InterviewSolution
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InterviewSolution
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A partical of mass `m` is driven by a machine that deleveres a constant power `k` watts. If the partical starts from rest the force on the partical at time `t` isA. `sqrt(mK)t^(-1//2)`B. `sqrt(2mK)t^(-1//2)`C. `(1)/(2)sqrt(mK)t^(-1//2)`D. `sqrt((mK)/(2))t^(-1//2)` |
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Answer» Correct Answer - D Here, power `= K wat t, time =t` as`P=Fupsilon=maupsilon, a=(d upsilon)/(dt)` `K=m (d upsilon)/(dt)` or `upsilond upsilon=(Kdt)/(m)` …(i) Integrating, we get `int upsilon dupsilon=int(Kdt)/(m)` or `(upsilon^(2))/(2)=(K)/(m)intdt` or `(upsilon^(2))/(2)=(K)/(m)t` or `upsilon=sqrt((2Kt)/(m))` Now, `a=(dupsilon)/(dt)=(d)/(dt)(sqrt((2Kt)/(m)))=sqrt((2K)/(m))((1)/(2)t^(-1//2))` `=(1)/(2)sqrt((2K)/(mt))` As `F=ma=mxx(1)/(2)sqrt((2K)/(mt))` or `F=sqrt((mK)/(2t))` or `F=sqrt((mK)/(2))t^(-1//2)` |
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