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A particle of mass \(m\) is moving in a circular path of constant radius \(r\) such that its centripetal acceleration \(a_c\) varies with time \(t\) as \(a_c = k^2 r t^2\), where \(k\) is a constant. The power delivered to the particle by the force acting on it is:

1. \(2\pi mk^2r^2\)
2. \(mk^2r^2t\)
3. \(4\sqrt{2}(mkrt)^3\)
4. Zero

Correct answer: \(mk^2r^2t\)

Solution

Given \(a_c = v^2/r = k^2rt^2\), we get \(v = krt\). The centripetal force is \(F = ma_c = mk^2rt^2\), so power is \(P = Fv = mk^2r^2t\).

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