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A seconds pendulum is suspended from rof of a vehicle that is moving along a circular track of radius `(10)/(sqrt(3))m` with speed `10m//s`. Its period of oscillation will be `(g = 10m//s^(2))`A. `sqrt(2)s`B. `2s`C. `1s`D. `0.5s` |
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Answer» Correct Answer - A `T = 2pi sqrt((l)/(sqrt(g^(2)+a^(2)))),a =(v^(2))/(R)` |
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