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Question: A wire of length \(L_{0}\) is supplied heat to raise its temperature by *T*. If *γ* is the coefficie...

A wire of length L0L_{0} is supplied heat to raise its temperature by T. If γ is the coefficient of volume expansion of the wire and Y is the Young’s modulus of the wire then the energy density stored in the wire is

A

12γ2T2Y\frac{1}{2}\gamma^{2}T^{2}Y

B

13γ2T2Y3\frac{1}{3}\gamma^{2}T^{2}Y^{3}

C

118γ2T2Y\frac{1}{18}\frac{\gamma^{2}T^{2}}{Y}

D

118γ2T2Y\frac{1}{18}\gamma^{2}T^{2}Y

Answer

118γ2T2Y\frac{1}{18}\gamma^{2}T^{2}Y

Explanation

Solution

Due to heating the length of the wire increases. ∴ Longitudinal strain is produced ⇒ ΔLL=α×ΔT\frac{\Delta L}{L} = \alpha \times \Delta T

Elastic potential energy per unit volume E = 12×Stress×Strain\frac{1}{2} \times \text{Stress} \times \text{Strain} = 12×Y×(Strain)2\frac{1}{2} \times Y \times (\text{Strain})^{2}

⇒ E = 12×Y×(ΔLL)2=12×Y×α2×ΔT2\frac{1}{2} \times Y \times \left( \frac{\Delta L}{L} \right)^{2} = \frac{1}{2} \times Y \times \alpha^{2} \times \Delta T^{2}or

E = 12×Y×(γ3)2×T2\frac{1}{2} \times Y \times \left( \frac{\gamma}{3} \right)^{2} \times T^{2}=118γ2YT2\frac{1}{18}\gamma^{2}YT^{2} [As γ=3α\gamma = 3\alpha and ∆T = T (given)]