Question

The area of cross-section of a wire of length $1.1$ meter is $1m{m}^2$. It is loaded with 1 kg. If Young's modulus of copper is $1.1\times {10}^{11}N/m^2$, then the increase in length will be (If $g=10m/s^2$) :

• A. $0.01mm$
• B. $0.075mm$
• C. $0.1mm$
• D. $0.15mm$

Answer 1

Answer: (C)

$0.1mm$

$y=\frac{F/A}{\frac{\Delta l}{l}}$

$\Delta l=\frac{Fl}{Ay}$

$\Delta l=\frac{Mgl}{Ay}$

$\Delta l=\frac{1\times 10\times 1.1}{1\times {10}^{-6}\times 1.1\times {10}^{11}}=1\times {10}^{-4}=0.1mm$