Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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

$\frac{\pi R G}{12 g}$

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

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

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

Formula for g: The acceleration due to gravity ( $g$ ) on the surface of the Earth is related to its mass ( $M$ ) and radius ( $R$ ) by the formula: $g = \frac{GM}{R^2}$
Mass in terms of Density: Assuming the Earth is a sphere of uniform mean density ( $\rho$ ), its mass is given by volume times density: $M = \frac{4}{3}\pi R^3 \rho$
Substitution and Solving: Substitute the expression for $M$ into the equation for $g$ : $g = \frac{G}{R^2} \left( \frac{4}{3}\pi R^3 \rho \right)$ $g = \frac{4}{3} \pi G R \rho$ Rearranging to solve for density ( $\rho$ ): $\rho = \frac{3g}{4 \pi G R}$

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