1.

Simple pendulum is executing simple harmonic motion with time period `T`. If the length of the pendulum is increased by `21%`, then the increase in the time period of the pendulum of the increased length is:A. 0.1 sB. 0.2 sC. 0.4 sD. 0.15 s

Answer» Correct Answer - B
`(T_(2))/(T_(1))=sqrt((l_(2))/(l_(1)))=sqrt(1.21)`
`therefore (T_(2))/(T_(1))=1.1`
`(T_(2)-T_(1))/(T_(1))=0.1`
`therefore T_(2)-T_(1)=0.1xx2=0.2s`


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