Solve this following


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):

  1. $\frac{\sqrt{5} R-R}{2}$

  2. $\frac{\sqrt{5}}{2} R-R$

  3. $\frac{\mathrm{R}}{2}$

  4. $\frac{\sqrt{3} R-R}{2}$

Correct Option: 1


- $\mathrm{M}=$ mass of earth

$\mathrm{M}_{1}=$ mass of shaded portion

$\mathrm{R}=$ Radius of earth

$\mathrm{M}_{1}=\frac{\mathrm{M}}{\frac{4}{3} \pi \mathrm{R}^{3}} \cdot \frac{4}{3} \pi(\mathrm{R}-\mathrm{h})^{3}$


- Weight of body is same at $\mathrm{P}$ and $\mathrm{Q}$

i.e. $m g_{P}=m g_{Q}$



$\frac{G M(R-h)^{3}}{(R-h)^{2} R^{3}}=\frac{G M}{(R+h)^{2}}$


$\mathrm{R}^{3}-\mathrm{h} \mathrm{R}^{2}-\mathrm{h}^{2} \mathrm{R}-\mathrm{h}^{3}+2 \mathrm{R}^{2} \mathrm{~h}-2 \mathrm{Rh}^{2}=\mathrm{R}^{3}$



$h^{2}+R h-R^{2}=0 \Rightarrow h=\frac{-R \pm \sqrt{R^{2}+4 R^{2}}}{2}$

ie $h=\frac{-R+\sqrt{5} R}{2}=\left(\frac{\sqrt{5}-1}{2}\right) R$


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