A simple pendulum is suspended in a lift which is going up with an acceleration of `5 m//s^(2)`. An electric field of magnitude `5 N//C` and direction vertically upward is also present in the lift. The charge of the is `1 mu C` and mass is `1 mg`. Talking `g=pi^(2)` and length of the simple pendulum 1m, find the time period of the simple pendulum (in sec).

Answer

2

Explanation

`T=2pisqrt(l/g_("eff"))`
`g_("eff")=g-(qE)/M+5=15-(1xx5xx10^(-6))/(1xx10^(-6))`
`g_("eff")=10=pi^(2)`
`T=2` sec