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(C) Let $T$ be the tension in the inclined string making an angle of $30^\circ$ with the horizontal. For the equilibrium of the knot,the vertical comp

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$2\sqrt{3}$

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(C) Let $T$ be the tension in the inclined string making an angle of $30^\circ$ with the horizontal. For the equilibrium of the knot,the vertical component of the tension $T$ must balance the weight of the body: $T \sin 30^\circ = 2 \, kg-wt$ $T 	imes (1/2) = 2 \, kg-wt$ $T = 4 \, kg-wt$ The horizontal component of the tension $T$ must balance the tension $T_1$ in the horizontal string: $T_1 = T \cos 30^\circ$ $T_1 = 4 	imes (\sqrt{3}/2)$ $T_1 = 2\sqrt{3} \, kg-wt$

$A$ body of weight $2 \, kg-wt$ is suspended as shown in the figure. The tension $T_1$ in the horizontal string (in $kg-wt$) is

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