JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2016
- Medium
- MCQ
A particle of mass \(M\) is moving in a circle of fixed radius \(R\) in such a way that its centripetal acceleration at time \(t\) is given by \(n^2Rt^2\) where \(n\) is a constant. The Power delivered to the particle by the force acting on it, is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(M\,{n^2}{R^2}{t^2}\)
Step-by-step Solution
Detailed explanation
\(v = \sqrt{a_c R}\)
\(v = \sqrt{(n^2Rt^2)R} = \sqrt{n^2R^2t^2} = nRt\)
\(a_t = \frac{dv}{dt} = \frac{d}{dt}(nRt) = nR\)
\(P = F_t v = (Ma_t)v = (MnR)(nRt)\)
\(P = Mn^2R^2t\)
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