A wire of length 1 , m and Young's modulus Y | vedclass

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(A) The thermal strain produced in the wire is given by $	ext{strain} = \frac{\Delta l}{L} = \alpha \Delta 	heta$.The energy density (energy per uni

Vedclass · Accepted Answer

$32.4 \, J$

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(A) The thermal strain produced in the wire is given by $	ext{strain} = \frac{\Delta l}{L} = \alpha \Delta 	heta$.The energy density (energy per unit volume) stored in the wire due to thermal stress is $u = \frac{1}{2} 	imes Y 	imes (	ext{strain})^2$.Total energy stored $U = u 	imes 	ext{Volume} = \frac{1}{2} 	imes Y 	imes (\alpha \Delta 	heta)^2 	imes (A 	imes L)$.Given values: $Y = 2.0 	imes 10^{11} \, N/m^2$,$\alpha = 18 	imes 10^{-6} \, ^\circ C^{-1}$,$\Delta 	heta = 100 ^\circ C$,$A = 1 \, cm^2 = 10^{-4} \, m^2$,$L = 1 \, m$.Substituting the values:$U = \frac{1}{2} 	imes (2.0 	imes 10^{11}) 	imes (18 	imes 10^{-6} 	imes 100)^2 	imes (10^{-4} 	imes 1)$$U = 10^{11} 	imes (18 	imes 10^{-4})^2 	imes 10^{-4}$$U = 10^{11} 	imes 324 	imes 10^{-8} 	imes 10^{-4}$$U = 324 	imes 10^{-1} = 32.4 \, J$.

$A$ wire of length $1 \, m$ and Young's modulus $Y = 2.0 \times 10^{11} \, N/m^2$ is fixed between two rigid supports. If its temperature is increased by $100 ^\circ C$,what is the energy stored in the wire? (Given: $\alpha = 18 \times 10^{-6} \, ^\circ C^{-1}$,$A = 1 \, cm^2$)

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