A particle of mass 10g moves along a circle of radius 6.4 cm with a constant tangential
acceleration. What is the magnitude of this acceleration, if the kinetic energy of the particle becomes equal to 8 X 10^-4 J by the end of the second revolution after the beginning of the motion?

Given, mass of particle, m = 0.01 kg
Radius of circle along which particle is moving, r = 6.4 cm
$\therefore$ Kinetic energy of particle, KE = 8 X l0^-4 J

$\Rightarrow \frac{1}{2}m{v}^2=8\times {10}^{-4} \mathrm{J}$

$\Rightarrow v^2=\frac{16\times {10}^{-4}}{0.01}=16\times {10}^{-2}$                   …(i)

As it is given that kinetic energy of particle is equal to  8 x 10^-4 J by the end of second revolution after the beginning
of motion of particle. It means, its initial velocity (u) is 0 ms^-1   at this moment.

$v^2=u^2+2a_{t}s$

$\Rightarrow v^2=2a_{t}s\text{ or }{v}^2=2a_t\left(4\pi r\right)$            ( $\because$ particle covers 2 revolutions)

$\Rightarrow a_t=\frac{v^2}{8\pi r}=\frac{16\times {10}^{-2}}{8\times 3.14\times 6.4\times {10}^{-2}}$                        [Using Eq. (i»)

$\therefore a_t\approx 0.1{ \mathrm{ms}}^{-2}$