + T_2}$$

Substituting $$T_1 = 24\text{ h}$$ and $$T_2 \approx 18.0\text{ h}$$:

$$T_{\text{rel}} = \frac{24 \times 18.0}{24 + 18.0} \text{ h} = \frac{432}{42} \text{ h} \approx 10.3 \text{ h}$$

This observed period is given in the problem as $$\dfrac{24}{p} \text{ h}$$, so

$$\frac{24}{p} = 10.3 \Longrightarrow p = \frac{24}{10.3} \approx 2.33$$

Thus the required value is

$$\boxed{p \approx 2.3 - 2.4}$$

$$T^2 \propto R^3 \implies T \propto R^{3/2}$$ (Kepler’s Third Law)

Since the increase in radius is small ($$R$$ to $$1.03R$$, a $$3\%$$ change), we can use the approximation method for small percentage changes.

The percentage change in the radius is:

$$\frac{\Delta R}{R} \times 100 = \frac{1.03R - R}{R} \times 100 = 3\%$$

$$\text{Percentage change in } T \approx \frac{3}{2} \times (\text{Percentage change in } R)$$

$$\text{Percentage change in } T \approx \frac{3}{2} \times 3\% = 4.5\%$$

According to Kepler’s second law, the line joining the Sun and any planet sweeps out equal areas in equal intervals of time. This statement is the same as saying that the areal velocity $$\frac{dA}{dt}$$ of the planet is constant.

In mechanics the areal velocity of a particle of mass $$m$$ moving with linear momentum $$\mathbf{p}=m\mathbf{v}$$ at an instantaneous position vector $$\mathbf{r}$$ from the origin is given by

$$\frac{dA}{dt}=\frac{1}{2m}\,|\mathbf{r}\times\mathbf{p}|$$

The quantity $$\mathbf{L}=\mathbf{r}\times\mathbf{p}$$ is the angular momentum of the particle about the origin. Hence

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

If a particle is subjected to a central force, the force vector $$\mathbf{F}$$ is always directed along the radius vector $$\mathbf{r}$$. Therefore the torque about the origin is

$$\boldsymbol{\tau}=\mathbf{r}\times\mathbf{F}=0$$

Zero torque implies $$\frac{d\mathbf{L}}{dt}=0$$, so the angular momentum $$\mathbf{L}$$ is conserved (constant in both magnitude and direction).

Because $$|\mathbf{L}|$$ is constant, the expression $$\frac{dA}{dt}=\frac{|\mathbf{L}|}{2m}$$ shows that the areal velocity $$\frac{dA}{dt}$$ must also be constant. This directly yields Kepler’s second law.

Thus:

  • Assertion (A) is correct: the radius vector sweeps out equal areas in equal times, so the areal velocity is constant.
  • Reason (R) is correct: in a central force field the angular momentum is conserved.
  • Reason (R) correctly explains why the areal velocity is constant, since constant angular momentum leads to constant $$\frac{dA}{dt}$$.

Therefore the appropriate choice is Option A: Both (A) and (R) are correct and (R) is the correct explanation of (A).

Mass of the projectile: $$m = 200 \text{ g} = 0.2 \text{ kg}$$

The drag force is linear: $$\vec F = -c\vec v,$$ with $$c = 0.1 \text{ kg s}^{-1}.$$
Hence for any component of velocity, $$m\dfrac{dv}{dt} = -cv \; \Rightarrow \; \dfrac{dv}{dt} = -\dfrac{c}{m}\,v.$$

Horizontal component
Initial horizontal speed: $$v_{x0} = u\cos 60^{\circ} = 270 \times 0.5 = 135 \text{ m s}^{-1}.$$

With $$\dfrac{c}{m} = \dfrac{0.1}{0.2} = 0.5 \text{ s}^{-1},$$ the differential equation becomes
$$\dfrac{dv_x}{dt} = -0.5\,v_x.$$

Solution for velocity:
$$v_x(t) = v_{x0}\,e^{-0.5t} = 135\,e^{-0.5t}.$$

Horizontal displacement up to time $$t$$:
$$x(t) = \int_0^{t} v_x(t')\,dt' = 135\int_0^{t} e^{-0.5t'}dt' = 135\left[\dfrac{-1}{0.5}e^{-0.5t'}\right]_0^{t}.$$
This simplifies to
$$x(t) = \dfrac{135}{0.5}\bigl(1-e^{-0.5t}\bigr) = 270\bigl(1-e^{-0.5t}\bigr).$$

The projectile hits the wall at $$t = 2 \text{ s}$$, so
$$x(2) = 270\bigl(1-e^{-1}\bigr).$$

Using the given value $$e = 2.7$$,
$$e^{-1} = \dfrac{1}{2.7} \approx 0.37037,$$
$$1 - e^{-1} \approx 1 - 0.37037 = 0.62963.$$

Therefore
$$x(2) = 270 \times 0.62963 \approx 170 \text{ m}.$$

Hence the vertical wall is about $$170 \text{ m}$$ from the point of projection, which lies within the accepted range 167 - 171 m.

The object is initially at rest at a height $$h = 3R$$ above Earth’s surface.
Therefore its initial distance from Earth’s centre is
$$r_0 = R + h = R + 3R = 4R$$.

To “never return to Earth” the object must just reach infinity with zero speed. Hence we apply conservation of mechanical energy between the initial point $$r_0$$ and infinity.

Initial mechanical energy
Kinetic: $$K_i = \frac12 m v^2$$ (where $$v$$ is the required minimum speed).
Gravitational potential: $$U_i = -\frac{GMm}{r_0} = -\frac{GMm}{4R}$$.

Final mechanical energy at infinity
Kinetic: $$K_f = 0$$ (minimum condition).
Potential: $$U_f = 0$$ (by convention).

Conservation of energy gives $$K_i + U_i = K_f + U_f$$ $$\frac12 m v^2 - \frac{GMm}{4R} = 0$$.

Solve for $$v$$:

$$\frac12 m v^2 = \frac{GMm}{4R}$$

$$v^2 = \frac{2GM}{4R} = \frac{GM}{2R}$$

$$v = \sqrt{\frac{GM}{2R}}$$.

The minimum speed required is $$\sqrt{\dfrac{GM}{2R}}$$, which corresponds to Option A.

Since the masses interact only via internal gravitational forces, there is no external torque, and the total angular momentum ($$\vec{L}$$) remains constant from $$t=0$$ until the point of collision.

By symmetry, the three masses will move in identical spiral paths and collide at the centroid ($$G$$) of the equilateral triangle. It is most convenient to calculate the angular momentum about this point.

For an equilateral triangle of side $$a$$, the distance from any vertex to the centroid ($$r$$) is $$r = \frac{a}{\sqrt{3}}$$

The velocity vector $$\vec{V}_A$$ is directed along the side $$AC$$. The angle between the position vector $$\vec{r}_A$$ (from $$G$$ to $$A$$) and the side $$AC$$ is $$30^\circ$$. Therefore, the angle $$\phi$$ between the position vector and the velocity vector is $$150^\circ$$

$$L_1 = mvr \sin \phi$$

$$L_1 = m V_0 \left( \frac{a}{\sqrt{3}} \right) \sin(150^\circ)$$

$$L_1 = \frac{maV_0}{2\sqrt{3}}$$

$$L_{net} = 3 \times L_1 = 3 \times \frac{maV_0}{2\sqrt{3}}$$

$$L_{net} = \frac{\sqrt{3}}{2} amV_0$$

For a point mass outside a uniform solid sphere, the entire mass $$M$$ acts as if it is concentrated at the center $$O$$.

$$F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$$

The cavity has a radius $$r = R/3$$. Since mass is proportional to volume ($$M \propto R^3$$),

Mass of cavity ($$m_c$$): $$M \cdot (\frac{R/3}{R})^3 = \frac{M}{27}$$

Distance to $$P$$ ($$d_c$$): The center of the cavity $$O'$$ is at a distance $$R - R/3 = 2R/3$$ from $$O$$. Thus, its distance from point $$P$$ is $$2R - \frac{2R}{3} = \frac{4R}{3}$$.

The force exerted by this removed mass is $$F_{cav} = \frac{G(M/27)m}{(4R/3)^2} = \frac{GMm}{27} \cdot \frac{9}{16R^2} = \frac{GMm}{48R^2}$$

The force from the remaining part ($$F_2$$) is $$F_2 = F_1 - F_{cav} = \frac{GMm}{4R^2} - \frac{GMm}{48R^2}$$

$$F_2 = \frac{12GMm - 1GMm}{48R^2} = \frac{11GMm}{48R^2}$$

$$\frac{F_1}{F_2} = \frac{\frac{GMm}{4R^2}}{\frac{11GMm}{48R^2}} = \frac{1}{4} \times \frac{48}{11} = \frac{12}{11}$$

Escape velocity: v_e = √(2GM/R). For planet: M_p = M_e/8, R_p = R_e/2.

v_p = √(2G(M/8)/(R/2)) = √(2GM/(4R)) = v_e/2 = 11.2/2 = 5.6 km/s.

The correct answer is Option 3: 5.6.

Assertion (A): A simple pendulum is taken to a planet with mass $$4M_E$$ and radius $$2R_E$$. The time period remains the same.

Reason (R): The mass of the pendulum remains unchanged.

Since gravitational acceleration is given by $$g = \frac{GM}{R^2}$$, on Earth we have $$g_E = \frac{GM_E}{R_E^2}$$ and on the planet

$$g_P = \frac{G \cdot 4M_E}{(2R_E)^2} = \frac{4GM_E}{4R_E^2} = \frac{GM_E}{R_E^2} = g_E\,. $$

From this it follows that $$g_P = g_E$$. Since the time period of a simple pendulum is $$T = 2\pi\sqrt{\frac{l}{g}}$$, the equality of gravitational acceleration implies that the time period remains the same on both Earth and the planet. Thus Assertion (A) is true.

The mass of the pendulum does remain unchanged, as mass is invariant under such a transfer, so Reason (R) is true.

However, the pendulum’s time period does not depend on its mass but only on $$g$$ and $$l$$, so the unchanged mass does not explain why the time period remains the same. Therefore, although Reason (R) is true, it does not correctly explain Assertion (A).

The correct answer is Option 4: Both (A) and (R) are true but (R) is NOT the correct explanation of (A).

For a body of mass $$m$$ projected from the earth’s surface (radius $$R$$) let the minimum speed required to reach infinity with zero speed be $$v_e$$ (escape velocity).

Gravitational potential energy at the earth’s surface is $$U_s = -\dfrac{GMm}{R}$$, while at infinity it is $$U_\infty = 0$$ (this is the highest possible value of gravitational potential energy).

Using conservation of mechanical energy, the total energy at the surface must equal the total energy at infinity:

$$\dfrac{1}{2} m v_e^2 + \left(-\dfrac{GMm}{R}\right) = 0 \quad -(1)$$

Solving $$(1)$$ gives the escape velocity:
$$v_e = \sqrt{\dfrac{2GM}{R}}$$

The kinetic energy that must be supplied is therefore
$$K = \dfrac{1}{2} m v_e^2 = \dfrac{1}{2} m \left(\dfrac{2GM}{R}\right) = \dfrac{GMm}{R}$$

Using $$g = \dfrac{GM}{R^2}$$, we rewrite this as
$$K = m g R$$

Assertion A claims the required kinetic energy is $$\dfrac{1}{2} m g R$$, which is not equal to the correct value $$m g R$$. Hence Assertion A is false.

Reason R states that “the maximum potential energy of a body is zero when it is projected to infinity from the earth’s surface.”
Because gravitational potential is defined to be zero at infinity and is negative everywhere else, this statement is true.

Therefore, A is false but R is true. The correct option is Option A.

We begin with a satellite orbiting Earth that is 9 times closer than the Moon and seek its time period.

We know from Kepler’s third law that $$T^2 \propto r^3$$.

Since the satellite is 9 times closer: $$r_s = \frac{r_m}{9}$$.

Substituting into the proportionality gives $$\frac{T_s^2}{T_m^2} = \frac{r_s^3}{r_m^3} = \left(\frac{1}{9}\right)^3 = \frac{1}{729}$$.

Taking the square root yields $$T_s = T_m \times \frac{1}{\sqrt{729}} = 27 \times \frac{1}{27} = 1$$ day.

The correct answer is Option 2: 1 day.

We need to find the ratio of electrostatic force to gravitational force between the two ions. The electrostatic force (Coulomb's law) is given by $$ F_e = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2} $$. The gravitational force is $$ F_g = \frac{Gm_1 m_2}{r^2} $$. Dividing these gives $$ \frac{F_e}{F_g} = \frac{q_1 q_2}{4\pi\epsilon_0 G m_1 m_2} $$.

The charges are $$q_1 = 6.67 \times 10^{-19}$$ C and $$q_2 = 9.6 \times 10^{-10}$$ C, and the masses are $$m_1 = 19.2 \times 10^{-27}$$ kg and $$m_2 = 9 \times 10^{-27}$$ kg. We also use $$\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}$$ Nm$$^2$$C$$^{-2}$$ and $$G = 6.67 \times 10^{-11}$$ Nm$$^2$$kg$$^{-2}$$.

Substituting these values into the ratio yields: $$ \frac{F_e}{F_g} = \frac{9 \times 10^{9} \times 6.67 \times 10^{-19} \times 9.6 \times 10^{-10}}{6.67 \times 10^{-11} \times 19.2 \times 10^{-27} \times 9 \times 10^{-27}} $$.

The numerator simplifies as $$ 9 \times 6.67 \times 9.6 \times 10^{9-19-10} = 9 \times 6.67 \times 9.6 \times 10^{-20} = 576.288 \times 10^{-20} = 5.76288 \times 10^{-18} $$. The denominator becomes $$ 6.67 \times 19.2 \times 9 \times 10^{-11-27-27} = 6.67 \times 19.2 \times 9 \times 10^{-65} = 1152.576 \times 10^{-65} = 1.152576 \times 10^{-62} $$.

Therefore, $$ \frac{F_e}{F_g} = \frac{5.76288 \times 10^{-18}}{1.152576 \times 10^{-62}} = 5 \times 10^{44} = 0.5 \times 10^{45} $$. Hence, $$P = 0.5$$ and $$10P = 5$$, giving the final answer as 5.

The only quantities that decide the collapse-time for three identical masses released from the corners of an equilateral triangle are

• the initial side length $$a$$,
• the mass of each sphere $$m$$,
• the universal constant of gravitation $$G$$.

Let the required time be $$T$$.  Assume a dependence of the form

$$T \propto G^{\,p}\,m^{\,q}\,a^{\,r}$$

Write dimensions (M ≡ mass, L ≡ length, T ≡ time):

$$[T] = T^{1},\quad [G]=L^{3}M^{-1}T^{-2},\quad [m]=M^{1},\quad [a]=L^{1}$$

Equating powers of the fundamental dimensions,

$$L: \; 0 = 3p + r$$ $$M: \; 0 = -p + q$$ $$T: \; 1 = -2p$$

Solving,

$$p = -\frac12,\quad q = -\frac12,\quad r = \frac32$$

Hence

$$T \propto G^{-1/2}\,m^{-1/2}\,a^{3/2} \;=\;k\,\sqrt{\dfrac{a^{3}}{G\,m}}$$

for some dimensionless constant $$k$$ that is the same for both experiments because the geometry is identical.

Initial situation: side $$=a$$, mass $$=m$$, time $$=T = 4\,$$s.

New situation: side $$=2a$$, mass $$=2m$$. Using the proportionality,

$$\dfrac{T'}{T}=\dfrac{\sqrt{(2a)^{3}/(G\,2m)}}{\sqrt{a^{3}/(G\,m)}}$$

$$\qquad=\,\dfrac{2^{3/2}}{2^{1/2}} = 2$$

Therefore $$T' = 2T = 2 \times 4\text{ s} = 8\text{ s}$$.

So the three spheres will collide after 8 seconds.

The angular momentum of a planet in circular orbit is $$L = mv r$$, where $$v$$ is the orbital velocity.

For a circular orbit we have $$\frac{GMm}{R^2} = \frac{mv^2}{R}$$, so $$v = \sqrt{\frac{GM}{R}}$$.

Substituting this into the expression for angular momentum gives $$L = mR\sqrt{\frac{GM}{R}} = m\sqrt{GMR}$$.

We then consider the ratio of the angular momenta of bodies B and A:

$$ \frac{L_B}{L_A} = \frac{m_B\sqrt{GMR_B}}{m_A\sqrt{GMR_A}} = \frac{m_B}{m_A}\sqrt{\frac{R_B}{R_A}} $$

Using the given relations $$m_B = 4\sqrt{2} \cdot m_A$$ and $$R_B = 2R_A$$, we find

$$ \frac{L_B}{L_A} = 4\sqrt{2} \times \sqrt{2} = 4\sqrt{2} \times \sqrt{2} = 4 \times 2 = 8 $$

The answer is 8.

A satellite of mass $$\frac{M}{2}$$ revolves around the Earth in a circular orbit at a height of $$\frac{R}{3}$$ from the Earth's surface. The angular momentum is given as $$M\sqrt{\frac{GMR}{x}}$$. We need to find $$x$$.

First we find the orbital radius, which is the distance from the center of the Earth to the satellite:

$$ r = R + \frac{R}{3} = \frac{3R + R}{3} = \frac{4R}{3} $$

Next we find the orbital velocity by equating the gravitational force to the centripetal force:

$$ \frac{GMm}{r^2} = \frac{mv^2}{r} $$

Here $$M$$ is Earth's mass, $$m = \frac{M}{2}$$ is the satellite mass, and $$r$$ is the orbital radius. Solving for $$v$$ gives

$$ v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{\frac{4R}{3}}} = \sqrt{\frac{3GM}{4R}} $$

Then the angular momentum of the satellite is $$L = mvr$$, where $$m = \frac{M}{2}$$, so

$$ L = \frac{M}{2} \times \sqrt{\frac{3GM}{4R}} \times \frac{4R}{3} $$

We simplify this expression step by step. First, multiply the constant factors:

$$ L = \frac{M}{2} \times \frac{4R}{3} \times \sqrt{\frac{3GM}{4R}} $$

$$ L = \frac{2MR}{3} \times \sqrt{\frac{3GM}{4R}} $$

Since

$$ \sqrt{\frac{3GM}{4R}} = \frac{\sqrt{3GM}}{2\sqrt{R}}, $$

substituting back gives

$$ L = \frac{2MR}{3} \times \frac{\sqrt{3GM}}{2\sqrt{R}} = \frac{MR}{3} \times \frac{\sqrt{3GM}}{\sqrt{R}} = \frac{M\sqrt{R} \times \sqrt{R}}{3} \times \frac{\sqrt{3GM}}{\sqrt{R}} $$

which simplifies to

$$ L = \frac{M\sqrt{R}}{3} \times \sqrt{3GM} = \frac{M}{3}\sqrt{3GMR}. $$

We now rewrite this result in the required form:

$$ L = \frac{M}{3}\sqrt{3GMR} = M \times \frac{\sqrt{3GMR}}{3} = M \times \sqrt{\frac{3GMR}{9}} = M\sqrt{\frac{GMR}{3}} $$

Here we used

$$\frac{\sqrt{3GMR}}{3} = \frac{\sqrt{3GMR}}{\sqrt{9}} = \sqrt{\frac{3GMR}{9}} = \sqrt{\frac{GMR}{3}}.$$

Comparing $$L = M\sqrt{\frac{GMR}{3}}$$ with $$L = M\sqrt{\frac{GMR}{x}}$$ shows

$$ x = 3 $$

Thus the value of $$x$$ is $$\boxed{3}$$.

If diameter reduced to 1/3, radius R/3, mass same. g' = GM/(R/3)² = 9GM/R² = 9g.

The answer is 9.

The gravitational potential energy due to the central mass $$M$$ is
$$V_g(r)= -\dfrac{G M m}{r}\,.$$

For a circular orbit of radius $$r_0$$ under gravity alone, the inward (centripetal) force balance is
$$\dfrac{m v_0^{\,2}}{r_0}= \dfrac{G M m}{r_0^{2}} \;\;\Longrightarrow\;\; v_0^{\,2}= \dfrac{G M}{r_0}\,.$$

The corresponding time period is therefore
$$T_0 =\dfrac{2\pi r_0}{v_0}=2\pi\sqrt{\dfrac{r_0^{3}}{G M}} \;\;\Longrightarrow\;\; T_0^{2}= \dfrac{4\pi^{2} r_0^{3}}{G M}\,.$$

Now an additional central potential is applied:
$$V_c(r)=\dfrac{m\alpha}{r^{3}}\qquad (\alpha\gt 0).$$

The radial force associated with this potential is obtained from $$F_r=-\dfrac{dV_c}{dr}$$:
$$F_c(r)=-\frac{d}{dr}\left(\dfrac{m\alpha}{r^{3}}\right)=+\,\dfrac{3m\alpha}{r^{4}}.$$
Since this force is positive in the outward radial direction, it is repulsive. Hence it opposes gravitation.

To keep the particle in the same circular orbit of radius $$r_0$$, the inward gravitational pull must now be reduced by this repulsion:
$$\dfrac{m v_1^{\,2}}{r_0}= \dfrac{G M m}{r_0^{2}}-\dfrac{3m\alpha}{r_0^{4}} \;\;\Longrightarrow\;\; v_1^{\,2}= \dfrac{G M}{r_0}-\dfrac{3\alpha}{r_0^{3}}.$$

The new time period becomes
$$T_1 =\dfrac{2\pi r_0}{v_1}\,,\qquad T_1^{2}= \dfrac{4\pi^{2} r_0^{2}}{v_1^{\,2}} =\dfrac{4\pi^{2} r_0^{3}} {\,G M-\dfrac{3\alpha}{r_0^{2}}}\;.$$ Define $$A=4\pi^{2}r_0^{3}$$ for brevity, so that
$$T_0^{2}= \dfrac{A}{G M},\qquad T_1^{2}= \dfrac{A}{\,G M-\dfrac{3\alpha}{r_0^{2}}}\;.$$

We now evaluate the required ratio:
$$\dfrac{T_1^{2}-T_0^{2}}{T_1^{2}} =1-\dfrac{T_0^{2}}{T_1^{2}} =1-\dfrac{\dfrac{A}{G M}} {\dfrac{A}{\,G M-\dfrac{3\alpha}{r_0^{2}}}} =1-\dfrac{G M-\dfrac{3\alpha}{r_0^{2}}}{G M} =\dfrac{3\alpha}{G M r_0^{2}}\;.$$

Thus
$$\boxed{\dfrac{T_1^{2}-T_0^{2}}{T_1^{2}}=\dfrac{3\alpha}{G M r_0^{2}}}\,.$$
Hence the correct option is:
Option A which is: $$\dfrac{3\alpha}{G M r_0^{2}}.$$

The given central force is $$F(r)=-kr$$, so the corresponding potential energy is $$V(r)=\dfrac{1}{2}kr^{2}$$.

For a stable circular orbit of radius $$r$$ and speed $$v$$, the required centripetal force is provided entirely by the central force:

$$\frac{mv^{2}}{r}=kr \qquad -(1)$$

From Bohr’s quantisation rule, the magnitude of the angular momentum is

$$L = n\hbar \qquad -(2)$$

But for circular motion, angular momentum is also

$$L = mvr \qquad -(3)$$

We now combine equations $$(1)$$ and $$(3)$$ to obtain relations among $$r$$, $$v$$, and $$L$$.

Step 1: Eliminate $$v$$ between $$(1)$$ and $$(3)$$
From $$(3)$$, $$v=\dfrac{L}{mr}$$. Substitute this in $$(1)$$:

$$\frac{m}{r}\left(\frac{L}{mr}\right)^{2}=kr \;\;\Longrightarrow\;\; \frac{L^{2}}{m r^{3}}=kr$$

$$\Longrightarrow\; L^{2}=km\,r^{4} \qquad -(4)$$

Step 2: Express $$r^{2}$$ in terms of $$L$$
From $$(4)$$,

$$r^{4}=\frac{L^{2}}{km}\;\;\Longrightarrow\;\; r^{2}=\frac{L}{\sqrt{km}} \qquad -(5)$$

Using $$(2)$$ ( $$L=n\hbar$$ ) in $$(5)$$ gives

$$r^{2}=n\hbar\sqrt{\frac{1}{mk}}$$

This is exactly Option A.

Step 3: Obtain $$v^{2}$$
Return to equation $$(1)$$: $$v^{2}=\frac{k}{m}r^{2}$$. Insert the value of $$r^{2}$$ from $$(5)$$:

$$v^{2}=\frac{k}{m}\left(n\hbar\sqrt{\frac{1}{mk}}\right) =n\hbar\sqrt{\frac{k}{m^{3}}}$$

This is Option B.

Step 4: Evaluate $$\dfrac{L}{mr^{2}}$$
Using $$(2)$$ and $$(5)$$,

$$\frac{L}{mr^{2}}=\frac{n\hbar}{m\left(n\hbar/\sqrt{km}\right)} =\frac{\sqrt{km}}{m} =\sqrt{\frac{k}{m}}$$

This matches Option C.

