Two pendulums of length 121cm and 100cm start vibrating in phase....

Time period of a pendulum,

$T ∝ \sqrt{l}$

$⇒ \dfrac{{T}_{1}}{{T}_{2}} = \sqrt{\dfrac{{l}_{1}}{{l}_{2}}}$

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,

$\dfrac{{T}_{1} \times \left(n - 1\right)}{{T}_{2} \times n} = 1 ⇒ \dfrac{\sqrt{121}\left(n - 1\right)}{\sqrt{100}\left(n\right)} = 1$

$⇒ 11n - 11 = 10n ⇒ n = 11$