JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A particle of mass \(m\) is moving in a circular path of constant radius \(I\) such that its centripetal acceleration \((a)\) is varying with time \(t\) as \(a=k^{2} rt^{2}\). where \(k\) is a constant. The 10.Power delivered to the particle by the force acting on it is given as
- A \(zero\)
- B \(mk ^{2} r ^{2} t ^{2}\)
- C \(mk ^{2} r ^{2} t\)
- D \(m k^{2} rt\)
Answer & Solution
Correct Answer
(C) \(mk ^{2} r ^{2} t\)
Step-by-step Solution
Detailed explanation
\(a = k ^{2} rt ^{2}=\frac{ V ^{2}}{ r }\) \(V = krt\) \(a _{ t }=\frac{ dv }{ dt }= kr\) \(F _{ t }= ma _{ t }= mkr\) \(P =\overrightarrow{ F }. \overrightarrow{ V }\) \(= F \cos \theta V = F _{ t } V = mkr ( krt )\) \(P = mk ^{2} r ^{2} t\)
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