1.

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

Answer»

`(omega^(2)R^(2))/(G)`
`(omega^(2)R^(2))/(2G)`
`(2 omega^(2)R^(2))/(g)`
`SQRT(RG)/(omega)`

Answer :A


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