Question

To increase the length by $0.5\, mm$ in a steel wire of length $2\, m$ and area of cross-section $2\, mm ^{2}$, the force required is $\left[ Y _{\text {steel }}=2.2 \times 10^{11}\, N / m ^{2}\right]$

• A. $1.1 \times 10^{5} N$
• B. $1.1 \times 10^{4} N$
• C. $1.1 \times 10^{3} N$
• D. $1.1 \times 10^{2} N$

Answer 1

Answer: (D)

$F=\frac{Y A e}{l}=\frac{2.2 \times 10^{11} \times 2 \times 10^{-6} \times 5 \times 10^{-4}}{2}$
$=11 \times 10^{1}=1.1 \times 10^{2} N$