A) 11
B) 10
C) 21
D) 20
Correct Answer: B
Solution :
Time period of simple pendulum \[T=2\pi \sqrt{\frac{l}{g}}\] \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{l}_{1}}}{{{l}_{2}}}}\] \[=\sqrt{\frac{121}{100}}\] \[=\frac{11}{10}\] Let after n oscillations, the pendulums are in same phase \[\therefore \] \[(n)\times 11=(n+1)\times 10\] \[\therefore \] \[n=10\]You need to login to perform this action.
You will be redirected in
3 sec