Two pendula of length 121 cm and 100 cm start vibrating. At some instant the two are in mean position in the same phase. After how many vibrations of the shorter pendulum the two will be in phase in the mean position

 Here, we have
$$\begin{gathered} {\mathrm{t}}_1=2\mathrm{\pi }\sqrt{\left(\frac{121}{\mathrm{g}}\right)}\text{ and }{\mathrm{t}}_2=2\mathrm{\pi }\sqrt{\left(\frac{100}{\mathrm{g}}\right)} \\ \mathrm{Let} \mathrm{shorterpendulum} \mathrm{male} \mathrm{n} \mathrm{vibmtions}. \mathrm{Then} \mathrm{the} \mathrm{longer} \mathrm{will} \mathrm{make} \mathrm{one} \mathrm{less} \mathrm{tian} \mathrm{n} \mathrm{to} \mathrm{come} \mathrm{in} \mathrm{phase} \mathrm{again}, \mathrm{i}.\mathrm{e}., \\ {\mathrm{nt}}_2=(\mathrm{n}-1){\mathrm{t}}_1 \\ \mathrm{or} \mathrm{n}\left[2\mathrm{\pi }\sqrt{\left(\frac{100}{\mathrm{g}}\right)}\right]=(\mathrm{n}-1)\left[2\mathrm{\pi }\sqrt{\left(\frac{121}{\mathrm{g}}\right)}\right] \\ \mathrm{n}\times 10=(\mathrm{n}-1)11 \\ \mathrm{Solving}, \mathrm{we} \mathrm{get} \mathrm{n} = 11 \end{gathered}$$