Question

What is the percentage increase in length of a wire of diameter 2.5 mm, stretched by a force of 100 kg wt? Young's modulus of elasticity of wire = 12.5 x ${10}^{11}dyne/c{m}^2$

• A. 0.16%
• B. 032%
• C. 0.18%
• D. 0.12%

Answer 1

Answer: (A)

0.16%

$\begin{array}{c}\text{Diameter}d=2.5mm=2.5\times {10}^{-3}m\\ \text{Force},f=100kgwt=100\times 9.8N=980N\\ Y=12.5\times {10}^{11}\text{dyne}/{cm}^2=12.5\times {10}^{10}N/m^2\end{array}$

$\begin{array}{cc}& \frac{\Delta L}{L}=\frac{F}{AY}=\frac{980}{12.5\times {10}^{10}\times \pi {\left(\frac{2.5\times {10}^{-3}}{2}\right)}^2}=0.0016\\ & \Rightarrow \frac{\Delta l}{l}\times 100=0.16\%\to \text{percentage increase in length.}\\ & \text{Ans : A}\end{array}$