A) \[3\,\,T\]
B) \[3/2\,\,T\]
C) \[4\,\,T\]
D) \[2\,\,T\]
Correct Answer: A
Solution :
The periodic-time of a simple pendulum is given by \[T=2\pi \sqrt{\frac{l}{g}}\] \[{{T}_{1}}=T,\,\,{{l}_{1}}=l,\,\,{{l}_{2}}=9l\] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{l}_{1}}}{{{l}_{2}}}}=\sqrt{\frac{1}{9}}=\frac{1}{3}\] \[{{T}_{2}}=3{{T}_{1}}=3T\]
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