MHT CET · Physics · Thermal Properties of Matter
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.)
- A \(\mathrm{YA} \alpha / \mathrm{T}(1+\alpha \mathrm{T})\)
- B \(\mathrm{YA} \alpha \mathrm{T} /(1-\alpha \mathrm{T})\)
- C \(\mathrm{YA} \alpha \mathrm{T} /(1+\alpha \mathrm{T})\)
- D \(\mathrm{YA} \alpha /(1-\alpha \mathrm{T})\)
Answer & Solution
Correct Answer
(C) \(\mathrm{YA} \alpha \mathrm{T} /(1+\alpha \mathrm{T})\)
Step-by-step Solution
Detailed explanation
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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