Two simple pendulums \( A \) and \( B \) are made to oscillate simultaneously and it is found that \( A \)
completes \( 10 \) oscillations in \( 20 \mathrm{sec} \) and \( \mathrm{B} \) completes \( 8 \) oscillations in \( 10 \mathrm{sec} \). The ratio of the
lengths of \( A \) and \( B \) is

Time period of oscillations of a simple pendulum is given as \( T=2 \sqrt{\frac{l}{g}} \)
where \( I \) is length of pendulum.
For pendulum A: Given, completes \( 10 \) oscillations is \( 20 \mathrm{~s} \).
Therefore, time period of single oscillation \( =\frac{20}{10}=2 \)
\(\Rightarrow T_{A}=2 \)
\(\Rightarrow 2 \pi \sqrt{\frac{I_{A}}{g}}=2 \rightarrow(1)\)
For pendulum B: Given, completes \( 8 \) oscillations in \( 10 \mathrm{~s} \).
Therefore, time period of single oscillation \( =\frac{10}{8}=\frac{5}{4} \)
\(\Rightarrow T_{B}=\frac{5}{4} \)
\(\Rightarrow 2 \Pi \sqrt{\frac{l_{B}}{g}}=\frac{5}{4} \rightarrow(2)\)
Divide Eq. (1) by Eq. (2), we get
\(\begin{aligned}& 2 \pi \sqrt{\frac{l_{A}}{g}} \\2 \pi & \sqrt{\frac{l_{B}}{g}}=\frac{\frac{2}{1}}{4} \\\Rightarrow & \sqrt{\frac{l_{A}}{l_{B}}}=\frac{2 \times 4}{5} \\\Rightarrow & \sqrt{\frac{l_{A}}{l_{B}}}=\frac{8}{5}\end{aligned}\)