The minimum speed for a particle at the lowest point of a vertical circle of radius R, to describe the circle is ‘v’. If the radius of the circle is reduced to one-fourth its value, the corresponding minimum speed will be

The correct option is v/2.

Step-by-step Explanation:
For a particle to successfully complete a vertical circle of radius $R$ under gravity $g$, the minimum speed required at the lowest point of the circle (often called the critical velocity) is given by the formula:
$v=\sqrt{5gR}$

Let the initial minimum speed be $v_1=v$ for a circle of radius $R_1=R$. Therefore:
$v=\sqrt{5gR}$

Now, the radius of the circle is reduced to one-fourth of its initial value. Let the new radius be $R_2$:
$R_2=\frac{R}{4}$

Let the corresponding new minimum speed at the lowest point be $v_2$:
$v_2=\sqrt{5g{R}_2}=\sqrt{5g\left(\frac{R}{4}\right)}$

By simplifying this expression, we can pull out the factor of $\frac{1}{4}$ from under the square root:
$v_2=\sqrt{\frac{1}{4}\cdot 5gR}=\frac{1}{2}\sqrt{5gR}$

Since $\sqrt{5gR}=v$, we substitute this back into the equation:
$v_2=\frac{v}{2}$

Thus, the corresponding minimum speed when the radius is reduced to one-fourth of its value will be v/2.