In the arrangement as shown,tension T_2 is .. | vedclass

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(B) The tension in the string supporting the $10\,kg$ block is $T = mg = 10 	imes 10 = 100\,N$.For the pulley,the downward force is $2T = 2 	imes 10

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$100$

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(B) The tension in the string supporting the $10\,kg$ block is $T = mg = 10 	imes 10 = 100\,N$.For the pulley,the downward force is $2T = 2 	imes 100 = 200\,N$. This force is balanced by the vertical components of $T_1$ and $T_2$.Let the angles with the horizontal be $60^{\circ}$ and $30^{\circ}$ respectively.For horizontal equilibrium: $T_1 \cos 60^{\circ} = T_2 \cos 30^{\circ} \implies T_1 	imes \frac{1}{2} = T_2 	imes \frac{\sqrt{3}}{2} \implies T_1 = \sqrt{3} T_2 \dots(i)$For vertical equilibrium: $T_1 \sin 60^{\circ} + T_2 \sin 30^{\circ} = 200\,N$.Substituting $(i)$ into the vertical equilibrium equation:$(\sqrt{3} T_2) \sin 60^{\circ} + T_2 \sin 30^{\circ} = 200$$(\sqrt{3} T_2) 	imes \frac{\sqrt{3}}{2} + T_2 	imes \frac{1}{2} = 200$$\frac{3}{2} T_2 + \frac{1}{2} T_2 = 200$$2 T_2 = 200 \implies T_2 = 100\,N$.

In the arrangement as shown,tension $T_2$ is ......... $N$ $\left(g=10\,m/s^2\right)$

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