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(D) For a conical pendulum,the forces acting on the mass $M$ are the tension $T$ in the string and its weight $Mg$.Resolving the tension $T$ into hori

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$16 \,ML$

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(D) For a conical pendulum,the forces acting on the mass $M$ are the tension $T$ in the string and its weight $Mg$.Resolving the tension $T$ into horizontal and vertical components:$T \sin 	heta = M \omega^2 R$ (Centripetal force equation) ... $(i)$$T \cos 	heta = Mg$ (Vertical equilibrium) ... (ii)From the geometry,the radius of the circular path is $R = L \sin 	heta$.Substituting $R$ into equation $(i)$:$T \sin 	heta = M \omega^2 (L \sin 	heta)$$T = M \omega^2 L$Given the frequency $n = 2/\pi$ revolutions per second,the angular velocity is $\omega = 2 \pi n = 2 \pi (2/\pi) = 4 	ext{ rad/s}$.Substituting $\omega = 4$ into the expression for tension:$T = M (4)^2 L = 16 ML$.

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

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