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

Option: 1
$zero$

Option: 2
$\mathrm{m k^{2} r^{2} t^{2}}$

Option: 3
$\mathrm{m k^{2} r^{2} t}$

Option: 4
$\mathrm{m k^{2} rt}$

Answers (1)

best_answer

$\mathrm{a=k^{2} r t^{2}}\\$

$\because \text{since centripetal acceleration}$

$\mathrm{a_{c}=\frac{v^{2}}{r} =k^{2} r t^{2}} \\$

$\mathrm{v^{2} =k^{2} r^{2} t^{2}} \\$

$\mathrm{v =k r t}$

Here speed is variable, therefore particle is performing non-uniform circular motion

$\mathrm{\text { Power }(P) =F_{\text {tangential }} \times \text { speed } }\\$

$\mathrm{=m a_{t} \times v} \\$

$\mathrm{a_{t}=\frac{d v}{d t} =\frac{d}{d t}(k r t)=k r} \\$

$\mathrm{p =m k r \times k r t} \\$

$\mathrm{p =m k^{2} r^{2} t}$

Hence the correct option is 3

Posted by

Devendra Khairwa

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