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(D) The value of acceleration due to gravity at a height $h$ above the surface is given by $g_h = g(1 - \frac{2h}{R})$,where $R$ is the radius of the

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(D) The value of acceleration due to gravity at a height $h$ above the surface is given by $g_h = g(1 - \frac{2h}{R})$,where $R$ is the radius of the earth.The value of acceleration due to gravity at a depth $d$ below the surface is given by $g_d = g(1 - \frac{d}{R})$.According to the problem,$g_h = g_d$.Therefore,$g(1 - \frac{2h}{R}) = g(1 - \frac{d}{R})$.Simplifying the equation,we get $\frac{2h}{R} = \frac{d}{R}$,which implies $d = 2h$.Given that $h = 10 \,km$,we have $d = 2 	imes 10 \,km = 20 \,km$.

At a height of $10 \,km$ above the surface of the earth,the value of acceleration due to gravity is the same as that at a particular depth below the surface of the earth. Assuming uniform mass density for the earth,the depth is ............. $km$.

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