Question

A wire of natural length $ l, $ Youngs modulus $ Y $ and area of cross-section $ A $ is extended by $ x $ . Then the energy stored in the wire is given by

• A. $ \frac{1}{2}\frac{YA}{l}{{x}^{2}} $
• B. $ \frac{1}{3}\frac{YA}{l}{{x}^{2}} $
• C. $ \frac{1}{2}\frac{Yl}{A}{{x}^{2}} $
• D. $ \frac{1}{2}\frac{YA}{{{l}^{2}}}{{x}^{2}} $
• E. $ \frac{1}{2}\frac{A}{Yl}{{x}^{2}} $

Answer 1

Answer: (A)

Energy stored in the wire
$ U=\frac{1}{2}Y\times {{(strain)}^{2}}\times volume $
Or $ U=\frac{1}{2}Y\times {{\left( \frac{x}{l} \right)}^{2}}\times Al $
Or $ U=\frac{1}{2}\frac{Y{{x}^{2}}}{l}\times A $
Or $ U=\frac{1}{2}\frac{YA}{l}{{x}^{2}} $