If the speed of revolution of a particle on the circumference of a circle and the speed gained in falling through a distance equal to half the radius are equal, then the centripetal acceleration will be

The correct option is g.

Let the radius of the circle be R and the speed of the particle revolving on the circumference of the circle be $v_1$.

The formula for centripetal acceleration ($a_c$) is given by:

$a_c=\frac{{v_1}^2}{R}$

Next, let's find the speed gained by a body falling freely from rest through a distance h. Using the third equation of motion:

${v_2}^2=u^2+2gh$

Since the body starts from rest, the initial velocity is $u=0$. The distance fallen is equal to half the radius of the circle, which means $h=\frac{R}{2}$.

Substituting these values, we get:

${v_2}^2=0+2g\left(\frac{R}{2}\right)$

${v_2}^2=gR$

Therefore, the speed gained is:

$v_2=\sqrt{gR}$

According to the given condition, the speed of revolution and the speed gained in falling are equal ($v_1=v_2$):

$v_1=\sqrt{gR}$

Squaring both sides gives:

${v_1}^2=gR$

Now, substituting this value back into the formula for centripetal acceleration:

$a_c=\frac{gR}{R}$

$a_c=g$

Thus, the centripetal acceleration of the particle is equal to the acceleration due to gravity, g.