Question

A steel wire of cross sectional area $0.05{\text{cm}}^2$ is under a tension of $15\text{N}$. Find the decrease in the cross-sectional area. Young's modulus of steel = $2\times {10}^{11}{\text{N/m}}^2$ and Poisson's ratio = $0.3$.

• A. $40\times {10}^{-8}{\text{cm}}^2$
• B. $50\times {10}^{-8}{\text{cm}}^2$
• C. $45\times {10}^{-8}{\text{cm}}^2$
• D. $22.5\times {10}^{-8}{\text{cm}}^2$

Answer 1

The correct option is C $45\times {10}^{-8}{\text{cm}}^2$

Let $d$ be the diameter of steel wire and $L$ be the length of wire.
So, Poisson ratio $\left(\nu \right)=0.3=\frac{\frac{\Delta d}{d}}{\frac{\Delta L}{L}}$
Now, Stress $\left(\sigma \right)=\frac{F}{A}=\frac{15}{0.05\times {10}^{-4}}=300\times {10}^4{\text{N/m}}^2$
Young's modulus $\left(Y\right)=\frac{\sigma }{\frac{\Delta L}{L}}$
$\Rightarrow \frac{\Delta L}{L}=\frac{\sigma }{Y}=\frac{300\times {10}^4}{2\times {10}^{11}}=1.5\times {10}^{-5}$
Now, $\nu =\frac{\frac{\Delta d}{d}}{\frac{\Delta L}{L}}=\frac{\frac{\Delta d}{d}}{1.5\times {10}^{-5}}$
$\Rightarrow 0.3=\frac{\frac{\Delta d}{d}}{1.5\times {10}^{-5}}$
$\Rightarrow \frac{\Delta d}{d}=0.3\times 1.5\times {10}^{-5}=0.45\times {10}^{-5}$
Now, $\frac{\Delta A}{A}=\frac{2\Delta d}{d}=0.9\times {10}^{-5}$
$\Rightarrow \Delta A=0.9\times {10}^{-5}\times A=0.9\times 0.05\times {10}^{-5}=0.045\times {10}^{-5}{\text{m}}^2$
$\Rightarrow \Delta A=45\times {10}^{-8}{\text{cm}}^2$