If $\mathrm{R}$ is the radius of the earth and $\mathrm{g}$ is the acceleration due to gravity on the earth surface. Then the mean density of the earth will be :

Question

$\frac{\pi \mathrm{RG}}{12 \mathrm{~g}}$

$\frac{3 \pi R}{4 g G}$

$\frac{3 g}{4 \pi R G}$

$\frac{4 \pi \mathrm{G}}{3 g R}$

Solution

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