Question

A work of $ 2\times {{10}^{-2}}J $ is done on a wire of length $ 50 $ cm and area of cross-section $ 0.5\text{ }m{{m}^{2}} $ . If the Youngs modulus of the material of the wire is $ 2\times {{10}^{10}}N{{m}^{-2}}, $ then the wire must be:

• A. elongated to 50.1414 cm
• B. contracted by 2.0 mm
• C. stretched by 0.707 mm
• D. of length changed to 49.293 cm
• E. of length changed to 50.2 cm

Answer 1

Answer: (C)

The work done by wire is stored as potential energy in the wire
$ U=\frac{1}{2}\,\times $ youngs modulus $ \times \,{{(strain)}^{2}} $ Given, $ y=2\times {{10}^{10}}N{{m}^{-2}},
$ Strain, $ \frac{l}{L}=\frac{l}{50\times {{10}^{-2}}}, $
$ U=2\times {{10}^{-2}}J $
$ \therefore $ $ 2\times {{10}^{-2}} $
= $ \frac{1}{2}\times 2\times {{10}^{10}}\times {{\left( \frac{1}{50\times {{10}^{-2}}} \right)}^{2}} $
$ \Rightarrow $ $ l\approx 0.707\,mm $ (stretched)