Q147 / 150 · MHT CET 2024 (15 May Shift 2) · Physics

The density of a new planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If $R$ is the radius of earth, then radius of the planet would be

  • A
    $\quad 4 \mathrm{R}$
  • Correct answer
    B
    $\mathrm{R} / 2$
  • C
    $\frac{R}{4}$
  • D
    $\quad 2 \mathrm{R}$

Solution

$\begin{aligned} & \text { Given, } \rho_{\mathrm{p}}=2 \rho_{\mathrm{e}}, \mathrm{g}_{\mathrm{p}}=\mathrm{g}_{\mathrm{e}} \\ & \mathrm{g}=\frac{4}{3} \pi \rho \mathrm{GR}\end{aligned}$ $\begin{array}{ll}\therefore & \frac{R_p}{R_e}=\left(\frac{g_p}{g_e}\right)\left(\frac{\rho_e}{\rho_p}\right)=(1) \times\left(\frac{1}{2}\right) \\ \therefore & R_p=\frac{R_e}{2}=\frac{R}{2}\end{array}$

Hence, the correct answer is option: 'B'.

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