A mass m attached to a string moves in a circle of radius R with velocity v. The string passes through a hole at the center. If the radius is reduced to R/4, what is the final kinetic energy?

Initially, the mass $m$ moves in a circle of radius $R_0$ with velocity $v_0$. The initial kinetic energy is $KE_0 = \frac{1}{2}mv_0^2$.
The string passes through a smooth hole, meaning the tension force acts radially and does no work on the mass. Therefore, the angular momentum of the mass about the center of the circle is conserved.

The angular momentum $L$ of a particle of mass $m$ moving with velocity $v$ at a radius $R$ is given by $L = mvR$.
Initial angular momentum: $L_0 = mv_0R_0$.
Final radius: $R_f = \frac{R_0}{4}$. Let the final velocity be $v_f$.
Final angular momentum: $L_f = mv_fR_f$.

By conservation of angular momentum:

Substitute $R_f = \frac{R_0}{4}$:

Cancel $m$ and $R_0$ from both sides:

Solve for $v_f$:

The final kinetic energy $KE_f$ is given by:

Substitute the expression for $v_f$ from Step 2:

Since $KE_0 = \frac{1}{2}mv_0^2$, we can write:

Comparing this with the given options, option (3) is the correct answer.

The final value of the kinetic energy is \text{8mv_0^2}.