Step 5: Compute the total mechanical energy $$E$$
Kinetic energy: $$T=\dfrac{1}{2}mv^{2}$$. Using $$(1)$$, $$v^{2}=\dfrac{k}{m}r^{2}$$ gives

$$T=\frac{1}{2}m\left(\frac{k}{m}r^{2}\right)=\frac{1}{2}kr^{2}$$

Potential energy: $$V=\dfrac{1}{2}kr^{2}$$.

Total energy: $$E=T+V=kr^{2}$$.

Substituting $$r^{2}=n\hbar/\sqrt{km}$$:

$$E=k\left(\frac{n\hbar}{\sqrt{km}}\right)=n\hbar\sqrt{\frac{k}{m}}$$

The factor is $$1$$, not $$\tfrac{1}{2}$$, so Option D is incorrect.

Therefore the correct statements are:
Option A, Option B, and Option C.

Radius of the ball, $$r = \frac{3}{2}\times 10^{-2}\;{\rm m}=0.015\;{\rm m}$$
Mass of the ball, $$m = \frac{22}{7}\times 10^{-3}\;{\rm kg}= \pi\times 10^{-3}\;{\rm kg}=3.1429\times 10^{-3}\;{\rm kg}$$

Volume of the ball
$$V = \frac{4}{3}\pi r^{3}$$
$$r^{3}=0.015^{3}=3.375\times 10^{-6}\,{\rm m^{3}}$$
$$V=\frac{4}{3}\pi(3.375\times 10^{-6})=4.18879\times3.375\times10^{-6}=1.414\times10^{-5}\;{\rm m^{3}}$$

Buoyant force in water
$$F_{b} = \rho_{\text w}gV = (1000)(10)(1.414\times10^{-5}) = 0.1414\;{\rm N}$$

Weight of the ball
$$W = mg = (3.1429\times10^{-3})(10)=0.03143\;{\rm N}$$

Net upward force (viscosity neglected)
$$F_{\text{net}} = F_{b}-W = 0.1414-0.03143 = 0.1100\;{\rm N}\approx 0.11\;{\rm N}$$

Option A  - Work done in pushing the ball down by $$d=0.7\;{\rm m}$$ (quasi-static)
To move slowly we must apply a downward force equal to the constant upward imbalance $$F_{b}-W$$. $$W_{\text{push}} = (F_{b}-W)\,d = (0.11)(0.7)=0.077\;{\rm J}$$
Hence Option A is correct.

Option B  - Speed $$v$$ at the water surface (viscosity neglected)
Acceleration inside water: $$a=\dfrac{F_{\text{net}}}{m}=\dfrac{0.11}{3.1429\times10^{-3}}\approx35\;{\rm m\,s^{-2}}$$
Using $$v^{2}=2as$$ with $$s=d=0.7\;{\rm m}$$:
$$v=\sqrt{2(35)(0.7)}=\sqrt{49}=7\;{\rm m\,s^{-1}}$$
Thus Option B is correct.

Option C  - Height $$H$$ above the surface
After emerging, only gravity acts. $$H=\dfrac{v^{2}}{2g}=\dfrac{49}{20}=2.45\;{\rm m}$$, not $$1.4\;{\rm m}$$. Therefore Option C is incorrect.

Option D  - Ratio of (net force) to (maximum viscous force)
Maximum possible viscous (Stokes) drag inside water occurs at speed $$v=7\;{\rm m\,s^{-1}}$$:
$$F_{\eta,\max}=6\pi\eta rv = 6\left(\frac{22}{7}\right)(1\times10^{-3})(0.015)(7)$$
$$(\frac{22}{7})\times7 = 22$$, so $$F_{\eta,\max}=6(22)(0.015)\times10^{-3}=1.98\times10^{-3}\;{\rm N}$$

Hence $$\dfrac{F_{\text{net}}}{F_{\eta,\max}}=\dfrac{0.11}{1.98\times10^{-3}}\approx55.56=\frac{500}{9}$$
Option D is therefore correct.

Final result: Option A, Option B and Option D are correct.

We need to find the dimensions of $$\sqrt{uG}$$ where $$u$$ is energy density and $$G$$ is the gravitational constant.

Dimensions of energy density $$u$$.

$$ [u] = \frac{\text{Energy}}{\text{Volume}} = \frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}] $$

Dimensions of gravitational constant $$G$$.

From $$F = \frac{Gm_1m_2}{r^2}$$:

$$ [G] = \frac{[F][r^2]}{[m^2]} = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}] $$

Dimensions of $$\sqrt{uG}$$.

$$ [uG] = [ML^{-1}T^{-2}] \times [M^{-1}L^3T^{-2}] = [M^0L^2T^{-4}] = [L^2T^{-4}] $$

$$ [\sqrt{uG}] = [LT^{-2}] $$

Identify the physical quantity.

$$[LT^{-2}]$$ is the dimension of acceleration, which is force per unit mass.

The correct answer is Option (4): Force per unit mass.

Satellite mass = $$10^3$$ kg, initial orbit radius = $$2R$$.

Total energy in circular orbit: $$E = -\frac{GMm}{2r}$$.

Initial energy: $$E_1 = -\frac{GM \times 10^3}{2 \times 2R} = -\frac{GM \times 10^3}{4R}$$.

Using $$GM = gR^2$$: $$E_1 = -\frac{gR \times 10^3}{4} = -\frac{10 \times R \times 10^3}{4} = -2500R$$.

Energy supplied: $$\frac{10^4R}{6}$$ J.

New energy: $$E_2 = E_1 + \frac{10^4R}{6} = -2500R + \frac{10000R}{6} = R(-2500 + \frac{5000}{3}) = R\frac{-7500+5000}{3} = \frac{-2500R}{3}$$.

$$E_2 = -\frac{GMm}{2r_2} = -\frac{gR^2 \times 10^3}{2r_2} = -\frac{10^4R^2}{2r_2}$$.

$$\frac{-2500R}{3} = \frac{-10^4R^2}{2r_2}$$

$$r_2 = \frac{10^4R^2 \times 3}{2 \times 2500R} = \frac{3 \times 10^4R}{5000} = 6R$$.

The correct answer is Option 4: $$6R$$.

A 90 kg body is placed at a height $$2R$$ above Earth’s surface and the gravitational pull on it is to be determined.

The acceleration due to gravity at a height $$h$$ above Earth’s surface is given by $$ g' = g\left(\frac{R}{R + h}\right)^2 $$. This follows from Newton’s law of gravitation, since $$g' = \frac{GM}{(R+h)^2}$$ and $$g = \frac{GM}{R^2}$$.

For $$h = 2R$$, the distance from Earth’s center is $$R + h = 3R$$, so substituting into the formula yields $$ g' = g\left(\frac{R}{3R}\right)^2 = g\times\frac{1}{9} = \frac{10}{9}\,\text{m/s}^2 $$.

The gravitational force on the body is then $$ F = m\,g' = 90 \times \frac{10}{9} = 100\,\text{N} $$.

The correct answer is Option A: 100 N.

A planet revolves around a star in a circular orbit of radius $$R$$ with period $$T$$, and the gravitational force is proportional to $$R^{-3/2}$$.

For a circular orbit, the gravitational force must provide the centripetal force, so we have $$F = \frac{mv^2}{R}$$.

Since the force varies as $$F \propto R^{-3/2}$$, it can be expressed as $$F = \frac{k}{R^{3/2}}$$ for some constant $$k$$.

Equating these expressions gives $$\frac{k}{R^{3/2}} = \frac{mv^2}{R}$$, which simplifies to $$v^2 = \frac{k}{m\sqrt{R}}$$ and hence $$v = \sqrt{\frac{k}{m}}\,R^{-1/4}$$.

The period $$T$$ is the orbital circumference divided by the speed, so $$T = \frac{2\pi R}{v} = 2\pi R \sqrt{\frac{m}{k}}\,R^{1/4} = 2\pi\sqrt{\frac{m}{k}}\,R^{5/4}$$.

It follows that $$T \propto R^{5/4}$$, and squaring this relationship yields $$T^2 \propto R^{5/2}$$.

The correct answer is $$T^2 \propto R^{5/2}$$ (Option A).

We need to find the new orbital period when the distance from the Sun is reduced to one-fourth.

According to Kepler’s Third Law, $$T^2 \propto r^3$$, where $$T$$ is the orbital period and $$r$$ is the orbital radius.

Applying this to the initial and new conditions, we get $$\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3} = \left(\frac{r_2}{r_1}\right)^3 = \left(\frac{1}{4}\right)^3 = \frac{1}{64}$$, which implies $$\frac{T_2}{T_1} = \frac{1}{8}$$.

Since the original period $$T_1$$ is 200 days, the new period is $$T_2 = \frac{T_1}{8} = \frac{200}{8} = 25$$ days.

The correct answer is Option (1): 25 days.

$$T_p=6$$h, $$M_p=M_E/4$$. For geostationary orbit: $$T²\propto r³/M$$.

$$r_p³=\frac{M_p}{M_E}\cdot\frac{T_p²}{T_E²}\cdot r_E³=\frac{1}{4}\cdot\frac{36}{576}\cdot r_E³=\frac{1}{4}\cdot\frac{1}{16}\cdot r_E³=\frac{r_E³}{64}$$.

$$r_p=r_E/4=4.2\times10^4/4=1.05\times10^4$$ km.

The answer is Option (2): $$1.05\times10^4$$ km.

The total mechanical energy on Earth's surface: $$E_1 = -\frac{GM_em}{R_e}$$ (taking KE = 0 on surface).

on Earth's surface, the ball is at rest, so $$KE = 0$$ and $$PE = -\frac{GM_em}{R_e}$$. Total $$E_1 = -\frac{GM_em}{R_e}$$.

In circular orbit at altitude $$h = 318.5$$ km:

$$r = R_e + h = 6370 + 318.5 = 6688.5 \text{ km}$$

$$\frac{r}{R_e} = \frac{6688.5}{6370} = \frac{6688.5}{6370}$$. Let me simplify: $$\frac{6688.5}{6370} = 1 + \frac{318.5}{6370} = 1 + \frac{1}{20} = \frac{21}{20}$$.

So $$r = \frac{21}{20}R_e$$.

Total energy in orbit: $$E_2 = -\frac{GM_em}{2r} = -\frac{GM_em}{2 \cdot \frac{21}{20}R_e} = -\frac{10GM_em}{21R_e}$$.

Change in total energy:

$$\Delta E = E_2 - E_1 = -\frac{10GM_em}{21R_e} + \frac{GM_em}{R_e} = GM_em\left(\frac{1}{R_e} - \frac{10}{21R_e}\right) = \frac{GM_em}{R_e}\left(\frac{21-10}{21}\right) = \frac{11GM_em}{21R_e}$$

Comparing with $$x\frac{GM_em}{21R_e}$$: $$x = 11$$.

The correct answer is Option 4: 11.

We need to find the weight of a body at a depth of $$R/4$$ below the Earth's surface, given that it weighs 300 N on the surface.

We know that the acceleration due to gravity at a depth $$d$$ below the Earth's surface (assuming uniform density) is $$g' = g\left(1 - \frac{d}{R}\right)$$, where $$g$$ is the acceleration due to gravity at the surface and $$R$$ is the radius of the Earth. This formula arises because at depth $$d$$, only the spherical shell of radius $$(R - d)$$ contributes to the gravitational field (by the shell theorem, the outer shell exerts no net force).

Substituting $$d = R/4$$ into this expression gives $$g' = g\left(1 - \frac{R/4}{R}\right) = g\left(1 - \frac{1}{4}\right) = g \times \frac{3}{4}$$.

Therefore, the weight of the body at this depth is $$W' = m \times g' = m \times \frac{3g}{4}$$.

Since the surface weight is $$W = mg = 300 \,\text{N}$$, it follows that $$W' = \frac{3}{4} \times W = \frac{3}{4} \times 300 = 225 \,\text{N}$$.

The correct answer is Option (4): 225 N.

At height $$h$$ above surface: $$g_h = g\frac{R^2}{(R+h)^2}$$.

At depth $$d$$ below surface: $$g_d = g\left(1 - \frac{d}{R}\right)$$.

Setting equal and using $$h = d$$:

$$\frac{R^2}{(R+h)^2} = 1 - \frac{h}{R} = \frac{R-h}{R}$$

$$R^3 = (R-h)(R+h)^2$$

Let $$x = h/R$$: $$1 = (1-x)(1+x)^2 = (1-x^2)(1+x) = 1+x-x^2-x^3$$

$$x + x^2(-1) - x^3 = 0$$... Actually: $$x - x^2 - x^3 = 0$$, $$x(1-x-x^2) = 0$$.

$$x^2 + x - 1 = 0$$, $$x = \frac{-1+\sqrt{5}}{2}$$.

$$h = \frac{(\sqrt{5}-1)R}{2} = \frac{\sqrt{5}R - R}{2}$$.

The answer corresponds to Option (4).

Derive the correct formula for the height of a satellite above Earth's surface.

For a satellite in circular orbit at distance $$r$$ from Earth's center, the gravitational force provides centripetal force:

$$ \frac{GMm}{r^2} = \frac{mv^2}{r} = m\omega^2 r $$

where $$\omega = \frac{2\pi}{T}$$.

$$ \frac{GM}{r^2} = \omega^2 r = \frac{4\pi^2}{T^2}r $$

$$ GM = \frac{4\pi^2 r^3}{T^2} $$

Since $$g = \frac{GM}{R^2}$$, we have $$GM = gR^2$$. Substituting:

$$ gR^2 = \frac{4\pi^2 r^3}{T^2} $$

$$ r^3 = \frac{gR^2T^2}{4\pi^2} = \frac{T^2R^2g}{4\pi^2} $$

$$ r = \left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3} $$

Since $$r = R + h$$:

$$ h = r - R = \left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3} - R $$

The correct answer is Option B: $$\left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3} - R$$.

Escape velocity: $$v_e = \sqrt{\frac{2GM}{R}}$$.

Given: $$R_p = \frac{R_e}{3}$$ and $$M_p = \frac{M_e}{6}$$.

$$\frac{v_p}{v_e} = \sqrt{\frac{M_p/R_p}{M_e/R_e}} = \sqrt{\frac{M_p \cdot R_e}{M_e \cdot R_p}} = \sqrt{\frac{(M_e/6) \cdot R_e}{M_e \cdot (R_e/3)}} = \sqrt{\frac{1/6}{1/3}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$

$$v_p = \frac{11.2}{\sqrt{2}} = \frac{11.2 \times \sqrt{2}}{2} = 5.6\sqrt{2} \approx 7.92 \text{ km/s}$$

The answer is Option (4): $$\boxed{7.9 \text{ km s}^{-1}}$$.

The angular speed ($$\omega$$) of an orbiting body is defined by its time period ($$T$$): $$\omega = \frac{2\pi}{T}$$

The Moon completes its orbit around the Earth in $$T_m \approx 27.3$$ days, whereas the Earth takes $$T_e \approx 365.25$$ days to orbit the Sun. Because the Moon has a much shorter time period ($$T_m < T_e$$), its angular speed is significantly higher: $$\omega_{\text{moon}} > \omega_{\text{earth}}$$

The Assertion is true, and the Reason provides the factual basis (shorter time period) that mathematically explains why the Moon's angular speed is greater.

Correct Option: (B)

The acceleration due to gravity on the surface of Earth is:

$$g = \frac{GM}{R^2}$$

If the diameter is halved, the radius becomes $$R' = R/2$$.

$$g' = \frac{GM}{(R/2)^2} = \frac{GM}{R^2/4} = \frac{4GM}{R^2} = 4g$$

The answer is $$4g$$, which corresponds to Option (4).

Given: Gravitational potential $$V = -5.12 \times 10^7 \text{ J kg}^{-1}$$, acceleration due to gravity $$g = 6.4 \text{ m s}^{-2}$$, and mean radius of Earth $$R = 6400 \text{ km}$$.

Key formulas:

At distance $$r$$ from the centre of Earth:

$$V = -\frac{GM}{r}$$ and $$g = \frac{GM}{r^2}$$

Find $$r$$ from the ratio $$V/g$$: $$\frac{|V|}{g} = \frac{GM/r}{GM/r^2} = r$$

$$r = \frac{5.12 \times 10^7}{6.4} = 8 \times 10^6 \text{ m} = 8000 \text{ km}$$

Find the height above the surface: $$h = r - R = 8000 - 6400 = 1600 \text{ km}$$

The correct answer is Option (1): $$1600 \text{ km}$$.

We need to find the escape velocity on the moon relative to the planet.

Since $$M_m = \frac{M_p}{144}$$, $$d_m = \frac{d_p}{16}$$, we have $$R_m = \frac{R_p}{16}$$ and the escape velocity on the planet is $$v$$.

We start with the escape velocity formula: $$v_e = \sqrt{\frac{2GM}{R}}$$

Next, the ratio of the escape velocity on the moon to that on the planet is given by $$\frac{v_m}{v_p} = \sqrt{\frac{M_m/R_m}{M_p/R_p}} = \sqrt{\frac{M_m \cdot R_p}{M_p \cdot R_m}}$$

Substituting the expressions for mass and radius yields $$= \sqrt{\frac{1}{144} \times 16} = \sqrt{\frac{16}{144}} = \frac{4}{12} = \frac{1}{3}$$

This gives $$v_m = \frac{v}{3}$$.

The correct answer is Option 1: $$\frac{v}{3}$$.

Two satellites A and B with orbital radii 4R and R. Speed of A = 3v. Find speed of B.

For a circular orbit, the orbital speed formula is $$v = \sqrt{\frac{GM}{r}}$$, so $$v \propto \frac{1}{\sqrt{r}}$$.

Using this relation, $$\frac{v_B}{v_A}=\sqrt{\frac{r_A}{r_B}}=\sqrt{\frac{4R}{R}}=2$$ and hence $$v_B=2v_A=2\times3v=6v$$.

The correct answer is Option (2): 6v.

A wire of length L and mass M is bent into a semicircular arc. The radius of the semicircle is determined by $$\pi R = L \Rightarrow R = \frac{L}{\pi}$$.

The linear mass density is $$\lambda = \frac{M}{L}$$. A particle of mass m is placed at the centre, and by symmetry the net force is directed toward the midpoint of the arc along the axis of symmetry.

Consider an element $$d\theta$$ at angle $$\theta$$ from the axis of symmetry. Its mass is $$dm = \lambda R\,d\theta = \frac{M}{L} \cdot \frac{L}{\pi}\,d\theta = \frac{M}{\pi}\,d\theta$$. The gravitational force from this element is $$dF = \frac{Gm\,dm}{R^2}$$, and the component along the axis of symmetry is $$dF\cos\theta$$.

Integrating from $$-\pi/2$$ to $$\pi/2$$ gives the total force:

$$ F = \int_{-\pi/2}^{\pi/2} \frac{Gm}{R^2} \cdot \frac{M}{\pi}\cos\theta\,d\theta = \frac{GmM}{\pi R^2}[\sin\theta]_{-\pi/2}^{\pi/2} = \frac{GmM}{\pi R^2}(2) = \frac{2GmM}{\pi R^2} $$

Substituting $$R = L/\pi$$:

$$ F = \frac{2GmM}{\pi(L/\pi)^2} = \frac{2GmM}{\pi \cdot L^2/\pi^2} = \frac{2GmM\pi}{L^2} $$

The correct answer is Option 4: $$\frac{2GmM\pi}{L^2}$$.

A simple pendulum at height $$R$$ above Earth's surface has time period $$T_1 = 4$$ s. We need $$T_2$$ at height $$2R$$.

The time period of a simple pendulum is $$T = 2\pi\sqrt{\frac{l}{g'}}$$ where $$g'$$ is the effective gravitational acceleration at height $$h$$:

$$ g' = \frac{g}{(1 + h/R)^2} $$

At height $$h = R$$:

$$ g_1 = \frac{g}{(1+1)^2} = \frac{g}{4} $$

$$ T_1 = 2\pi\sqrt{\frac{l}{g/4}} = 2\pi\sqrt{\frac{4l}{g}} $$

At height $$h = 2R$$:

$$ g_2 = \frac{g}{(1+2)^2} = \frac{g}{9} $$

$$ T_2 = 2\pi\sqrt{\frac{l}{g/9}} = 2\pi\sqrt{\frac{9l}{g}} $$

Dividing to find the relation between $$T_2$$ and $$T_1$$:

$$ \frac{T_2}{T_1} = \frac{\sqrt{9}}{\sqrt{4}} = \frac{3}{2} $$

Therefore,

$$ T_2 = \frac{3}{2}T_1 $$

Multiplying both sides by 2:

$$ 2T_2 = 3T_1 $$

This corresponds to $$3T_1 = 2T_2$$.

The correct answer is Option (4): $$3T_1 = 2T_2$$.

Find the length of a seconds pendulum at height $$h = 2R$$ where $$g = \pi^2$$ m/s$$^2$$.

At height $$h=2R$$ above Earth’s surface the acceleration due to gravity becomes $$g' = g\left(\frac{R}{R+h}\right)^2 = \pi^2 \left(\frac{R}{R+2R}\right)^2 = \pi^2 \times \frac{1}{9} = \frac{\pi^2}{9}$$ m/s$$^2$$.

A seconds pendulum has a time period $$T = 2$$ s and its period satisfies $$T = 2\pi\sqrt{\frac{l}{g'}}$$. Substituting $$T=2$$ and $$g'=\pi^2/9$$ gives $$2 = 2\pi\sqrt{\frac{l}{\pi^2/9}}$$. Inside the square root one finds $$\sqrt{\frac{l}{\pi^2/9}} = \sqrt{\frac{9l}{\pi^2}} = \frac{3\sqrt{l}}{\pi}$$. Hence $$2 = 2\pi\times\frac{3\sqrt{l}}{\pi} = 6\sqrt{l}$$, which leads to $$6\sqrt{l} = 2$$, so $$\sqrt{l} = \frac{2}{6} = \frac{1}{3}$$ and therefore $$l = \frac{1}{9}$$ m.

The correct answer is Option B: $$\frac{1}{9}$$ m.

Four identical particles of mass $$m$$ are at the corners of a square of side $$a$$. The net gravitational force on one particle due to the other three is given as $$\frac{(2\sqrt{2}+1)}{32}\frac{Gm^2}{L^2}$$. We need to find $$a$$ in terms of $$L$$.

Consider the particle at one corner. The other three particles are two adjacent particles at distance $$a$$ (along the sides) and one diagonal particle at distance $$a\sqrt{2}$$. Each adjacent particle exerts a force: $$F_{adj} = \frac{Gm^2}{a^2}$$. The two adjacent forces are perpendicular to each other along the two sides of the square, so by the Pythagorean theorem the resultant is $$F_{resultant,adj} = \sqrt{F_{adj}^2 + F_{adj}^2} = F_{adj}\sqrt{2} = \frac{\sqrt{2}\cdot Gm^2}{a^2}$$. This resultant points along the diagonal toward the diagonally opposite corner.

The diagonal particle exerts a force $$F_{diag} = \frac{Gm^2}{(a\sqrt{2})^2} = \frac{Gm^2}{2a^2}$$, which also points along the same diagonal direction. Since all forces are along the diagonal, the total net force is $$F_{net} = \frac{\sqrt{2}\cdot Gm^2}{a^2} + \frac{Gm^2}{2a^2} = \frac{Gm^2}{a^2}\left(\sqrt{2} + \tfrac{1}{2}\right) = \frac{(2\sqrt{2}+1)}{2}\cdot\frac{Gm^2}{a^2}$$. Equating this to the given expression $$\frac{(2\sqrt{2}+1)}{2}\cdot\frac{Gm^2}{a^2} = \frac{(2\sqrt{2}+1)}{32}\cdot\frac{Gm^2}{L^2}$$, the common factor $$(2\sqrt{2}+1)\,Gm^2$$ cancels, giving $$\frac{1}{2a^2} = \frac{1}{32L^2}$$, and hence $$2a^2 = 32L^2$$, $$a^2 = 16L^2$$, $$a = 4L$$.

The correct answer is Option (2): $$4L$$.

The positions of the two identical masses just before the collision are

$$x_1(t)=\bigl(x_0+d\bigr)+a\sin\omega t,\qquad x_2(t)=\bigl(x_0-d\bigr)-a\sin\omega t.$$

Differentiating, their velocities are

$$v_1(t)=a\omega\cos\omega t,\qquad v_2(t)=-a\omega\cos\omega t.$$

The collision takes place at $$t_0=0$$, so

$$v_{1i}=v_1(0)=a\omega,\qquad v_{2i}=v_2(0)=-a\omega.$$

Particle 3 (mass $$m$$) approaches from the left with speed

$$u_0=\tfrac{a\omega}{2}$$

towards the right (&plus;ve $$x$$ direction). All three masses are equal, so a one-dimensional perfectly elastic collision simply makes the two colliding particles exchange their velocities.

