Gravitation

JEE Main level, single correct, medium

84 of 223

Question 84 of 223

The acceleration due to gravity on the earth's surface at the poles is gg and angular velocity of the earth about the axis passing through the pole is ω\omega. An object is weighed at the equator and at a height hh above the poles by using a spring balance. If the weights are found to be same, then hh is: (h≪Rh≪R, where RR is the radius of the earth)

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Answer: (A) R2ω22g\dfrac{{R}^{2}{\omega }^{2}}{2g}

Acceleration due to gravity at height hh above the pole
is given by
gh=g(1−2hR)...(1){g}_{h}=g\left(1-\dfrac{2h}{R}\right)...(1)
Acceleration due to gravity at the equator due to rotation of the earth is given by
g′=g−ω2Rcos⁡2λ...(2)g'=g-{\omega }^{2}R{\cos }^{2}\lambda ...(2)
At the equator, λ=0\lambda =0
⇒g′=g−ω2R...(3)\Rightarrow g'=g-{\omega }^{2}R...(3)
As both weights are equal, i.e.; gh=g′{g}_{h}={g}^{'}
From equation (1)(1) and (2)(2), we get,
⇒ω2R=2gRh\Rightarrow {\omega }^{2}R=\dfrac{2g}{R}h
∴h=R2ω22g∴h=\dfrac{{R}^{2}{\omega }^{2}}{2g}

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