Question

A force of $100$ N produces a change of $0.1$% in the length of a wire of area of cross section $1.0$ $m{m}^2$. Young's modulus of the material of the wire is :

• A. ${10}^5$ $N/m^2$
• B. ${10}^4$ $N/m^2$
• C. ${10}^{11}$ $N/m^2$
• D. ${10}^{12}$ $N/m^2$

Answer 1

The correct option is C ${10}^{11}$ $N/m^2$

$\begin{array}{c}\text{Given;}\\ \text{Force}\left(F\right)=100N\\ \text{Change in length}(\frac{\Delta L}{L}\times 100)=0.1\%\\ \text{Cross section area}\left(A\right)=1.0{mm}^2\end{array}$

$\begin{array}{c}y=\frac{\text{stress}}{\text{we know}}\\ y=(\frac{F}{A)\left(\frac{\Delta L}{L}\right)}\\ y=\frac{FL}{A\Delta L}\end{array}$

$\begin{array}{c}\text{Putting know value from above data}\\ y=\frac{100x}{\left(\frac{1}{{10}^6}\right)\left({10}^{-3}\right)}\end{array}$

$y={10}^{11}N/m^2$