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MHT CET 2024 (11 May, Shift 1) · Acceleration Due to Gravity

Acceleration Due to Gravity question from MHT CET 2024 (11 May, Shift 1)

Q24MHT CET 2024 · 11 MAY · SHIFT 1Easy

Acceleration Due to Gravity

The acceleration due to gravity at the surface of the planet is same as that at the surface of the earth, but the density of planet is thrice that of the earth. If 'R' is the radius of the earth, the radius of the planet will be

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Answer: (b)

$g = \frac{4}{3}\pi G\rho R$ , so equal $g$ needs equal $\rho R$ .
$3\rho \cdot R_p = \rho R \Rightarrow R_p = \frac{R}{3}$
So the answer is (b) $\frac{R}{3}$ .

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