A) 10
B) 11
C) 20
D) 21
Correct Answer: A
Solution :
Let the two pendulums are in same phase, after n vibrations of the longer pendulum. In this time the shorter pendulum will complete \[(n+1)\]vibrations. \[\therefore \]\[n\times 2\pi \sqrt{\frac{{{l}_{2}}}{g}}=(n+1)2\pi \sqrt{\frac{{{l}_{1}}}{g}}\] \[n\times 2\pi \sqrt{\frac{121}{g}}=(n+1)2\pi \sqrt{\frac{100}{g}}\] \[11n=10(n+1)\] \[n=10\]You need to login to perform this action.
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