JEE Advanced · Physics · 12. Thermal Properties
Steel wire of length \(L\) at \(40^{\circ} \mathrm{C}\) is suspended from the ceiling and then a mass \(m\) is hung from its free end. The wire is cooled down from \(40^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\) to regain its original length \(L\). The coefficient of linear thermal expansion of the steel is \(10^{-5} /{ }^{\circ} \mathrm{C}\), Young's modulus of steel is \(10^{11} \mathrm{~N} / \mathrm{m}^2\) and radius of the wire is \(1 \mathrm{~mm}\). Assume that \(L>\) diameter of the wire. Then, the value of \(m\) in \(\mathrm{kg}\) is nearly
- A 1
- B 2
- C 4
- D 3
Answer & Solution
Correct Answer
(D) 3
Step-by-step Solution
Detailed explanation
\(\Delta l_1=\frac{F L}{A Y}=\frac{m g L}{\pi r^2 Y}=\) Increase in length \(\Delta l_2=L \alpha \Delta \theta=\) Decrease in length To regain its original length,
\(\Delta l_1 =\Delta l_2 \)
\( \therefore \frac{m g L}{\pi r^2 Y} =L \alpha \Delta \theta \)
\( \therefore m =\left(\frac{r^2 Y \alpha \Delta \theta}{g}\right)\)
Substituting the values, we get
\(
m \simeq 3 \mathrm{~kg}
\)
\(\therefore\) Answer is 3 .
Analysis of Question
Question is very simple. In my opinion any student who have gone through the syllabus once can solve this problem easily.
\(\Delta l_1 =\Delta l_2 \)
\( \therefore \frac{m g L}{\pi r^2 Y} =L \alpha \Delta \theta \)
\( \therefore m =\left(\frac{r^2 Y \alpha \Delta \theta}{g}\right)\)
Substituting the values, we get
\(
m \simeq 3 \mathrm{~kg}
\)
\(\therefore\) Answer is 3 .
Analysis of Question
Question is very simple. In my opinion any student who have gone through the syllabus once can solve this problem easily.
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