1.In a stationary lift, time period of a simple pendulum is TTT. The lift starts accelerating downwards with acceleration g/4g/4g/4, then the time period of the pendulum will bea.(3/2) T(\sqrt{3}/2)\,T(3/2)Tb.(2/3) T(2/\sqrt{3})\,T(2/3)Tc.(3/4) T(3/4)\,T(3/4)Td.(4/3) T(4/3)\,T(4/3)TLogin to continueOnly logged in users canattempt or see the solution.