Question

A $5m$ long cylindrical steel wire with radius $2\times {10}^{-3}$ m is suspended vertically from a rigid support and carrics a bob of mass 100 kg at the other end. If the bob gets snapped, calculate the change in temperature of the wire ignoring radiation losses. $(Takeg=10m/s^2)$(For the steel wire: Young's $modulus=2.1\times {10}^{11}N/m^2$; Density $=7860kg/m^3$ Specific heat= $420J/k{g}^{}C)$

Answer 1

Given : $L=5m$ $r=2\times {10}^{-3}m$ $S=420J/k{g}^{o}C$

Mass of the bob $m=100kg$

Elastic potential energy stored in the wire $U=\frac{1}{2}\times \frac{F}{A}\times \frac{\Delta L}{L}\times AL$ where $\Delta L=\frac{FL}{AY}$

$⟹$ $U=\frac{F^{2}L}{2AY}$

Mass of the wire $M=\rho \left(\pi r^{2}L\right)$

$\therefore$ $MS\Delta T=\frac{F^{2}L}{2AY}$ where $F=mg$

OR $\rho \left(\pi r^{2}L\right)S\Delta T=\frac{m^2{g}^{2}L}{2\left(\pi r^2\right)Y}$ $⟹$ $\Delta T=\frac{m^2{g}^2}{2{\pi }^2{r}^4\rho YS}$

$\therefore$ $\Delta T=\frac{(100{)}^2\times (10{)}^2}{2{\pi }^2(2\times {10}^{-3}{)}^4\times 7860\times (2.1\times {10}^{11}\left)\right(420)}=4.54\times {10}^{-3}$ ${}^{o}C$