As shown in the figure,two strings of length | vedclass

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(D) Let $	heta$ be the angle that each string makes with the vertical axis. Since the triangle $ABC$ is equilateral (all sides are $l$),$	heta = 60^

Vedclass · Accepted Answer

$T_1 - T_2 = 2mg$

Vedclass · Answer

(D) Let $	heta$ be the angle that each string makes with the vertical axis. Since the triangle $ABC$ is equilateral (all sides are $l$),$	heta = 60^\circ$.Resolving forces on mass $m$:Vertical direction: $(T_1 - T_2) \cos 	heta = mg \implies T_1 - T_2 = \frac{mg}{\cos 60^\circ} = 2mg$.Horizontal direction: $(T_1 + T_2) \sin 	heta = m \omega^2 r$,where $r = l \sin 60^\circ = l \frac{\sqrt{3}}{2}$.$(T_1 + T_2) \frac{\sqrt{3}}{2} = m \omega^2 l \frac{\sqrt{3}}{2} \implies T_1 + T_2 = m \omega^2 l$.Solving these equations: $T_1 = \frac{m \omega^2 l}{2} + mg$ and $T_2 = \frac{m \omega^2 l}{2} - mg$.For string $BC$ to remain taut,$T_2 \geq 0 \implies \frac{m \omega^2 l}{2} \geq mg \implies \omega \geq \sqrt{2g/l}$.Thus,option $D$ is correct.

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