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 $1mm$, Assume that $L>>$ diameter of the wire Then the valure of $m$ in $kg$ is nearly

Answer 1

Answer: 3

$\Delta l_1=\frac{FL}{AY}=\frac{mgL}{\pi r^{2}Y}=$ In crease in length
$\Delta l_2=L\alpha \Delta \theta =$ Decrease in length.
To regain its original length, $\Delta l_1=\Delta l_2$
$\therefore \frac{mgL}{\pi r^{2}Y}=L\alpha \Delta \theta$
$\Rightarrow m=\left(\frac{r^{2}Y\alpha \Delta \theta }{g}\right)$
Substituting the values we get, $m=3kg$