MHT CET · Physics · Thermal Properties of Matter
A metal rod of Young's modulus 'Y' and coefficient of linear expansion ' \(\propto^{\prime}\) has its
temperature raised by ' \(\Delta \theta^{\prime} .\) The linear stress to prevent the expansion of rod is
(L and \(\ell\) is original length of rod and expansion respectively)
- A \(\mathrm{Y} \frac{\mathrm{L}}{\ell}\)
- B \(\frac{Y \propto}{\Delta \theta}\)
- C \(\mathrm{Y} \propto \Delta \theta\)
- D \(\mathrm{Y}\left(\frac{\ell}{\mathrm{L}}\right)^{2}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{Y} \propto \Delta \theta\)
Step-by-step Solution
Detailed explanation
For thermal expansion
\(\Delta \mathrm{L}=\mathrm{L} \Delta \theta\)
If rod is compress by \(\Delta \mathrm{L}\) then
\(\begin{aligned}
\Delta \mathrm{L}=& \frac{\mathrm{FL}}{\mathrm{AY}} \\
\therefore \frac{\mathrm{FL}}{\mathrm{AY}} &=\mathrm{L} \propto \Delta \theta \quad \therefore \frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y} \propto \Delta \theta
\end{aligned}\)
\(\Delta \mathrm{L}=\mathrm{L} \Delta \theta\)
If rod is compress by \(\Delta \mathrm{L}\) then
\(\begin{aligned}
\Delta \mathrm{L}=& \frac{\mathrm{FL}}{\mathrm{AY}} \\
\therefore \frac{\mathrm{FL}}{\mathrm{AY}} &=\mathrm{L} \propto \Delta \theta \quad \therefore \frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y} \propto \Delta \theta
\end{aligned}\)
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