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In order that a body of 15 kg weighs zer...

In order that a body of 15 kg weighs zero at the equator, then the angular speed of earth is :
`(g=10ms^(-1))`

A

`(1)/(80) " rad s"^(-1)`

B

`(1)/(400)" rad s"^(-1)`

C

`(1)/(800)" rad s"^(-1)`

D

`(1)/(1600)" rad s"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C

`mg.=0 rArr mg - mRomega^(2)=0`
`rArr omega= sqrt((g)/(R ))=sqrt((1.0)/(6.4xx10^(6)))=(1)/(800)" rad sec"^(-1)`
Thus correct choice is (c ).
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Knowledge Check

  • With what velocity should a body be thrown up so that it rises to a height equal to the radius of the earth ( g=10ms^(-2) at the surface) ?

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