The length of a second's pendulum on the surface of the Earth is `0.9m`. The length of the same pendulum of surface of planet X such that the acceleration of planet X is n times greater than the Earth is :

Options

A. `0.9 n`
B. `(0.9)/(n)m`
C. `0.9 n^(2)m`
D. `(0.9)/(n^(2))`

Answer

A

Explanation

`T= 2pi sqrt((l)/(g)),`
`l=0.9m, T = 2pi sqrt((l_(x))/(g_(x)))`
`g_(x)=ng`
`l_(x)=0.9 n`