Q134 / 180 · KCET 2017 · Physics
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
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
- A\( \frac{8}{5} \)
- Correct answerB\( \frac{64}{25} \)
- C\( \frac{5}{4} \)
- D\( \frac{25}{64} \)
Solution
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 \( =\frac{20}{10}=2 \)
\[ \begin{array}{l} \Rightarrow T_{A}=2 \\ \Rightarrow 2 \pi \sqrt{\frac{I_{A}}{g}}=2 \rightarrow(1) \end{array} \]
For pendulum B: Given, completes \( 8 \) oscillations in \( 10 \mathrm{~s} \).
Therefore, time period of single oscillation \( =\frac{10}{8}=\frac{5}{4} \)
\[ \begin{array}{l} \Rightarrow T_{B}=\frac{5}{4} \\ \Rightarrow 2 \Pi \sqrt{\frac{l_{B}}{g}}=\frac{5}{4} \rightarrow(2) \end{array} \]
Divide Eq. (1) by Eq. (2), we get
\[ \begin{aligned} & 2 \pi \sqrt{\frac{l_{A}}{g}} \\ 2 \pi & \sqrt{\frac{l_{B}}{g}}=\frac{\frac{2}{1}}{4} \\ \Rightarrow & \sqrt{\frac{l_{A}}{l_{B}}}=\frac{2 \times 4}{5} \\ \Rightarrow & \sqrt{\frac{l_{A}}{l_{B}}}=\frac{8}{5} \end{aligned} \]
where \( I \) is length of pendulum.
For pendulum A: Given, completes \( 10 \) oscillations is \( 20 \mathrm{~s} \).
Therefore, time period of single oscillation \( =\frac{20}{10}=2 \)
\[ \begin{array}{l} \Rightarrow T_{A}=2 \\ \Rightarrow 2 \pi \sqrt{\frac{I_{A}}{g}}=2 \rightarrow(1) \end{array} \]
For pendulum B: Given, completes \( 8 \) oscillations in \( 10 \mathrm{~s} \).
Therefore, time period of single oscillation \( =\frac{10}{8}=\frac{5}{4} \)
\[ \begin{array}{l} \Rightarrow T_{B}=\frac{5}{4} \\ \Rightarrow 2 \Pi \sqrt{\frac{l_{B}}{g}}=\frac{5}{4} \rightarrow(2) \end{array} \]
Divide Eq. (1) by Eq. (2), we get
\[ \begin{aligned} & 2 \pi \sqrt{\frac{l_{A}}{g}} \\ 2 \pi & \sqrt{\frac{l_{B}}{g}}=\frac{\frac{2}{1}}{4} \\ \Rightarrow & \sqrt{\frac{l_{A}}{l_{B}}}=\frac{2 \times 4}{5} \\ \Rightarrow & \sqrt{\frac{l_{A}}{l_{B}}}=\frac{8}{5} \end{aligned} \]
Hence, the correct answer is option: 'B'.