A steel wire of length 3.2 , m (Y_{S} = 2.0 t | vedclass

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(D) The total elongation $\Delta \ell$ is the sum of the individual elongations of the steel wire $(\Delta \ell_{S})$ and the copper wire $(\Delta \el

Vedclass · Accepted Answer

$154$

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(D) The total elongation $\Delta \ell$ is the sum of the individual elongations of the steel wire $(\Delta \ell_{S})$ and the copper wire $(\Delta \ell_{C})$:$\Delta \ell = \Delta \ell_{S} + \Delta \ell_{C}$Using the formula for elongation $\Delta \ell = \frac{F \ell}{AY}$,we have:$\Delta \ell = \frac{F \ell_{S}}{A Y_{S}} + \frac{F \ell_{C}}{A Y_{C}} = \frac{F}{A} \left( \frac{\ell_{S}}{Y_{S}} + \frac{\ell_{C}}{Y_{C}} ight)$Given $r = 1.4 \, mm = 1.4 	imes 10^{-3} \, m$,the cross-sectional area $A = \pi r^{2} = \frac{22}{7} 	imes (1.4 	imes 10^{-3})^{2} = \frac{22}{7} 	imes 1.96 	imes 10^{-6} = 6.16 	imes 10^{-6} \, m^{2}$.Substituting the values:$1.4 	imes 10^{-3} = \frac{F}{6.16 	imes 10^{-6}} \left( \frac{3.2}{2.0 	imes 10^{11}} + \frac{4.4}{1.1 	imes 10^{11}} ight)$$1.4 	imes 10^{-3} = \frac{F}{6.16 	imes 10^{-6}} \left( 1.6 	imes 10^{-11} + 4.0 	imes 10^{-11} ight)$$1.4 	imes 10^{-3} = \frac{F}{6.16 	imes 10^{-6}} 	imes 5.6 	imes 10^{-11}$$F = \frac{1.4 	imes 10^{-3} 	imes 6.16 	imes 10^{-6}}{5.6 	imes 10^{-11}} = \frac{8.624 	imes 10^{-9}}{5.6 	imes 10^{-11}} = 1.54 	imes 10^{2} = 154 \, N$.

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