Question

On increasing the length by $0.5 \,mm$ in a steel wire of length $4 \,m$ and area of cross section $5 \,mm ^{2},$ the force required is $\left(Y=11.90 \times 10^{12} N / m ^{2}\right)$

• A. $4 \times 10^{3} N$
• B. $2 \times 10^{3} N$
• C. $7.4 \times 10^{3} N$
• D. $10^{3} N$

Answer 1

Answer: (C)

$Y=\frac{F \times L}{A \times \Delta L}$
$\Rightarrow F=\frac{Y A \Delta L}{L}$
$\therefore F=\frac{11.9 \times 10^{12} \times 5 \times 10^{-6} \times 5 \times 10^{-4}}{4}$
$=\frac{297.5 \times 10^{2}}{4}$
$=74.375 \times 10^{2} N$
$F =7.4 \times 10^{+3} N$