A) 21
B) 11
C) 10.5
D) 10
Correct Answer: B
Solution :
The time period (T) of pendulum is given by \[T=2\pi \sqrt{\frac{l}{g}}\] where \[l\] is length and g the acceleration due to gravity. \[\Rightarrow \] \[{{T}^{2}}\propto l\] \[\therefore \] \[\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{l}{{{l}_{2}}}\] \[\Rightarrow \] \[\frac{T_{2}^{2}-T_{1}^{2}}{T_{1}^{2}}=\frac{{{l}_{2}}-{{l}_{1}}}{{{l}_{1}}}\] \[\Rightarrow \] \[\frac{1.1\times \Delta T}{T}=\frac{\Delta l}{l}\]. Hence, increase in length \[\frac{\Delta l}{l}\times 100=1.1\times \frac{\Delta T}{T}\times 100\] \[=1.1\times \frac{10}{100}\times 100=11%\]
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