A) 5
B) 1
C) 2
D) 3
Correct Answer: C
Solution :
Key Idea: For the pendulum to be again in the same phase, there should be difference of 1 complete oscillation. If smaller pendulum completes n oscillations the larger pendulum will complete \[(n-1)\] oscillations, so Time period of n oscillations of first = Time period of \[(n-1)\] oscillations of second i.e., \[n{{T}_{1}}=(n-1){{T}_{2}}\] or \[n\,2\,\pi \,\sqrt{\frac{{{L}_{1}}}{g}}=(n-1)\,2\,\pi \sqrt{\frac{{{L}_{2}}}{g}}\] or \[n\sqrt{{{L}_{1}}}=(n-1)\,\sqrt{{{L}_{2}}}\] or \[\frac{n}{n-1}=\sqrt{\frac{{{L}_{2}}}{{{L}_{1}}}}=\sqrt{\frac{2.0}{0.5}}\] or \[\frac{n}{n-1}=2\] or \[n=2n-2\] \[\therefore \] n = 2
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