Q106 / 125 · BITSAT 2010 · Physics
A simple pendulum has time period 't'. Its time period in a lift which is moving upwards with acceleration $3 \mathrm{~ms}^{-2}$ is
- Correct answerA$t \sqrt{\frac{9.8}{12.8}}$
- B$t \sqrt{\frac{12.8}{9.8}}$
- C$\mathrm{t} \sqrt{\frac{9.8}{6.8}}$
- D$\mathrm{t} \sqrt{\frac{6.8}{9.8}}$
Solution
$\mathrm{t} \propto \frac{1}{\sqrt{9.8}}, \mathrm{t}^{\prime} \propto \frac{1}{\sqrt{12.8}}$
$\left(\because \mathrm{g}^{\prime}=9.8+3=12.8\right)$
$\therefore \frac{\mathrm{t}^{\prime}}{\mathrm{t}}=\sqrt{\frac{9.8}{12.8}} \Rightarrow \mathrm{t}^{\prime}=\sqrt{\frac{9.8}{12.8}} \mathrm{t}$
Hence, the correct answer is option: 'A'.