Hence, immediately after the collision,

$$v_{3f}=-a\omega,\qquad v_{2f}=u_0=\tfrac{a\omega}{2},\qquad v_{1f}=v_{1i}=a\omega \;(\text{particle 1 is untouched}).$$

The centre-of-mass velocity of particles 1 and 2 after the collision is

$$v_{cm}=\frac{v_{1f}+v_{2f}}{2}=\frac{a\omega+\tfrac{a\omega}{2}}{2} =\frac{3a\omega/2}{2}= \frac{3}{4}\,a\omega.$$

Therefore,

$$\frac{v_{cm}}{a\omega}=\frac{3}{4}=0.75.$$

The required value is 0.75.

For a uniform solid sphere of mass M and radius R:

Gravitational potential at the surface:

$$V_{surface} = -\frac{GM}{R}$$

Gravitational potential at the centre:

$$V_{centre} = -\frac{3GM}{2R}$$

Since $$V = V_{surface} = -\frac{GM}{R}$$:

$$V_{centre} = -\frac{3GM}{2R} = \frac{3}{2}\left(-\frac{GM}{R}\right) = \frac{3}{2}V$$

The gravitational potential at the centre is $$\frac{3V}{2}$$.

The impulse $$\vec J_0 = J_0\,\hat x$$ is applied at a point situated a vertical distance $$h$$ above the centre of mass (COM) of the annular disc. Choose the co-ordinate system   •  origin at the COM,   •  $$x$$-axis along the direction of the impulse (and of the subsequent translation),   •  $$y$$-axis horizontal and perpendicular to the plane of the disc (i.e. along the axle),   •  $$z$$-axis vertically upward.

1. Linear impulse-momentum relation
The impulse imparts a COM velocity $$v = \frac{J_0}{M}$$ along the $$+x$$-direction.

2. Angular impulse-momentum relation
The position vector of the point of application is $$\vec r = h\,\hat z$$. The impulse produces an angular impulse $$\vec \tau \,\Delta t = \vec r \times \vec J_0 = hJ_0\,\hat y.$$ Hence the disc acquires an angular momentum about the $$y$$-axis equal to $$hJ_0$$. For an annular disc the moment of inertia about a diameter (the $$y$$-axis here) is $$I_y = \frac{1}{2}M(a^2+b^2)\;.$$ Therefore the initial angular velocity about the $$y$$-axis is $$\omega = \frac{hJ_0}{I_y} = \frac{2hJ_0}{M\,(a^2+b^2)}.$$

3. Condition for instantaneous pure rolling
For rolling without slipping, the velocity of the point of contact with the floor must be zero. The lowest point is at a distance $$b$$ below the COM, i.e. $$\vec r_c = -\,b\,\hat z$$. Its velocity relative to the ground is $$\vec v_c = \vec v\;+\;\vec\omega \times \vec r_c = \frac{J_0}{M}\,\hat x \;+\;\omega\,\hat y \times (-\,b\,\hat z) = \left(\frac{J_0}{M} - \omega b\right)\hat x.$$ The rolling condition $$\vec v_c = 0$$ therefore gives $$\frac{J_0}{M} = \omega b.$$ Substituting $$\omega$$ from step 2, $$\frac{J_0}{M} = \frac{2hJ_0}{M(a^2+b^2)}\,b \quad\Longrightarrow\quad h = \frac{a^2+b^2}{2b}\;.$$

4. Optimum height
Thus the unique height at which the impulse must be delivered so that the disc starts pure rolling immediately is $$h_m = \frac{a^2+b^2}{2b}\;.$$

5. Verification of the given options
•  Take the limiting case $$a \to 0$$ (a solid disc). Then $$h_m \to \dfrac{b^2}{2b} = \dfrac{b}{2}\;,$$ which coincides with Option A. ✅

Hence, the correct statement is:

Option A which is: For $$\mu \neq 0$$ and $$a \to 0$$, $$h_m = b/2$$.

The acceleration due to gravity at a distance $$r$$ from the centre of the Earth is

$$g(r)=\frac{GM}{r^{2}}$$

where $$G$$ is the gravitational constant and $$M$$ is the mass of the Earth.

For a satellite at height $$h$$ above the Earth’s surface (Earth’s radius $$R$$), the distance from the centre of the Earth is

$$r = R + h$$

Hence, for satellites P and Q we have

$$g_P = \frac{GM}{(R+h_P)^{2}}, \qquad g_Q = \frac{GM}{(R+h_Q)^{2}}$$

Taking their ratio, the common factor $$GM$$ cancels:

$$\frac{g_P}{g_Q} = \frac{(R+h_Q)^{2}}{(R+h_P)^{2}}$$ $$-(1)$$

Given data:

• $$h_P = \frac{R}{3} \;\;\Rightarrow\;\; R+h_P = R + \frac{R}{3} = \frac{4R}{3}$$
• $$\dfrac{g_P}{g_Q} = \dfrac{36}{25}$$

Substitute in equation $$(1)$$:

$$\frac{36}{25} = \frac{(R+h_Q)^{2}}{\left(\dfrac{4R}{3}\right)^{2}}$$

Solve for $$R+h_Q$$:

$$R+h_Q = \frac{6}{5}\left(\frac{4R}{3}\right) = \frac{24R}{15} = \frac{8R}{5}$$

Therefore,

$$h_Q = \frac{8R}{5} - R = \frac{8R}{5} - \frac{5R}{5} = \frac{3R}{5}$$

Hence, the height of satellite Q above the Earth’s surface is $$\displaystyle \frac{3R}{5}$$.

Option A which is: $$3R/5$$

Statement I: True. Earth's rotation affects g: $$g' = g - \omega^2 R\cos^2\lambda$$.

Statement II: The effect is maximum at the equator ($$\cos^2 0° = 1$$) and minimum (zero) at the poles. So Statement II says the opposite, which is false.

The correct answer is Option 4: Statement I is true but Statement II is false.

The escape velocities of planets A and B are in ratio 1:2, and their radii are in ratio 1:3. We need the ratio of their gravitational accelerations.

To begin,

$$ v_e = \sqrt{2gR} $$

Squaring: $$v_e^2 = 2gR$$, which gives $$g = \frac{v_e^2}{2R}$$.

Next,

$$ \frac{v_{eA}^2}{v_{eB}^2} = \frac{g_A R_A}{g_B R_B} $$

Given $$\frac{v_{eA}}{v_{eB}} = \frac{1}{2}$$ and $$\frac{R_A}{R_B} = \frac{1}{3}$$:

$$ \left(\frac{1}{2}\right)^2 = \frac{g_A}{g_B} \times \frac{1}{3} $$

$$ \frac{1}{4} = \frac{g_A}{3g_B} $$

$$ \frac{g_A}{g_B} = \frac{3}{4} $$

The ratio of acceleration due to gravity is $$\dfrac{3}{4}$$.

The correct answer is Option 4: $$\dfrac{3}{4}$$.

A body is released from height $$h = R$$ above the Earth's surface. Using conservation of energy:

$$-\frac{GMm}{R+R} = \frac{1}{2}mv^2 - \frac{GMm}{R}$$

$$\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$$

$$v^2 = \frac{GM}{R}$$

Since $$g = \frac{GM}{R^2}$$, we have $$GM = gR^2$$:

$$v^2 = \frac{gR^2}{R} = gR$$

$$v = \sqrt{gR}$$

This matches option 2.

A spaceship in a circular orbit close to Earth's surface needs additional velocity to escape. We need to find this additional velocity.

First, calculate the orbital velocity for a circular orbit close to Earth's surface.

For a circular orbit at radius $$R$$ (Earth's radius, since the orbit is close to the surface), the gravitational force provides the centripetal force:

$$ \frac{mv_o^2}{R} = \frac{mg}{1} \implies v_o = \sqrt{gR} $$

Substituting $$g = 10$$ m/s$$^2$$ and $$R = 6400$$ km = $$6.4 \times 10^6$$ m:

$$ v_o = \sqrt{10 \times 6.4 \times 10^6} = \sqrt{64 \times 10^6} = 8000\;\text{m/s} = 8\;\text{km/s} $$

Next, calculate the escape velocity from this orbit.

The escape velocity from the surface of Earth is:

$$ v_e = \sqrt{2gR} = \sqrt{2} \times \sqrt{gR} = \sqrt{2} \times v_o $$

$$ v_e = 8\sqrt{2}\;\text{km/s} $$

Now, find the additional velocity needed.

The spaceship is already moving at orbital velocity $$v_o$$. To escape, it needs to reach escape velocity $$v_e$$. The additional velocity required is:

$$ \Delta v = v_e - v_o = 8\sqrt{2} - 8 = 8(\sqrt{2} - 1)\;\text{km/s} $$

Note: The mass of the spaceship ($$2 \times 10^4$$ kg) is not needed for this calculation, as the velocities are independent of mass.

The correct answer is Option 2: $$8(\sqrt{2} - 1)$$ km s$$^{-1}$$.

We need to find the depth $$d$$ below the Earth's surface where the gravitational acceleration equals four times its value at height $$3R$$ above the surface.

To begin,

At a height $$h$$ above the Earth's surface, the acceleration due to gravity is:

$$ g_h = \frac{g}{\left(1 + \frac{h}{R}\right)^2} $$

This follows from Newton's law of gravitation: $$g_h = \frac{GM}{(R+h)^2} = \frac{g R^2}{(R+h)^2}$$, where $$g = GM/R^2$$ is the surface value.

Next,

$$ g_h = \frac{g}{\left(1 + \frac{3R}{R}\right)^2} = \frac{g}{(1+3)^2} = \frac{g}{16} $$

From this,

At depth $$d$$, assuming uniform density of the Earth:

$$ g_d = g\left(1 - \frac{d}{R}\right) $$

This is because only the mass in the sphere of radius $$(R-d)$$ contributes to gravity at that depth (by the shell theorem), and that mass is proportional to $$(R-d)^3$$ while the distance from the centre is $$(R-d)$$, giving $$g_d = g(R-d)/R$$.

Continuing,

$$ g\left(1 - \frac{d}{R}\right) = 4 \times \frac{g}{16} $$

$$ g\left(1 - \frac{d}{R}\right) = \frac{g}{4} $$

Dividing both sides by $$g$$:

$$ 1 - \frac{d}{R} = \frac{1}{4} $$

$$ \frac{d}{R} = 1 - \frac{1}{4} = \frac{3}{4} $$

Now,

$$ d = \frac{3}{4} \times R = \frac{3}{4} \times 6400 = 4800\;\text{km} $$

The depth is 4800 km.

The correct answer is Option 4: 4800 km.

Statement I: Acceleration due to gravity decreases as you go 'up' or 'down' from earth's surface.

Above the surface: $$g_h = g\left(\frac{R}{R+h}\right)^2$$ which decreases with height $$h$$.

Below the surface: $$g_d = g\left(1 - \frac{d}{R}\right)$$ which decreases with depth $$d$$.

So Statement I is correct.

Statement II: Acceleration due to gravity is same at height $$h$$ and depth $$d$$ if $$h = d$$.

At height $$h$$: $$g_h = g\left(1 - \frac{2h}{R}\right)$$ (for small $$h$$)

At depth $$d$$: $$g_d = g\left(1 - \frac{d}{R}\right)$$

For $$g_h = g_d$$: $$1 - \frac{2h}{R} = 1 - \frac{d}{R}$$, which gives $$d = 2h$$.

So $$g$$ is the same when $$d = 2h$$, not when $$h = d$$. Statement II is incorrect.

The correct answer is Option 3: Statement I is correct but Statement II is incorrect.

We are given that Earth has mass $$M_e = 9M_p$$ and radius $$R_e = 2R_p$$, where $$M_p$$ and $$R_p$$ are the mass and radius of planet $$P$$.

The escape velocity on Earth is $$v_e = \sqrt{\frac{2GM_e}{R_e}}$$, and on planet $$P$$ it is $$v_p = \sqrt{\frac{2GM_p}{R_p}}$$.

From the given relations, $$M_p = \frac{M_e}{9}$$ and $$R_p = \frac{R_e}{2}$$. Substituting these in:

$$v_p = \sqrt{\frac{2G \cdot \frac{M_e}{9}}{\frac{R_e}{2}}} = \sqrt{\frac{2GM_e \cdot 2}{9R_e}} = \sqrt{\frac{2}{9}} \cdot \sqrt{\frac{2GM_e}{R_e}}$$

So $$v_p = \frac{\sqrt{2}}{3} \cdot v_e = \frac{v_e}{3}\sqrt{2}$$.

Comparing with the given form $$\frac{v_e}{3}\sqrt{x}$$, we get $$x = 2$$. Hence, the correct answer is $$2$$.

For a satellite in circular orbit, the orbital angular momentum is $$L = mvr$$, where $$v$$ is the orbital velocity and $$r$$ is the distance from the Earth's centre.

For a circular orbit, gravitational force provides centripetal force, so

$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$

which gives $$v = \sqrt{\frac{GM}{r}}$$.

Substituting into the angular momentum expression:

$$L = mr\sqrt{\frac{GM}{r}} = m\sqrt{GMr}$$

So $$L \propto \sqrt{r}$$. Now, the problem says the distance is increased by eight times to its initial value, meaning the new distance is $$r' = r + 8r = 9r$$ (increased by eight times the original).

The new angular momentum is then

$$\frac{L'}{L} = \sqrt{\frac{9r}{r}} = \sqrt{9} = 3$$

So $$L' = 3L$$. Hence, the correct answer is Option 4.

The acceleration due to gravity at the surface of a planet is given by:

$$g = \frac{GM}{R^2}$$

Since $$M = \frac{4}{3}\pi R^3 \rho$$, we get:

$$g = \frac{G \times \frac{4}{3}\pi R^3 \rho}{R^2} = \frac{4}{3}\pi G R \rho$$

So $$g \propto R\rho$$.

For Planet A: Radius = $$R$$, Density = $$\rho$$

$$g_A = \frac{4}{3}\pi G R \rho$$

For Planet B: Radius = $$4R$$, Density = $$\frac{\rho}{3}$$

$$g_B = \frac{4}{3}\pi G (4R) \left(\frac{\rho}{3}\right) = \frac{4}{3}\pi G \times \frac{4R\rho}{3}$$

The ratio:

$$\frac{g_A}{g_B} = \frac{R\rho}{\frac{4R\rho}{3}} = \frac{R\rho \times 3}{4R\rho} = \frac{3}{4}$$

Therefore, $$g_A : g_B = 3 : 4$$.

We need to find the time period of a satellite revolving at a height $$h = R$$ above the Earth's surface, where $$R$$ is the radius of the Earth and $$g = \pi^2$$ m/s$$^2$$. The satellite is at height $$h = R$$ above the Earth's surface so the orbital radius (distance from the center of the Earth) is
$$r = R + h = R + R = 2R$$.

For a satellite of mass $$m$$ in a circular orbit the gravitational force is balanced by the centripetal force:
$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$ which gives the orbital velocity $$v = \sqrt{\frac{GM}{r}}$$. The time period is the circumference divided by velocity:
$$T = \frac{2\pi r}{v} = 2\pi r \cdot \sqrt{\frac{r}{GM}} = 2\pi\sqrt{\frac{r^3}{GM}}$$. At the Earth's surface the gravitational acceleration satisfies $$g = \frac{GM}{R^2}$$, which gives $$GM = gR^2$$. Substituting this into the time period formula yields
$$T = 2\pi\sqrt{\frac{r^3}{gR^2}}$$.

Substituting $$r = 2R$$ gives
$$T = 2\pi\sqrt{\frac{(2R)^3}{gR^2}} = 2\pi\sqrt{\frac{8R^3}{gR^2}} = 2\pi\sqrt{\frac{8R}{g}}$$. With $$g = \pi^2$$ we get
$$T = 2\pi\sqrt{\frac{8R}{\pi^2}} = 2\pi \cdot \frac{\sqrt{8R}}{\pi} = 2\sqrt{8R}$$. Since $$\sqrt{8} = 2\sqrt{2}$$, it follows that
$$T = 2 \cdot 2\sqrt{2R} = 4\sqrt{2R}$$ which can also be written as
$$T = \sqrt{16 \cdot 2R} = \sqrt{32R}$$.

The correct answer is Option 3: $$\sqrt{32R}$$.

We need to find the weight of a body at an altitude of 3200 km, given that its weight at the earth's surface is 18 N.

The acceleration due to gravity at a height $$h$$ above the earth's surface is: $$g' = \frac{g}{\left(1 + \frac{h}{R_e}\right)^2}$$ where $$g$$ is the surface gravity and $$R_e$$ is the radius of the earth.

At $$h = 3200$$ km and $$R_e = 6400$$ km, $$\frac{h}{R_e} = \frac{3200}{6400} = \frac{1}{2}$$, so $$g' = \frac{g}{\left(1 + \frac{1}{2}\right)^2} = \frac{g}{\left(\frac{3}{2}\right)^2} = \frac{g}{\frac{9}{4}} = \frac{4g}{9}$$.

Since weight is proportional to $$g$$, $$W' = W \times \frac{g'}{g} = 18 \times \frac{4}{9} = 8 \text{ N}$$.

The correct answer is Option (4): 8 N.

The weight of a body on the surface of the earth is $$100$$ N. We need to find the gravitational force at height $$h = \dfrac{R}{4}$$ from the surface.

$$g' = g \cdot \dfrac{R^2}{(R+h)^2}$$

$$g' = g \cdot \dfrac{R^2}{(R + R/4)^2} = g \cdot \dfrac{R^2}{(5R/4)^2} = g \cdot \dfrac{R^2}{25R^2/16} = g \cdot \dfrac{16}{25}$$

$$W' = mg' = mg \cdot \dfrac{16}{25} = 100 \times \dfrac{16}{25} = 64 \text{ N}$$

The gravitational force at height $$R/4$$ is $$64$$ N, which corresponds to Option A.

Two particles of equal mass $$m$$ move in a circle of radius $$r$$ under their mutual gravitational attraction. They are diametrically opposite on the circle, so the distance between them is $$2r$$.

$$F = \frac{Gm^2}{(2r)^2} = \frac{Gm^2}{4r^2}$$

This gravitational force provides the centripetal force for each particle moving in a circle of radius $$r$$:

$$\frac{Gm^2}{4r^2} = \frac{mv^2}{r}$$

$$\frac{Gm}{4r^2} = \frac{v^2}{r}$$

$$v^2 = \frac{Gm}{4r}$$

$$v = \sqrt{\frac{Gm}{4r}}$$

The correct answer is Option D: $$\sqrt{\dfrac{Gm}{4r}}$$.

Two satellites of masses $$m$$ and $$3m$$ revolve in circular orbits of radii $$r$$ and $$3r$$ respectively. We need to find the ratio of their orbital speeds.

First, derive the orbital speed formula.

For a satellite in circular orbit, the gravitational force provides the centripetal force:

$$ \frac{GMm_s}{r_s^2} = \frac{m_s v^2}{r_s} $$

where $$M$$ is Earth's mass, $$m_s$$ is the satellite mass, and $$r_s$$ is the orbital radius. Notice that the satellite mass $$m_s$$ cancels:

$$ v = \sqrt{\frac{GM}{r_s}} $$

This is a crucial result: orbital speed depends only on the orbital radius, not on the satellite's mass.

Next, calculate the ratio of orbital speeds.

For satellite 1 (orbit radius $$r$$): $$v_1 = \sqrt{\frac{GM}{r}}$$

For satellite 2 (orbit radius $$3r$$): $$v_2 = \sqrt{\frac{GM}{3r}}$$

$$ \frac{v_1}{v_2} = \sqrt{\frac{GM/r}{GM/3r}} = \sqrt{\frac{3r}{r}} = \sqrt{3} $$

The ratio of orbital speeds is $$\sqrt{3} : 1$$.

The correct answer is Option 1: $$\sqrt{3} : 1$$.

A planet has mass $$M_p = 2M_e$$ and average density $$\rho_p = \rho_e$$. We need the weight of an object on this planet.

Since density is the same: $$\rho = \dfrac{M}{\frac{4}{3}\pi R^3}$$.

$$\dfrac{M_p}{\frac{4}{3}\pi R_p^3} = \dfrac{M_e}{\frac{4}{3}\pi R_e^3}$$.

$$\dfrac{R_p^3}{R_e^3} = \dfrac{M_p}{M_e} = 2 \implies R_p = 2^{1/3}\,R_e$$.

$$\dfrac{g_p}{g_e} = \dfrac{M_p}{M_e} \cdot \dfrac{R_e^2}{R_p^2} = 2 \cdot \dfrac{R_e^2}{2^{2/3}\,R_e^2} = 2 \cdot 2^{-2/3} = 2^{1/3}$$.

$$W_p = \dfrac{g_p}{g_e} \cdot W = 2^{1/3}\,W$$.

The correct answer is Option A: $$2^{1/3}W$$.

We need to find the weight of a body at depth $$d = R/2$$ below Earth's surface, given its surface weight is 200 N.

To begin,

Assuming Earth has uniform mass density, the acceleration due to gravity at depth $$d$$ below the surface is:

$$ g_d = g\left(1 - \frac{d}{R}\right) $$

This is because, by the shell theorem, only the spherical mass at radii less than $$(R - d)$$ contributes to gravity. This inner mass is proportional to $$(R-d)^3$$, and the gravitational acceleration at distance $$(R-d)$$ from the centre is $$g_d = g(R-d)/R$$.

Next,

$$ W' = W \times \frac{g_d}{g} = W\left(1 - \frac{d}{R}\right) = 200\left(1 - \frac{R/2}{R}\right) = 200\left(1 - \frac{1}{2}\right) = 200 \times \frac{1}{2} = 100\;\text{N} $$

The weight at depth $$R/2$$ is 100 N.

The correct answer is Option 2: 100 N.

We need to identify the incorrect statement from the given options.

Option A: The speed of satellite in a given circular orbit remains constant.

This is correct. In a circular orbit, the gravitational force is always perpendicular to the velocity, so it only changes the direction, not the magnitude of velocity. The speed remains constant.

Option B: For a planet revolving around the sun in an elliptical orbit, the total energy of the planet remains constant.

This is correct. The gravitational force is conservative, so the total mechanical energy (kinetic + potential) is conserved in any orbit.

Option C: The linear speed of a planet revolving around the sun remains constant.

This is INCORRECT. A planet moves in an elliptical orbit (Kepler's First Law). By Kepler's Second Law (equal areas in equal times), the planet moves faster at perihelion (closest to the sun) and slower at aphelion (farthest from the sun). The linear speed varies throughout the orbit.

Option D: When a body falls towards earth, the displacement of earth towards the body is negligible.

This is correct. By Newton's Third Law, the earth also accelerates toward the body, but since the earth's mass is enormous, its acceleration (and hence displacement) is negligibly small.

The incorrect statement is Option C.

The assertion states that a pendulum clock when taken to Mount Everest becomes fast.

It is reasoned that the value of g (acceleration due to gravity) is less at Mount Everest than on the surface of the earth.

Upon analysis, at higher altitude such as Mount Everest the gravitational acceleration decreases, so $$g$$ is indeed less there. Therefore, the reason is correct.

Analyzing the assertion further, we recall that the time period of a simple pendulum is:

$$T = 2\pi\sqrt{\frac{L}{g}}$$

Since $$g$$ decreases at Mount Everest, the time period $$T$$ increases (the pendulum oscillates more slowly). A longer time period means each swing takes more time, so the clock runs slow rather than fast, making the assertion incorrect.

The assertion is incorrect but the reason is correct, so the correct answer is Option D.

Consider Statement I, which says acceleration due to gravity is different at different places on the surface of Earth. This is true because the Earth is not a perfect sphere (it is an oblate spheroid), so the radius varies with latitude. Additionally, the rotational effect of the Earth reduces effective gravity at the equator compared to the poles. The effective gravity is given by $$g' = g - R\omega^2\cos^2\lambda$$, where $$\lambda$$ is the latitude. So Statement I is correct.

Now consider Statement II, which says acceleration due to gravity increases as we go down below the Earth's surface. Below the surface, the acceleration due to gravity is given by

$$g_d = g\left(1 - \frac{d}{R}\right)$$

where $$d$$ is the depth. As depth increases, $$g_d$$ actually decreases (reaching zero at the center of the Earth). So Statement II is false.

Hence, Statement I is true but Statement II is false.

We need to evaluate two statements about satellite energy.

Recall the energy relations for a satellite in orbit. For a satellite of mass $$m$$ orbiting Earth at radius $$r$$:

- Kinetic Energy: $$KE = \frac{GMm}{2r}$$

- Potential Energy: $$PE = -\frac{GMm}{r}$$

- Total Energy: $$E = KE + PE = -\frac{GMm}{2r}$$

Note that $$E$$ is negative.

Check Statement I — "PE = E/2": $$E/2 = -\frac{GMm}{4r}$$

But $$PE = -\frac{GMm}{r}$$

Since $$PE \neq E/2$$ (in fact, $$PE = 2E$$), Statement I is incorrect.

Check Statement II — "KE = |E|/2": $$|E| = \frac{GMm}{2r}$$

$$|E|/2 = \frac{GMm}{4r}$$

But $$KE = \frac{GMm}{2r} = |E|$$

Since $$KE = |E| \neq |E|/2$$, Statement II is incorrect.

