Question

The Youngs modulus of the material of a wire is $2\times {10}^{10}N{m}^{-2}$ .If the elongation strain is 1%, then the energy stored in the wire per unit: volume in $J{m}^{-3}$ is

• A. ${10}^6$
• B. ${10}^8$
• C. $2\times {10}^6$
• D. $2\times {10}^8$
• E. $0.5\times {10}^6$

Answer 1

Answer: (A)

Per unit volume energy stored
$=\frac{1}{2}\times Y\times {(strain)}^2$
$=\frac{1}{2}\times Y\times {\left(\frac{l}{L}\right)}^2$
Given $l=L\times 1$
Or $l=\frac{L}{100}$
Stored energy $Y=\frac{1}{2}\times 2\times {10}^{10}\times {\left(\frac{L}{100L}\right)}^2$
$={10}^{6}J{m}^{-3}$