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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