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The length of a simple pendulum executing simple harmonic motion is increased by $21 \%$. The percentage increase in the time period of the pendulum of increased length is
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Verified Answer
The correct answer is:
$10 \%$
$10 \%$
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{l}}{\mathrm{g}}} ; \log \mathrm{T}=\log (2 \pi)+\frac{1}{2} \log \left(\frac{\mathrm{l}}{\mathrm{g}}\right) \Rightarrow \log \mathrm{T}=\log (2 \pi)+\frac{1}{2} \log (1)-\frac{1}{2} \log (\mathrm{g})$
Differentiating
$\frac{\Delta \mathrm{T}}{\mathrm{T}}=0+\frac{1}{2} \times \frac{\Delta \mathrm{l}}{\mathrm{l}}-0 \Rightarrow \frac{\Delta \mathrm{T}}{\mathrm{T}} \times 100=\frac{1}{2} \times \frac{\Delta \mathrm{l}}{\mathrm{l}} \times 100$ $=\frac{1}{2} \times 21=10.5 \approx 10 \%$
Note: In this method, the $\%$ error obtained is an approximate value on the higher side. Exact value is less than the obtained one.
Differentiating
$\frac{\Delta \mathrm{T}}{\mathrm{T}}=0+\frac{1}{2} \times \frac{\Delta \mathrm{l}}{\mathrm{l}}-0 \Rightarrow \frac{\Delta \mathrm{T}}{\mathrm{T}} \times 100=\frac{1}{2} \times \frac{\Delta \mathrm{l}}{\mathrm{l}} \times 100$ $=\frac{1}{2} \times 21=10.5 \approx 10 \%$
Note: In this method, the $\%$ error obtained is an approximate value on the higher side. Exact value is less than the obtained one.
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