Question

The area of cross-section of a wire of length 1.1 m is $ 1m{{m}^{2}} $ . It is leaded with 1 kg of Youngs 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. 0.01 mm
• B. 0.075 mm
• C. 0.1 mm
• D. 0.15mm

Answer 1

Answer: (C)

Increase in length $ l=\frac{mgL}{AY} $ $ =\frac{1\times 10\times 1.1}{1.1\times {{10}^{11}}\times {{10}^{-6}}}m=0.1\,mm $ ?(i)