Q143 / 150 · MHT CET 2020 (19 Oct Shift 2) · Physics

The escape velocity of a body from any planet whose mass is six times the mass of earth and radius is twice the radius of earth will be $\left(\mathrm{V}_{e}=\right.$ escape velocity of a body from the earth's surface)

Correct answer
A
$\sqrt{3} \mathrm{~V}_{\mathrm{e}}$
B
$2 \mathrm{~V}_{\mathrm{e}}$
C
$\frac{3}{2} \mathrm{~V}_{\mathrm{e}}$
D
$2 \sqrt{2} \mathrm{~V}_{\mathrm{e}}$

Solution

(D) $\begin{aligned} \mathrm{V}_{\mathrm{e}} &=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\ \mathrm{V}_{\mathrm{e}}^{\prime} &=\sqrt{\frac{2 \mathrm{G} \times 6 \mathrm{M}}{2 \mathrm{R}}} \\ &=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}} \times 3} \\ &=\sqrt{3} \mathrm{~V}_{\mathrm{e}} \end{aligned}$

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

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