1.

Use the assumptions of the previous question. An object weighted by a spring balance at the equator gives the same reading as a reading taken at a depth d below the earth's surface at a pole (d lt lt R). The value of d is

Answer»

`(OMEGA^(2) R^(2))/(g)`
`(omega^(2) R^(2))/(2G)`
`(2omega^(2) R^(2))/(g)`
`(sqrt(Rg))/(omega)`

ANSWER :B


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