A string of length 'L' fixed at one end carries a mass 'm' at the other end. The string makes $\dfrac{3}{2\pi}$ revolutions per second around the vertical axis through fixed end. The tension in the string is

Let $T$ be the tension in the string and $\theta$ be the angle it makes with the vertical.

The radius of the circular path is $r = L \sin\theta$.

The horizontal component of the tension provides the necessary centripetal force:

$T \sin\theta = m \omega^2 r$

Substituting $r = L \sin\theta$:

$T \sin\theta = m \omega^2 L \sin\theta$

$T = m \omega^2 L$

Given the frequency of revolution $f = \dfrac{3}{2\pi}$ Hz.

The angular velocity $\omega = 2\pi f = 2\pi \left(\dfrac{3}{2\pi}\right) = 3$ rad/s.

Substituting the value of $\omega$ in the expression for tension:

$T = m (3)^2 L = 9mL$

Answer: $9\,mL$