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}} $