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BHU PMT BHU PMT (Screening) Solved Paper-2008

  • question_answer
    A particle of mass m is moving in a circular path of constant radius r such that centripetal acceleration\[{{a}_{c}}\]varying with time is\[{{a}_{c}}={{k}^{2}}r{{t}^{2}}\]where k is a constant. What is the power delivered to the particle by the force acting on it?

    A)  \[2mk{{r}^{2}}t\]                            

    B) \[mk{{r}^{2}}{{t}^{2}}\]

    C)  \[m{{k}^{2}}{{r}^{2}}t\]                               

    D) \[m{{k}^{2}}r{{t}^{2}}\]

    Correct Answer: C

    Solution :

                     \[{{a}_{c}}=\frac{{{v}^{2}}}{r}={{k}^{2}}r{{t}^{2}}\] \[\Rightarrow \] \[{{v}^{2}}={{k}^{2}}{{r}^{2}}{{t}^{2}}\] \[KE=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{k}^{2}}{{r}^{2}}{{t}^{2}}\] According to work-energy theorem, change in kinetic energy is equal to work done. \[\therefore \]  \[W=\frac{1}{2}m{{k}^{2}}{{r}^{2}}{{t}^{2}}\] Thus, power delivered to the particle, \[P=\frac{dW}{dt}=m{{k}^{2}}{{r}^{2}}t\]


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