Question

A work of $2\times {10}^{-2}J$ is done on a wire of length $50$ cm and area of cross-section $0.5m{m}^2$ . 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, $y=2\times {10}^{10}N{m}^{-2},<br>$ Strain, $\frac{l}{L}=\frac{l}{50\times {10}^{-2}},$
$U=2\times {10}^{-2}J$
$\therefore$ $2\times {10}^{-2}$
= $\frac{1}{2}\times 2\times {10}^{10}\times {\left(\frac{1}{50\times {10}^{-2}}\right)}^2$
$\Rightarrow$ $l\approx 0.707mm$ (stretched)