Question:
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 $kg$ in nearly.
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 $kg$ in nearly.
Updated On: Aug 15, 2022
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The Correct Option is C
Solution and Explanation
Youngs modulus
We know that
$E=\frac{F / A}{\Delta L / L}$
$ \Rightarrow \frac{\Delta L}{L}=\frac{F}{A E} \ldots $ (i)
Also
$\frac{\Delta L }{ L }=\alpha \Delta T\ldots$ (ii)
$\frac{F}{A E}=\alpha \Delta T$
$\Rightarrow m g=(\alpha \Delta T) A E$
$\Rightarrow m=\frac{\alpha \Delta T}{g} A E$
$=\frac{10^{-5} \times 10 \times \pi \times 10^{-6} \times 10^{11}}{10}$
$=\pi \approx 3$
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