The depth 'd' at which the value of acce...

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

A

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

B

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

C

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

D

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

Verified by Experts

The correct Answer is:

C

Similar Questions

Explore conceptually related problems

The value of acceleration due to gravity is zero at

The value of acceleration due to gravity is maximum at

The value of acceleration due to gravity is maximum at____

The value of the acceleration due to gravity g on earth depends upon

The value of acceleration due to gravity at a depth of 1600 km is equal to

The acceleration due to gravity on the surface of earth varies

The value of acceleration due to gravity will be 1% of its value at the surface of earth at a height of (R_(e )=6400 km)

As we go above the earth's surface , the value of acceleration due to gravity increases.

Find the value of acceleration due to gravity in a mine at a depth of 80 km from the surface of the earth . Radius of the earth = 6400 km .

Derive an expression for the acceleration due to gravity at a depth d below the earth'ssurface