Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their…

Time period of a pendulum,

$T\propto \sqrt{l}$

$\Rightarrow \frac{T_1}{T_2}=\sqrt{\frac{l_1}{l_2}}$

If $n$ is the number of vibrations after which the pendulums are again in phase, the number of vibration of the longer pendulum will be $\left(n-1\right)$. Therefore,

$\frac{T_1\times \left(n-1\right)}{T_2\times n}=1$ $\Rightarrow \frac{\sqrt{121}\left(n-1\right)}{\sqrt{100}\left(n\right)}=1$

$\Rightarrow 11n-11=10n$ $\Rightarrow n=11$