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The depth d, at which the value of accel...

The depth `d`, at which the value of acceleration due to gravity becomes 1/n times the value at the surface is (R = radius of the earth)

A

`d=R((n)/(n-1))`

B

`d=R((n-1)/(2n))`

C

`d=R((n-1)/(n))`

D

`d R^(2)((n-1)/(n))`

Text Solution

Verified by Experts

The correct Answer is:
C
A dept d we have `g'=g(1-d/R)`
`g'=g/n`
`g/n=g (1-d/R)`
`therefore 1/n=1-d/R`
`therefore d/R=1-1/n=(n-1)/(n)`
`d=R((n-1)/(n))`
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