Question

A uniform steel wire of length 4 m and area of cross-section $3\times {10}^{-6}{m}^2$ is extended by 1 mm by the application of a force. If Young's modulus of steel is $2\times {10}^{11}N{m}^{-2}$, the energy stored in the wire is

• A. 0.025 J
• B. 0.050 J
• C. 0.075 J
• D. 0.100 J

Answer 1

The correct option is C

0.075 J

Strain, $ϵ=\frac{l}{L}=\frac{1mm}{4m}=\frac{1\times {10}^{-3}m}{4m}=2.5\times {10}^{-4}$
Now, energy stored per unit volume$=\frac{1}{2}Y\times {ϵ}^2=\frac{1}{2}\times 2\times {10}^{11}\times (2.5\times {10}^{-4}{)}^2=6.25\times {10}^{3}J{m}^{-3}$
Volume of the wire, $V=AL=3\times {10}^{-6}\times 4=1.2\times {10}^{-5}{m}^3$
$\therefore$ Energy stored in the wire $=6.25\times {10}^3\times 1.2\times {10}^{-5}=7.5\times {10}^{-2}J=0.075J$
Hence, the correct choice is (c).