Question

The area of cross-section of a wire of length 1.1 m is 1 $mm^2$. It is leaded with 1 kg of Young's modulus of copper is $1.1\times10^i Nm^\circ$ , then the increase in length will be $(ifg = 10{ms^-2})$

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

Answer 1

Answer: (C)

0.1 mm

Answer 2

Increase in length $l=\frac{mg^2}{AY}$ $=\frac{1\times10\times1.1}{1.1\times10^{11}\times10^{-6}}m=0.1 mm$