If R is the radius of the Earth and g is the | vedclass

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(C) The mass of the Earth $M$ can be expressed in terms of its volume $V$ and mean density $\rho$ as: $M = V \rho = \frac{4}{3} \pi R^3 \rho$ ... $(i)

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

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(C) The mass of the Earth $M$ can be expressed in terms of its volume $V$ and mean density $ho$ as: $M = V ho = \frac{4}{3} \pi R^3 ho$ ... $(i)$ The acceleration due to gravity $g$ on the surface of the Earth is given by: $g = \frac{GM}{R^2}$ ... $(ii)$ Substituting the expression for $M$ from equation $(i)$ into equation $(ii)$: $g = \frac{G}{R^2} 	imes \left( \frac{4}{3} \pi R^3 ho ight)$ $g = \frac{4}{3} \pi R G ho$ Rearranging the equation to solve for the mean density $ho$: $ho = \frac{3 g}{4 \pi R G}$

If $R$ is the radius of the Earth and $g$ is the acceleration due to gravity on the Earth's surface,then the mean density of the Earth is:

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