UPSEE 2019 Solved Paper

According to the question, we draw the following situationAs, the point of suspension is moving upward according to the relation $y =\lambda t^{2}$ $\frac{d y}{d t} =2 \lambda t$ Further, $\frac{d^{2} y}{d t^{2}}=2 \lambda=$ upward acceleration $a$ Here, $2 \lambda$ is the acceleration of point of suspension. The bob experiences a downward reaction force. Hence, the new time period $T'=2 \pi \sqrt{\frac{L}{g+a}}$ .......(i) Whereas the time period of simple pendulum, $T=2 \pi \sqrt{\frac{L}{g}}$ ......(ii) By comparing Eqs. (i) and (ii), we get $T' < T$ Hence, the new time period is less than $T$.