Two stones of masses m and 3 m are whirled in horizontal circles, the heavier one in a radius $\left(\frac{\mathrm{r}}{3}\right)$ and lighter one in a radius r . The tangential speed of lighter stone is ' $n$ ' times the value of heavier stone. When the magnitude of centripetal force becomes equal the value of $n$ is

The centripetal forces are equal: $F_L = F_H$. For stones with masses $m_L = m$, $m_H = 3m$ and radii $r_L = r$, $r_H = \frac{r}{3}$, with tangential speeds related by $v_L = n v_H$, the force equality becomes:

$\frac{m (n v_H)^2}{r} = \frac{3m v_H^2}{r/3}$

Simplifying both sides yields:

$\frac{m n^2 v_H^2}{r} = \frac{9m v_H^2}{r}$

Canceling common factors ($m$, $v_H^2$, and $r$) from both sides gives $n^2 = 9$, so $n = 3$.

Answer: $\boxed{3}$