A) 50%
B) 30%
C) 21 %
D) 10 %
Correct Answer: D
Solution :
Key Idea: As length of a pendulum increases, its time period increases. The time period T, of a simple pendulum of length\[l,\] and gravity g is given by \[T=2\pi \sqrt{\frac{l}{g}}\] \[\Rightarrow \] \[T\propto \sqrt{l}\] \[\therefore \] \[\frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\frac{{{l}_{2}}}{{{l}_{1}}}}=\sqrt{\frac{121\,l}{100\,l}}\] \[{{T}_{2}}=1.1\,{{T}_{1}}\] Therefore, increase in time period is \[=1.1{{T}_{1}}-{{T}_{1}}=0.1{{T}_{1}}=10%{{T}_{1}}\] Alternative: Time period of simple pendulum of length \[l\] and gravity g is given by \[T=2\pi \sqrt{\frac{l}{g}}\] \[\Rightarrow \]\[T\propto \sqrt{l}\] \[\Rightarrow \] \[\frac{\Delta T}{T}=\frac{1}{2}\frac{\Delta l}{l}\] \[\therefore \] \[\frac{\Delta \Tau }{T}=\frac{1}{2}\times 21%\approx 10%\] As \[l\] increases, time period of pendulum will also increase.
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