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

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(C) The acceleration due to gravity on the surface of the earth is given by $g = \frac{GM}{R^2}$,where $M$ is the mass of the earth and $G$ is the uni

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$3g/4\pi RG$

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(C) The acceleration due to gravity on the surface of the earth is given by $g = \frac{GM}{R^2}$,where $M$ is the mass of the earth and $G$ is the universal gravitational constant.The mass of the earth $M$ can be expressed in terms of its mean density $D$ and radius $R$ as $M = \frac{4}{3}\pi R^3 D$.Substituting the expression for $M$ into the formula for $g$,we get:$g = \frac{G}{R^2} 	imes (\frac{4}{3}\pi R^3 D)$Simplifying the equation:$g = \frac{4}{3} \pi R G D$Solving for the mean density $D$:$D = \frac{3g}{4\pi RG}$

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

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