A steel wire of length 3 m and a copper wire | vedclass

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Question

(A) The formula for Young's modulus is $Y = \frac{F \cdot L}{A \cdot \Delta L}$, which gives elongation $\Delta L = \frac{F \cdot L}{Y \cdot A}$.Given

Vedclass · Accepted Answer

$180$

Vedclass · Answer

(A) The formula for Young's modulus is $Y = \frac{F \cdot L}{A \cdot \Delta L}$, which gives elongation $\Delta L = \frac{F \cdot L}{Y \cdot A}$.Given total elongation $\Delta L_{total} = \Delta L_s + \Delta L_c = 1.05 	imes 10^{-3} \ m$.Substituting the values: $\frac{F \cdot L_s}{Y_s \cdot A} + \frac{F \cdot L_c}{Y_c \cdot A} = 1.05 	imes 10^{-3}$.$F \left( \frac{3}{2 	imes 10^{11} 	imes 6 	imes 10^{-6}} + \frac{2.2}{1.1 	imes 10^{11} 	imes 6 	imes 10^{-6}} ight) = 1.05 	imes 10^{-3}$.$F \left( \frac{3}{12 	imes 10^5} + \frac{2.2}{6.6 	imes 10^5} ight) = 1.05 	imes 10^{-3}$.$F \left( \frac{0.25}{10^5} + \frac{0.333}{10^5} ight) = 1.05 	imes 10^{-3}$.$F \left( \frac{0.5833}{10^5} ight) = 1.05 	imes 10^{-3}$.Calculating precisely: $F \left( \frac{3}{1.2 	imes 10^6} + \frac{2.2}{0.66 	imes 10^6} ight) = 1.05 	imes 10^{-3} \Rightarrow F \left( 2.5 	imes 10^{-6} + 3.33 	imes 10^{-6} ight) = 1.05 	imes 10^{-3}$.Using fractions: $F \left( \frac{3}{12 	imes 10^5} + \frac{2.2}{6.6 	imes 10^5} ight) = F \left( \frac{1}{4 	imes 10^5} + \frac{1}{3 	imes 10^5} ight) = F \left( \frac{3+4}{12 	imes 10^5} ight) = F \left( \frac{7}{12 	imes 10^5} ight) = 1.05 	imes 10^{-3}$.$F = \frac{1.05 	imes 10^{-3} 	imes 12 	imes 10^5}{7} = \frac{1.26 	imes 10^3}{7} = 180 \ N$.

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