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 $40^{\circ} C$ to $30^{\circ} C$ to regain its original length ' $L$ '. The coefficient of linear thermal expansion of the steel is $10^{-5} /{ }^{\circ} C$, Young's modulus of steel is $10^{11} N / m ^2$ and radius of the wire is $1 \ mm$. Assume that $L \gg$ diameter of the wire. Then the value of ' $m$ ' in $kg$ is nearly
IIT 2011, Advanced
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We know that
$E =\frac{\frac{ F }{ A }}{\frac{\Delta L }{ L }} \Rightarrow \frac{\Delta L }{ L }=\frac{ F }{ AE } \ldots . . \text {. (i) }$
$\text { Also }$
$\frac{\Delta L }{ L } \propto \alpha \Delta T . \ldots . \text { (ii) }$
$\text { from (i) and (ii) }$
$\frac{ F }{ AE }=\alpha \Delta T$
$\Rightarrow mg =(a \Delta T ) AE$
$\Rightarrow m =g a \Delta TAE$
$=\frac{10^{-5} \times 10 \times \pi \times 10^{-6} \times 10^{11}}{10}$
$=\pi \approx 3$
So correct answer is a
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