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(A) The thermal stress $F/A$ developed in a wire due to a temperature change $\Delta T$ is given by $F/A = Y \alpha \Delta T$.The speed of a transvers

Vedclass · Accepted Answer

$64.6$

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(A) The thermal stress $F/A$ developed in a wire due to a temperature change $\Delta T$ is given by $F/A = Y \alpha \Delta T$.The speed of a transverse wave on a stretched wire is $v = \sqrt{T/\mu}$,where $T$ is the tension and $\mu$ is the linear mass density.Here,the tension $T = F = Y A \alpha \Delta T$ and the linear mass density $\mu = ho A$.Substituting these into the velocity formula: $v = \sqrt{\frac{Y A \alpha \Delta T}{ho A}} = \sqrt{\frac{Y \alpha \Delta T}{ho}}$.Given: $Y = 1.2 	imes 10^{11} \, N/m^2$,$\alpha = 1.6 	imes 10^{-5} /^{\circ} C$,$ho = 9.2 	imes 10^3 \, kg/m^3$,and $\Delta T = 50^{\circ} C - 30^{\circ} C = 20^{\circ} C$.Calculating the speed: $v = \sqrt{\frac{1.2 	imes 10^{11} 	imes 1.6 	imes 10^{-5} 	imes 20}{9.2 	imes 10^3}}$.$v = \sqrt{\frac{3.84 	imes 10^7}{9.2 	imes 10^3}} = \sqrt{4173.9} \approx 64.6 \, m/s$.

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