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MHT CET 2023 (9 May, Shift 2) · Acceleration Due to Gravity

Acceleration Due to Gravity question from MHT CET 2023 (9 May, Shift 2)

Q45MHT CET 2023 · 9 MAY · SHIFT 2Easy

Acceleration Due to Gravity

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 ' $g_n$ ' on the surface of planet is $g_n = x$ . $g$ where $g$ is acceleration due to gravity on surface of earth. The value of ' $x$ ' is

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Answer: (b)

$g = \frac{4}{3}\pi G\rho R$ ; same $\rho$ , so $g \propto R$ .
$g_n = 3g \Rightarrow x = 3$
So the answer is (b) 3.

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