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Previous Years JEE Advanced Questions
Q. Steel wire of length ‘L’ at $40^{\circ}$ C is suspended from the ceiling and then a mass ‘m’ is hung from its free end. The wire is cooled down from $40^{\circ}$ to $30^{\circ}$ 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 1011 N/m2 and radius of the wire is 1 mm. Assume that L>> diameter of the wire. Then the value of ‘m’ in kg is nearly -
[JEE 2011]
Ans. $\Delta L_{1}=\frac{F L}{A Y}=\frac{m g L}{\pi r^{2} Y}=$Increase in length
$\Delta L_{2}=L \alpha \Delta T=$ Decrease in length
$\Delta L_{1}=\Delta L_{2}$
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