Question

The area of cross-section of a steel wire $(Y=2.0\times {10}^{11}N/m^2)$ is $0.1{cm}^2$. The force required to double its length will be :

• A. $2\times {10}^{12}N$
• B. $2\times {10}^{11}N$
• C. $2\times {10}^{10}N$
• D. $2\times {10}^{6}N$

Answer 1

Answer: (D)

$2\times {10}^{6}N$

Young's modulus of elasticity

$Y=\frac{FL}{Al}$

To double the length $l=L$

$\therefore Y=\frac{F}{A}\Rightarrow F=YA$

$=2\times {10}^{11}\times 0.1\times {10}^{-4}=2\times {10}^{6}N$