JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 6)
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)
$${R \over 2}$$
$${{\sqrt 5 R - R} \over 2}$$
$${{\sqrt 3 R - R} \over 2}$$
$${{\sqrt 5 } \over 2}R - R$$
Explanation
_2nd_September_Evening_Slot_en_6_2.png)
M = mass of earth
M1 = mass of shaded portion
Re = Radius of earth
M1 = $${M \over {{4 \over 3}\pi {R^3}}}.{4 \over 3}\pi {\left( {R - h} \right)^3}$$
= $${{M{{\left( {R - h} \right)}^3}} \over R^3}$$
Given, Weight of body is same at P and Q
$$ \therefore $$ mgP = mgQ
$$ \Rightarrow $$ gP = gQ
$$ \Rightarrow $$ $${{G{M_1}} \over {{{\left( {R - h} \right)}^2}}} = {{GM} \over {{{\left( {R + h} \right)}^2}}}$$
$$ \Rightarrow $$ $${{GM{{\left( {R - h} \right)}^3}} \over {{R^3}{{\left( {R - h} \right)}^2}}} = {{GM} \over {{{\left( {R + h} \right)}^2}}}$$
$$ \Rightarrow $$ (R – h) (R + h)2 = R3
$$ \Rightarrow $$ R3 – hR2 + h2R – h3 + 2R2h – 2Rh2 = R3
$$ \Rightarrow $$ R2h – Rh2 – h3 = 0
$$ \Rightarrow $$ R2 – Rh – h2 = 0
$$ \Rightarrow $$ h2 + Rh – R2 = 0
$$ \Rightarrow $$ h = $${{ - R \pm \sqrt {{R^2} + 4{R^2}} } \over 2}$$
$$ \Rightarrow $$ h = $${{ - R + \sqrt 5 R} \over 2}$$
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