A ball of mass 0.1 kg is whirled in a horizontal circle of radius 1 m by means of a string at an initial speed of 10 r.p.m. Keeping the radius constant, the tension in the string is reduced to one quarter of its initial value. The new speed is

Given data:
Mass of the ball, _$m$ = 0.1 kg
Radius of the horizontal circle, _$r$ = 1 m
Initial speed, _{$N_1$} = 10 r.p.m.
Let the initial tension in the string be _{$T_1$}.
The tension in the string is reduced to one-quarter of its initial value, so the new tension is:
$T_2=\frac{1}{4}{T}_1$

For a ball whirled in a horizontal circle, the tension in the string provides the necessary centripetal force.
The centripetal force in terms of angular speed _$\omega$ (or rotational speed _$N$ in r.p.m., since _{$\omega \propto N$}) is given by:
$T=m{\omega }^{2}r$

Since the mass _$m$ of the ball and the radius _$r$ of the circle are kept constant:
$T\propto {\omega }^2$
And because the rotational speed _$N$ is directly proportional to _$\omega$, we have:
$T\propto N^2$

We can write the ratio of the tensions as:
$\frac{T_2}{T_1}={\left(\frac{N_2}{N_1}\right)}^2$

Taking the square root on both sides:
$\frac{N_2}{N_1}=\sqrt{\frac{T_2}{T_1}}$

Substitute the given relationship _{$T_2/T_1=1/4$} into the equation:
$\frac{N_2}{N_1}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

Now, solve for the new speed _{$N_2$}:
$N_2=\frac{1}{2}{N}_1$
$N_2=\frac{1}{2}\times 10\text{r.p.m.}$
$N_2=5\text{r.p.m.}$

Thus, the new speed of the ball is 5 r.p.m.