Question

A $5\text{m}$ long cylindrical steel wire with radius $2\times {10}^{-3}\text{m}$ is suspended vertically from a rigid support and carries a bob of mass $100\text{kg}$ at the other end. If the bob gets snapped,The change in temperature of the wire is ${(x\times {10}^{-3})}^{\circ }C$ (ignoring radiation losses). Find the value of $x$.

(For the steel wire: Young's modulus $=2.1\times {10}^{11}{\text{N/m}}^2$); Density $=7860{\text{kg/m}}^3$; Specific heat $=420{\text{J/kg-}}^{\circ }\text{C}$

Answer 1

Given that,
Length of the wire, $l=5\text{m}$
Radius of the wire, $r=2\times {10}^{-3}\text{m}$
Density of wire, $\rho =7860{\text{kg/m}}^3$
Mass of bob, $M=100\text{kg}$



Young's modulus, $Y=2.1\times {10}^{11}{\text{N/m}}^2$
Specific heat, $s=420\text{J/kg-K}$
Mass of wire, $m=\left(\text{density}\right)\left(\text{volume}\right)$
$\Rightarrow m=\left(\rho \right)\left(\pi r^{2}l\right)$
$\Rightarrow m=\left(7860\right)\times \pi \times (2\times {10}^{-3}{)}^2\times 5$
$\Rightarrow m=0.494\text{kg}$

Let the change in length of wire be $\Delta l$ due to mass of bob.
By using young's modulus
$Y=\frac{\frac{Mg}{\pi r^2}}{\frac{\Delta l}{l}}$
$\Rightarrow \Delta l=\frac{Mgl}{\pi r^{2}Y}$

The elastic potential energy stored in the wire,
$U=\frac{1}{2}\left(\text{stress}\right)\times \left(\text{strain}\right)\times \left(\text{volume}\right)$
$\Rightarrow U=\frac{1}{2}\left(\frac{Mg}{\pi r^2}\right)\times \left(\frac{\Delta l}{l}\right)\times \left(\pi r^{2}l\right)$
$\Rightarrow U=\frac{1}{2}\left(Mg\right)\times \Delta l$
$\Rightarrow U=\frac{1}{2}\left(Mg\right)\times \frac{\left(Mgl\right)}{\left(\pi r^2\right)Y}$
$\Rightarrow U=\frac{1}{2}\frac{M^2{g}^{2}l}{\pi r^{2}Y}$

Substituting the values, we have
$\Rightarrow U=\frac{1}{2}\times \frac{\left(100{)}^2\right(10{)}^2\left(5\right)}{\left(3.14\right)(2\times {10}^{-3}{)}^2(2.1\times {10}^{11})}$
$\therefore U=0.9478\text{J}$

When the bob gets snapped, this energy is utilized in raising the temperature of the wire by $\Delta \theta$.
So, $U=ms\Delta \theta$
$\therefore \Delta \theta =\frac{U}{ms}={\frac{0.9478}{0.494\times \left(420\right)}}^{\circ }\text{C}$
$\Delta \theta =4.568\times {{10}^{-3}}^{\circ }\text{C}$

Accepted answers $4.568,4.57,4.6$