Question

The length of a wire of cross-sectional area $1\times$ 10${}^{-6}{m}^2$ is 10m. The young's modulus of the material of the wire is 25 G.pa. When the wire is subjected to a tensile force of $100N$, the elongation produced in $mm$ is:

• A. $0.04$
• B. $0.4$
• C. $4$
• D. $40$

Answer 1

Answer: (D)

$40$

$\Delta L=\frac{100\times 10}{{10}^{-6}\times 25\times {10}^9}$$=0.04m$$=40mm$ (using standard result $Y=\frac{\left(\frac{F}{A}\right)}{\left(\frac{\text{change in length}}{\text{actual length}}\right)}$)