A) 1
B) 3
C) 2
D) 5
Correct Answer: C
Solution :
Time period of simple pendulum For 1st pendulum \[{{T}_{1}}=2\pi \sqrt{\frac{5}{g}}\] \[=2\pi \frac{\sqrt{5}}{9.8}=\sqrt{20}\,\sec \] For 2nd pendulum \[{{T}_{2}}=2\pi \frac{\sqrt{20}}{g}\] \[=2\pi \sqrt{\frac{20}{9.8}}=2\sqrt{20}\,\sec \] Suppose the time will be in same phase is \[t\] then \[\frac{1}{t}=\frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}}=\frac{1}{\sqrt{20}}-\frac{1}{2\sqrt{20}}=\frac{1}{2\sqrt{20}}\] So \[t=2\sqrt{20}\] Hence, the number of oscillation of simple pendulum of shorter length \[=\frac{2\sqrt{20}}{\sqrt{20}}=2\]
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