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  • The value of the acceleration due to gravity is  at a height <img alt="h=\frac{R}{2}" src="http://entrancecorner.codecogs.com/gif.latex?h%3D%5Cfrac%7BR%7D%7B

The value of the acceleration due to gravity is g_{1} at a height h=\frac{R}{2} (R= radius of the earth) from the surface of the earth. It is again equal to g_{1} at a depth d below the suface of the earth. The ratio \left ( \frac{d}{R} \right ) equals :
Option: 1 \frac{4}{9}
Option: 2 \frac{5}{9}
Option: 3 \frac{1}{3}
Option: 4 \frac{7}{9}

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g_{1}=\frac{G M}{\left(R+\frac{R}{2}\right)^{2}} \ldots (1)

\begin{aligned} &g_{2}=\frac{G M(R-d)}{R^{3}} \ldots (2) \\ &g_{1}=g_{2}\\ &\frac{\mathrm{GM}}{\left(\frac{3 \mathrm{R}}{2}\right)^{2}}=\frac{\operatorname{GM}(\mathrm{R}-\mathrm{d})}{\mathrm{R}^{3}}\\ &\Rightarrow \frac{4}{9}=\frac{(\mathrm{R}-\mathrm{d})}{\mathrm{R}}\\ &4 R=9 R-9 d\\ &5 \mathrm{R}=9 \mathrm{~d} \Rightarrow \frac{\mathrm{d}}{\mathrm{R}}=\frac{5}{9} \end{aligned}

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Deependra Verma

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