The correct relationships are: $$PE = 2E$$ and $$KE = -E = |E|$$.

Since both statements are incorrect, the correct answer is Option D: Both Statement I and Statement II are incorrect.

We need to find the gravitational potential at $$r = 3$$ cm, given the gravitational field $$g = -\frac{K}{r^2}$$ and $$V(r=2) = 10$$ J/kg.

Relation between field and potential:

$$V(r) - V(r_0) = -\int_{r_0}^{r} g \, dr$$

Calculating the potential difference:

$$V(3) - V(2) = -\int_{2}^{3} \left(-\frac{K}{r^2}\right) dr = K\int_{2}^{3} \frac{dr}{r^2}$$

$$= K\left[-\frac{1}{r}\right]_{2}^{3} = K\left(-\frac{1}{3} + \frac{1}{2}\right) = K \times \frac{1}{6}$$

Substituting values:

$$K = 6$$ J cm/kg (note: distances are in cm)

$$V(3) = V(2) + \frac{K}{6} = 10 + \frac{6}{6} = 10 + 1 = 11$$ J/kg

The gravitational potential at $$r = 3$$ cm is $$11$$ J/kg.

The correct answer is Option 2: $$11$$.

We need to find the time period of a simple pendulum when taken to a height $$R$$ (equal to Earth's radius) above the Earth's surface.

Find the acceleration due to gravity at height $$h = R$$.

At height $$h$$ above the Earth's surface:

$$ g' = g \cdot \frac{R^2}{(R + h)^2} $$

At $$h = R$$:

$$ g' = g \cdot \frac{R^2}{(R + R)^2} = g \cdot \frac{R^2}{4R^2} = \frac{g}{4} $$

Find the new time period.

The time period of a simple pendulum is $$T = 2\pi\sqrt{\frac{L}{g}}$$.

At height $$R$$:

$$ T' = 2\pi\sqrt{\frac{L}{g'}} = 2\pi\sqrt{\frac{L}{g/4}} = 2\pi\sqrt{\frac{4L}{g}} = 2 \cdot 2\pi\sqrt{\frac{L}{g}} = 2T $$

Find the value of $$x$$.

Since $$T' = xT$$ and $$T' = 2T$$:

$$ x = 2 $$

The correct answer is Option B: $$2$$.

Escape velocity: $$v_e = \sqrt{\frac{2GM}{R}}$$

$$\frac{v_p}{v_e} = \sqrt{\frac{M_p}{M_e} \times \frac{R_e}{R_p}} = \sqrt{\frac{16}{1} \times \frac{1}{4}} = \sqrt{4} = 2$$

Ratio = $$2 : 1$$

Two identical particles of mass $$m$$ go round a circle of radius $$a$$. They are diametrically opposite, so the distance between them is $$2a$$.

The gravitational force between them provides the centripetal force:

$$\frac{Gm^2}{(2a)^2} = m\omega^2 a$$

$$\frac{Gm}{4a^2} = \omega^2 a$$

$$\omega^2 = \frac{Gm}{4a^3}$$

$$\omega = \sqrt{\frac{Gm}{4a^3}}$$

This matches option 3.

Assertion A: Earth has atmosphere whereas moon doesn't have any atmosphere.

This is TRUE. Earth has a significant atmosphere while the moon has virtually no atmosphere.

Reason R: The escape velocity on moon is very small as compared to that on earth.

This is TRUE. The escape velocity on the moon is about $$2.38\;\text{km/s}$$, while on Earth it is about $$11.2\;\text{km/s}$$.

The escape velocity determines whether gas molecules can escape the gravitational pull of a celestial body. Since the moon's escape velocity is very low, gas molecules (which have thermal velocities comparable to or exceeding the moon's escape velocity) easily escape from the moon's surface. On Earth, the higher escape velocity retains the atmosphere.

Therefore, R is the correct explanation of A.

The correct answer is Both A and R are correct and R is the correct explanation of A.

Distance of Earth from Sun = $$1.5 \times 10^6$$ km and the period of the imaginary planet is 2.83 years.

According to Kepler’s Third Law, $$T^2 \propto R^3 \implies \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{R_2}{R_1}\right)^3$$

Substituting Earth’s values $$T_1 = 1$$ year and $$R_1 = 1.5 \times 10^6$$ km into this relation gives $$\left(\frac{2.83}{1}\right)^2 = \left(\frac{R_2}{1.5 \times 10^6}\right)^3$$

Noting that $$2.83 \approx \sqrt{8}$$ so $$(2.83)^2 \approx 8$$ leads to $$8 = \left(\frac{R_2}{1.5 \times 10^6}\right)^3$$ and hence $$\frac{R_2}{1.5 \times 10^6} = \sqrt[3]{8} = 2$$

It follows that $$R_2 = 2 \times 1.5 \times 10^6 = 3 \times 10^6 \text{ km}$$

The correct answer is Option C: $$3 \times 10^6$$ km.

For a uniform solid sphere of mass $$M$$ and radius $$R$$ the escape speed from its surface is
$$v = \sqrt{\dfrac{2GM}{R}} \; \; -(1)$$

Initial data
Both stars have the same initial radius $$R$$.
Masses are related by $$M_B = 2M_A$$.
Since mass $$M = \dfrac{4}{3}\pi R^{3}\rho$$ for a sphere of density $$\rho$$, we have
$$\dfrac{\rho_B}{\rho_A} = \dfrac{M_B}{M_A} = 2 \;\;\; \Rightarrow \;\;\; \rho_B = 2\rho_A$$.

1. Final parameters of star A
Its radius is halved: $$R_A' = \dfrac{R}{2}$$.
Its density is still $$\rho_A$$, hence its final mass is
$$M_A' = \dfrac{4}{3}\pi \left(\dfrac{R}{2}\right)^3 \rho_A = \dfrac{1}{8}\left(\dfrac{4}{3}\pi R^{3}\rho_A\right)=\dfrac{M_A}{8}$$.
Mass lost by A
$$\Delta M = M_A - M_A' = M_A\left(1-\dfrac{1}{8}\right)=\dfrac{7M_A}{8}$$.

2. Final parameters of star B
The whole mass $$\Delta M$$ forms a spherical shell of density $$\rho_A$$ around B.
Let the outer radius after coating be $$R_B'$$.
Mass of the shell:
$$\dfrac{4}{3}\pi\bigl(R_B'^{3}-R^{3}\bigr)\rho_A = \dfrac{7M_A}{8} = \dfrac{7}{8}\left(\dfrac{4}{3}\pi R^{3}\rho_A\right).$$
Cancelling the common factor $$\dfrac{4}{3}\pi\rho_A$$ gives
$$R_B'^{3}-R^{3}= \dfrac{7}{8}R^{3}\;\;\;\Rightarrow\;\;\;R_B'^{3} =\left(1+\dfrac{7}{8}\right)R^{3}= \dfrac{15}{8}R^{3}.$$

Therefore
$$R_B' = R\left(\dfrac{15}{8}\right)^{1/3} \;\;\; -(2)$$
Final mass of B:
$$M_B' = M_B + \Delta M = 2M_A + \dfrac{7M_A}{8}= \dfrac{23M_A}{8} \;\;\; -(3)$$

3. Escape speeds after interaction
From (1):
Escape speed from A:
$$v_A = \sqrt{\dfrac{2G M_A'}{R_A'}} = \sqrt{\dfrac{2G\left(\dfrac{M_A}{8}\right)}{R/2}} = \sqrt{\dfrac{G M_A}{2R}} \;\;\; -(4)$$
Escape speed from B:
$$v_B = \sqrt{\dfrac{2G M_B'}{R_B'}} = \sqrt{\dfrac{2G\left(\dfrac{23M_A}{8}\right)} {R\left( \dfrac{15}{8}\right)^{1/3}}} \;\;\; -(5)$$

4. Required ratio
From (4) and (5):
$$\left(\dfrac{v_B}{v_A}\right)^2 = \dfrac{\,\dfrac{2G(23M_A/8)}{R(15/8)^{1/3}}\,}{\,\dfrac{G M_A}{2R}\,} = \left(\dfrac{2\cdot23}{8}\right)\cdot2\cdot\left(\dfrac{8}{15}\right)^{1/3} = \dfrac{23}{2}\left(\dfrac{8}{15}\right)^{1/3}.$$

Note that $$\left(\dfrac{8}{15}\right)^{1/3} = \dfrac{2^{3/3}}{15^{1/3}}= \dfrac{2}{15^{1/3}}.$$ Hence
$$\left(\dfrac{v_B}{v_A}\right)^2 = \dfrac{23}{2}\cdot\dfrac{2}{15^{1/3}} = \dfrac{23}{15^{1/3}}.$$

Taking the square root,
$$\dfrac{v_B}{v_A}= \dfrac{\sqrt{23}}{15^{1/6}} = \dfrac{\sqrt{10\,n}}{15^{1/6}},$$ which matches the given form $$\sqrt{\dfrac{10n}{15^{1/3}}}$$ because $$1/15^{1/6} = \sqrt{1/15^{1/3}}.$$

Equating the numerators:
$$\sqrt{10\,n} = \sqrt{23}\;\;\;\Rightarrow\;\;\;10n = 23 \;\;\;\Rightarrow\;\;\;n = 2.3.$$

The required value is 2.30.

We need to find $$\alpha$$ such that the acceleration due to gravity at height $$h$$ above the Earth's surface equals the acceleration due to gravity at depth $$\alpha h$$ below the surface, given $$h \ll R_E$$.

The acceleration due to gravity at a height $$h$$ above the surface (for $$h \ll R_E$$) is approximately $$g_h = g\left(1 - \dfrac{2h}{R_E}\right)$$.

The acceleration due to gravity at a depth $$d$$ below the surface is $$g_d = g\left(1 - \dfrac{d}{R_E}\right)$$.

Setting these equal with $$d = \alpha h$$: $$g\left(1 - \dfrac{2h}{R_E}\right) = g\left(1 - \dfrac{\alpha h}{R_E}\right)$$.

Cancelling $$g$$ and simplifying: $$-\dfrac{2h}{R_E} = -\dfrac{\alpha h}{R_E}$$, which gives $$\alpha = 2$$.

Hence, the correct answer is 2.

The radii of the circular orbits of the two satellites are given by $$R_1 = 3200$$ km for satellite $$S_1$$ and $$R_2 = 800$$ km for satellite $$S_2$$.

Since the orbital speed of a satellite in a circular orbit of radius $$R$$ is given by $$v = \sqrt{\frac{GM}{R}}$$, where $$M$$ is the mass of the planet, we can derive the ratio of their speeds.

Substituting the given radii into the speed formula yields $$\frac{v_1}{v_2} = \sqrt{\frac{R_2}{R_1}} = \sqrt{\frac{800}{3200}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\,.$$

Comparing this result with the form $$\frac{1}{x}$$, we have $$\frac{v_1}{v_2} = \frac{1}{2}$$, which implies $$x = 2$$. Therefore, the value of $$x$$ is 2.

A body is projected vertically upwards with velocity $$v = \dfrac{v_e}{3}$$ (one-third of escape velocity), and we need to find the maximum height attained. The escape velocity from Earth's surface is $$v_e = \sqrt{2gR}$$ where $$R = 6400 \text{ km}$$ and $$g = 10 \text{ m s}^{-2}$$.

At the surface, the total energy equals kinetic plus potential energy: $$\dfrac{1}{2}mv^2 - \dfrac{GMm}{R} = -\dfrac{GMm}{R+h}$$.

Since $$v = \dfrac{v_e}{3} = \dfrac{\sqrt{2gR}}{3}$$, substituting into the energy equation yields $$\dfrac{1}{2}m \cdot \dfrac{2gR}{9} = \dfrac{GMm}{R} - \dfrac{GMm}{R+h}$$.

Using $$GM = gR^2$$, this gives $$\dfrac{mgR}{9} = mgR\left(1 - \dfrac{R}{R+h}\right) = mgR \cdot \dfrac{h}{R+h}$$. From this, $$\dfrac{1}{9} = \dfrac{h}{R+h}$$ so that $$R + h = 9h$$ and $$R = 8h$$, which leads to $$h = \dfrac{R}{8} = \dfrac{6400}{8} = 800 \text{ km}$$.

The correct answer is Option A: $$800 \text{ km}$$.

We know that the time period of a simple pendulum is $$T = 2\pi\sqrt{\frac{l}{g'}}$$, where $$g'$$ is the effective gravitational acceleration at that location. Since both clocks use identical pendulums (same length $$l$$), the ratio of their time periods depends only on the ratio of gravitational accelerations.

On the Earth's surface, $$g_{\text{earth}} = g$$, and at height $$h$$ above the surface, $$g_h = g\left(\frac{R_E}{R_E + h}\right)^2$$.

The ratio of time periods gives us $$\frac{T_2}{T_1} = \sqrt{\frac{g}{g_h}} = \frac{R_E + h}{R_E}$$.

Substituting $$T_1 = 4$$ s and $$T_2 = 6$$ s, we get $$\frac{6}{4} = \frac{R_E + h}{R_E}$$, which gives $$\frac{3}{2} = \frac{R_E + h}{R_E}$$.

So $$R_E + h = \frac{3}{2}R_E$$, which means $$h = \frac{1}{2}R_E = \frac{1}{2} \times 6400 = 3200$$ km.

Hence, the correct answer is Option C.

Let us analyze both the Assertion and Reason carefully.

Assertion A: "If we move from poles to equator, the direction of acceleration due to gravity of earth always points towards the center of earth without any variation in its magnitude."

Analysis of Assertion A:

The effective gravitational acceleration at a latitude $$\lambda$$ on the rotating Earth is:

$$g_{eff} = g - R\omega^2\cos^2\lambda$$

where $$g$$ is the gravitational acceleration without rotation, $$R$$ is the radius of the Earth, and $$\omega$$ is the angular velocity of Earth's rotation.

Variation in magnitude: At the poles ($$\lambda = 90°$$), $$g_{eff} = g$$. At the equator ($$\lambda = 0°$$), $$g_{eff} = g - R\omega^2$$, which is smaller. So the magnitude clearly varies as we move from poles to equator. This part of the assertion is false.

Variation in direction: The centrifugal acceleration at latitude $$\lambda$$ is directed perpendicular to Earth's rotation axis (i.e., radially outward from the axis, not from the center of Earth). Only at the equator and at the poles does the centrifugal acceleration align radially with respect to Earth's center. At intermediate latitudes, the centrifugal force has a component tangential to the Earth's surface, causing the net effective gravity to deviate slightly from pointing towards the center. So the direction also varies, making the assertion false.

Reason R: "At the equator, the direction of acceleration due to gravity is towards the center of earth."

Analysis of Reason R:

At the equator, the centrifugal acceleration $$R\omega^2$$ is directed radially outward from Earth's center (since at the equator, the direction perpendicular to the rotation axis coincides with the radial direction from Earth's center). The gravitational pull is directed radially inward towards the center. Since both are along the same line (radial direction), the resultant effective gravity still points towards the center of Earth. Only the magnitude is reduced. This makes Reason R true.

Conclusion:

The Assertion is false (both magnitude and direction vary) but the Reason is true (at the equator specifically, the direction is towards the center).

Hence, the correct answer is Option D.

We need to find the percentage change in acceleration due to gravity when the radius of Earth shrinks by 2% while mass remains constant. Using the relation

$$g = \frac{GM}{R^2}$$

and taking its differential yields

$$\frac{\Delta g}{g} = -2\frac{\Delta R}{R}.$$

Since the radius decreases by 2%, i.e., $$\frac{\Delta R}{R} = -2\%,$$ it follows that

$$\frac{\Delta g}{g} = -2 \times (-2\%) = +4\%.$$

Equivalently, because $$g \propto \frac{1}{R^2},$$ a reduction in R increases g, so a 2% decrease in the radius brings about approximately a 4% increase in g. Hence, the correct answer is Option D: increase by 4%.

We need to find the acceleration due to gravity at point $$P$$ which is at a height equal to the diameter of the Earth above the surface. The height of point $$P$$ above the surface is the diameter of the Earth, $$2R$$, where $$R$$ is the radius of the Earth, so the distance from the center of the Earth is $$d = R + h = R + 2R = 3R$$.

The acceleration due to gravity at a distance $$d$$ from the center (when $$d > R$$) is given by $$g' = \frac{g R^2}{d^2}$$. Substituting $$d = 3R$$ yields $$g' = \frac{g R^2}{(3R)^2} = \frac{g R^2}{9R^2} = \frac{g}{9}$$.

The correct answer is Option D: $$\frac{g}{9}$$.

We need to find the percentage decrease in weight when a rocket is taken to a height of $$h = 32 \text{ km}$$ above the Earth's surface. The radius of Earth is $$R = 6400 \text{ km}$$.

We start by expressing the acceleration due to gravity at height $$h$$ above the surface as:

$$g' = g\left(\frac{R}{R + h}\right)^2$$

Since $$h \ll R$$ ($$32 \text{ km} \ll 6400 \text{ km}$$), we can use the binomial approximation:

$$g' \approx g\left(1 - \frac{2h}{R}\right)$$

Next, the fractional decrease in gravitational acceleration is given by:

$$\frac{\Delta g}{g} = \frac{g - g'}{g} = \frac{2h}{R}$$

Substituting the values, we get $$\frac{\Delta g}{g} = \frac{2 \times 32}{6400} = \frac{64}{6400} = \frac{1}{100} = 0.01$$

Finally, converting this to a percentage gives:

$$\text{Percentage decrease} = \frac{\Delta g}{g} \times 100 = 0.01 \times 100 = 1\%$$

Since weight is directly proportional to $$g$$, the percentage decrease in weight is also $$1\%$$.

The correct answer is Option A: $$1\%$$.

We are given two planets A and B with equal mass, revolving in circular orbits with periods $$T_A = 2T_B$$. According to Kepler’s third law, $$T^2 \propto r^3$$, or more precisely $$T^2 = \frac{4\pi^2}{GM}r^3$$, so for the two planets we have $$\frac{T_A^2}{T_B^2} = \frac{r_A^3}{r_B^3}$$.

Substituting $$T_A = 2T_B$$ into this relation gives $$\frac{(2T_B)^2}{T_B^2} = \frac{r_A^3}{r_B^3}$$, which simplifies to $$\frac{4T_B^2}{T_B^2} = \frac{r_A^3}{r_B^3}$$, or $$4 = \frac{r_A^3}{r_B^3}$$. Therefore, $$r_A^3 = 4r_B^3$$. The correct answer is Option C.

We need to find the ratio of total mechanical energy of satellites A and B. The total mechanical energy of a satellite of mass $$m$$ in a circular orbit of radius $$r$$ around Earth (mass $$M$$) is given by $$E = -\frac{GMm}{2r}$$.

For satellite A, which has mass $$4m_0$$ and orbits at radius $$3r$$, the energy becomes $$E_A = -\frac{GM(4m_0)}{2(3r)} = -\frac{4GMm_0}{6r} = -\frac{2GMm_0}{3r}$$. Similarly, for satellite B with mass $$3m_0$$ at radius $$4r$$, one finds $$E_B = -\frac{GM(3m_0)}{2(4r)} = -\frac{3GMm_0}{8r}$$.

Taking the ratio yields $$\frac{E_A}{E_B} = \frac{-\frac{2GMm_0}{3r}}{-\frac{3GMm_0}{8r}} = \frac{2}{3} \times \frac{8}{3} = \frac{16}{9}$$, so the correct answer is Option B: $$16:9$$.

A body of mass $$m$$ is projected vertically upward with velocity $$\lambda v_e$$ (where $$v_e$$ is escape velocity and $$\lambda < 1$$). We need to determine its maximum height measured from the centre of the earth.

Since energy is conserved, at the surface (distance $$R$$ from the centre) the sum of kinetic and gravitational potential energy equals the total energy at the maximum height $$h$$ (where the velocity becomes zero):

$$\frac{1}{2}m(\lambda v_e)^2 - \frac{GMm}{R} = 0 - \frac{GMm}{h}$$

We know that $$v_e = \sqrt{\frac{2GM}{R}}$$, so $$v_e^2 = \frac{2GM}{R}$$, which implies $$GM = \frac{v_e^2 R}{2}$$.

Substituting this expression for $$GM$$ into the energy equation gives

$$\frac{1}{2}m\lambda^2 v_e^2 - \frac{GMm}{R} = -\frac{GMm}{h}$$

Next, dividing through by $$m$$ yields

$$\frac{1}{2}\lambda^2 v_e^2 - \frac{GM}{R} = -\frac{GM}{h}$$

Substituting $$GM = \frac{v_e^2 R}{2}$$ once more into this result gives

$$\frac{1}{2}\lambda^2 v_e^2 - \frac{v_e^2}{2} = -\frac{v_e^2 R}{2h}$$

Dividing both sides by $$\frac{v_e^2}{2}$$ leads to

$$\lambda^2 - 1 = -\frac{R}{h}$$

Rewriting this expression gives

$$1 - \lambda^2 = \frac{R}{h}$$

Therefore, solving for $$h$$ yields

$$h = \frac{R}{1 - \lambda^2}$$

This represents the maximum distance from the centre of the earth.

Answer: Option B: $$\dfrac{R}{1-\lambda^2}$$

We need to find the height $$h$$ from the surface of the Earth where the weight becomes $$\frac{1}{3}$$ of its weight on the surface.

Write the formula for gravitational acceleration at height $$h$$: $$ g' = \frac{g}{\left(1 + \frac{h}{R}\right)^2} $$

Set up the equation: for weight to become $$\frac{1}{3}$$:

$$ \frac{g}{3} = \frac{g}{\left(1 + \frac{h}{R}\right)^2} $$

$$ \left(1 + \frac{h}{R}\right)^2 = 3 $$

$$ 1 + \frac{h}{R} = \sqrt{3} $$

Solve for $$h$$: $$ \frac{h}{R} = \sqrt{3} - 1 $$

$$ h = R(\sqrt{3} - 1) = 6400 \times (1.732 - 1) $$

$$ h = 6400 \times 0.732 = 4684.8 \approx 4685 \text{ km} $$

Therefore, the correct answer is Option B.

We use Kepler’s Third Law, which states that the square of the orbital period is proportional to the cube of the semi-major axis (orbital radius): $$T^2 \propto R^3$$.

For the original distance $$R$$ with period $$T_1 = 1$$ year: $$T_1^2 \propto R^3$$. For the new distance $$3R$$ with period $$T_2$$: $$T_2^2 \propto (3R)^3$$.

Taking the ratio gives $$\frac{T_2^2}{T_1^2} = \frac{(3R)^3}{R^3} = 27$$, so $$T_2^2 = 27 \times T_1^2 = 27$$ and hence $$T_2 = \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} \text{ years}$$.

The duration of the year when the distance becomes $$3R$$ is $$3\sqrt{3}$$ years. The correct answer is Option B.

We need to find the length of a seconds pendulum at height $$h = 2R$$ from Earth's surface.

Find the acceleration due to gravity at height h = 2R.

Use the time period formula for a seconds pendulum.

A seconds pendulum has time period $$T = 2 \text{ s}$$:

Substitute $$g = \pi^2 \text{ m s}^{-2}$$.

The correct answer is Option D: $$\dfrac{1}{9} \text{ m}$$.

The time period of a satellite is 7 hours. The radius is increased to 3 times its previous value. Find the new time period.

$$T^2 \propto r^3$$, so:

$$\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3} = \left(\frac{r_2}{r_1}\right)^3 = 3^3 = 27$$

$$T_2 = T_1\sqrt{27} = 7 \times 3\sqrt{3} = 21\sqrt{3} \approx 21 \times 1.732 \approx 36.37 \text{ hours}$$

The approximate new time period is 36 hours.

Hence, the correct answer is Option A: 36 hours.

Three identical particles A, B, and C each of mass $$100 \text{ kg}$$ are placed in a straight line with $$AB = BC = 13 \text{ m}$$. Particle P of the same mass is placed at $$13 \text{ m}$$ from B on the perpendicular bisector of AC.

We place B at the origin, A at $$(-13, 0)$$, C at $$(13, 0)$$, and P at $$(0, 13)$$.

The distances from P to each particle are found to be $$BP = 13 \text{ m}$$, $$AP = \sqrt{13^2 + 13^2} = 13\sqrt{2} \text{ m}$$, and $$CP = \sqrt{13^2 + 13^2} = 13\sqrt{2} \text{ m}$$.

The gravitational force exerted on P by B is given by $$F_B = \frac{G \times 100 \times 100}{13^2} = \frac{10000G}{169}$$, which acts along BP (vertically upward in this setup).

The gravitational force exerted on P by A is $$F_A = \frac{G \times 100 \times 100}{(13\sqrt{2})^2} = \frac{10000G}{338}$$.

Since AP makes an angle of $$45°$$ with BP, the component of this force along BP is $$F_A \cos 45° = \frac{10000G}{338} \times \frac{1}{\sqrt{2}} = \frac{10000G}{338\sqrt{2}}$$.

