Kinematics

JEE Main level, single correct, medium

298 of 308

Question 298 of 308

If the angular velocity of earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately :
(Take : g=10ms−2g=10{ms}^{-2}, the radius of earth, R=6400×103mR=6400\times {10}^{3}m, Take π=3.14\pi =3.14)

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Answer: (D) 8484 minutes

For objects to float,
mg=mω2Rmg=m{\omega }^{2}R
ω=\omega = angular velocity of the earth.
R=R= Radius of earth
ω=gR...(1)\omega =\sqrt{\dfrac{g}{R}}...\left(1\right)
Duration of day =T=T
T=2πω...(2)T=\dfrac{2\pi }{\omega }...\left(2\right)
⇒T=2πRg\Rightarrow T=2\pi \sqrt{\dfrac{R}{g}}
=2π6400×10310=2\pi \sqrt{\dfrac{6400\times {10}^{3}}{10}}
⇒T60=83.775\Rightarrow \dfrac{T}{60}=83.775 minutes
≃84≃84 minutes

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