Q57 / 75 · JEE Main 2026 (22 January Shift 1) · Physics
The escape velocity from a spherical planet $A$ is $10 \mathrm{~km} / \mathrm{s}$. The escape velocity from another planet $B$ whose density and radius are $10 \%$ of those of planet $A$, is $\_\_\_\_$ $\mathrm{m} / \mathrm{s}$.
- A1000
- B$1000 \sqrt{2}$
- C$200 \sqrt{5}$
- Correct answerD$100 \sqrt{10}$
Solution
Escape velocity $v_e = R\sqrt{\frac{8\pi G\rho}{3}} \propto R\sqrt{\rho}$.
Planet B: $\rho_B = 0.1\rho_A$, $R_B = 0.1R_A$.
$\frac{v_B}{v_A} = \frac{0.1R_A\sqrt{0.1\rho_A}}{R_A\sqrt{\rho_A}} = 0.1\sqrt{0.1} = \frac{1}{10\sqrt{10}}$.
$v_B = 10 \times \frac{1}{10\sqrt{10}} = \frac{1}{\sqrt{10}}$ km/s $= \frac{1000}{\sqrt{10}} = 100\sqrt{10}$ m/s.
Planet B: $\rho_B = 0.1\rho_A$, $R_B = 0.1R_A$.
$\frac{v_B}{v_A} = \frac{0.1R_A\sqrt{0.1\rho_A}}{R_A\sqrt{\rho_A}} = 0.1\sqrt{0.1} = \frac{1}{10\sqrt{10}}$.
$v_B = 10 \times \frac{1}{10\sqrt{10}} = \frac{1}{\sqrt{10}}$ km/s $= \frac{1000}{\sqrt{10}} = 100\sqrt{10}$ m/s.
Hence, the correct answer is option: 'D'.