MCQ
The time period of a simple pendulum measured inside a stationary lift is found to be $T$. If the lift starts accelerating upwards with an acceleration $g/3$, the time period is
- A$T\sqrt 3 $
- ✓$T\sqrt 3 /2$
- C$T/\sqrt 3 $
- D$T/3$
$[As\;g' = g + a = g + \frac{g}{3} = \frac{{4g}}{3}$]
$T' = \frac{{\sqrt 3 }}{2}T$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.


