Question

The area of cross section of a steel wire of Young's modulus $Y=2.0\times {10}^{11}{\text{N/m}}^2$ is $0.1{\text{cm}}^2$. Find the force required to increase its length by $1$%.
(Assume area of cross section remains same)

• A. $2\times {10}^{12}\text{N}$
• B. $2\times {10}^{11}\text{N}$
• C. $2\times {10}^{10}\text{N}$
• D. $2\times {10}^4\text{N}$

Answer 1

The correct option is D $2\times {10}^4\text{N}$

Given,
Young's modulus $\left(Y\right)=2.0\times {10}^{11}{\text{N/m}}^2$
Area of cross-section$\left(A\right)=0.1{\text{cm}}^2$
Let $l$ be the length of steel wire
When the length of wire increases by 1%, then strain $=\frac{\Delta l}{l}=\frac{1}{100}$
Stress = Y $\times$ strain
$\frac{F}{A}=Y\frac{\Delta l}{l}$
$F=AY\frac{\Delta l}{l}=0.1\times {10}^{-4}\times 2\times {10}^{11}\times \left(\frac{1}{100}\right)$
$F=2\times {10}^4\text{N}$