By symmetry, the force from C has the same component along BP, and the horizontal components of forces from A and C cancel out, so the net force along BP is: $$F = F_B + 2F_A\cos 45°$$ $$F = \frac{10000G}{169} + 2 \times \frac{10000G}{338\sqrt{2}}$$ $$F = \frac{10000G}{169} + \frac{10000G}{169\sqrt{2}}$$ $$F = \frac{10000G}{169}\left(1 + \frac{1}{\sqrt{2}}\right)$$ $$F = \frac{10000G}{169} \times 1.707$$ $$F \approx 59.2 \times 1.707 \times G$$ $$F \approx 101G \approx 100G$$

Hence, the correct answer is Option B.

Two objects of equal mass $$m$$ are placed at distance $$d$$ apart. The initial gravitational force is $$F = \frac{Gm^2}{d^2}$$.

Find the new masses after transferring one-third of one object's mass.

One-third mass of one object ($$m/3$$) is transferred to the other.

New mass of first object: $$m - \frac{m}{3} = \frac{2m}{3}$$

New mass of second object: $$m + \frac{m}{3} = \frac{4m}{3}$$

Calculate the new gravitational force.

$$F' = \frac{G \times \frac{2m}{3} \times \frac{4m}{3}}{d^2} = \frac{G \times \frac{8m^2}{9}}{d^2} = \frac{8}{9} \times \frac{Gm^2}{d^2} = \frac{8}{9}F$$

The correct answer is Option C.

An object is at a distance $$\frac{5}{4}R$$ from the centre of the Earth, i.e., at height $$h = \frac{5R}{4} - R = \frac{R}{4}$$ above the surface.

Find the acceleration due to gravity at this height.

Since the object is above the surface, at distance $$r = \frac{5R}{4}$$ from the centre:

Calculate the percentage decrease in weight.

The correct answer is Option A: $$36\%$$.

We have an object of mass $$m = 1$$ kg taken from the surface of the Earth to a height $$h = 3R$$, where $$R = 6400$$ km is the radius of the Earth.

The gravitational potential energy at a distance $$r$$ from the centre of the Earth is $$U = -\frac{GMm}{r}$$. At the surface ($$r = R$$), the potential energy is $$U_i = -\frac{GMm}{R}$$. At height $$3R$$ above the surface ($$r = 4R$$), the potential energy is $$U_f = -\frac{GMm}{4R}$$.

The gain in potential energy is: $$\Delta U = U_f - U_i = -\frac{GMm}{4R} + \frac{GMm}{R} = \frac{GMm}{R}\left(1 - \frac{1}{4}\right) = \frac{3GMm}{4R}$$

Now, since $$g = \frac{GM}{R^2}$$, we have $$GM = gR^2$$. Substituting: $$\Delta U = \frac{3gR^2 m}{4R} = \frac{3}{4}mgR$$

Plugging in values: $$\Delta U = \frac{3}{4} \times 1 \times 10 \times 6400 \times 10^3 = \frac{3}{4} \times 64 \times 10^6 = 48 \times 10^6 \text{ J} = 48 \text{ MJ}$$.

Hence, the correct answer is Option 1.

Concept:
Total gravitational potential energy = sum of energies of all interacting pairs:

$$U=-\sum_{ }^{ }\frac{Gm_im_j}{r_{ij}}$$

Step 1: Energy between corner masses (m)

• 4 sides of square (distance d):

$$U_1=-4\cdot\frac{Gm^2}{d}$$

• 2 diagonals (distance $$\sqrt{2}d$$):

$$U_2=-2\cdot\frac{Gm^2}{\sqrt{2}d}$$

Step 2: Energy between centre mass M and each corner mass

Distance from centre to corner:

$$r=\frac{d}{\sqrt{2}}$$

For 4 pairs:

$$U_3=-4\cdot\frac{GmM}{d/\sqrt{2}}=-4\sqrt{2}\frac{GmM}{d}$$

Step 3: Total energy

$$U=U_1+U_2+U_3$$

$$U=-\frac{Gm^2}{d}\left(4+\sqrt{2}\right)-\frac{4\sqrt{2}GmM}{d}$$

Final Answer:

$$U=-\frac{Gm^2}{d}\left(4+\sqrt{2}\right)-\frac{4\sqrt{2}GmM}{d}$$

Statement I: The law of gravitation holds good for any pair of bodies in the universe.

Newton's law of universal gravitation states that every particle in the universe attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the distance between them. This is a universal law and holds for any pair of bodies. So, Statement I is true.

Statement II: The weight of any person becomes zero when the person is at the centre of the earth.

At the centre of the earth, the gravitational field is zero because the gravitational pull from all directions cancels out. The acceleration due to gravity at the centre is $$g = 0$$.

Since weight $$W = mg$$, and $$g = 0$$ at the centre:

$$W = m \times 0 = 0$$

So, Statement II is also true.

Both statements are true.

The correct answer is Option A.

At the equator, bodies start floating when the centripetal acceleration equals the gravitational acceleration, i.e., $$\omega^2 R = g$$. This gives $$\omega = \sqrt{\frac{g}{R}}$$.

The duration of the day is $$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{R}{g}}$$.

Substituting the given values: $$T = 2 \times 3.14 \times \sqrt{\frac{6400 \times 10^3}{10}} = 6.28 \times \sqrt{6.4 \times 10^5} = 6.28 \times 800 = 5024$$ seconds.

Converting to minutes: $$T = \frac{5024}{60} \approx 84$$ minutes.

We use conservation of energy for a body projected from Earth's surface to height $$10R$$. At the surface, kinetic energy is $$\frac{1}{2}mv_i^2$$ and gravitational potential energy is $$-\frac{GMm}{R}$$. At height $$10R$$ from the surface (i.e., distance $$11R$$ from Earth's centre), the velocity is zero and potential energy is $$-\frac{GMm}{11R}$$.

Applying conservation of energy: $$\frac{1}{2}mv_i^2 - \frac{GMm}{R} = -\frac{GMm}{11R}$$

This gives $$\frac{1}{2}mv_i^2 = \frac{GMm}{R} - \frac{GMm}{11R} = \frac{GMm}{R}\left(1 - \frac{1}{11}\right) = \frac{GMm}{R} \cdot \frac{10}{11}$$

So $$v_i^2 = \frac{2GM}{R} \cdot \frac{10}{11}$$. We know the escape velocity is $$v_e = \sqrt{\frac{2GM}{R}}$$, so $$v_e^2 = \frac{2GM}{R}$$.

Substituting: $$v_i^2 = \frac{10}{11} v_e^2$$, which gives $$v_i = \sqrt{\frac{10}{11}} \times v_e$$.

Comparing with $$v_i = \sqrt{\frac{x}{y}} \times v_e$$, we get $$x = 10$$ and $$y = 11$$. Therefore, the value of $$x$$ is $$10$$.

We need to find the compressed radius $$R'$$ of the Earth such that the escape velocity becomes 10 times its current value, while the mass remains the same.

The escape velocity from the surface of a planet is given by $$v_e = \sqrt{\frac{2GM}{R}}$$, where $$M$$ is the mass and $$R$$ is the radius.

Let the new escape velocity be $$v_e' = 10 v_e$$. Since $$v_e' = \sqrt{\frac{2GM}{R'}}$$ and $$v_e = \sqrt{\frac{2GM}{R}}$$, we get $$\frac{v_e'}{v_e} = \sqrt{\frac{R}{R'}}$$.

Substituting $$v_e' = 10 v_e$$, we have $$10 = \sqrt{\frac{R}{R'}}$$, which gives $$100 = \frac{R}{R'}$$, so $$R' = \frac{R}{100}$$.

With $$R = 6400$$ km, the compressed radius is $$R' = \frac{6400}{100} = 64$$ km.

Therefore, the required compressed radius is $$\boxed{64}$$ km.

The two satellites move in uniform circular motion round the same planet and both turn anticlockwise. Let us name the nearer satellite as 1 and the farther satellite as 2. We are given

$$T_1 = 1 \text{ h}, \qquad T_2 = 8 \text{ h}, \qquad r_1 = 2 \times 10^{3}\ \text{km}$$

First we write their angular speeds about the planet. For uniform circular motion the relation is

$$\omega = \frac{2\pi}{T}.$$

So we have

$$\omega_1 = \frac{2\pi}{T_1} = \frac{2\pi}{1} = 2\pi \ \text{rad h}^{-1},$$

$$\omega_2 = \frac{2\pi}{T_2} = \frac{2\pi}{8} = \frac{\pi}{4} \ \text{rad h}^{-1}.$$

The radius of the outer orbit, $$r_2,$$ is not given directly, but the satellites circle the same planet, so Kepler’s third law applies:

$$\frac{T^2}{r^3} = \text{constant for the same planet}.$$

Hence

$$\frac{T_2^2}{r_2^3} = \frac{T_1^2}{r_1^3}\; \Longrightarrow\; r_2 = r_1\biggl(\frac{T_2}{T_1}\biggr)^{2/3}.$$

Substituting $$T_2 = 8 \text{ h},\; T_1 = 1 \text{ h}$$ and $$r_1 = 2 \times 10^{3} \text{ km}$$ gives

$$r_2 = 2 \times 10^{3}\ \text{km} \;\bigl(8\bigr)^{2/3}.$$

Because $$8 = 2^3,$$ we get $$8^{2/3} = (2^3)^{2/3} = 2^{2} = 4,$$ therefore

$$r_2 = 2 \times 10^{3}\ \text{km} \times 4 = 8 \times 10^{3}\ \text{km}.$$

To find how fast the farther satellite seems to move across the sky of an observer on the nearer satellite, we look at the instantaneous rate of change of the angle that the line joining the two satellites makes. Let us place the planet at the origin. At the instant of closest approach (both satellites on the same straight line through the planet) put that line along the $$x$$-axis. A little time $$t$$ later their angular displacements about the planet are $$\theta_1 = \omega_1 t$$ and $$\theta_2 = \omega_2 t.$$ The position vectors are

$$\vec r_1 = r_1(\cos\theta_1\,\hat i + \sin\theta_1\,\hat j),$$

$$\vec r_2 = r_2(\cos\theta_2\,\hat i + \sin\theta_2\,\hat j).$$

The vector from satellite 1 to satellite 2 is

$$\vec R = \vec r_2 - \vec r_1.$$

For small $$t$$ we expand $$\cos\theta \approx 1-\theta^2/2$$ and $$\sin\theta \approx \theta.$$ Then to first order in $$t$$

$$\vec R \approx \bigl(r_2 - r_1\bigr)\hat i + \bigl(r_2\omega_2 - r_1\omega_1\bigr)t\,\hat j.$$

The $$x$$-component is essentially constant, while the $$y$$-component grows linearly. The apparent angle $$\phi$$ measured from the initial line is therefore

$$\tan\phi \approx \frac{(r_2\omega_2 - r_1\omega_1)t}{r_2 - r_1}\; \Longrightarrow\; \phi \approx \frac{(r_2\omega_2 - r_1\omega_1)t}{r_2 - r_1}.$$

Hence the instantaneous angular speed of satellite 2 as seen from satellite 1 is

$$\frac{d\phi}{dt} = \frac{r_2\omega_2 - r_1\omega_1}{\,r_2 - r_1\,}.$$

Substituting the numerical values:

$$r_2\omega_2 = (8 \times 10^{3}\ \text{km})\left(\frac{\pi}{4}\ \text{rad h}^{-1}\right) = 2000\pi\ \text{km h}^{-1},$$

$$r_1\omega_1 = (2 \times 10^{3}\ \text{km})(2\pi\ \text{rad h}^{-1}) = 4000\pi\ \text{km h}^{-1},$$

$$r_2 - r_1 = 8 \times 10^{3}\ \text{km} - 2 \times 10^{3}\ \text{km} = 6 \times 10^{3}\ \text{km}.$$

Therefore

$$\left|\frac{d\phi}{dt}\right| = \left|\frac{2000\pi - 4000\pi}{6 \times 10^{3}}\right| = \left|\frac{-2000\pi}{6000}\right| = \frac{\pi}{3}\ \text{rad h}^{-1}.$$

This matches the form $$\dfrac{\pi}{x}\ \text{rad h}^{-1},$$ so we identify $$x = 3.$$

Hence, the correct answer is Option 3.

At the north pole, the body weighs 49 N. Using $$W = mg$$, the mass of the body is $$m = \frac{49}{9.8} = 5$$ kg.

At the equator, the effective acceleration due to gravity is reduced because of Earth's rotation. The effective gravity at the equator is $$g' = g - R\omega^2$$, where $$R = 6400 \times 10^3$$ m is the radius of Earth and $$\omega = \frac{2\pi}{T}$$ is the angular velocity of Earth with $$T = 24 \times 3600 = 86400$$ s.

Calculating $$\omega = \frac{2\pi}{86400} = 7.27 \times 10^{-5}$$ rad/s. Then $$R\omega^2 = 6.4 \times 10^6 \times (7.27 \times 10^{-5})^2 = 6.4 \times 10^6 \times 5.285 \times 10^{-9} = 0.0338$$ m/s$$^2$$.

So the effective gravity at the equator is $$g' = 9.8 - 0.0338 = 9.766$$ m/s$$^2$$.

The weight at the equator is $$W' = mg' = 5 \times 9.766 = 48.83$$ N.

The correct answer is 48.83 N.

The angular velocity of a satellite is related to its period of revolution by $$\omega = \frac{2\pi}{T}$$.

For satellite $$S_1$$, the period is $$T_1 = 1$$ hr, so $$\omega_1 = \frac{2\pi}{1}$$.

For satellite $$S_2$$, the period is $$T_2 = 8$$ hr, so $$\omega_2 = \frac{2\pi}{8}$$.

Now the ratio of angular velocities is $$\frac{\omega_1}{\omega_2} = \frac{T_2}{T_1} = \frac{8}{1}$$.

So $$\omega_1 : \omega_2 = 8 : 1$$.

Hence, the correct answer is Option A.

We recall the shell theorem, which states two key results for a thin, uniform, spherical shell of total mass $$M$$ and radius $$R$$:

1. For any internal point whose distance from the centre is $$r<R$$, the gravitational field (intensity) is zero:

$$\vec g(r)=0.$$

2. The gravitational potential at an internal point equals the potential at the surface and is therefore a constant given by

$$V(r)=V(R)=-\dfrac{GM}{R},$$

where $$G$$ is the universal gravitational constant. The minus sign shows the usual convention that potential is taken to be zero at infinity.

Now we examine each statement in the question:

    (a) “The gravitational field is zero.”  We have just established $$\vec g(r)=0,$$ so statement (a) is true.

    (b) “The gravitational potential is zero.”  From $$V(r)=-\dfrac{GM}{R},$$ the value is generally a non-zero negative constant (unless the mass itself is zero). Hence statement (b) is false.

    (c) “The gravitational field is the same everywhere.”  Because the field is identically zero at every internal point, every location experiences the same value (namely zero). So statement (c) is true.

    (d) “The gravitational potential is the same everywhere.”  We have $$V(r)=$$ constant for all internal points, so statement (d) is true.

    (e) “All the above.”  Since statement (b) is false, “all” cannot be correct, so (e) is false.

Combining the valid statements, only (a), (c) and (d) are correct. These correspond to Option A.

Hence, the correct answer is Option A.

We are given a comet orbiting the Sun with maximum distance (aphelion) $$r_{\text{max}} = 1.6 \times 10^{12} \text{ m}$$, minimum distance (perihelion) $$r_{\text{min}} = 8.0 \times 10^{10} \text{ m}$$, and speed at the nearest point $$v_{\text{near}} = 6 \times 10^4 \text{ m/s}$$.

By conservation of angular momentum about the Sun, at the nearest and farthest points the velocity is perpendicular to the radius vector, so $$m \, v_{\text{near}} \, r_{\text{min}} = m \, v_{\text{far}} \, r_{\text{max}}$$.

Solving for $$v_{\text{far}}$$: $$v_{\text{far}} = \frac{v_{\text{near}} \, r_{\text{min}}}{r_{\text{max}}} = \frac{6 \times 10^4 \times 8.0 \times 10^{10}}{1.6 \times 10^{12}}$$.

Computing: $$v_{\text{far}} = \frac{48 \times 10^{14}}{1.6 \times 10^{12}} = 30 \times 10^{2} = 3.0 \times 10^3 \text{ m/s}$$.

For a geostationary satellite orbiting at a height of $$11R$$ above the surface, the orbital radius is $$r_1 = R + 11R = 12R$$ and its time period is $$T_1 = 24$$ hours.

For the second satellite at a height of $$2R$$, the orbital radius is $$r_2 = R + 2R = 3R$$.

By Kepler's third law, $$\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}$$, which gives $$T_2 = T_1 \left(\frac{r_2}{r_1}\right)^{3/2} = 24 \left(\frac{3R}{12R}\right)^{3/2} = 24 \left(\frac{1}{4}\right)^{3/2}$$.

Now $$\left(\frac{1}{4}\right)^{3/2} = \frac{1}{4\sqrt{4}} = \frac{1}{8}$$, so $$T_2 = \frac{24}{8} = 3$$ hours.

As the passenger travels from Earth to Mars, their weight $$W = mg$$ changes due to the variation in gravitational acceleration. On Earth's surface, $$W = 100 \times 10 = 1000$$ N. On Mars' surface, $$W = 100 \times 4 = 400$$ N.

As the spacecraft moves away from Earth, the gravitational pull of Earth decreases as $$\frac{1}{r^2}$$, so the weight decreases. At some point between Earth and Mars — the neutral point where Earth's gravitational pull equals Mars' gravitational pull — the weight becomes zero (the passenger is weightless).

After passing the neutral point, Mars' gravity becomes dominant and the weight starts increasing, eventually reaching 400 N at Mars' surface. Since Earth's surface gravity is stronger than Mars', the neutral point is closer to Mars, so the weight drops to zero more gradually and rises more quickly.

The correct curve starts at 1000 N, continuously decreases to zero at the neutral point, and then increases back to 400 N — corresponding to curve (c).

Each particle experiences gravitational forces from the other three.

Net Gravitational Force ($$F_{net}$$)

A single particle is pulled by two adjacent particles at a distance $$d = \sqrt{2}R$$ and one opposite particle at a distance $$D = 2R$$. The resultant force acting towards the center is the sum of the components of these forces:

$$F_{net} = 2 \left( \frac{Gm^2}{(\sqrt{2}R)^2} \right) \cos 45^\circ + \frac{Gm^2}{(2R)^2}$$

$$F_{net} = \frac{Gm^2}{R^2} \left( \frac{1}{\sqrt{2}} + \frac{1}{4} \right) = \frac{Gm^2}{R^2} \left( \frac{2\sqrt{2} + 1}{4} \right)$$

This mutual attraction provides the centripetal force required for the circular motion of each particle:

$$\frac{mv^2}{R} = \frac{Gm^2}{R^2} \left( \frac{1 + 2\sqrt{2}}{4} \right)$$

$$v^2 = G \left( \frac{1 + 2\sqrt{2}}{4} \right)$$

$$v = \frac{\sqrt{G(1 + 2\sqrt{2})}}{2}$$

We have a planet of mass $$m$$ revolving around the Sun under the action of the gravitational force. Because this force is always directed towards the Sun (the centre of force), it is a central force. For motion under a central force, the angular momentum about the centre remains conserved. We therefore begin with the statement of the law of conservation of angular momentum:

$$L \;=\; mvr \;=\; \text{constant}.$$

Here $$v$$ is the linear speed of the planet at any instant and $$r$$ is the corresponding Sun-planet distance. Since the mass $$m$$ of the planet does not change, the product $$vr$$ must stay the same at every point along the orbit.

The orbit is elliptical. Let

$$x_1 = \text{minimum distance (perihelion)},$$

$$x_2 = \text{maximum distance (aphelion)}.$$

At aphelion the distance is greatest, so the speed is least; this minimum speed is given to be $$v_0$$. At perihelion the distance is least, so the speed is greatest; let us call this unknown maximum speed $$v_{\max}$$. Using the conserved quantity $$mvr$$ at the two extreme positions:

$$m\,v_{\max}\,x_1 \;=\; m\,v_0\,x_2.$$

Now we cancel the common factor $$m$$ from both sides:

$$v_{\max}\,x_1 \;=\; v_0\,x_2.$$

To isolate $$v_{\max}$$ we divide both sides by $$x_1$$:

$$v_{\max} \;=\; \dfrac{v_0\,x_2}{x_1}.$$

This is the required expression for the maximum speed of the planet. Comparing with the given options, we notice that this matches option D.

Hence, the correct answer is Option D.

A body projected with escape velocity satisfies the energy equation at any height $$r$$ from the Earth's center: $$\frac{1}{2}mv^2 - \frac{GMm}{r} = 0$$ so $$v = \sqrt{\frac{2GM}{r}}$$.

Using $$GM = gR_e^2$$, we get $$v = \frac{dr}{dt} = \sqrt{\frac{2gR_e^2}{r}}$$.

Separating variables: $$\sqrt{r}\, dr = \sqrt{2gR_e^2}\, dt = R_e\sqrt{2g}\, dt$$

Integrating from $$r = R_e$$ to $$r = R_e + h$$: $$\int_{R_e}^{R_e+h} r^{1/2}\, dr = R_e\sqrt{2g} \int_0^t dt$$

$$\left[\frac{2}{3}r^{3/2}\right]_{R_e}^{R_e+h} = R_e\sqrt{2g}\cdot t$$

$$\frac{2}{3}\left[(R_e+h)^{3/2} - R_e^{3/2}\right] = R_e\sqrt{2g}\cdot t$$

$$t = \frac{2}{3R_e\sqrt{2g}}\left[(R_e+h)^{3/2} - R_e^{3/2}\right]$$

Factoring out $$R_e^{3/2}$$: $$t = \frac{2R_e^{3/2}}{3R_e\sqrt{2g}}\left[\left(1+\frac{h}{R_e}\right)^{3/2} - 1\right] = \frac{2\sqrt{R_e}}{3\sqrt{2g}}\left[\left(1+\frac{h}{R_e}\right)^{3/2} - 1\right]$$

Simplifying: $$t = \frac{1}{3}\sqrt{\frac{2R_e}{g}}\left[\left(1+\frac{h}{R_e}\right)^{3/2} - 1\right]$$

This matches option 4.

The escape velocity from the surface of a planet of mass $$M$$ and radius $$R$$ is given by $$v_e = \sqrt{\frac{2GM}{R}}$$.

For the escape velocities of planets A and B to be equal, we need $$\frac{M_1}{R_1} = \frac{M_2}{R_2}$$. This condition allows planets of unequal mass to have the same escape velocity, as long as the ratio of mass to radius is the same. For example, if planet A has twice the mass and twice the radius of planet B, they will have the same escape velocity. Therefore, Assertion A is correct.

The Reason states that $$M_1R_1 = M_2R_2$$. However, from the escape velocity formula, the actual requirement is $$\frac{M_1}{R_1} = \frac{M_2}{R_2}$$, which gives $$M_1R_2 = M_2R_1$$, not $$M_1R_1 = M_2R_2$$. The condition given in Reason R is incorrect.

Therefore, A is correct but R is not correct, which corresponds to Option (4).

We have to find the ratio between the acceleration due to gravity at a depth $$r$$ below the Earth’s surface and at a height $$r$$ above the surface, where $$r \lt R_E$$ and $$R_E$$ is the Earth’s radius.

First recall the standard results for the variation of acceleration due to gravity with depth and with height:

1. Variation with depth. For a depth $$r$$ (or $$d$$) inside the Earth, the formula is stated as $$g_d = g\left(1 - \frac{r}{R_E}\right)$$ because only the mass enclosed within radius $$R_E - r$$ contributes to the gravitational pull, leading to a linear decrease.

2. Variation with height. For a height $$r$$ (or $$h$$) above the surface, the formula is stated as $$g_h = \frac{g}{\left(1 + \frac{r}{R_E}\right)^2}$$ because the field outside the Earth follows the inverse-square law while the distance from the centre becomes $$R_E + r$$.

Now we are asked for the ratio $$\dfrac{g_d}{g_h}$$. Substituting the two formulae, we write

$$ \frac{g_d}{g_h} = \frac{g\left(1 - \frac{r}{R_E}\right)} {\displaystyle \frac{g}{\left(1 + \frac{r}{R_E}\right)^2}} = \left(1 - \frac{r}{R_E}\right)\left(1 + \frac{r}{R_E}\right)^2 . $$

The factor $$g$$ cancels out, as expected. To make the algebra clearer, set

$$x = \frac{r}{R_E}, \qquad \text{with } x \lt 1.$$

With this substitution the ratio becomes

$$ \frac{g_d}{g_h} = (1 - x)(1 + x)^2 . $$

Now expand every factor step by step. First expand the square:

$$ (1 + x)^2 = 1 + 2x + x^2 . $$

Next multiply this result by the remaining factor $$(1 - x)$$:

$$ (1 - x)(1 + 2x + x^2) = 1 + 2x + x^2 \;-\; x - 2x^2 - x^3 . $$

Combine like terms:

$$ 1 + 2x + x^2 - x - 2x^2 - x^3 = 1 + (2x - x) + (x^2 - 2x^2) - x^3 = 1 + x - x^2 - x^3 . $$

Finally replace $$x$$ by $$\dfrac{r}{R_E}$$ to return to the original variables:

$$ \frac{g_d}{g_h} = 1 + \frac{r}{R_E} - \frac{r^2}{R_E^2} - \frac{r^3}{R_E^3}. $$

This result matches exactly the expression given in Option A.

