A) 11
B) 10
C) 21
D) 20
Correct Answer: B
Solution :
The time period of a simple pendulum is \[T=2\,\pi \sqrt{\frac{l}{g}}\] where Us length. \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{l}_{1}}}{{{l}_{2}}}}\] Given, \[{{l}_{1}}=100\,cm,\,{{l}_{2}}=121\,cm\] \[\therefore \] \[\frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\frac{121}{100}}=\frac{11}{10}\]. In a-given time shorter pendulum will make one oscillation more than the longer. Let after n oscillations, the pendulums are in same phase, then \[n\times 11=(n+1)\times 10\] \[\Rightarrow \] \[n=10\].
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