A brass wire 1.8; m long at 27;^{circ}C is he | vedclass

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Question

(A) Initial temperature $T_{1} = 27\;^{\circ}C$. Final temperature $T_{2} = -39\;^{\circ}C$. Change in temperature $\Delta T = T_{2} - T_{1} = -39 - 2

Vedclass · Accepted Answer

$3.8 	imes 10^{2}\; N$

Vedclass · Answer

(A) Initial temperature $T_{1} = 27\;^{\circ}C$. Final temperature $T_{2} = -39\;^{\circ}C$. Change in temperature $\Delta T = T_{2} - T_{1} = -39 - 27 = -66\;^{\circ}C$ (or $K$).Length $L = 1.8\; m$. Diameter $d = 2.0 	imes 10^{-3}\; m$. Radius $r = 1.0 	imes 10^{-3}\; m$.Area of cross-section $A = \pi r^{2} = 3.14 	imes (1.0 	imes 10^{-3})^{2} = 3.14 	imes 10^{-6}\; m^{2}$.Thermal strain is given by $\frac{\Delta L}{L} = \alpha \Delta T$.From Hooke's Law,$Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta L/L}$.Therefore,$F = Y A \alpha \Delta T$.Substituting the values:$F = (0.91 	imes 10^{11}) 	imes (3.14 	imes 10^{-6}) 	imes (2.0 	imes 10^{-5}) 	imes (-66)$.$F = 0.91 	imes 3.14 	imes 2.0 	imes 66 	imes 10^{11-6-5} = 5.71228 	imes 66 \approx 377\; N$.The magnitude of the tension developed is approximately $3.8 	imes 10^{2}\; N$.

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