Question

The area of cross-section of a steel wire $\left(Y=2.0 \times 10^{11} N / m ^{2}\right)$ 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)

Young's modulus of elasticity
$Y=\frac{F L}{A l}$
To double the length $l=L$
$\therefore Y =\frac{F}{A} \Rightarrow F=Y A$
$=2 \times 10^{11} \times 0.1 \times 10^{-4}$
$=2 \times 10^{6} N$