The time period of a simple pendulum measured inside a ststionary lift is found to be T. IF the lift starts accelarating upwards with an acceslaration of `g//3`, the time period of the pendulum will be.

Options

A. `sqrt3T`
B. `sqrt3/2T`
C. `T//sqrt3`
D. `T//3`

Answer

b

Explanation

`T=2pisqrt(l)/(g)`
`implies Talpha 1/sqrt(g)`
`implies T_1/T=sqrt(g)/(g_1)`
`g_1=g+g//3`(Lift accesleration upward)
`g_1=4g//3`
`therefore T_1=sqrt(3)/(2)T`