A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. if the lift accelerates upwards with an acceleration g/4, then the period of the pendulum will be

Options

A. T
B. `(T)/(4)`
C. `(2T)/(sqrt(5))`
D. `2T sqrt(5)`

Answer

C

Explanation

`(T_(2))/(T_(1))=sqrt((g_(1))/(g_(2)))=sqrt((9)/(9+(9)/(4)))=sqrt((4)/(5))`
`therefore T_(2)=(2)/(sqrt(5))T`.