A metal rod of length $\mathrm{L}$ and cross-sectional area $\mathrm{A}$ is heated through…

A metal rod of length $\mathrm{L}$ and cross-sectional area $\mathrm{A}$ is heated through $\mathrm{T}^{\circ} \mathrm{C}$. What is the force required to prevent the expansion of the rod lengthwise? (Y= Young's modulus of material of the rod, $\alpha=$ coefficient of linear expansion of the rod.)
  1. $\mathrm{YA} \alpha / \mathrm{T}(1+\alpha \mathrm{T})$
  2. $\mathrm{YA} \alpha \mathrm{T} /(1-\alpha \mathrm{T})$
  3. $\mathrm{YA} \alpha \mathrm{T} /(1+\alpha \mathrm{T})$
  4. $\mathrm{YA} \alpha /(1-\alpha \mathrm{T})$

Solution

The increase in length due to heating is given by $\mathrm{L}=\mathrm{L}_{0}(1+\alpha \mathrm{T}) \quad$ or $\quad \mathrm{L}-\mathrm{L}_{0}=\Delta \mathrm{L}=\mathrm{L}_{0} \alpha \mathrm{T}$ To compress the rod by $\Delta \mathrm{L}$, the force required is given by $\mathrm{F}=\frac{\mathrm{YA} \Delta \mathrm{L}}{\mathrm{L}}$ Substituting the value of $\mathrm{L}$ and $\Delta \mathrm{L}$ we get $\mathrm{F}=\frac{\mathrm{YA} \times \mathrm{L}_{0} \alpha \mathrm{T}}{\mathrm{L}_{\mathrm{o}}(1+\alpha \mathrm{T})}=\frac{\mathrm{YA} \alpha \mathrm{T}}{1+\alpha \mathrm{T}}$

Asked in: MHT CET 2020 (14 Oct Shift 2)

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