JEE Main 2016 · 10 Apr · Shift 1 · Physics Q2

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PhysicsSingle correct+4 / -1Question 2 of 90

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 :

Mn^2R^2t

MnR^2t

MnR^2t^2

\dfrac12Mn^2R^2t^2

Answer and worked solution

Official answer: A

$\frac{v^2}R=n^2Rt^2$ gives $v=nRt$ , so the tangential force is $M\frac{dv}{dt}=MnR$ . The centripetal force does no work, so
$P=F_tv=MnR\cdot nRt=Mn^2R^2t.$

Topics:Circular motion dynamics Power

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JEE Main 2016 · 10 Apr · Shift 1, Physics question 2