A body weighs 300 N on the surface of the ea | vedclass

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(D) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula: $g_d = g(1 - \frac{d}{R})$.Given that the d

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$150$

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(D) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula: $g_d = g(1 - \frac{d}{R})$.Given that the depth $d = \frac{R}{2}$,we substitute this value into the equation:$g_d = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.The weight of the body at the surface is $W = mg = 300 \ N$.The weight of the body at depth $d$ is $W_d = mg_d$.Substituting $g_d = \frac{g}{2}$,we get $W_d = m(\frac{g}{2}) = \frac{1}{2} 	imes mg = \frac{1}{2} 	imes 300 \ N = 150 \ N$.

$A$ body weighs $300 \ N$ on the surface of the earth. How much will it weigh at a distance $\frac{R}{2}$ below the surface of the earth (in $N$)? ($R$ is the radius of the earth.)

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