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MHT CET 2024 (4 May, Shift 1) · Acceleration Due to Gravity

Acceleration Due to Gravity question from MHT CET 2024 (4 May, Shift 1)

Q27MHT CET 2024 · 4 MAY · SHIFT 1Easy

Acceleration Due to Gravity

If ' $R$ ' is the radius of earth \& ' $g$ ' is acceleration due to gravity on earth's surface, then mean density of earth is

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Answer: (c)

$g = \frac{GM}{R^2}$ and $M = \rho \cdot \frac{4}{3}\pi R^3$ .
So $g = \frac{4}{3}\pi G \rho R$ .
$\rho = \frac{3g}{4\pi RG}$
So the answer is (c) $\frac{3g}{4\pi RG}$ .

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