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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:  
Option: 1 \frac{R}{2}
Option: 2 \frac{\sqrt{5}R-R}{2}
Option: 3 \frac{\sqrt{3}R-R}{2}
Option: 4 \frac{\sqrt{5}R-2R}{2}

Answers (1)



weight at hieght h  = weight at depth d

So, mgh = mgd

So, gh = gd


\frac{g_o R^2}{(R+h)^2} = \frac{g_o (R-d)}{R}


go = acceleration due to gravity at the surface of earth

h= height

d = Depth

R = Radius of the earth

On solving,

\frac{ R^2}{(R+h)^2} = \frac{ (R-d)}{R}

\Rightarrow R^3 = (R^2 + h^2 +2Rh)(R-d)

On Solving this we get - 

h^2 +hR-R^2 = 0


\Rightarrow h = \frac{\sqrt{5}R - R}{2}

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