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**Hint:**In this question first we need to find the total distance travelled by particle after two revolutions in a circular path, by finding the circumference of the circular path and then multiplying it by two since the particle made two revolutions.

**Complete step by step answer:**Mass of the particle \[m = 10g = 10 \times {10^{ - 3}}kg\]

The radius of the circle \[r = 6.4cm = 6.4 \times {10^{ - 2}}m\]

Kinetic energy =\[8 \times {10^{ - 4}}J\]

Where kinetic energy is given by the formula\[KE = \dfrac{1}{2}m{v^2} - - (i)\]

Hence we can write

\[

\dfrac{1}{2}m{v^2} = 8 \times {10^{ - 4}}J \\

\dfrac{1}{2}\left( {10 \times {{10}^{ - 3}}} \right){v^2} = 8 \times {10^{ - 4}}J \\

\]

By further solving we get

\[

{v^2} = \dfrac{{2 \times 8 \times {{10}^{ - 4}}}}{{10 \times {{10}^{ - 3}}}} \\

{v^2} = 16 \times {10^{ - 2}} \\

v = 4 \times {10^{ - 1}} \\

v = 0.4{\text{ m/s}} - - (ii) \\

\]

Now find the total distance covered by the particle after two revolutions by using

\[S = 2\left( {2\pi r} \right) - - (iii)\]

By solving we get

\[

S = 4\pi r \\

= 4 \times 3.14 \times 6.4 \times {10^{ - 2}} \\

= 0.8m \\

\]

Now use the third equation of newton’s law of motion which is given as

\[{v^2} = {u^2} + 2aS - - (iii)\]

Now since the particle started from the rest, hence its initial velocity will be

\[u = 0\]

Now substitute the value of distance S and the velocity v in equation (iii) to find the acceleration; hence we get

\[

{v^2} = {u^2} + 2aS - - (iii) \\

{\left( {0.4} \right)^2} = 0 + 2a(0.8) \\

0.16 = 2a(0.8) \\

a = 0.1{\text{ m/}}{{\text{s}}^2} \\

\]

Therefore, the magnitude of this acceleration by the end of the second revolution \[ = 0.1{\text{ m/}}{{\text{s}}^2}\]

Option A is correct

**Note:**The newton’s third equation of motion establishes a relation between the initial velocity, final velocity, acceleration and the displacement of the particle which is given as \[{v^2} = {u^2} + 2aS\], where S is the total distance travelled by the particle, u is the initial velocity, v is the final velocity.

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