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A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. if the lift accelerates upwards with an acceleration g/4, then the period of the pendulum will beA. TB. `(T)/(4)`C. `(2T)/(sqrt(5))`D. `2T sqrt(5)` |
Answer» Correct Answer - C `(T_(2))/(T_(1))=sqrt((g_(1))/(g_(2)))=sqrt((9)/(9+(9)/(4)))=sqrt((4)/(5))` `therefore T_(2)=(2)/(sqrt(5))T`. |
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