In the provided figure of the Earth,the value | vedclass

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(C) Let $R$ be the radius of the Earth. Point $A$ is at a distance $r$ from the center $O$ inside the Earth. Point $C$ is at a height $h = 3200 	ext{

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$4$

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(C) Let $R$ be the radius of the Earth. Point $A$ is at a distance $r$ from the center $O$ inside the Earth. Point $C$ is at a height $h = 3200 	ext{ km} = R/2$ above the surface $B$.The acceleration due to gravity at point $A$ (inside the Earth) is given by $g_A = \frac{G M r}{R^3}$.The acceleration due to gravity at point $C$ (outside the Earth) is given by $g_C = \frac{G M}{(R + h)^2} = \frac{G M}{(R + R/2)^2} = \frac{G M}{(3R/2)^2} = \frac{4 G M}{9 R^2}$.Given that $g_A = g_C$,we equate the expressions:$\frac{G M r}{R^3} = \frac{4 G M}{9 R^2}$Solving for $r$:$r = \frac{4 R}{9}$.Thus,$OA = r = \frac{4 R}{9}$.The distance $AB = R - r = R - \frac{4 R}{9} = \frac{5 R}{9}$.Therefore,the ratio $OA : AB = \frac{4 R}{9} : \frac{5 R}{9} = 4 : 5$.Since the ratio is $x : y = 4 : 5$,the value of $x$ is $4$.

In the provided figure of the Earth,the value of acceleration due to gravity is the same at points $A$ and $C$,but it is smaller than its value at point $B$ (surface of the Earth). The value of $OA : AB$ is $x : y$. The value of $x$ is $\ldots \ldots \ldots$

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