Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is :

The longer pendulum (121 cm) has the larger time period. For them to be in phase again, the shorter one must complete one more oscillation: $n\,T_\ell = (n + 1)T_s$ $n \times 2\pi\sqrt{\frac{1.21}{g}} = (n + 1) \times 2\pi\sqrt{\frac{1}{g}}$ $1.1n = n + 1 \Rightarrow n = 10$ Number of oscillations of the shorter pendulum $= n + 1 = 11$