Hence, the correct answer is Option A.

We begin by recalling the expression for the gravitational potential energy of interaction between two point masses. The formula is stated first:

$$U \;=\; -\dfrac{G\,M\,m}{d}$$

Here $$G$$ is the universal gravitational constant, $$M$$ is the source mass, $$m$$ is the test mass and $$d$$ is the separation of their centres. The negative sign tells us that gravity is an attractive interaction whose potential energy decreases (becomes more negative) as the distance decreases.

In the present problem two large bodies are involved. Their masses and radii are $$M_1,\,R_1$$ (Earth) and $$M_2,\,R_2$$ (Moon). The distance between their centres is given to be $$r$$. A particle of mass $$m$$ is released from the exact mid-point of the line joining the two centres, so its distance from each of the two bodies is clearly $$\dfrac{r}{2}$$.

We write the total initial gravitational potential energy of the particle due to both masses by adding the individual contributions:

$$U_{\text{initial}} \;=\; -\dfrac{G\,M_1\,m}{\dfrac{r}{2}} \;-\; \dfrac{G\,M_2\,m}{\dfrac{r}{2}}$$

Because the denominator $$\dfrac{r}{2}$$ appears in both terms we simplify each fraction by inverting and multiplying:

$$ U_{\text{initial}} = -\,G\,M_1\,m\left(\dfrac{2}{r}\right) -\,G\,M_2\,m\left(\dfrac{2}{r}\right) = -\,\dfrac{2Gm}{r}\left(M_1+M_2\right) $$

Thus the combined potential energy at the mid-point is

$$U_{\text{initial}} = -\dfrac{2Gm\,(M_1+M_2)}{r}. $$

Next we invoke the principle of conservation of mechanical energy. Let the particle be projected from the mid-point with a speed $$v_e$$ just sufficient to escape to infinity. “Just sufficient” (minimum escape condition) means that when it finally reaches infinity its speed falls to zero. We therefore have for the final state at infinity:

$$K_{\text{final}} \;=\; 0, \qquad U_{\text{final}} \;=\; 0.$$

The mechanical energy at the starting point equals that at infinity:

$$ K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}}. $$

Substituting the known expressions, we write

$$ \dfrac{1}{2}m\,v_e^{\,2} + \left(-\dfrac{2Gm\,(M_1+M_2)}{r}\right) = 0 + 0. $$

Now we isolate the kinetic term and solve for $$v_e^{\,2}$$ step by step:

$$ \dfrac{1}{2}m\,v_e^{\,2} = \dfrac{2Gm\,(M_1+M_2)}{r}. $$

We cancel the common factor $$m$$ from both sides:

$$ \dfrac{1}{2}\,v_e^{\,2} = \dfrac{2G\,(M_1+M_2)}{r}. $$

Multiplying both sides by $$2$$ gives

$$ v_e^{\,2} = \dfrac{4G\,(M_1+M_2)}{r}. $$

Finally, taking the square root, we obtain the minimum escape velocity:

$$ v_e = \sqrt{\dfrac{4G\,(M_1+M_2)}{r}}. $$

This expression matches Option A exactly.

Hence, the correct answer is Option A.

Let us consider the two identical particles, each of mass $$m = 1\text{ kg}$$. Because the particles are identical, their centre of mass lies exactly at the midpoint of the line joining them. We are told that each particle moves in a circle of radius $$R$$ around this midpoint, so the distance between the two particles is twice this radius:

$$d = 2R.$$

First, we write the universal law of gravitation:

Gravitational force between two point masses is $$F = \dfrac{G\,m_1 m_2}{d^2}.$$

Substituting $$m_1 = m_2 = 1\text{ kg}$$ and $$d = 2R$$, we get

$$F = \dfrac{G \,(1)(1)}{(2R)^2} = \dfrac{G}{4R^2}.$$

This gravitational force is the only force acting on each particle, and it points along the line joining the particles, which also passes through the centre of the circular path. Therefore, this same force provides the necessary centripetal force for uniform circular motion.

The formula for centripetal force is

$$F_{\text{centripetal}} = m\,\omega^2\,r,$$

where $$\omega$$ is the angular speed and $$r$$ is the radius of the circular path. For each particle, $$m = 1\text{ kg}$$ and $$r = R$$, so

$$F_{\text{centripetal}} = (1)\,\omega^2\,R = \omega^2 R.$$

Because the gravitational force supplies the required centripetal force, we equate the two:

$$\dfrac{G}{4R^2} = \omega^2 R.$$

We now solve this equation for $$\omega$$ step by step.

First, isolate $$\omega^2$$ by dividing both sides by $$R$$:

$$\omega^2 = \dfrac{G}{4R^2} \cdot \dfrac{1}{R} = \dfrac{G}{4R^3}.$$

Now take the positive square root to obtain $$\omega$$ (angular speed is taken as positive):

$$\omega = \sqrt{\dfrac{G}{4R^3}} = \dfrac{1}{2}\sqrt{\dfrac{G}{R^3}}.$$

This matches Option B.

Hence, the correct answer is Option B.

We have two stars of masses $$m$$ and $$2m$$ separated by a distance $$d$$, rotating about their common centre of mass.

The centre of mass divides the line joining them in the inverse ratio of their masses. So the distance of mass $$m$$ from the centre of mass is $$r_1 = \frac{2m \cdot d}{m + 2m} = \frac{2d}{3}$$, and the distance of mass $$2m$$ from the centre of mass is $$r_2 = \frac{m \cdot d}{3m} = \frac{d}{3}$$.

The gravitational force between them provides the centripetal force. For the star of mass $$m$$ orbiting at radius $$r_1 = \frac{2d}{3}$$, we write $$m\omega^2 r_1 = \frac{G \cdot m \cdot 2m}{d^2}$$.

Substituting $$r_1 = \frac{2d}{3}$$, we get $$m\omega^2 \cdot \frac{2d}{3} = \frac{2Gm^2}{d^2}$$.

Simplifying, $$\omega^2 = \frac{2Gm}{d^2} \times \frac{3}{2d} = \frac{3Gm}{d^3}$$.

The period of revolution is $$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{d^3}{3Gm}}$$.

Hence, the correct answer is Option A.

By Kepler's third law, $$T \propto R^{3/2}$$. If the first satellite has orbital radius $$R$$ with period $$T$$, and the second has radius $$1.02R$$, then $$T' = T(1.02)^{3/2}$$.

The percentage difference is $$\frac{T' - T}{T} \times 100 = \left[(1.02)^{3/2} - 1\right] \times 100$$.

Using the binomial approximation $$(1+x)^n \approx 1 + nx$$ for small $$x$$: $$(1.02)^{3/2} \approx 1 + \frac{3}{2}(0.02) = 1 + 0.03 = 1.03$$.

Therefore the percentage difference is approximately $$3.0\%$$.

First, recall that the weight of an object on any planet is given by the familiar relation $$W = m\,g,$$ where $$m$$ is the mass of the object and $$g$$ is the acceleration due to gravity on that planet. For two different planets we can compare the weights by comparing their respective values of $$g.$$

The acceleration due to gravity at the surface of a spherical planet is obtained from Newton’s law of gravitation:

$$g = \dfrac{G\,M}{R^{2}},$$

where $$G$$ is the universal gravitational constant, $$M$$ is the mass of the planet, and $$R$$ is its radius. Thus, if we know how the mass and radius of another planet relate to the Earth’s, we can get the ratio of the two $$g$$’s.

We are told that the new planet has a mass which is double the mass of the Earth. Writing Earth’s mass as $$M_E,$$ we have

$$M_P = 2\,M_E.$$

Next, we are told that the density of the new planet is equal to the average density of the Earth. Density is defined as mass divided by volume. For a sphere, volume is $$\dfrac{4}{3}\pi R^{3},$$ so

$$\rho = \dfrac{M}{\frac{4}{3}\pi R^{3}} \quad\Longrightarrow\quad M = \rho\,\frac{4}{3}\pi R^{3}.$$

If the densities of the two planets are equal, we can set up the equality

$$\rho_P = \rho_E \quad\Longrightarrow\quad \dfrac{M_P}{\frac{4}{3}\pi R_P^{3}} = \dfrac{M_E}{\frac{4}{3}\pi R_E^{3}}.$$

Cancelling the common factors $$\frac{4}{3}\pi$$ on both sides gives

$$\dfrac{M_P}{R_P^{3}} = \dfrac{M_E}{R_E^{3}}.$$

Substituting $$M_P = 2\,M_E$$ into the above relation, we get

$$\dfrac{2\,M_E}{R_P^{3}} = \dfrac{M_E}{R_E^{3}}.$$

Dividing both sides by $$M_E$$ and then cross-multiplying, we have

$$2\,R_E^{3} = R_P^{3}.$$

Taking the cube root of both sides, we obtain the radius of the planet in terms of Earth’s radius:

$$R_P = 2^{\frac{1}{3}}\,R_E.$$

Now we can find the ratio of the gravitational accelerations. Using the formula $$g = \dfrac{G\,M}{R^{2}},$$ we write

$$\dfrac{g_P}{g_E} = \dfrac{\dfrac{G\,M_P}{R_P^{2}}}{\dfrac{G\,M_E}{R_E^{2}}}.$$

The constant $$G$$ cancels out, so

$$\dfrac{g_P}{g_E} = \dfrac{M_P}{M_E}\,\dfrac{R_E^{2}}{R_P^{2}}.$$

Substituting $$M_P = 2\,M_E$$ and $$R_P = 2^{\frac{1}{3}}\,R_E,$$ we get

$$\dfrac{g_P}{g_E} = 2 \times \dfrac{R_E^{2}}{\left(2^{\frac{1}{3}}\,R_E\right)^{2}}.$$

Simplifying the expression inside the denominator first,

$$(2^{\tfrac{1}{3}}\,R_E)^{2} = 2^{\tfrac{2}{3}}\,R_E^{2}.$$

Therefore,

$$\dfrac{g_P}{g_E} = 2 \times \dfrac{R_E^{2}}{2^{\tfrac{2}{3}}\,R_E^{2}}.$$

The factor $$R_E^{2}$$ cancels, giving

$$\dfrac{g_P}{g_E} = \dfrac{2}{2^{\tfrac{2}{3}}} = 2^{1 - \tfrac{2}{3}} = 2^{\tfrac{1}{3}}.$$

Finally, the weight of the object on the new planet will be

$$W_P = m\,g_P = m\,g_E \left(\dfrac{g_P}{g_E}\right) = W \times 2^{\tfrac{1}{3}}.$$

So the weight becomes $$2^{\tfrac{1}{3}}\,W.$$

Hence, the correct answer is Option C.

The angular momentum of the planet about the sun is $$L = M v_\perp r$$, where $$v_\perp$$ is the component of velocity perpendicular to the radius vector and $$r$$ is the distance from the sun.

The areal velocity is the rate at which the radius vector sweeps out area: $$\frac{dA}{dt} = \frac{1}{2} r \, v_\perp$$.

Since $$L = M r \, v_\perp$$, we get $$r \, v_\perp = \frac{L}{M}$$. Substituting into the areal velocity expression: $$\frac{dA}{dt} = \frac{1}{2} \cdot \frac{L}{M} = \frac{L}{2M}$$.

We are told that one of Mars’ moons has an orbital period of 7 hours 30 minutes and an orbital radius of $$9.0 \times 10^3$$ km. First we convert all the given data into SI units.

The period is 7 hours 30 minutes. Since 30 minutes is half an hour, this is a total of 7.5 hours. Converting hours to seconds we use the relation $$1 \text{ hour} = 3600 \text{ s}.$$ So $$T = 7.5 \times 3600 \text{ s} = 27000 \text{ s} = 2.7 \times 10^4 \text{ s}.$$

The radius is $$9.0 \times 10^3$$ km. Knowing that $$1 \text{ km} = 1000 \text{ m},$$ we get $$r = 9.0 \times 10^3 \times 10^3 \text{ m} = 9.0 \times 10^6 \text{ m}.$$

For a satellite orbiting a planet, Kepler’s third law (derived from Newton’s law of gravitation) states $$T^2 = \frac{4\pi^2}{G M}\,r^3,$$ where $$M$$ is the mass of the planet. We are asked to find $$M$$, so we solve this equation for $$M$$. Multiplying both sides by $$G M$$ and then dividing by $$T^2$$ gives

$$M = \frac{4\pi^2 r^3}{G T^2}.$$

The problem supplies the convenient constant $$\frac{4\pi^2}{G} = 6 \times 10^{11}\;\text{N}^{-1}\text{ m}^{-2}\text{ kg}^2.$$ Using this, we can rewrite the expression for the mass as

$$M = \left(\frac{4\pi^2}{G}\right)\frac{r^3}{T^2}.$$

Now we compute each part step by step. First, the cube of the radius:

$$r^3 = \left(9.0 \times 10^6\right)^3 = 9^3 \times 10^{18} = 729 \times 10^{18} = 7.29 \times 10^{20} \text{ m}^3.$$

Next, the square of the period:

$$T^2 = \left(2.7 \times 10^4\right)^2 = 2.7^2 \times 10^{8} = 7.29 \times 10^{8} \text{ s}^2.$$

We now form the ratio $$\dfrac{r^3}{T^2}$$:

$$\frac{r^3}{T^2} = \frac{7.29 \times 10^{20}}{7.29 \times 10^{8}} = 10^{12} \text{ m}^3\text{ s}^{-2}.$$

Finally, substituting this result into the mass formula, we get

$$M = \left(6 \times 10^{11}\right)\left(10^{12}\right) \text{ kg} = 6 \times 10^{23} \text{ kg}.$$

Hence, the correct answer is Option D.

We are asked to find the gravitational potential at a point that lies inside a thin, uniform spherical shell of mass $$100\ \text{kg}$$ and radius $$50\ \text{m}$$, while a point mass of $$50\ \text{kg}$$ sits exactly at the common centre. The point of interest is situated at a distance $$25\ \text{m}$$ from this centre.

We recall two standard results from Newtonian gravitation:

1. Potential due to a point mass. For a point mass $$m$$, the gravitational potential at a distance $$r$$ is given by the formula $$\displaystyle V_{\text{point}} = -\,\frac{G\,m}{r}$$ where $$G$$ is the universal gravitational constant.

2. Potential inside a uniform spherical shell. At any interior point (including the very centre) of a thin spherical shell of total mass $$M$$ and radius $$R$$, the potential has the same constant value $$\displaystyle V_{\text{shell (inside)}} = -\,\frac{G\,M}{R}.$$ It is independent of how deep inside we are, as long as we remain inside the hollow shell.

Because potential is a scalar quantity, the net potential is simply the algebraic sum of the individual potentials contributed by each mass.

We first compute the shell’s contribution. Here $$M = 100\ \text{kg}, \qquad R = 50\ \text{m}.$$ Using the stated formula, we have $$V_{\text{shell}} = -\,\frac{G\,M}{R} = -\,\frac{G \times 100}{50} = -\,\frac{100}{50}\,G = -\,2\,G.$$

Next we compute the point mass’s contribution. For the point mass at the centre, $$m = 50\ \text{kg}, \qquad r = 25\ \text{m}.$$ Using the point-mass formula, $$V_{\text{point}} = -\,\frac{G\,m}{r} = -\,\frac{G \times 50}{25} = -\,\frac{50}{25}\,G = -\,2\,G.$$

Now we add the two potentials:

$$\begin{aligned} V_{\text{total}} & = V_{\text{shell}} + V_{\text{point}} \\ & = (-\,2\,G) + (-\,2\,G) \\ & = -\,4\,G. \end{aligned}$$

Thus the gravitational potential at the given point is

$$V = -\,4\,G.$$

Hence, the correct answer is Option C.

In a binary star system, both stars revolve around their common centre of mass. Crucially, both stars complete one full orbit in the same time — they always remain on opposite sides of the centre of mass. This is a direct consequence of Newton's third law: the gravitational force each exerts on the other is equal and opposite, and since they orbit the same centre of mass with the same angular velocity, their periods must be identical.

Although the orbital radii $$r_A$$ and $$r_B$$ may differ (depending on the mass ratio $$m_A r_A = m_B r_B$$), and $$r_A \neq r_B$$ in general, the angular velocity $$\omega$$ is the same for both, so $$T_A = T_B$$.

Therefore $$T_A = T_B$$.

We use conservation of mechanical energy, because the gravitational force is conservative and the atmosphere is neglected.

Let $$R$$ be the radius of the earth, $$M$$ its mass and $$G$$ the universal gravitational constant. The asteroid has

initial distance from the earth’s centre $$r_1 = 10R$$, initial speed $$v_1 = 12\ {\rm km\,s^{-1}}$$, final distance when it just touches the surface $$r_2 = R$$, and final speed $$v_2$$ (to be found).

The mechanical energy at any position is the sum of kinetic and gravitational potential energies. For a mass $$m$$ at a distance $$r$$ from the earth’s centre, the energies are

$$\text{K.E.} = \tfrac12 m v^2,\qquad \text{P.E.} = -\dfrac{G M m}{r}.$$

Conservation of energy gives

$$\tfrac12 m v_1^2 - \dfrac{G M m}{r_1} \;=\; \tfrac12 m v_2^2 - \dfrac{G M m}{r_2}.$$

Dividing by $$m$$ and multiplying by $$2$$ to remove the fractions, we get

$$v_1^2 - \dfrac{2GM}{r_1} \;=\; v_2^2 - \dfrac{2GM}{r_2}.$$

Re-arranging for $$v_2^2$$:

$$v_2^2 \;=\; v_1^2 - \dfrac{2GM}{r_1} + \dfrac{2GM}{r_2}.$$

Now substitute $$r_1 = 10R$$ and $$r_2 = R$$:

$$v_2^2 \;=\; v_1^2 - \dfrac{2GM}{10R} + \dfrac{2GM}{R}.$$

Combine the potential-energy terms:

$$-\dfrac{2GM}{10R} + \dfrac{2GM}{R} \;=\; \dfrac{2GM}{R}\Bigl(1 - \tfrac1{10}\Bigr) \;=\; \dfrac{2GM}{R}\,\dfrac{9}{10}.$$

Thus

$$v_2^2 \;=\; v_1^2 + \dfrac{9}{10}\,\dfrac{2GM}{R}.$$

The escape velocity from the earth is given in the question as $$11.2\ {\rm km\,s^{-1}}$$. By definition, escape velocity $$v_\text{esc}$$ satisfies

$$v_\text{esc} = \sqrt{\dfrac{2GM}{R}}.$$

Therefore

$$\dfrac{2GM}{R} = v_\text{esc}^{\,2} = (11.2)^2 = 125.44.$$

Insert this and $$v_1 = 12\ {\rm km\,s^{-1}}$$ into the expression for $$v_2^2$$:

$$v_2^2 = (12)^2 + \dfrac{9}{10}\,(125.44) = 144 + 0.9 \times 125.44 = 144 + 112.896 = 256.896.$$

Taking the square root,

$$v_2 = \sqrt{256.896}\ {\rm km\,s^{-1}} \;\approx\; 16.03\ {\rm km\,s^{-1}}.$$

Rounding to the nearest integer, the speed is $$16\ {\rm km\,s^{-1}}.$$

So, the answer is $$16\ {\rm km\,s^{-1}}.$$

We start with the satellite in a very low, almost circular orbit whose radius is practically the earth’s radius itself. Hence the initial orbital radius is $$r_1 = R_e$$.

For a circular orbit, the standard formula for orbital speed is

$$v_c = \sqrt{\frac{GM}{r}},$$

where $$G$$ is the universal gravitational constant and $$M$$ is the mass of the earth. Substituting $$r = R_e$$ gives the initial speed

$$v_1 = \sqrt{\frac{GM}{R_e}}.$$

Immediately after the rockets are fired, the speed is increased tangentially to

$$v_2 = \sqrt{\frac{3}{2}}\;v_1 = \sqrt{\frac{3}{2}}\;\sqrt{\frac{GM}{R_e}} = \sqrt{\frac{3GM}{2R_e}}.$$

The moment after the impulse, the satellite is still at the same position $$r = R_e$$ but now with the higher speed $$v_2$$. Its subsequent path is no longer circular; it becomes an ellipse with the present point as the perigee (closest point). We shall find the apogee distance $$R$$ using conservation laws.

1. Specific mechanical energy (total energy per unit mass)
The formula is

$$\varepsilon = \frac{v^2}{2} - \frac{GM}{r}.$$

At perigee just after firing, we substitute $$v = v_2$$ and $$r = R_e$$:

$$\varepsilon = \frac{1}{2}\left(\frac{3GM}{2R_e}\right) - \frac{GM}{R_e} = \frac{3GM}{4R_e} - \frac{GM}{R_e} = \frac{3GM}{4R_e} - \frac{4GM}{4R_e} = -\frac{GM}{4R_e}.$$

For an elliptical orbit, the same energy is also expressed as

$$\varepsilon = -\frac{GM}{2a},$$

where $$a$$ is the semi-major axis. Equating the two expressions, we get

$$-\frac{GM}{2a} = -\frac{GM}{4R_e} \;\;\Longrightarrow\;\; 2a = 4R_e \;\;\Longrightarrow\;\; a = 2R_e.$$

2. Relation between perigee, apogee, and semi-major axis
For any ellipse,

$$r_p + r_a = 2a,$$

where $$r_p$$ is the perigee distance and $$r_a$$ is the apogee distance. Here

$$r_p = R_e,\qquad r_a = R.$$

Substituting these values and $$a = 2R_e$$, we obtain

$$R_e + R = 2a = 4R_e \;\;\Longrightarrow\;\; R = 4R_e - R_e = 3R_e.$$

So the farthest distance from the earth’s centre reached by the satellite is three times the earth’s radius.

Hence, the correct answer is Option C.

Let the acceleration due to gravity on the surface of the Earth be denoted by $$g$$. We have two different positions to consider for the same body:

1. At a height $$h$$ above the surface. 2. At a depth $$h$$ below the surface.

According to Newton’s law of gravitation, the magnitude of $$g$$ varies with distance from the centre of the Earth. First we state the two standard formulae that connect $$g$$ with height and depth (rotation neglected):

• For a height $$h$$ above the surface, the distance from the centre becomes $$R+h$$.  The formula is $$g_h = g\left(\dfrac{R}{R+h}\right)^2$$.

• For a depth $$h$$ below the surface, only the mass enclosed within the radius $$R-h$$ contributes.  For a uniform Earth, $$g_d = g\left(1-\dfrac{h}{R}\right)$$.

The problem states that the weight (and hence the gravitational acceleration) is the same in both cases, so we equate the two expressions:

$$g\left(\dfrac{R}{R+h}\right)^2 \;=\; g\left(1-\dfrac{h}{R}\right).$$

Because the factor $$g$$ appears on both sides, it cancels out, leaving

$$\left(\dfrac{R}{R+h}\right)^2 = 1-\dfrac{h}{R}.$$

To simplify the algebra, we introduce the dimensionless variable

$$x = \dfrac{h}{R}.$$

Substituting $$h = xR$$ into the equation gives

$$\left(\dfrac{R}{R + xR}\right)^2 = 1 - x.$$

Inside the brackets, factor out $$R$$:

$$\left(\dfrac{R}{R(1 + x)}\right)^2 = 1 - x \;\;\Longrightarrow\;\; \left(\dfrac{1}{1 + x}\right)^2 = 1 - x.$$

Remove the fraction by multiplying both sides by $$(1 + x)^2$$:

$$(1 + x)^2\,(1 - x) = 1.$$

Now expand the left-hand side step by step. First, square the binomial:

$$(1 + x)^2 = 1 + 2x + x^2.$$

Next, multiply this result by $$(1 - x)$$:

$$(1 + 2x + x^2)(1 - x) = 1(1 - x) + 2x(1 - x) + x^2(1 - x).$$

Carrying out each product:

$$1 - x + 2x - 2x^2 + x^2 - x^3.$$

Combine like terms:

$$1 + x - x^2 - x^3.$$

Therefore the equation becomes

$$1 + x - x^2 - x^3 = 1.$$

Subtract $$1$$ from both sides to bring all terms to one side:

$$x - x^2 - x^3 = 0.$$

Factor out the common factor $$x$$:

$$x\bigl(1 - x - x^2\bigr) = 0.$$

One solution is $$x = 0$$, corresponding to $$h = 0$$, which is trivial. For a non-zero height we require

$$1 - x - x^2 = 0.$$

This is a quadratic in $$x$$. Rewriting it in standard form:

$$x^2 + x - 1 = 0.$$

We now apply the quadratic formula $$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a = 1,\; b = 1,\; c = -1$$:

$$x = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \dfrac{-1 \pm \sqrt{1 + 4}}{2} = \dfrac{-1 \pm \sqrt{5}}{2}.$$

The negative root gives a negative height, which is not meaningful here, so we choose the positive root:

$$x = \dfrac{-1 + \sqrt{5}}{2}.$$

Now revert to the original variable $$h$$ using $$x = \dfrac{h}{R}$$:

$$\dfrac{h}{R} = \dfrac{\sqrt{5} - 1}{2} \;\;\Longrightarrow\;\; h = \dfrac{\sqrt{5}\,R - R}{2}.$$

This expression exactly matches Option C.

