A) 6 s
B) 2 s
C) 8 s
D) 4 s
Correct Answer: D
Solution :
The time period of simple pendulum is given by \[T=2\pi \sqrt{\frac{l}{g}}\] (Given: \[{{l}_{2}}=4{{l}_{1}},{{T}_{1}}=2s\]) \[T\propto \sqrt{l}\] Hence, \[\frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\frac{{{l}_{2}}}{{{l}_{1}}}}\] \[{{T}_{2}}=\sqrt{\frac{4{{l}_{1}}}{{{l}_{1}}}}\times {{T}_{1}}=2{{T}_{1}}\] \[=2\times 2=4s\]
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