A) 16 s
B) 12 s
C) 8 s
D) 4 s
Correct Answer: D
Solution :
Time period of simple pendulum \[T=2\pi \frac{\sqrt{l}}{g}\] i.e., \[T\propto \sqrt{l}\] Hence, \[\frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\frac{{{l}_{2}}}{{{l}_{1}}}}\] Given, \[{{l}_{2}}=4{{l}_{1}},\,{{T}_{1}}=2s\] Substituting the values in Eq. (i), we get \[{{T}_{2}}=\sqrt{\frac{4{{l}_{1}}}{{{l}_{1}}}}\times 2=2\times 2=4\,s\]
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