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(C) Let $T_S$ be the tension in the steel wire and $T_C$ be the tension in the copper wire.Resolving forces in the horizontal $(x)$ direction:$T_C \co

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Let $T_S$ be the tension in the steel wire and $T_C$ be the tension in the copper wire.Resolving forces in the horizontal $(x)$ direction:$T_C \cos 30^{\circ} = T_S \cos 60^{\circ}$$T_C 	imes \frac{\sqrt{3}}{2} = T_S 	imes \frac{1}{2}$$T_S = \sqrt{3} T_C \quad \dots (i)$Resolving forces in the vertical $(y)$ direction:$T_C \sin 30^{\circ} + T_S \sin 60^{\circ} = 100$$\frac{T_C}{2} + \frac{T_S \sqrt{3}}{2} = 100 \quad \dots (ii)$Substituting $(i)$ into $(ii)$:$\frac{T_C}{2} + \frac{(\sqrt{3} T_C) \sqrt{3}}{2} = 100$$\frac{T_C}{2} + \frac{3 T_C}{2} = 100 \implies 2 T_C = 100 \implies T_C = 50 \ N$$T_S = \sqrt{3} 	imes 50 = 50\sqrt{3} \ N$Using the formula for elongation $\Delta \ell = \frac{FL}{AY}$:$\frac{\Delta \ell_C}{\Delta \ell_S} = \frac{T_C L_C}{A_C Y_C} 	imes \frac{A_S Y_S}{T_S L_S}$Given $A_C = A_S = 0.5 \ cm^2$,$L_C = \sqrt{3} \ m$,$L_S = 1 \ m$,$Y_C = 1 	imes 10^{11} \ N/m^2$,$Y_S = 2 	imes 10^{11} \ N/m^2$:$\frac{\Delta \ell_C}{\Delta \ell_S} = \left( \frac{50 	imes \sqrt{3}}{0.5 	imes 10^{11}} ight) 	imes \left( \frac{0.5 	imes 2 	imes 10^{11}}{50\sqrt{3} 	imes 1} ight) = \frac{50\sqrt{3}}{50\sqrt{3}} 	imes \frac{2}{1} = 2$Thus,the ratio is $2$.

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