A particle does uniform circular motion in a horizontal plane. The radius of the circle is 20 cm. The centripetal force acting on the particle is 10 N. It’s kinetic energy is

To find the kinetic energy of the particle, we can establish a relationship between the centripetal force and the kinetic energy.

First, let us identify the given values from the problem:
Radius of the circular path, $r=20\text{cm}=0.2\text{m}$
Centripetal force acting on the particle, $F_c=10\text{N}$

The formula for centripetal force ($F_c$) acting on a particle of mass $m$ moving with velocity $v$ in a circle of radius $r$ is given by:
$F_c=\frac{m{v}^2}{r}$

We can rearrange this formula to express $m{v}^2$ in terms of force and radius:
$m{v}^2=F_c\cdot r$

The kinetic energy ($K$) of a moving particle is given by the standard formula:
$K=\frac{1}{2}m{v}^2$

Substitute the expression $m{v}^2=F_c\cdot r$ into the kinetic energy equation:
$K=\frac{1}{2}(F_c\cdot r)$

Now, substitute the given numerical values into the equation:
$K=\frac{1}{2}\cdot 10\text{N}\cdot 0.2\text{m}$
$K=5\cdot 0.2$
$K=1.0\text{Joule}$

Therefore, the kinetic energy of the particle is 1.0 Joule.