Q141 / 180 · COMEDK 2024 (Memory Based) · Physics
In an experiment to determine the period of a simple pendulum of length 100 cm , it is attached to different spherical bobs have uniform mass distribution. If the relative difference in the periods is found to be $5 \times 10^{-4} \mathrm{~s}$, the difference in radii $\left|r_1-r_2\right|$ is best given by
- A$10^{-2} \mathrm{~m}$
- Correct answerB$10^{-3} \mathrm{~m}$
- C$10^{-1} \mathrm{~m}$
- D$10^{-4} \mathrm{~m}$
Solution
Given, $\Delta T=5 \times 10^{-4} \mathrm{~s}$
We know that,
$\begin{array}{cc} <br/>& T=2 \pi \sqrt{\frac{l}{g}} \\<br/>\Rightarrow & T \propto \sqrt{l} \\<br/>\therefore & \Delta T=\frac{1}{2} \Delta l \\<br/>\Rightarrow & 5 \times 10^{-4}=\frac{\left|r_1-r_2\right|}{2} \\<br/>\Rightarrow & \left|r_1-r_2\right|=2 \times 5 \times 10^{-4}=10^{-3} \mathrm{~m}<br/>\end{array}$
We know that,
$\begin{array}{cc} <br/>& T=2 \pi \sqrt{\frac{l}{g}} \\<br/>\Rightarrow & T \propto \sqrt{l} \\<br/>\therefore & \Delta T=\frac{1}{2} \Delta l \\<br/>\Rightarrow & 5 \times 10^{-4}=\frac{\left|r_1-r_2\right|}{2} \\<br/>\Rightarrow & \left|r_1-r_2\right|=2 \times 5 \times 10^{-4}=10^{-3} \mathrm{~m}<br/>\end{array}$
Hence, the correct answer is option: 'B'.