The kinetic energy is given as = 8 × 10-4 J
or, 1/2 mv² = 8 × 10-4
or, 1/2 × 10 × 10-3 v² = 8 × 10-4
or, v² = 16 × 10-2 => v = 0.4 m/s
initial velocity of particle, u = 0 m/s
we have to find Tangential acceleration at the end of 2nd revolution.
total distance covered, s = 2(2πr) = 4πr
so, v² = 2as
a = v²/2s = (0.4)²/2(4πr)
= 16 × 10^-2/(8 × 3.14 × 6.4 × 10-2)
= 0.0995 m/s² ≈ 0.1 m/s²