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7.Gravitation
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The height at which the weight of the body become $\frac{1}{9}^{th}$. Its weight on the surface of earth (radius of earth $R$)
A
$h = 3R$
B
$h = R$
C
$h = \frac{R}{2}$
D
$h = 2R$
Solution
$\mathrm{g}^{\prime}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}=\mathrm{g}^{\prime}=\frac{\mathrm{g}}{9}$
$\frac{\mathrm{g}}{9}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}} \Rightarrow 1+\frac{\mathrm{h}}{\mathrm{R}}=3$
$\quad \mathrm{h}=2 \mathrm{R}$
Standard 11
Physics
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