To solve this problem, we need to determine the tangential acceleration of a particle moving in a circle.$\\$Given:$\\$• Mass of the particle $m = 10$ g$= 0.01$ kg$\\$• Radius of the circle $r = 6.4$ cm$= 0.064$ m$\\$• Kinetic energy after two revolutions $K = 8 \times 10^{-4}$ J$\\$We need to find the tangential acceleration $a_t. \\$First, let's find the velocity of the particle after two revolutions.$\\$The kinetic energy is given by:$\\ K = \frac{1}{2} m v^2 \\$Rearranging for $v: \\ v = \sqrt{\frac{2K}{m}} \\$Substitute the known values:$\\ v = \sqrt{\frac{2 \times 8 \times 10^{-4}}{0.01}} \\ v = \sqrt{\frac{16 \times 10^{-4}}{0.01}} \\ v = \sqrt{1.6} \\ v = 1.2649$ m/s$\\$Now, let's find the tangential acceleration.$\\$The distance covered in two revolutions is:$\\ s = 2 \times 2\pi r = 4\pi r \\ s = 4\pi \times 0.064 \\ s = 0.256\pi$ m$\\$Using the kinematic equation for constant acceleration:$\\ v^2 = u^2 + 2a_t s \\$Since the initial velocity $u = 0,$ we have:$\\ v^2 = 2a_t s \\$Substitute the known values:$\\ (1.2649)^2 = 2a_t \times 0.256\pi \\ 1.6 = 2a_t \times 0.256\pi \\ a_t = \frac{1.6}{2 \times 0.256\pi} \\ a_t = \frac{1.6}{0.512\pi} \\ a_t = \frac{1.6}{1.608} \\ a_t \approx 0.1$ m/s$^2 \\$Therefore, the magnitude of the tangential acceleration is $0.1$ m/s$^2. \\$This corresponds to Option 3.