Q128 / 150 · MHT CET 2024 (10 May Shift 1) · Physics
Let ' $l_1$ ' be the length of simple pendulum. Its length changes to ' $l_2$ ' to increase the periodic time by $20 \%$. The ratio $\frac{l_2}{l_1}=$
- A1.22
- B1.33
- Correct answerC1.44
- D1.55
Solution
$\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$
As the periodic time increases by $20 \%$, $\begin{aligned} & \mathrm{T}_2=\mathrm{T}_1+\frac{20}{100}\mathrm{T}_1=\frac{120}{100} \mathrm{~T}_1 \\ & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\sqrt{\frac{l_2}{l_1}} \\ & \left(\frac{\mathrm{~T}_2}{\mathrm{~T}_1}\right)^2=\frac{l_2}{l_1} \\ & \left(\frac{120 \mathrm{~T}_1}{100 \times \mathrm{T}_1}\right)^2=\frac{l_2}{l_1} \\ & \frac{36}{25}=\frac{l_2}{l_1} \\ & \frac{l_2}{l_1}=1.44 \end{aligned}$
As the periodic time increases by $20 \%$, $\begin{aligned} & \mathrm{T}_2=\mathrm{T}_1+\frac{20}{100}\mathrm{T}_1=\frac{120}{100} \mathrm{~T}_1 \\ & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\sqrt{\frac{l_2}{l_1}} \\ & \left(\frac{\mathrm{~T}_2}{\mathrm{~T}_1}\right)^2=\frac{l_2}{l_1} \\ & \left(\frac{120 \mathrm{~T}_1}{100 \times \mathrm{T}_1}\right)^2=\frac{l_2}{l_1} \\ & \frac{36}{25}=\frac{l_2}{l_1} \\ & \frac{l_2}{l_1}=1.44 \end{aligned}$
Hence, the correct answer is option: 'C'.