Question

Two pendulums whose lengths differ by $11\text{cm}$ oscillate at the same place so that one of them makes $30$ oscillations and the other $36$ oscillations during same time. Find the lengths of the pendulum.

• A. $36\text{cm}\text{and}25\text{cm}$
• B. $37\text{cm}\text{and}26\text{cm}$
• C. $25\text{cm}\text{and}14\text{cm}$
• D. $30\text{cm}\text{and}19\text{cm}$

Answer 1

The correct option is A $36\text{cm}\text{and}25\text{cm}$

Given,
$l_1-l_2=11\text{cm}...\left(1\right)$
Pendulum $1$ makes $30$ oscillations and pendulum $2$ makes $36$ oscillations during same time.
Let $T_1$ and $T_2$ are the time period of pendulum $\left(1\right)$ and $\left(2\right)$ respectively.
$\Rightarrow 30\times T_1=36\times T_2$
$\Rightarrow 30\times 2\pi \sqrt{\frac{l_1}{g}}=36\times 2\pi \sqrt{\frac{l_2}{g}}$
$\Rightarrow \frac{l_1}{l_2}=\frac{36\times 36}{30\times 30}=\frac{36}{25}$
from $\left(1\right)$
$\frac{l_1}{l_2}-1=\frac{36}{25}-1$
$\Rightarrow \frac{l_1-l_2}{l_2}=\frac{11}{25}$
$\Rightarrow \frac{11}{l_2}=\frac{11}{25}$
$\Rightarrow l_2=25\text{cm}$
$\Rightarrow l_1=36\text{cm}$