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$A$ particle is moving along a circular path of radius $R$ in such a way that at any instant magnitude of radial acceleration & tangential acceleration are equal. If at $t = 0$ velocity of particle is $V_0$, the time period of first revolution of the particle is
$\frac{R}{{{V_0}}} e^{-2 \pi}$
$\frac{R}{{{V_0}}}$ $(e^{2\pi} {-1})$
$\frac{R}{{{V_0}}}$
$\frac{R}{{{V_0}}}$ $(1 - e^{-2 \pi} )$
Solution
$\because$Given $\frac{d v}{d t}=\frac{v^{2}}{R}$$…(1)$
$\Rightarrow \int_{v_{0}}^{v} \frac{d v}{v^{2}}=\int_{0}^{t} \frac{1}{R} d t \Rightarrow R\left(\frac{1}{v_{0}}-\frac{1}{v}\right)=t \ldots(2)$
Again from eqn $( 1 )$
$\because \frac{d v}{d s} \cdot \frac{d s}{d t}=\frac{v^{2}}{R} \Rightarrow v \frac{d v}{d s}=\frac{v^{2}}{R}$
$\int_{-6}^{v} \frac{d v}{v}=\int_{0}^{2 \pi R} \frac{d s}{R} \Rightarrow v=v_{0} e^{-2 \pi} \ldots(3)$
from $( 2)$ and $(3), t=\frac{R}{v_{0}}\left(1-e^{-2 \pi}\right)$