The height ' $h$ ' at which the weight of a body will be the same as that at the same depth 'h' from the surface of the earth is (Radius of the earth is $R$ and effect of the rotation of the earth is neglected):
Correct Option: 1
- $\mathrm{M}=$ mass of earth
$\mathrm{M}_{1}=$ mass of shaded portion
$\mathrm{R}=$ Radius of earth
$\mathrm{M}_{1}=\frac{\mathrm{M}}{\frac{4}{3} \pi \mathrm{R}^{3}} \cdot \frac{4}{3} \pi(\mathrm{R}-\mathrm{h})^{3}$
$=\frac{\mathrm{M}(\mathrm{R}-\mathrm{h})^{3}}{\mathrm{R}}$
- Weight of body is same at $\mathrm{P}$ and $\mathrm{Q}$
i.e. $m g_{P}=m g_{Q}$
$\mathrm{g}_{\mathrm{P}}=\mathrm{g}_{\mathrm{Q}}$
$\frac{\mathrm{GM}_{1}}{(\mathrm{R}-\mathrm{h})^{2}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}$
$\frac{G M(R-h)^{3}}{(R-h)^{2} R^{3}}=\frac{G M}{(R+h)^{2}}$
$(\mathrm{R}-\mathrm{h})(\mathrm{R}+\mathrm{h})^{2}=\mathrm{R}^{3}$
$\mathrm{R}^{3}-\mathrm{h} \mathrm{R}^{2}-\mathrm{h}^{2} \mathrm{R}-\mathrm{h}^{3}+2 \mathrm{R}^{2} \mathrm{~h}-2 \mathrm{Rh}^{2}=\mathrm{R}^{3}$
$\mathrm{R}^{2}-\mathrm{Rh}^{2}-\mathrm{h}^{3}=0$
$\mathrm{R}^{2}-\mathrm{Rh}-\mathrm{h}^{2}=0$
$h^{2}+R h-R^{2}=0 \Rightarrow h=\frac{-R \pm \sqrt{R^{2}+4 R^{2}}}{2}$
ie $h=\frac{-R+\sqrt{5} R}{2}=\left(\frac{\sqrt{5}-1}{2}\right) R$
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