Discussion Forum

Steel wire of length L at $40^{0}$ C is suspended from the ceiling and then a mass m is hung from its free end. The wire is cooled down from $40^{0}$C to $30^{0}$C to regain its original length L. The coefficient of linear thermal expansion of the steel is$10^{-5}/^{0}C$,  Young's modulus of steel is $10^{11}$ N/m² and the radius of the wire is 1 mm.
Assume that L >> the diameter of the wire. Then, the value of m in kg is nearly

Answer:

Explanation:

$\triangle l_{1}=\frac{FL}{AY}=\frac{mgL}{\pi r^{2}Y}$ =increase in length

 $\triangle l_{2}=L \alpha \triangle \theta$ = Decrease in length 

 To regain is original length

 $\triangle l_{1}= \triangle l_{2}$

 $\therefore$  $\frac{mgL}{\pi r^{2} Y}=L \propto \triangle \theta$

 $\therefore$    $m=\left(\frac{r^{2} Y \alpha \triangle \theta}{g}\right)$

Substituting  the values , we get

 m= 3kg

 $\therefore$ Answer is 3.