MHT CET · Physics · Oscillations
Let ' \(l_1\) ' be the length of simple pendulum. Its length changes to ' \(l_2\) ' to increase the periodic time by \(20 \%\). The ratio \(\frac{l_2}{l_1}=\)
- A 1.22
- B 1.33
- C 1.44
- D 1.55
Answer & Solution
Correct Answer
(C) 1.44
Step-by-step Solution
Detailed explanation
\(\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\)
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 \(20 \%\),
\(\begin{aligned}
& \mathrm{T}_2=\mathrm{T}_1+\frac{20}{100}\mathrm{T}_1=\frac{120}{100} \mathrm{~T}_1 \\
& \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\sqrt{\frac{l_2}{l_1}} \\
& \left(\frac{\mathrm{~T}_2}{\mathrm{~T}_1}\right)^2=\frac{l_2}{l_1} \\
& \left(\frac{120 \mathrm{~T}_1}{100 \times \mathrm{T}_1}\right)^2=\frac{l_2}{l_1} \\
& \frac{36}{25}=\frac{l_2}{l_1} \\
& \frac{l_2}{l_1}=1.44
\end{aligned}\)
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