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 modules of steel is $10^{11} N / m^{2}$ and radius of the wire is $1 \,mm$, Assume that $L > >$ diameter of the wire Then the valure of $m$ in $kg$ is nearly

Answer 1

$\Delta l_{1}=\frac{F L}{A Y}=\frac{m g L}{\pi r^{2} Y}=$ In crease in length
$\Delta l_{2}=L \alpha \Delta \theta=$ Decrease in length.
To regain its original length, $\Delta l_{1}=\Delta l_{2} $
$\therefore \frac{m g L}{\pi r^{2} Y}=L \alpha \Delta \theta$
$ \Rightarrow m=\left(\frac{r^{2} Y \alpha \Delta \theta}{g}\right)$
Substituting the values we get, $m =3 \,kg$