The average density of the earth is [ $\mathrm{g}$ is acceleration due to gravity]
The average density of the earth is [ $\mathrm{g}$ is acceleration due to gravity]
- inversely proportional to $\mathrm{g}^2$
- directly proportional to $g$
- inversely proportional to $\mathrm{g}$
- directly proportional to $\mathrm{g}^2$
Solution
$\begin{aligned} & \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}=\frac{\mathrm{G}}{\mathrm{R}^2} \cdot \frac{4}{3} \pi \mathrm{R}^3 \cdot \rho=\mathrm{G} \frac{4}{3} \cdot \pi \mathrm{R} \rho \\ & \therefore \mathrm{g} \propto \rho\end{aligned}$
Asked in: MHT CET 2021 (24 Sep Shift 2)
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