The elongation of the copper wire of cross-se | vedclass

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Question

(A) The tension in the copper wire is due to the weight of the $7 \,kg$ mass hanging from it.$T = m 	imes g = 7 \,kg 	imes 10 \,m/s^2 = 70 \,N$.The

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$10^{-4} \,m$

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(A) The tension in the copper wire is due to the weight of the $7 \,kg$ mass hanging from it.$T = m 	imes g = 7 \,kg 	imes 10 \,m/s^2 = 70 \,N$.The formula for elongation $\Delta l$ is given by $\Delta l = \frac{T 	imes l}{Y 	imes A}$.Given:$T = 70 \,N$$l = 0.5 \,m$$Y = 10 	imes 10^{10} \,N/m^2$$A = 3.5 \,mm^2 = 3.5 	imes 10^{-6} \,m^2$Substituting the values:$\Delta l = \frac{70 	imes 0.5}{10 	imes 10^{10} 	imes 3.5 	imes 10^{-6}}$$\Delta l = \frac{35}{35 	imes 10^4} = 10^{-4} \,m$.

The elongation of the copper wire of cross-sectional area $3.5 \,mm^2$, as shown in the figure, is (Given: $Y_{\text{copper}} = 10 \times 10^{10} \,N/m^2$ and $g = 10 \,m/s^2$).

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