A mass of 10 , kg is suspended vertically by | vedclass

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(D) Let the tension in the upper half of the rope be $T_1$ and the tension in the lower half be $T_2$. The lower half of the rope supports the mass $m

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let the tension in the upper half of the rope be $T_1$ and the tension in the lower half be $T_2$. The lower half of the rope supports the mass $m = 10 \, kg$,so $T_2 = mg = 10 	imes 10 = 100 \, N$.At the middle point where the horizontal force $F = 30 \, N$ is applied,we consider the equilibrium of forces in the horizontal and vertical directions.For horizontal equilibrium: $T_1 \sin 	heta = F = 30 \, N$.For vertical equilibrium: $T_1 \cos 	heta = T_2 = 100 \, N$.Dividing the two equations: $\frac{T_1 \sin 	heta}{T_1 \cos 	heta} = \frac{30}{100} \Rightarrow 	an 	heta = 0.3$.Given $	heta = 	an^{-1}(x 	imes 10^{-1})$,we have $	an 	heta = x 	imes 10^{-1} = \frac{x}{10}$.Equating the two expressions for $	an 	heta$: $\frac{x}{10} = 0.3 \Rightarrow x = 3$.

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