Gravitation

JEE Main level, single correct, medium

74 of 223

Question 74 of 223

An object is taken to a height above the surface of earth at a distance 54R\dfrac{5}{4}R from the centre of the earth. Where radius of earth, R=6400kmR=6400km. The percentage decrease in the weight of the object will be

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Answer: (A) 3636%

Acceleration due to gravity at height hh is given by, g′=g(1+hR)2{g}^{'}=\dfrac{g}{{\left(1+\dfrac{h}{R}\right)}^{2}}
For distance from the centre r=54Rr=\dfrac{5}{4}R height h=R4h=\dfrac{R}{4},
⇒g′=g(1+14)2=1625g\Rightarrow {g}^{'}=\dfrac{g}{{\left(1+\dfrac{1}{4}\right)}^{2}}=\dfrac{16}{25}g
Change in weight
∆W=mg−mg′⇒∆W=mg(1−1625)∆W=mg-m{g}^{'} \Rightarrow ∆W=mg\left(1-\dfrac{16}{25}\right)
⇒∆W=925mg\Rightarrow ∆W=\dfrac{9}{25}mg
Percentage change =925×100=\dfrac{9}{25}\times 100%
=36=36%

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