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JEE Advanced 2011Paper 1
Exam
JEE Advanced
Year
2011
Paper
Paper 1
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Advanced 2011 · Paper 1 · Physics · Thermal Properties of Matter

Steel wire of length L at 40^ C is suspended from the ceiling and...

Steel wire of length LL at 40∘C40^{\circ} \mathrm{C} is suspended from the ceiling and then a mass mm is hung from its free end. The wire is cooled down from 40∘C40^{\circ} \mathrm{C} to 30∘C30^{\circ} \mathrm{C} to regain its original length LL. The coefficient of linear thermal expansion of the steel is 10−5/∘C10^{-5} /{ }^{\circ} \mathrm{C}, Young's modulus of steel is 1011 N/m210^{11} \mathrm{~N} / \mathrm{m}^2 and radius of the wire is 1 mm1 \mathrm{~mm}. Assume that L>L> diameter of the wire. Then, the value of mm in kg\mathrm{kg} is nearly

Correct answer

0

Step-by-step solution

Δl1=FLAY=mgLπr2Y=\Delta l_1=\frac{F L}{A Y}=\frac{m g L}{\pi r^2 Y}= Increase in length Δl2=LαΔθ=\Delta l_2=L \alpha \Delta \theta= Decrease in length To regain its original length, Δl1=Δl2∴mgLπr2Y=LαΔθ∴m=(r2YαΔθg)\begin{aligned}& \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) \end{aligned}Substituting the values,wegetm≃3values, 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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