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(D) At the surface of the earth, the weight of the body is given by $W_s = mg_s = 300 \,N$.The acceleration due to gravity at a depth $d$ below the su

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$225$

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(D) At the surface of the earth, the weight of the body is given by $W_s = mg_s = 300 \,N$.The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula $g_d = g_s \left(1 - \frac{d}{R}ight)$, where $g_s$ is the acceleration due to gravity at the surface and $R$ is the radius of the earth.Given the depth $d = R/4$, we substitute this into the formula:$g_d = g_s \left(1 - \frac{R/4}{R}ight)$$g_d = g_s \left(1 - \frac{1}{4}ight)$$g_d = g_s \left(\frac{3}{4}ight)$The weight of the body at depth $d$ is $W_d = mg_d$.Substituting $g_d$ into the weight equation:$W_d = m 	imes \left(\frac{3}{4} g_sight)$$W_d = \frac{3}{4} 	imes (mg_s)$Since $mg_s = 300 \,N$, we have:$W_d = \frac{3}{4} 	imes 300 \,N = 225 \,N$.

Assuming the earth to be a sphere of uniform mass density, a body weighs $300 \,N$ on the surface of the earth. How much would it weigh at a depth of $R/4$ below the surface of the earth (in $\,N$)?

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