Question

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
A. T
B. `(T)/(sqrt(2))`
C. `sqrt(2)T`
D. `(T)/(2^(1//4))`

Answer 1

Correct Answer - D
`(T_(2))/(T_(1))=sqrt((g)/(sqrt(g^(2)+g^(2))))=sqrt((g)/(gsqrt(2)))`
`T_(2)=(T)/(sqrt(2))=(T)/(2^(1//4))`