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

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(C) The acceleration due to gravity $g$ at the Earth's surface is given by the formula:$g = \frac{G M}{R^2}$where $M$ is the mass of the Earth and $G$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The acceleration due to gravity $g$ at the Earth's surface is given by the formula:$g = \frac{G M}{R^2}$where $M$ is the mass of the Earth and $G$ is the universal gravitational constant.The mass $M$ of the Earth can be expressed in terms of its mean density $ho$ and radius $R$ as:$M = 	ext{Volume} 	imes 	ext{Density} = \left( \frac{4}{3} \pi R^3 ight) ho$Substituting the expression for $M$ into the formula for $g$:$g = \frac{G}{R^2} 	imes \left( \frac{4}{3} \pi R^3 ho ight)$Simplifying the equation:$g = \frac{4}{3} \pi G R ho$Solving for the density $ho$:$ho = \frac{3 g}{4 \pi G R}$Thus,the correct option is $C$.

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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