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A pendulum suspended from the ceiling of the train has a time period of two seconds when the train is at rest, then the time period of the pendulum, if the train accelerates 10 `m//s^(2)` will be `(g=10m//s^(2))`A. 2sB. `2 sqrt(2) s`C. `2//sqrt(2) s`D. `2^(3//4)` |
Answer» Correct Answer - D The net acceleration = `sqrt((a^(2)+g^(2)))` `=sqrt(100+100)=sqrt(200)` `T_(1)=2pisqrt((l)/(g))and T_(2)=2pisqrt((l)/(a_(R)))` `(T_(2))/(T_(1))=sqrt((10)/(10sqrt(2)))therefore T_(2)=(2)/(sqrt(2))` |
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