Question

A wire is stretched by 0.01 m by a certain force 'F' another wire of same material whose diameter and lengths are double to original wire is stretched b the same force then its elongation will be-

• A. 0.005 m
• B. 0.01m
• C. 0.02 m
• D. 0.04 m

Answer 1

The correct option is A 0.005 m

$\begin{array}{c}\text{Given, Force applied = F}\\ \text{Let, length}=L\\ \text{Area of cross section}=A=\pi r^2\\ \text{Young's modulus}=Y\end{array}$

$\begin{array}{c}\text{Case1) Given elongation=}0.01m\\ \text{elongation formula,}e=\frac{FL}{Ay}\\ \Rightarrow e=\frac{FL}{\pi r^{2}y}=0.01\end{array}$

$\begin{array}{c}\text{Case 2)}\text{Length and diameter are doubled.}\\ \text{That means radius}P\text{s also doubled.}\\ \text{Now elongation}{e}^{\prime }=\frac{F\left(2L\right)}{\pi (2r{)}^{2}y}\end{array}$

$\begin{array}{cc}{e}^{\prime }& =\frac{2FL}{4\pi r^{2}Y}=\frac{e}{2}\\ e^{\prime }& =\frac{0.01}{2}\\ \Rightarrow & e^{\prime }=0.005m\end{array}$