A) 3T
B) 3/2T
C) 4T
D) 2T
Correct Answer: A
Solution :
The periodic-time of a simple pendulum is given by \[T=2\pi \sqrt{\frac{l}{g}}\] where\[l\]is length of pendulum, g is acceleration due to gravity. Given, \[{{T}_{1}}=T,{{l}_{1}}=l,{{l}_{2}}=9l\] \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{l}_{1}}}{{{l}_{2}}}}=\sqrt{\frac{1}{9}}=\frac{1}{3}\] \[\Rightarrow \] \[{{T}_{2}}=3{{T}_{1}}=3T\] Note: The periodic-time of the pendulum does not depend upon the mass of the bob.
You need to login to perform this action.
You will be redirected in
3 sec
