Question

A wire of area of cross-section $3.0m{m}^2$ and natural length $50cm$ is fixed at one end and a mass of $2.1kg$ is hung from the other end. Find the elastic potential energy stored in the wire in steady state. Young’s modulus of the material of the wire $=1.9\times {10}^{11}N{m}^{-2}$. Take $g=10m{s}^{-2}$.

• A. $1.9\times {10}^{-4}J$
• B. $2.9\times {10}^{-4}J$
• C. $3.9\times {10}^{-4}J$
• D. $4.9\times {10}^{-4}J$

Answer 1

The correct option is A $1.9\times {10}^{-4}J$

Area = $3\times {10}^{-6}{m}^{2}l=0.50m\text{Volume}=3\times {10}^{-6}\times 0.50=1.5\times {10}^{-6}{m}^3$


$T=mg=2.1\times 10=21N\text{Stress}=\frac{T}{A}=\frac{21}{3\times {10}^{-6}}=7\times {10}^{6}N{m}^{-2}$
Strain = $\frac{\text{Stress}}{\text{Y}}=\frac{7\times {10}^6}{1.9\times {10}^{11}}=3.7\times {10}^{-5}$
Elastic potential energy, $\text{U}=\frac{1}{2}\times \text{Stress}\times \text{Strain}\times \text{Volume}=\frac{1}{2}\times 7\times {10}^6\times 3.7\times {10}^{-5}\times 1.5\times {10}^{-6}=1.9\times {10}^{-4}J$