Mathematics in Physics

JEE Main level, single correct, medium

120 of 121

Question 120 of 121

The length of a simple pendulum executing simple harmonic motion is increased by 21%21 \%. The percentage increase in the time period of the pendulum of increased length is

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Answer: (D) 10%10 \%

T=2πlg;log⁡T=log⁡(2π)+12log⁡(lg)⇒log⁡T=log⁡(2π)+12log⁡(1)−12log⁡(g)\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

ΔTT=0+12×Δll−0⇒ΔTT×100=12×Δll×100\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 =12×21=10.5≈10%=\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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