Q136 / 150 · MHT CET 2023 (09 May Shift 2) · Physics
Consider a planet whose density is same as that of the earth but whose radius is three times the radius ' $R$ ' of the earth. The acceleration due to gravity ' $\mathrm{g}_{\mathrm{n}}$ ' on the surface of planet is $g_n=x$. $g$ where $g$ is acceleration due to gravity on surface, of earth. The value of ' $\mathrm{x}$ ' is
- A$9$
- Correct answerB$3$
- C$\frac{1}{3}$
- D$\frac{1}{9}$
Solution
$\begin{array}{ll}
& \mathrm{As}, \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2} \text { and } \mathrm{M}=\rho \mathrm{V} \\
\therefore \quad & \mathrm{g}=\frac{\mathrm{G} \rho \mathrm{V}}{\mathrm{R}^2}=\frac{\mathrm{G} \rho \frac{4}{3} \pi \mathrm{R}^3}{\mathrm{R}^2} \\
\therefore \quad & \mathrm{g} \propto \mathrm{R}
\end{array}$
For the planet: Radius $\mathrm{R}=3 \mathrm{R}$
$\begin{array}{ll}
\therefore \quad & \mathrm{g}_{\text {planet }}=\frac{\mathrm{G} \mathrm{V}_{\text {planet }}}{(3 \mathrm{R})^2} \\
& \text { where } \mathrm{V}_{\text {planet }}=\frac{4}{3} \pi(3 \mathrm{R})^3 \\
\therefore \quad & \mathrm{g}_{\text {planet }}=\frac{\mathrm{G} \rho \frac{4}{3} \pi(3 \mathrm{R})^3}{(3 \mathrm{R})^2} \\
\therefore \quad & \mathrm{g}_{\text {planet }} \propto 3 \mathrm{R}
\end{array}$
$\therefore \quad \frac{\mathrm{g}}{\mathrm{g}_{\text {planet }}}=\frac{\mathrm{R}}{3 \mathrm{R}} \quad$....(since $\rho$ is constant)
$\begin{array}{ll}\therefore & \mathrm{g}_{\text {planet }}=3 \mathrm{~g} \\ \therefore & \mathrm{x}=3\end{array}$
Hence, the correct answer is option: 'B'.