Question

A wire of length $10 \, m$ is subjected to a force of $100 \, N$ along its length and hence the lateral strain produced is $0.01\times 10^{- 3} \, m$ . The Poisson's ratio was found to be $0.4$ . If the area of cross-section of wire is $0.025 \, m^{2}$ , its Young's modulus will be

• A. $1.6\times 10^{8 \, }N m^{- 2}$
• B. $2.5\times 10^{10 \, }N m^{- 2}$
• C. $1.25\times 10^{11}N m^{- 2}$
• D. $16\times 10^{9}N m^{- 2}$

Answer 1

Answer: (A)

Poisson's ratio $=\frac{L a t e r a l \, s t r a i n}{L o n g i t u d i n a l \, s t r a i n}$
$ie, \, \, \, 0.4=\frac{0.01 \times 10^{- 3}}{\frac{l}{L}}$
or $\frac{L}{l}=\frac{0.4}{0.01 \times 10^{- 3}}=4\times 10^{4}$
Young's modulus
$Y=\frac{F L}{A l}$
$=\frac{100}{0.025}\times 4\times 10^{4}=1.6\times 10^{8}N m^{- 2}$