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A $90 \mathrm{~kg}$ body placed at $2 \mathrm{R}$ distance from surface of earth experiences gravitational pull of : $\text { ( } \mathrm{R}=\text { Radius of earth, } \mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2} \text { ) }$
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The correct answer is:
$100 \mathrm{~N}$
Here $g_s=$ gravitational acceleration at surface
$\begin{aligned}
& \text {Value of } g=g_5\left(1+\frac{h}{R}\right)^{-2} \\
& =g_5(1+2)^{-2}=\frac{g_5}{9}
\end{aligned}$
$\text {Force }=\mathrm{mg}=90 \times \frac{\mathrm{g}_5}{9}=100 \mathrm{~N}$
$\begin{aligned}
& \text {Value of } g=g_5\left(1+\frac{h}{R}\right)^{-2} \\
& =g_5(1+2)^{-2}=\frac{g_5}{9}
\end{aligned}$
$\text {Force }=\mathrm{mg}=90 \times \frac{\mathrm{g}_5}{9}=100 \mathrm{~N}$
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