Question

A block of mass $500\, kg$ is suspended by wire of length $70\, cm$. The area of cross-section of wire is $10 mm^{2}$. When the load is removed, the wire contracts by $0.5\, cm$. The Young’s modulus of the material of wire will be

• A. $10 \times 10^{14} N / m ^{2}$
• B. $4 \times 10^{14} N / m ^{2}$
• C. $8 \times 10^{11} N / m ^{2}$
• D. $7 \times 10^{10} N / m ^{2}$

Answer 1

Answer: (D)

$Y=\frac{F \times L}{A \times \Delta L}$
$=\frac{500 \times 10 \times 0.7}{10 \times 10^{-6} \times 0.5 \times 10^{-2}}$
$=\frac{3500}{5 \times 10^{-8}}=7 \times 10^{10} N / m ^{2}$