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A particle starts moving along a circle of radius `(20//pi)m` with constant tangential acceleration. If the velocity of the parthcle is `50 m//s` at the end of the second revolution after motion has began, the tangential acceleration in `m//s^(2)` is :A. `1.6`B. `4`C. `15.6`D. `13.2` |
Answer» Correct Answer - C Given `r=20/pi m` Angular velocity after second revolution `omega=v/r=(50pi)/20=(5pi)/2` `omega_("final")^(2)=omega_("initial")^(2)+2alpha theta` `25/4 pi^(2)=2alpha(4pi) rArr alpha=(25 pi)/32` `a_(t)=alphar=(25pi)/32xx20/pi=15.6` |
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