A block of sqrt{3} , kg is attached to a stri | vedclass

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(A) Let the mass of the block be $m = \sqrt{3} \, kg$. The weight of the block acting downwards is $W = mg = \sqrt{3} 	imes 10 = 10\sqrt{3} \, N$.Con

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$20$

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(A) Let the mass of the block be $m = \sqrt{3} \, kg$. The weight of the block acting downwards is $W = mg = \sqrt{3} 	imes 10 = 10\sqrt{3} \, N$.Consider the equilibrium of the point where the string,the force $F$,and the weight are connected. The tension $T$ in the string makes an angle of $30^{\circ}$ with the vertical wall.Resolving the tension $T$ into components:Vertical component: $T \cos 30^{\circ}$ (acting upwards)Horizontal component: $T \sin 30^{\circ}$ (acting towards the wall)For the system to be in equilibrium:$1$. The vertical forces must balance: $T \cos 30^{\circ} = mg$$T 	imes \frac{\sqrt{3}}{2} = 10\sqrt{3}$$T = 10 	imes 2 = 20 \, N$Thus,the tension $T$ is $20 \, N$.

$A$ block of $\sqrt{3} \, kg$ is attached to a string whose other end is attached to the wall. An unknown force $F$ is applied so that the string makes an angle of $30^{\circ}$ with the wall. The tension $T$ is $........... \, N$ : (Given $g = 10 \, ms^{-2}$)

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