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(C) Given: Frequency $f = \frac{\pi}{2} 	ext{ rev/s}$.Angular velocity $\omega = 2 \pi f = 2 \pi \left( \frac{\pi}{2} ight) = \pi^2 	ext{ rad/s}$.

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$16 M L$

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(C) Given: Frequency $f = \frac{\pi}{2} 	ext{ rev/s}$.Angular velocity $\omega = 2 \pi f = 2 \pi \left( \frac{\pi}{2} ight) = \pi^2 	ext{ rad/s}$.For a conical pendulum,the forces acting on the mass $M$ are the tension $T$ in the string and the gravitational force $Mg$.The horizontal component of tension provides the centripetal force: $T \sin 	heta = M R \omega^2$.From the geometry of the figure,the radius of the circular path is $R = L \sin 	heta$.Substituting $R$ into the force equation: $T \sin 	heta = M (L \sin 	heta) \omega^2$.Simplifying,we get $T = M L \omega^2$.Substituting the value of $\omega = \pi^2 	ext{ rad/s}$:$T = M L (\pi^2)^2 = M L \pi^4$.Note: If the frequency was given as $\frac{2}{\pi} 	ext{ rev/s}$,then $\omega = 2 \pi (\frac{2}{\pi}) = 4 	ext{ rad/s}$,leading to $T = M L (4)^2 = 16 M L$. Given the options,the intended frequency is $\frac{2}{\pi} 	ext{ rev/s}$.

$A$ string of length $L$ is fixed at one end and carries a mass $M$ at the other end. The string makes $\frac{\pi}{2}$ revolutions per second around the vertical axis through the fixed end as shown in the figure. The tension in the string is:

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