A string of length $L$ is fixed at one end and carries a mass of $M$ at the other end. The mass makes $\left(\frac{3}{\pi}\right)$ rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is $\ldots\ldots$ ML.

Solution

$\begin{aligned} & \omega=\frac{3}{\pi} \times 2 \pi=6 \mathrm{rad} / \mathrm{s} \\ & R=L \sin \theta\end{aligned}$
and $T=M \sqrt{g^2+\omega^4 R^2}$ Also, $T \sin \theta=M \omega^2 \cdot L \sin \theta$ $\begin{aligned} & \Rightarrow \quad T=M(36) \mathrm{L} \\ & \Rightarrow \quad T=36 \mathrm{ML} \end{aligned}$

Hence, the correct answer is 36.