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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)/(sqrt(2))`C. `sqrt(2)T`D. `(T)/(2^(1//4))` |
Answer» Correct Answer - D `(T_(2))/(T_(1))=sqrt((g)/(sqrt(g^(2)+g^(2))))=sqrt((g)/(gsqrt(2)))` `T_(2)=(T)/(sqrt(2))=(T)/(2^(1//4))` |
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