JEE Physics 2023 Question with Solution

Given: The escape velocities of planets A and B are in the ratio $1:2$ and their radii are in the ratio $1:3$.

Find: The ratio $\frac{g_A}{g_B}$.

Using the relation for escape velocity,

$$V_e = \sqrt{\frac{2GM}{R}}$$

and with $M = \rho \times \frac{4}{3}\pi R^3$, we get

$$V_e \propto \sqrt{\rho R^2} = R\sqrt{\rho}$$

So,

$$\frac{V_{eA}}{V_{eB}} = \frac{R_A\sqrt{\rho_A}}{R_B\sqrt{\rho_B}} = \frac{1}{2}$$

Given $\frac{R_A}{R_B} = \frac{1}{3}$,

$$\frac{1}{3}\sqrt{\frac{\rho_A}{\rho_B}} = \frac{1}{2}$$ $$\sqrt{\frac{\rho_A}{\rho_B}} = \frac{3}{2}$$ $$\frac{\rho_A}{\rho_B} = \frac{9}{4}$$

Now acceleration due to gravity is

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

Using $M = \frac{4}{3}\pi R^3\rho$,

$$g \propto \rho R$$

Hence,

$$\frac{g_A}{g_B} = \frac{\rho_A R_A}{\rho_B R_B} = \frac{9}{4} \times \frac{1}{3} = \frac{3}{4}$$

Therefore, the ratio of acceleration due to gravity is $\frac{3}{4}$. The solution concludes that the correct option is C, but the listed options show $\frac{3}{4}$ as option D. The defensible answer by value is D.

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

Find: $\frac{g_A}{g_B}$.

From

$$V_e = \sqrt{\frac{2GM}{R}}$$

we write

$$V_e^2 \propto \frac{M}{R}$$

Since $M \propto \rho R^3$,

$$V_e^2 \propto \rho R^2$$

Therefore,

$$\left(\frac{V_{eA}}{V_{eB}}\right)^2 = \frac{\rho_A R_A^2}{\rho_B R_B^2}$$

Substitute the given ratio:

$$\frac{1}{4} = \frac{\rho_A}{\rho_B} \times \frac{1}{9}$$ $$\frac{\rho_A}{\rho_B} = \frac{9}{4}$$

Now,

$$g = \frac{GM}{R^2} \propto \frac{\rho R^3}{R^2} = \rho R$$

So,

$$\frac{g_A}{g_B} = \frac{\rho_A}{\rho_B} \times \frac{R_A}{R_B} = \frac{9}{4} \times \frac{1}{3} = \frac{3}{4}$$

Thus, the required ratio is $\frac{3}{4}$.