JEE MAIN - Physics (2024 - 27th January Morning Shift - No. 5)
The acceleration due to gravity on the surface of earth is $$\mathrm{g}$$. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be :
g/4
2g
g/2
4g
Explanation
$$\begin{aligned}
& g=\frac{G M}{R^2} \Rightarrow g \propto \frac{1}{R^2} \\
& \frac{g_2}{g_1}=\frac{R_1^2}{R_2^2} \\
& g_2=4 g_1\left(R_2=\frac{R_1}{2}\right)
\end{aligned}$$
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