Hence, the correct answer is Option C.

Let the mass of the object be $$m$$. Whenever a spring balance is used, it actually registers the normal reaction on the object, which we call the apparent weight.

First we consider the object kept on the equator at the earth’s surface. At the equator the object is moving in a horizontal circle of radius $$R$$ with the earth, so it experiences a centrifugal acceleration directed radially outward. The magnitude of this acceleration is $$\omega^{2}R$$, where $$\omega$$ is the earth’s angular velocity.

Hence the effective or apparent acceleration acting downward on the object at the equator is the vector difference of the true gravitational acceleration $$g$$ and the centrifugal acceleration $$\omega^{2}R$$. Therefore the apparent weight at the equator is

$$W_{e}=m\left(g-\omega^{2}R\right).$$

Now we take the same object to a point directly above the pole at a small height $$h$$ (with $$h\ll R$$). Exactly at the pole the distance of the object from the earth’s axis is zero, so it has no circular motion about the axis and hence no centrifugal effect. The only change comes from the fact that we are now at a larger distance $$R+h$$ from the earth’s centre.

The law of gravitation gives the gravitational acceleration at distance $$r$$ as $$g(r)=\dfrac{GM}{r^{2}}$$, where $$G$$ is the gravitational constant and $$M$$ is the earth’s mass. Putting $$r=R+h$$ we get

$$g_{p}=g\left(\dfrac{R}{R+h}\right)^{2}.$$

Because $$h\ll R$$ we expand by the binomial theorem. For any small $$x$$, $$(1+x)^{-2}\approx1-2x$$. Taking $$x=\dfrac{h}{R}$$ we have

$$\left(1+\dfrac{h}{R}\right)^{-2}\approx1-\dfrac{2h}{R}.$$

So

$$g_{p}\approx g\left(1-\dfrac{2h}{R}\right).$$

Thus the apparent weight at the height $$h$$ above the pole is simply

$$W_{p}=m\,g_{p}=m\,g\left(1-\dfrac{2h}{R}\right).$$

The problem states that the two spring-balance readings are the same, i.e.

$$W_{e}=W_{p}.$$

Substituting the expressions we have obtained:

$$m\left(g-\omega^{2}R\right)=m\,g\left(1-\dfrac{2h}{R}\right).$$

We cancel the common factor $$m$$ on both sides, giving

$$g-\omega^{2}R=g\left(1-\dfrac{2h}{R}\right).$$

Expanding the right-hand side:

$$g-\omega^{2}R=g-\dfrac{2gh}{R}.$$

Now we subtract $$g$$ from both sides to isolate the terms that contain $$h$$:

$$-\omega^{2}R=-\dfrac{2gh}{R}.$$

Multiplying by $$-1$$ removes the minus signs:

$$\omega^{2}R=\dfrac{2gh}{R}.$$

Finally we solve for $$h$$ by multiplying both sides by $$\dfrac{R}{2g}$$:

$$h=\dfrac{\omega^{2}R^{2}}{2g}.$$

This matches Option A.

Hence, the correct answer is Option A.

Let the mass of the planet be $$M$$. At the instant just before the collision body $$A$$ of mass $$m$$ is moving in a circular path of radius $$R$$ with speed $$v$$. For a circular orbit we first recall the condition obtained by equating the required centripetal force to the gravitational attraction

$$\frac{m v^{2}}{R}= \frac{G M m}{R^{2}}\; ,\qquad\Rightarrow\qquad v^{2}= \frac{G M}{R}$$

The gravitational potential energy of $$A$$ in that orbit is

$$U_{A}= -\,\frac{G M m}{R}= -m v^{2}$$

and its kinetic energy is

$$K_{A}= \frac12 m v^{2}$$

so the total mechanical energy of $$A$$ is

$$E_{A}=K_{A}+U_{A}=\frac12 m v^{2}-m v^{2}= -\frac12 m v^{2}$$

Now a second body $$B$$ of mass $$\dfrac{m}{2}$$ approaches and collides head-on (along the same tangential direction) with $$A$$ with speed $$\dfrac{v}{2}$$. The collision is completely inelastic, hence the two bodies stick together. During the very short collision the only significant forces are their mutual internal forces; the external gravitational force acts radially and therefore exerts no impulse in the tangential direction. Consequently the linear (tangential) momentum is conserved.

Initial tangential momentum:

$$p_{\text{initial}} \;=\; m\,v \;+\;\frac{m}{2}\,\frac{v}{2}\;=\;m v+\frac14 m v \;=\;\frac54\,m v$$

Total mass after collision:

$$m_{\text{tot}} = m+\frac{m}{2}= \frac32\,m$$

If $$V$$ is the common speed immediately after sticking together, conservation of momentum gives

$$\frac54\,m v = \frac32\,m\,V \quad\Rightarrow\quad V = \frac{\frac54}{\frac32}\,v = \frac{5}{4}\times\frac{2}{3}\,v = \frac56\,v$$

Thus the speed of the combined body has fallen to $$\dfrac56\,v$$.

Its kinetic energy right after the collision is

$$K' = \frac12\Bigl(\frac32\,m\Bigr) V^{2} =\frac34\,m\,\Bigl(\frac56\,v\Bigr)^{2} =\frac34\,m\,\frac{25}{36}\,v^{2} =\frac{25}{48}\,m v^{2}$$

The collision occurs at the same distance $$R$$ from the planet, so the gravitational potential energy of the combined body is

$$U' = -\,\frac{G M}{R}\Bigl(\frac32\,m\Bigr) =-\frac32\,m\,v^{2}$$ (because $$v^{2}= \dfrac{G M}{R}$$)

Hence the total mechanical energy after collision is

$$E' = K' + U' = \frac{25}{48}\,m v^{2} \;-\;\frac32\,m v^{2} = \frac{25}{48}\,m v^{2}-\frac{72}{48}\,m v^{2} = -\frac{47}{48}\,m v^{2}$$

We immediately notice that $$E'$$ is negative, so the system is still gravitationally bound; therefore it cannot escape (escape requires $$E\ge 0$$).

Next, compare the new speed $$V=\dfrac56\,v$$ with the speed required for a circular orbit at radius $$R$$. The circular-orbit speed does not depend on the mass of the moving body, and we have already found it to be $$v$$. Because

$$V=\frac56\,v \;\lt \; v$$

the combined body has less speed than needed for a circular path, so it cannot continue in the same circular orbit. However, its angular momentum about the planet is not zero:

$$L' = m_{\text{tot}} V R = \Bigl(\frac32\,m\Bigr)\Bigl(\frac56\,v\Bigr)R = \frac54\,m v R \ne 0$$

Because angular momentum is non-zero the body cannot fall straight towards the planet; instead it must follow another bound orbit. For a gravitational potential, bound non-circular orbits are elliptical. Hence the new path is an ellipse with the point of collision lying on that ellipse (specifically at the apogee, since the speed there is lower than the circular value).

Therefore the combined body starts moving in an elliptical orbit around the planet.

Hence, the correct answer is Option 4.

At the north pole the spring balance reads the true gravitational weight, so we have

$$W_{\text{pole}} = m\,g$$

The reading is given as $$W_{\text{pole}} = 196\ \text{N}$$ and the acceleration due to gravity at the pole is $$g = 10\ \text{m s}^{-2}$$. Substituting these values we can find the mass of the box:

$$m = \frac{W_{\text{pole}}}{g} = \frac{196}{10} = 19.6\ \text{kg}$$

When the box is taken to the equator it revolves with the Earth, so a centrifugal force acts outward. A spring balance measures the normal reaction, which equals the effective weight. The effective acceleration $$g'$$ at the equator is therefore reduced by the centrifugal term. The relation is

$$g' = g - \omega^{2}R$$

Here $$\omega$$ is the angular speed of the Earth and $$R$$ is the radius of the Earth.

First, we calculate $$\omega$$. The Earth completes one rotation in 24 hours:

$$\omega = \frac{2\pi}{T}, \qquad T = 24 \times 3600\ \text{s}$$

$$\omega = \frac{2\pi}{24 \times 3600} = 7.27 \times 10^{-5}\ \text{rad s}^{-1}$$

Next, we compute $$\omega^{2}R$$. The radius is given as $$R = 6400\ \text{km} = 6.4 \times 10^{6}\ \text{m}$$.

$$\omega^{2} = (7.27 \times 10^{-5})^{2} = 5.29 \times 10^{-9}\ \text{s}^{-2}$$

Multiplying by the radius,

$$\omega^{2}R = 5.29 \times 10^{-9}\ \text{s}^{-2} \times 6.4 \times 10^{6}\ \text{m} = 33.856 \times 10^{-3}\ \text{m s}^{-2} = 0.033856\ \text{m s}^{-2}$$

Now we obtain the effective acceleration at the equator:

$$g' = 10 - 0.033856 = 9.966144\ \text{m s}^{-2}$$

The spring balance reading at the equator is

$$W_{\text{equator}} = m\,g' = 19.6\ \text{kg} \times 9.966144\ \text{m s}^{-2}$$

Performing the multiplication,

$$W_{\text{equator}} = 195.329\ \text{N} \approx 195.32\ \text{N}$$

Among the given choices, this value matches Option D.

Hence, the correct answer is Option D.

We have a satellite of mass $$M$$ that is fired vertically from the surface of the earth (radius $$R$$) with speed $$u$$. Its total mechanical energy just after launch is

$$E_0=\frac12Mu^{2}-\frac{GM_eM}{R}.$$

When it rises to a height $$R$$ above the surface its distance from the earth’s centre is $$r=R+R=2R$$. At that instant let its speed be $$v$$. Applying conservation of mechanical energy between the point of launch and the point $$r=2R$$, we write

$$\frac12Mu^{2}-\frac{GM_eM}{R}= \frac12Mv^{2}-\frac{GM_eM}{2R}.$$

Simplifying,

$$\frac12Mu^{2}-\frac{GM_eM}{R}+ \frac{GM_eM}{2R}= \frac12Mv^{2}

\;\Longrightarrow\;

\frac12Mv^{2}= \frac12Mu^{2}-\frac{GM_eM}{2R}.$$

Dividing by $$\tfrac12M$$ we get

$$v^{2}=u^{2}-\frac{GM_e}{R}. \quad -(1)$$

Immediately afterwards the satellite ejects a small rocket of mass $$\dfrac{M}{10}$$, so the remaining mass of the satellite becomes

$$M_s=\frac{9M}{10}.$$

The problem states that the satellite (of mass $$\dfrac{9M}{10}$$) now moves in a circular orbit of radius $$2R$$. The speed required for a circular orbit of radius $$r$$ is obtained from the centripetal-gravitational balance

$$\frac{v_c^{2}}{r}=\frac{GM_e}{r^{2}}

\;\Longrightarrow\;

v_c=\sqrt{\frac{GM_e}{r}}

=\sqrt{\frac{GM_e}{2R}}. \quad -(2)$$

During the extremely short interval of ejection, the only external force is gravity, which is radial; its impulse is therefore negligible. Hence the linear momentum of the system (satellite + rocket) is conserved separately in the radial and tangential directions.

Radial direction. Before ejection the entire mass $$M$$ is moving radially outward with speed $$v$$, so the initial radial momentum is

$$P_{r,\,\text{initial}}=Mv.$$

After ejection, the satellite is moving purely tangentially (no radial component), so all the radial momentum must be carried by the rocket (mass $$\dfrac{M}{10}$$). Let $$v_{r,\parallel}$$ be the radial component of the rocket’s velocity. Conservation gives

$$Mv=\frac{M}{10}\,v_{r,\parallel}

\;\Longrightarrow\;

v_{r,\parallel}=10v. \quad -(3)$$

Tangential direction. Initially the tangential momentum is zero (vertical launch). After ejection the satellite (mass $$\dfrac{9M}{10}$$) moves tangentially with speed $$v_c$$. Let $$v_{r,\perp}$$ be the tangential component of the rocket’s velocity. Conservation of tangential momentum gives

$$0=\frac{9M}{10}\,v_c+\frac{M}{10}\,v_{r,\perp}

\;\Longrightarrow\;

v_{r,\perp}=-9v_c. \quad -(4)$$

(The negative sign shows that the rocket’s tangential motion is opposite to that of the satellite.)

Magnitude of the rocket’s velocity. Because the radial and tangential components are perpendicular, the speed of the rocket is

$$v_r=\sqrt{v_{r,\parallel}^{2}+v_{r,\perp}^{2}}

=\sqrt{(10v)^{2}+(-9v_c)^{2}}

=\sqrt{100v^{2}+81v_c^{2}}. \quad -(5)$$

Kinetic energy of the rocket. Using $$K=\tfrac12mv^{2}$$,

$$K_r=\frac12\left(\frac{M}{10}\right)v_r^{2}

=\frac{M}{20}\bigl(100v^{2}+81v_c^{2}\bigr). \quad -(6)$$

Now substitute from (1) and (2):

$$v^{2}=u^{2}-\frac{GM_e}{R}, \qquad

v_c^{2}=\frac{GM_e}{2R}.$$

Putting these into (6),

$$

\begin{aligned}

K_r

&=\frac{M}{20}\Bigl[100\Bigl(u^{2}-\frac{GM_e}{R}\Bigr)

+81\Bigl(\frac{GM_e}{2R}\Bigr)\Bigr] \\

&=\frac{M}{20}\Bigl[100u^{2}-100\frac{GM_e}{R}

+\frac{81}{2}\frac{GM_e}{R}\Bigr] \\

&=\frac{M}{20}\Bigl[100u^{2}-100\frac{GM_e}{R}

+40.5\frac{GM_e}{R}\Bigr] \\

&=\frac{M}{20}\Bigl[100u^{2}-59.5\frac{GM_e}{R}\Bigr].

\end{aligned}

$$

Factorising the common factor $$100$$ inside the bracket,

$$

K_r

=\frac{M}{20}\cdot100\Bigl[u^{2}-0.595\frac{GM_e}{R}\Bigr]

=5M\left(u^{2}-\frac{119}{200}\frac{GM_e}{R}\right). \quad -(7)

$$

This matches Option B exactly.

Hence, the correct answer is Option 2.

We are told that on the $$x$$-axis the gravitational field (which we may denote by $$\vec g(x)$$) has only an $$x$$-component and is given by

$$\vec g(x)=\frac{A x}{(x^2+a^2)^{3/2}}\;\hat i.$$

For a one-dimensional situation along the $$x$$-axis, the relation between gravitational field and gravitational potential $$\phi(x)$$ is the well-known formula

$$\vec g(x)= -\,\frac{d\phi}{dx}\;\hat i.$$

In scalar form (the direction is already along $$\hat i$$) this becomes

$$\frac{d\phi}{dx}= -\,\frac{A x}{(x^2+a^2)^{3/2}}.$$

We now integrate both sides with respect to $$x$$. Taking the potential to be zero at infinity, we write

$$\phi(x)-\phi(\infty)=\int_{\infty}^{x}\frac{d\phi}{dx'}\,dx'=-\int_{\infty}^{x}\frac{A x'}{(x'^2+a^2)^{3/2}}\,dx'.$$

Because $$\phi(\infty)=0$$, this simplifies to

$$\phi(x)= -A\int_{\infty}^{x}\frac{x'}{(x'^2+a^2)^{3/2}}\;dx'.$$

To evaluate the integral, we first consider the indefinite form

$$\int \frac{x}{(x^2+a^2)^{3/2}}\;dx.$$

We observe that

$$\frac{d}{dx}\left(\frac{1}{\sqrt{x^2+a^2}}\right)=\frac{d}{dx}\left((x^2+a^2)^{-1/2}\right)= -\frac{1}{2}(x^2+a^2)^{-3/2}\cdot 2x=-\frac{x}{(x^2+a^2)^{3/2}}.$$

Thus,

$$\int \frac{x}{(x^2+a^2)^{3/2}}\;dx = -\frac{1}{\sqrt{x^2+a^2}}+C.$$

Substituting this result back into the definite integral we have

$$\phi(x)= -A\Bigl[-\frac{1}{\sqrt{x'^2+a^2}}\Bigr]_{\infty}^{x}=A\Bigl[\frac{1}{\sqrt{x'^2+a^2}}\Bigr]_{\infty}^{x}.$$

Now we evaluate the limits. At the upper limit $$x' = x$$ we obtain $$\dfrac{1}{\sqrt{x^2+a^2}}$$, while at the lower limit $$x' \to \infty$$ we have $$\dfrac{1}{\sqrt{\infty}}=0$$. Hence

$$\phi(x)=\frac{A}{\sqrt{x^2+a^2}}.$$

Therefore the magnitude of the gravitational potential on the $$x$$-axis at a distance $$x$$ is

$$\boxed{\displaystyle \frac{A}{\sqrt{x^2+a^2}}}.$$

Hence, the correct answer is Option A.

Let $$g$$ denote the acceleration due to gravity at the surface of the Earth. We need the expressions for gravity when we go up to a height and when we go down to a depth.

First, the standard relation for acceleration due to gravity at a height $$h$$ above the surface is stated as $$g_h = g\left(\frac{R}{R+h}\right)^2.$$ This follows from Newton’s law of gravitation, because the distance of the mass from the centre becomes $$R+h$$ and the force is inversely proportional to the square of that distance.

Second, the standard relation for acceleration due to gravity at a depth $$d$$ below the surface is stated as $$g_d = g\left(1-\frac{d}{R}\right).$$ Inside the Earth the gravitational field falls off linearly with distance from the centre, giving this simple linear reduction factor.

According to the question we are given that the value of gravity at height $$h = \dfrac{R}{2}$$ equals the same value at some depth $$d$$. So we write $$g_h = g_d.$$

Substituting the formulae, we have $$g\left(\frac{R}{R+\dfrac{R}{2}}\right)^2 = g\left(1-\frac{d}{R}\right).$$

Now we simplify the left-hand side step by step. The denominator inside the parentheses is $$R + \frac{R}{2} = \frac{3R}{2}.$$ So inside the square we get $$\frac{R}{\dfrac{3R}{2}} = \frac{R}{1}\times \frac{2}{3R} = \frac{2}{3}.$$

Squaring this fraction, $$\left(\frac{2}{3}\right)^2 = \frac{4}{9}.$$ Hence the left-hand side becomes $$g \times \frac{4}{9}.$$

The equation thus reduces to $$g\left(\frac{4}{9}\right) = g\left(1-\frac{d}{R}\right).$$

We can cancel the common factor $$g$$ on both sides, giving $$\frac{4}{9} = 1-\frac{d}{R}.$$

Now we isolate the ratio $$\dfrac{d}{R}$$. Transposing the fraction yields $$\frac{d}{R} = 1 - \frac{4}{9}.$$

Writing the right-hand side with a common denominator, $$1 = \frac{9}{9},$$ so $$\frac{d}{R} = \frac{9}{9} - \frac{4}{9} = \frac{5}{9}.$$

Thus the required ratio is $$\boxed{\dfrac{d}{R} = \dfrac{5}{9}}.$$

Hence, the correct answer is Option B.

For a satellite moving under the central gravitational force of a planet, there is no external torque about the planet. Therefore, the angular momentum of the satellite about the planet remains conserved throughout the orbit. The statement of conservation of angular momentum is:

$$m\,r\,v = \text{constant}$$

Here $$m$$ is the mass of the satellite, $$r$$ is the instantaneous distance from the planet, and $$v$$ is the corresponding linear speed. We compare two special points of the elliptical orbit:

• Closest point (periapsis): distance $$r_c$$ and speed $$v_c$$.
• Farthest point (apoapsis): distance $$r_f$$ and speed $$v_f$$.

Applying the conservation law between these two points, we have

$$m\,r_c\,v_c \;=\; m\,r_f\,v_f.$$

Because the satellite’s mass $$m$$ is common to both terms, it cancels:

$$r_c\,v_c \;=\; r_f\,v_f.$$

The problem states that when the satellite is farthest from the planet, its speed is six times smaller than when it is closest. Expressing this in symbols,

$$v_f \;=\;\dfrac{v_c}{6}.$$

Substituting this value of $$v_f$$ into the conserved‐angular‐momentum equation gives

$$r_c\,v_c \;=\; r_f\left(\dfrac{v_c}{6}\right).$$

Now the speed $$v_c$$ appears on both sides, so it divides out:

$$r_c \;=\;\dfrac{r_f}{6}.$$

Rearranging, the ratio of distances becomes

$$\dfrac{r_c}{r_f} \;=\;\dfrac{1}{6}.$$

Thus, the distance of the satellite from the planet at the closest point is one‐sixth of its distance at the farthest point. Writing this as a ratio closest : farthest, we get

$$r_c : r_f \;=\; 1 : 6.$$

Hence, the correct answer is Option A.

We are told that the mass-density inside the planet depends on the distance from the centre according to

$$\rho(r)=\rho_0\!\left(1-\frac{r^{2}}{R^{2}}\right).$$

To find the point where the gravitational field $$g(r)$$ is maximum we must first obtain an explicit expression for $$g(r)$$ inside the planet and then differentiate it with respect to $$r$$.

The gravitational field at a point situated at a distance $$r<R$$ from the centre depends only on the mass contained within the sphere of radius $$r$$. By Gauss’s law for gravitation (equivalent to Newton’s law), we have

$$g(r)=\frac{G\,M(r)}{r^{2}},$$

where $$M(r)$$ is the mass enclosed within radius $$r$$.

To calculate $$M(r)$$, we integrate the density over the volume from the centre up to $$r$$:

$$M(r)=\int_{0}^{r} 4\pi r'^{2}\,\rho(r')\,dr'.$$

Substituting $$\rho(r')=\rho_0\!\left(1-\dfrac{r'^{2}}{R^{2}}\right)$$, we get

$$M(r)=4\pi\rho_0\!\int_{0}^{r}\!\left(1-\frac{r'^{2}}{R^{2}}\right)r'^{2}dr'.$$

Expanding inside the integral,

$$M(r)=4\pi\rho_0\!\left[\int_{0}^{r}r'^{2}dr'-\frac{1}{R^{2}}\int_{0}^{r}r'^{4}dr'\right].$$

Using the power-integral formulas $$\displaystyle\int r'^{2}dr'=\frac{r'^{3}}{3}$$ and $$\displaystyle\int r'^{4}dr'=\frac{r'^{5}}{5},$$ we obtain

$$M(r)=4\pi\rho_0\!\left[\frac{r^{3}}{3}-\frac{r^{5}}{5R^{2}}\right].$$

Now we substitute this result into $$g(r)=\dfrac{G\,M(r)}{r^{2}}$$:

$$g(r)=\frac{G}{r^{2}}\;4\pi\rho_0\!\left[\frac{r^{3}}{3}-\frac{r^{5}}{5R^{2}}\right].$$

Simplifying the factors of $$r$$,

$$g(r)=4\pi G\rho_0\left[\frac{r}{3}-\frac{r^{3}}{5R^{2}}\right].$$

It is convenient to pull out a common constant $$k=4\pi G\rho_0$$, so

$$g(r)=k\,r\!\left(\frac13-\frac{r^{2}}{5R^{2}}\right).$$

To locate the maximum of $$g(r)$$ for $$0<r<R$$ we differentiate with respect to $$r$$ and set the derivative to zero. Let

$$f(r)=r\!\left(\frac13-\frac{r^{2}}{5R^{2}}\right) \quad\text{so that}\quad g(r)=k\,f(r).$$

Differentiating term by term, we have

$$\frac{df}{dr}=\left(\frac13-\frac{r^{2}}{5R^{2}}\right)+r\!\left(-\frac{2r}{5R^{2}}\right).$$

The two terms give

$$\frac{df}{dr}=\frac13-\frac{r^{2}}{5R^{2}}-\frac{2r^{2}}{5R^{2}}=\frac13-\frac{3r^{2}}{5R^{2}}.$$

Setting $$\dfrac{df}{dr}=0$$ for an extremum,

$$\frac13-\frac{3r^{2}}{5R^{2}}=0.$$ $$\frac{3r^{2}}{5R^{2}}=\frac13.$$ $$r^{2}=R^{2}\left(\frac13\cdot\frac{5}{3}\right)=\frac{5}{9}R^{2}.$$

Taking the positive square root (since $$r$$ is a radial distance),

$$r=\sqrt{\frac{5}{9}}\;R=\frac{\sqrt5}{3}\,R.$$

This value indeed lies within the planet ($$r<R$$) and represents the point where $$g(r)$$ attains its maximum. Comparing with the given choices, this corresponds to Option D.

Hence, the correct answer is Option D.

We begin by recalling that the speed of a satellite in a circular orbit of radius $$r$$ around a planet of mass $$M$$ is obtained by equating the gravitational attraction to the centripetal force.

