Question

Find the increment in the length of a steel wire of length $5m$ and radius $6mm$ under its own weight.Density of steel $=8000kg/m^3$ and Young's modulus of steel $=2\times {10}^{11}N/m^2$. What is the energy stored in the wire? $(Takeg=9.8m/s^2).$

Answer 1

Given : Radius of the wire $R=6mm=0.006m$ $L=5m$

$\therefore$ Cross-sectional area $A=\pi R^2=3.14(0.006{)}^2=1.13\times {10}^{-4}{m}^2$

Tensile force $F=\left(AL\right)\rho g$ $⟹\frac{F}{A}=L\rho g$

$\therefore$ $\frac{F}{A}=5\times 8000\times 9.8=392000N/m^2$

Increment in the length $\Delta L=\frac{FL}{2AY}=392000\times \frac{5}{2(2\times {10}^{11})}=4.9\times {10}^{-6}m$

Energy stored $E=\frac{YA(\Delta L{)}^2}{2L}=\frac{2\times {10}^{11}\times 1.13\times {10}^{-4}\times (4.9\times {10}^{-6}{)}^2}{2\times 5}=5.43\times {10}^{-5}$ J