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 double its length.
(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 is doubled, then strain $=\frac{\Delta l}{l}=1$

$\therefore Y=\frac{Stress}{strain}=\frac{F}{A}$

$\therefore \text{Force}=Y\times A=2\times {10}^{11}\times 0.1\times {10}^{-4}=2\times {10}^6\text{N}$