Gravitational force: $$F_g=\dfrac{G M m}{r^{2}}.$$

Centripetal force needed for circular motion: $$F_c=\dfrac{m v_{\text{orbit}}^{2}}{r}.$$

Setting $$F_g=F_c$$ gives

$$\dfrac{G M m}{r^{2}}=\dfrac{m v_{\text{orbit}}^{2}}{r}.$$

The mass $$m$$ of the body cancels out, and one factor of $$r$$ in the denominator cancels as well, so we have

$$\dfrac{G M}{r}=v_{\text{orbit}}^{2}.$$

Taking the square root on both sides, we obtain the orbital speed,

$$v_{\text{orbit}}=\sqrt{\dfrac{G M}{r}}.$$

In this problem the orbit is said to be “low,” meaning its radius is practically the planet’s radius itself, $$r\approx R.$$ Substituting $$r=R$$ we get

$$v_{\text{orbit}}=\sqrt{\dfrac{G M}{R}}.$$

Next, we state the formula for escape velocity. A body will just escape the planet’s gravitational field when its kinetic energy at the surface equals the magnitude of its gravitational potential energy (taking zero potential energy at infinity):

$$\dfrac{1}{2}m v_{\text{escape}}^{2}=\dfrac{G M m}{R}.$$

Again $$m$$ cancels, and solving for $$v_{\text{escape}}$$ we obtain

$$v_{\text{escape}}=\sqrt{\dfrac{2 G M}{R}}.$$

We are asked for the ratio $$\dfrac{v_{\text{orbit}}}{v_{\text{escape}}}.$$ Substituting the two expressions derived above:

$$\dfrac{v_{\text{orbit}}}{v_{\text{escape}}}=\dfrac{\sqrt{\dfrac{G M}{R}}}{\sqrt{\dfrac{2 G M}{R}}}.$$

The common factor $$\sqrt{\dfrac{G M}{R}}$$ in numerator and denominator simplifies, leaving

$$\dfrac{v_{\text{orbit}}}{v_{\text{escape}}}=\dfrac{1}{\sqrt{2}}.$$

Hence, the correct answer is Option A.

We know that the escape-velocity from the surface of a spherical planet of mass $$M$$ and radius $$R$$ is given by the standard formula

$$v_{\text{esc}} \;=\;\sqrt{\dfrac{2\,G\,M}{R}}$$

where $$G$$ is the universal gravitational constant.

First we apply this formula to Planet $$A$$. For Planet $$A$$ the mass is $$M_A = M$$ and the radius is $$R_A = R$$, so

$$v_A \;=\;\sqrt{\dfrac{2\,G\,M_A}{R_A}} \;=\;\sqrt{\dfrac{2\,G\,M}{R}}.$$

Now we consider Planet $$B$$. By statement of the question, Planet $$B$$ has half the mass and half the radius of Planet $$A$$, that is

$$M_B = \dfrac{M}{2}, \qquad R_B = \dfrac{R}{2}.$$

Substituting these values into the escape-velocity formula gives

$$\begin{aligned} v_B &= \sqrt{\dfrac{2\,G\,M_B}{R_B}} \\ &= \sqrt{\dfrac{2\,G\,\left(\dfrac{M}{2}\right)}{\dfrac{R}{2}}}. \end{aligned}$$

We simplify the expression inside the square root step by step:

$$\dfrac{2\,G\,\left(\dfrac{M}{2}\right)}{\dfrac{R}{2}} = \dfrac{G\,M}{\dfrac{R}{2}} = G\,M \times \dfrac{2}{R} = \dfrac{2\,G\,M}{R}.$$

Therefore

$$v_B = \sqrt{\dfrac{2\,G\,M}{R}}.$$

We can now compare the two escape velocities:

$$v_A = \sqrt{\dfrac{2\,G\,M}{R}}, \qquad v_B = \sqrt{\dfrac{2\,G\,M}{R}}.$$

Clearly, these two quantities are equal, so

$$\dfrac{v_A}{v_B} = 1.$$

The question states that this ratio can be written as $$\dfrac{v_A}{v_B} = \dfrac{n}{4}.$$ Equating the two expressions for the ratio, we have

$$\dfrac{n}{4} = 1 \;\;\Longrightarrow\;\; n = 4.$$

Hence, the correct answer is Option A.

Let the mass of the earth be denoted by $$M_E$$ and the universal gravitational constant by $$G$$. For a satellite of mass $$M$$ moving in a circular orbit of radius $$R$$ the velocity is obtained from the condition for uniform circular motion

$$\displaystyle \frac{G\,M_E\,M}{R^2}= \frac{M v^2}{R}\;\;\Longrightarrow\;\;v=\sqrt{\frac{G\,M_E}{R}}\;.$$

The meteorite has the same mass $$M$$ and, just before impact, is given to have the same speed $$v$$. At the instant of collision the two velocity directions are orthogonal: the satellite velocity is tangential while the meteorite velocity is radially inward. Hence the angle between $$\vec v_{\!s}$$ (satellite) and $$\vec v_{\!m}$$ (meteorite) is $$90^\circ\;.$$

The collision is perfectly inelastic, so the two bodies stick together to form a single body of mass $$2M$$. Because external forces on the system are purely central (gravity through the earth’s centre), the angular momentum about the earth’s centre is conserved during the collision.

Initially, only the satellite possesses angular momentum because the meteorite is moving radially. Using $$\vec L=M\;\vec r\times\vec v$$ with $$r=R$$ we have

$$L_i = M\,R\,v\;.$$

After the collision the linear momenta add vectorially:

$$\vec p_f = M\vec v_{\!s}+M\vec v_{\!m} = M(\vec v_{\!s}+\vec v_{\!m}).$$

Since the two velocities are at right angles and equal in magnitude, their vector sum has magnitude

$$|\vec v_{\!s}+\vec v_{\!m}| = v\sqrt{2}\;.$$

The velocity of the fused body (mass $$2M$$) is therefore

$$\vec v'=\frac{\vec p_f}{2M}= \frac{\vec v_{\!s}+\vec v_{\!m}}{2},\qquad |\vec v'|=\frac{v\sqrt{2}}{2}= \frac{v}{\sqrt2}\;.$$

Because $$\vec v'$$ bisects the right angle between the original velocities, it makes $$45^\circ$$ with the tangent. Its tangential and radial components are each

$$v_t=\frac{v}{2},\qquad v_r=\frac{v}{2}\;.$$

The conserved angular momentum after collision equals

$$L_f = 2M\,R\,v_t = 2M\,R\left(\frac{v}{2}\right)=M\,R\,v=L_i\;,$$

so angular momentum conservation is satisfied.

The specific (per-unit-mass) angular momentum of the combined body is

$$h=\frac{L_f}{2M}=R\,v_t = R\left(\frac{v}{2}\right)=\frac{R\,v}{2}\;.$$

Next we evaluate the specific mechanical energy $$\varepsilon=\dfrac{v'^2}{2}-\dfrac{G\,M_E}{R}\;.$$ Using $$v'^2=\dfrac{v^2}{2}$$ and $$v^2=\dfrac{G\,M_E}{R}$$ we obtain

$$\varepsilon=\frac{1}{2}\left(\frac{v^2}{2}\right)-\frac{v^2}{1}= \frac{v^2}{4}-v^2=-\frac{3\,v^2}{4}\;.$$

The energy is negative, indicating a bound orbit.

To decide whether the bound orbit is circular or elliptical we employ the standard formula for the eccentricity $$e$$ of a Keplerian orbit expressed in terms of specific energy $$\varepsilon$$ and specific angular momentum $$h$$ :

$$e=\sqrt{\,1+\frac{2\varepsilon h^{2}}{(G\,M_E)^{2}}\,}\;.$$

Because $$v^2=\dfrac{G\,M_E}{R}$$ we write $$G\,M_E = v^2 R\;.$$ Now compute the term inside the square root:

$$\frac{2\varepsilon h^{2}}{(G\,M_E)^{2}} =\frac{2\left(-\dfrac{3v^{2}}{4}\right)\left(\dfrac{R^{2}v^{2}}{4}\right)}{(v^{2}R)^{2}} =\frac{-3R^{2}v^{4}/8}{R^{2}v^{4}} =-\frac{3}{8}\;.$$

Therefore

$$e^{2}=1-\frac{3}{8}=\frac{5}{8}\quad\Longrightarrow\quad e=\sqrt{\frac{5}{8}}<1\;.$$

An eccentricity less than unity but greater than zero corresponds to an ellipse that is not a circle. Thus the fused body of mass $$2M$$ moves in an elliptical orbit whose apogee is at the collision point (radius $$R$$) and whose perigee lies closer to the earth.

Hence, the correct answer is Option A.

We know that the acceleration due to gravity at any distance from the centre of Earth is given by the Newtonian formula $$g \;=\; \dfrac{G\,M}{r^{\,2}}$$ where $$G$$ is the gravitational constant, $$M$$ is the mass of Earth and $$r$$ is the distance of the point from the centre of Earth.

At Earth’s surface the distance from the centre is simply the radius $$R$$ of Earth, so the surface value is

$$g \;=\; \dfrac{G\,M}{R^{\,2}} \;=\; 9.8 \text{ m s}^{-2}.$$

If we go to an altitude $$h$$ above the surface, the distance from the centre becomes $$R+h$$ and the acceleration due to gravity there (call it $$g'$$) is

$$g' \;=\; \dfrac{G\,M}{(R+h)^{2}}.$$

To find the relation between $$g'$$ and $$g$$, we divide the two expressions:

$$\dfrac{g'}{g} \;=\; \dfrac{\dfrac{G\,M}{(R+h)^{2}}}{\dfrac{G\,M}{R^{\,2}}} \;=\; \dfrac{R^{\,2}}{(R+h)^{2}}.$$

Simplifying the right-hand side, we get the useful relation

$$g' \;=\; g\;\left(\dfrac{R}{R+h}\right)^{2}.$$

According to the question, the value at altitude must be one-half of the surface value, i.e.

$$g' = 4.9 \text{ m s}^{-2} = \dfrac{1}{2}\,g.$$

Substituting this condition into the previous relation, we have

$$\dfrac{1}{2}\,g \;=\; g\;\left(\dfrac{R}{R+h}\right)^{2}.$$

Now we cancel the common factor $$g$$ on both sides:

$$\dfrac{1}{2} \;=\; \left(\dfrac{R}{R+h}\right)^{2}.$$

Taking the square root of both sides (and keeping only the positive root because distances are positive), we obtain

$$\sqrt{\dfrac{1}{2}} \;=\; \dfrac{R}{R+h},$$

so

$$\dfrac{1}{\sqrt{2}} \;=\; \dfrac{R}{R+h}.$$

Cross-multiplying gives

$$R \;=\; \dfrac{R+h}{\sqrt{2}}.$$

Rewriting this in a clearer form, multiply both sides by $$\sqrt{2}$$:

$$\sqrt{2}\,R \;=\; R + h.$$

Now we isolate $$h$$ by subtracting $$R$$ from both sides:

$$h \;=\; (\sqrt{2} - 1)\,R.$$

We substitute the numerical value of Earth’s radius $$R = 6.4 \times 10^{6} \text{ m}$$ and the approximate value $$\sqrt{2} \approx 1.414$$:

$$h \;=\; (1.414 - 1)\,(6.4 \times 10^{6}) \text{ m}.$$

This difference $$1.414 - 1 = 0.414$$, so

$$h \;=\; 0.414 \times 6.4 \times 10^{6} \text{ m}.$$

Multiplying $$0.414$$ and $$6.4$$ step by step:

$$0.414 \times 6 = 2.484,$$

$$0.414 \times 0.4 = 0.1656,$$

Adding, $$2.484 + 0.1656 = 2.6496.$$

Thus

$$h \approx 2.65 \times 10^{6} \text{ m}.$$

Among the given options the closest numerical value is $$2.6 \times 10^{6} \text{ m}$$, which corresponds to Option B.

Hence, the correct answer is Option B.

We have a point mass of magnitude $$m$$ placed at the origin $$x = 0$$. A thin straight rod lies along the positive $$x$$-axis, beginning at $$x = a$$ and ending at $$x = a + L$$. Its mass per unit length (linear mass density) varies with position according to

$$\lambda(x)=A + Bx^{2}.$$

Our aim is to find the total gravitational force (magnitude) exerted by every small element of the rod on the mass at the origin. Throughout we shall use Newton’s law of gravitation, which in differential form for two small masses separated by a distance $$r$$ is

$$dF=\frac{G\,dm_1\,dm_2}{r^{2}}.$$

Here $$dm_1 = m$$ (the point mass) and $$dm_2 = dm$$ (an infinitesimal element of the rod). The separation between the element located at position $$x$$ and the origin is simply $$r = x$$. Substituting these facts, we get

$$dF = \frac{G\,m\,dm}{x^{2}}.$$

Because each element lies on the positive $$x$$-axis while the point mass is at the origin, every elemental force acts along the negative $$x$$-direction. We shall calculate its magnitude first and later attach the direction if required.

The infinitesimal mass element of the rod is given by the product of its linear density and an infinitesimal length $$dx$$:

$$dm = \lambda(x)\,dx = (A + Bx^{2})\,dx.$$

Placing this expression for $$dm$$ into the formula for $$dF$$, we obtain

$$dF = \frac{G\,m\,(A + Bx^{2})}{x^{2}}\;dx.$$

Now we separate the fraction:

$$dF = Gm\left(\frac{A}{x^{2}} + B\right)\,dx.$$

To find the total force, we integrate $$dF$$ from the near end of the rod at $$x = a$$ to the far end at $$x = a + L$$:

$$F = Gm \int_{a}^{a+L} \left(\frac{A}{x^{2}} + B\right)\,dx.$$

We evaluate the integral term by term.

First term: for $$\displaystyle \int \frac{A}{x^{2}}\,dx$$, we recall the standard result

$$\int x^{-2}\,dx = -x^{-1} + C,$$

so

$$\int \frac{A}{x^{2}}\,dx = A\int x^{-2}\,dx = -\frac{A}{x}.$$

Second term: for $$\displaystyle \int B\,dx$$, $$B$$ is a constant, hence

$$\int B\,dx = Bx.$$

Combining these results, the definite integral becomes

$$F = Gm\left[\,-\frac{A}{x} + Bx\;\right]_{x=a}^{x=a+L}.$$

We substitute the limits. First, the upper limit $$x = a + L$$:

Upper contribution  $$= -\frac{A}{a + L} + B(a + L).$$

Next, the lower limit $$x = a$$:

Lower contribution  $$= -\frac{A}{a} + Ba.$$

Subtracting the lower contribution from the upper gives the net value of the bracketed expression:

$$\begin{aligned} F &= Gm\Bigg[\left(-\frac{A}{a + L} + B(a + L)\right) \;-\; \left(-\frac{A}{a} + Ba\right)\Bigg] \\ &= Gm\Bigg[-\frac{A}{a + L} + B(a + L) + \frac{A}{a} - Ba\Bigg]. \end{aligned}$$

The two $$Ba$$ terms cancel:

$$B(a + L) - Ba \;=\; BL.$$

Hence the expression inside the brackets simplifies to

$$-\frac{A}{a + L} + \frac{A}{a} + BL.$$

Reordering the terms more conventionally, we write

$$\frac{A}{a} - \frac{A}{a + L} + BL.$$

Therefore the magnitude of the gravitational force is

$$F = Gm\left[A\!\left(\frac{1}{a} - \frac{1}{a + L}\right) + BL\right].$$

This matches the expression given in Option D.

Hence, the correct answer is Option D.

We place the four identical particles at the corners of a square ABCD of side $$a$$. Each particle has mass $$M$$ and all of them are made to revolve together, keeping the square rigid, about the centre $$O$$ of the square. Because they keep their relative positions unchanged, every particle describes a circle whose centre is $$O$$ and whose radius is the distance from a corner to the centre.

The diagonal of the square is $$a\sqrt 2$$, so half of this diagonal gives the radius of the circular path,

$$R=\dfrac{a\sqrt 2}{2}=\dfrac{a}{\sqrt 2}.$$

We now pick one particle, say the one at corner A. It experiences gravitational attraction from the three remaining particles situated at B, C and D. According to Newton’s law of gravitation, the magnitude of the gravitational force exerted by a particle of mass $$M$$ on another particle of mass $$M$$ separated by a distance $$r$$ is

$$F=\dfrac{G M^2}{r^2},$$

where $$G$$ is the universal gravitational constant.

1. Attraction by the two adjacent particles B and D    • Distance AB = AD = $$a$$    • Magnitude of each force

$$F_1=\dfrac{G M^2}{a^2}.$$

These two forces act along AB and AD, which are perpendicular to each other. The direction OA (from A towards the centre O) bisects the right angle at A, therefore it makes an angle of $$45^{\circ}$$ with each side. Resolving each adjacent force towards the centre, we obtain the radial component

$$F_{1r}=F_1\cos45^{\circ}=F_1\dfrac{1}{\sqrt 2}.$$

Because there are two identical adjacent forces, the total radial pull supplied by them is

$$F_{adj}=2F_{1r}=2F_1\dfrac{1}{\sqrt 2}=\sqrt 2\,F_1.$$

(The tangential components cancel each other because they are equal in magnitude and opposite in direction.)

2. Attraction by the particle at the opposite corner C    • Distance AC (full diagonal) $$=a\sqrt 2$$    • Magnitude of the force

$$F_2=\dfrac{G M^2}{(a\sqrt 2)^2}=\dfrac{G M^2}{2a^2}=\dfrac{F_1}{2}.$$

This force acts exactly along the diagonal AC, i.e. along AO, hence it is already directed towards the centre and needs no resolution.

Therefore the net gravitational force on the particle at A directed towards the centre is

$$F_{\text{net}}=F_{adj}+F_2=\sqrt 2\,F_1+\dfrac{F_1}{2}=\left(\sqrt 2+\dfrac12\right)F_1.$$

Substituting $$F_1=\dfrac{G M^2}{a^2}$$ gives

$$F_{\text{net}}=\left(\sqrt 2+\dfrac12\right)\dfrac{G M^2}{a^2}.$$

For uniform circular motion, the required centripetal force is provided exactly by this net inward force. The centripetal force needed for a particle of mass $$M$$ moving with speed $$v$$ in a circle of radius $$R$$ is

$$F_c=M\dfrac{v^2}{R}.$$

Equating the two forces,

$$M\dfrac{v^2}{R}=\left(\sqrt 2+\dfrac12\right)\dfrac{G M^2}{a^2}.$$

We cancel one factor of $$M$$ from both sides and substitute $$R=\dfrac{a}{\sqrt 2}$$:

$$\dfrac{v^2}{a/\sqrt 2}=\left(\sqrt 2+\dfrac12\right)\dfrac{G M}{a^2}.$$

Multiplying both sides by $$\dfrac{a}{\sqrt 2}$$ gives

$$v^2=\dfrac{1}{\sqrt 2}\left(\sqrt 2+\dfrac12\right)\dfrac{G M}{a}.$$

Simplify the numerical factor:

$$\dfrac{1}{\sqrt 2}\left(\sqrt 2+\dfrac12\right)=\dfrac{\sqrt 2}{\sqrt 2}+\dfrac{1}{2\sqrt 2}=1+\dfrac{1}{2\sqrt 2}.$$

The term $$\dfrac{1}{2\sqrt 2}\approx0.3536$$, so the bracket equals $$1.3536$$. Hence

$$v^2\approx1.3536\,\dfrac{G M}{a}.$$

Taking the square root,

$$v\approx1.16\,\sqrt{\dfrac{G M}{a}}.$$

This numerical coefficient (1.16 rounded to two decimal places) matches option D.

Hence, the correct answer is Option D.

Let the mass of the satellite be $$m$$ and the universal gravitational constant be $$G$$. The radius of the Earth is given as $$R = 6.4 \times 10^3 \text{ km}$$.

First, recall the formula for gravitational potential energy of a body of mass $$m$$ at a distance $$r$$ from the centre of the Earth:

$$U = -\dfrac{G M m}{r},$$

where $$M$$ is the mass of the Earth. The negative sign shows that the potential energy is taken to be zero at infinity.

The energy required to lift the satellite from the Earth’s surface (where the distance from the centre is $$R$$) to a height $$h$$ (where the distance from the centre is $$R + h$$) is simply the increase in potential energy. We therefore write

$$E_1 = U_{\text{final}} - U_{\text{initial}}.$$

Substituting the expressions for the potentials, we have

$$E_1 = \Bigl(-\dfrac{G M m}{R + h}\Bigr) \;-\; \Bigl(-\dfrac{G M m}{R}\Bigr) = G M m\Bigl(\dfrac{1}{R} - \dfrac{1}{R + h}\Bigr).$$

Now, to keep the satellite moving in a circular orbit of radius $$R + h$$, it must possess some kinetic energy. For a circular orbit, the necessary speed $$v$$ is found from the equality of centripetal force and gravitational attraction:

$$\dfrac{m v^2}{R + h} = \dfrac{G M m}{(R + h)^2}.$$

Simplifying, we get the orbital speed

$$v^2 = \dfrac{G M}{R + h}.$$

The kinetic energy corresponding to this speed is

$$E_2 = \dfrac{1}{2} m v^2 = \dfrac{1}{2} m \Bigl(\dfrac{G M}{R + h}\Bigr) = \dfrac{G M m}{2\,(R + h)}.$$

We are asked to find the height $$h$$ for which the lifting energy $$E_1$$ equals the orbital kinetic energy $$E_2$$. Hence we equate them:

$$G M m\Bigl(\dfrac{1}{R} - \dfrac{1}{R + h}\Bigr) = \dfrac{G M m}{2\,(R + h)}.$$

We notice that the common factor $$G M m$$ appears on both sides, so it can be cancelled:

$$\dfrac{1}{R} - \dfrac{1}{R + h} = \dfrac{1}{2\,(R + h)}.$$

To combine the left-hand side into a single fraction, we bring it to a common denominator:

$$\dfrac{(R + h) - R}{R\,(R + h)} = \dfrac{h}{R\,(R + h)}.$$

So the equality becomes

$$\dfrac{h}{R\,(R + h)} = \dfrac{1}{2\,(R + h)}.$$

Because the factor $$R + h$$ occurs in both denominators, it can again be cancelled from numerator and denominator on each side, yielding

$$\dfrac{h}{R} = \dfrac{1}{2}.$$

Now we solve for $$h$$:

$$h = \dfrac{R}{2}.$$

The radius of the Earth is $$R = 6.4 \times 10^3 \text{ km}$$, so substituting, we have

$$h = \dfrac{6.4 \times 10^3 \text{ km}}{2} = 3.2 \times 10^3 \text{ km}.$$

Hence, the correct answer is Option C.

For a simple pendulum the time‐period is given by the well-known formula

$$T = 2\pi\sqrt{\dfrac{l}{g}}\; ,$$

where $$l$$ is the length of the pendulum and $$g$$ is the acceleration due to gravity at the place of observation.

On Earth the period is provided as $$T_E = 2\ \text{s}.$$ So, on the Earth we have

$$2 = 2\pi\sqrt{\dfrac{l}{g_E}} \; ,$$

which we shall keep for later comparison. The length $$l$$ of the pendulum is the same everywhere; only $$g$$ changes from one celestial body to another.

Now we consider the given planet. Its mass is three times that of Earth, so

$$M_P = 3M_E.$$

Its diameter is also three times that of Earth, therefore its radius is three times Earth’s radius:

$$R_P = 3R_E.$$

The surface acceleration due to gravity is determined by Newton’s universal-gravitation expression

$$g = \dfrac{GM}{R^{2}},$$

where $$G$$ is the gravitational constant, $$M$$ is the mass of the body, and $$R$$ is its radius.

Applying this to the planet we write

$$g_P = \dfrac{G\,M_P}{R_P^{2}}.$$

Substituting $$M_P = 3M_E$$ and $$R_P = 3R_E$$ gives

$$g_P = \dfrac{G\,(3M_E)}{(3R_E)^{2}} = \dfrac{3G M_E}{9R_E^{2}} = \dfrac{1}{3}\,\dfrac{G M_E}{R_E^{2}} = \dfrac{g_E}{3}.$$

Thus the planet’s gravitational acceleration is one-third that of Earth:

$$g_P = \dfrac{g_E}{3}.$$

We now compute the pendulum’s period on the planet. Using the same length $$l$$ in the original period formula, we have

$$T_P = 2\pi\sqrt{\dfrac{l}{g_P}} = 2\pi\sqrt{\dfrac{l}{g_E/3}} = 2\pi\sqrt{\dfrac{3l}{g_E}} = \sqrt{3}\,\bigl(2\pi\sqrt{\dfrac{l}{g_E}}\bigr).$$

The expression in large parentheses is exactly the Earth period $$T_E$$, so

$$T_P = \sqrt{3}\,T_E.$$

Given $$T_E = 2\ \text{s},$$ we substitute:

$$T_P = \sqrt{3}\times 2 = 2\sqrt{3}\ \text{s}.$$

Hence, the correct answer is Option D.