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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.)
Options:
(Y= Young's modulus of material of the rod, $\alpha=$ coefficient of linear expansion of
the rod.)
Solution:
1060 Upvotes
Verified Answer
The correct answer is:
$\mathrm{YA} \alpha \mathrm{T} /(1+\alpha \mathrm{T})$
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
$\mathrm{F}=\frac{\mathrm{YA} \times \mathrm{L}_{0} \alpha \mathrm{T}}{\mathrm{L}_{\mathrm{o}}(1+\alpha \mathrm{T})}=\frac{\mathrm{YA} \alpha \mathrm{T}}{1+\alpha \mathrm{T}}$
$\mathrm{L}=\mathrm{L}_{0}(1+\alpha \mathrm{T}) \quad$ 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
$\mathrm{F}=\frac{\mathrm{YA} \times \mathrm{L}_{0} \alpha \mathrm{T}}{\mathrm{L}_{\mathrm{o}}(1+\alpha \mathrm{T})}=\frac{\mathrm{YA} \alpha \mathrm{T}}{1+\alpha \mathrm{T}}$
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