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(B) Let $l$ be the length of the thread. When the disk is released from the horizontal position,it moves to the lowest point of its circular path on t

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(B) Let $l$ be the length of the thread. When the disk is released from the horizontal position,it moves to the lowest point of its circular path on the inclined plane.In the lowest position,the disk has descended a vertical height $h = l \sin 30^\circ = l/2$.Applying the work-energy theorem,the change in kinetic energy equals the work done by gravity:$\frac{1}{2}mv^2 = mg(l \sin 30^\circ) = mg(l/2) = \frac{mgl}{2}$.Thus,$mv^2 = mgl$.At the lowest point,the forces acting along the radial direction are the tension $T$ (towards the center) and the component of gravity $mg \sin 30^\circ$ (away from the center).The net centripetal force is $T - mg \sin 30^\circ = \frac{mv^2}{l}$.Substituting $mv^2 = mgl$ and $\sin 30^\circ = 0.5$:$T - mg(0.5) = \frac{mgl}{l} = mg$.$T = mg + 0.5mg = 1.5mg$.Therefore,the tension is $1.5mg$.

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