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

Answer

B

Explanation

In a stationary lift, time period of simple pendulum is `T=2pisqrt((l)/(g))" "...(i)`
where l is the length of the simple pendulum.
When the lift accelerates upwards with an acceleration `g//4`, the effective acceleration on the bob of pendulum is
`g.=g+(g)/(4)=(5g)/(4)`
`therefore T.=2pisqrt((l)/(g.))=2pisqrt((l)/(((5g)/(4))))=2pisqrt((4)/(5)((l)/(g)))=(2T)/(sqrt(5))` (Using (i))