Gravitation

JEE Main level, single correct, medium

79 of 223

Question 79 of 223

The approximate height from the surface of earth at which the weight of the body becomes 13\dfrac{1}{3} of its weight on the surface of earth is :
[Radius of earth R=6400kmR=6400km and 3=1.732\sqrt{3}=1.732]

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Answer: (B) 4685km4685km

The variation of the acceleration due to gravity with height hh is given by,
g′=g(1+hR)2{g}^{'}=\dfrac{g}{{\left(1+\dfrac{h}{R}\right)}^{2}}.
Since the weight at given height is 13\dfrac{1}{3} of the weight at the surface
⇒g′=g3⇒g(1+hR)2=g3⇒(1+hR)=3⇒h=R(3−1)=6400×(1.732−1)=4685km\Rightarrow {g}^{'}=\dfrac{g}{3} \Rightarrow \dfrac{g}{{\left(1+\dfrac{h}{R}\right)}^{2}}=\dfrac{g}{3} \Rightarrow \left(1+\dfrac{h}{R}\right)=\sqrt{3} \Rightarrow h=R\left(\sqrt{3}-1\right)=6400\times \left(1.732-1\right)=4685km

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