Question

A stress of $3.18 \times 10^{8} Nm ^{-2}$ is applied to a steel rod of length $1\, m$ along its length. Its Young's modulus is $2 \times 10^{11} N-m^{-2}$. Then, the elongation produced in the rod (in $mm$ ) is

• A. 3.18
• B. 6.36
• C. 5.18
• D. 1.59

Answer 1

Answer: (D)

1.59

Answer 2

$Y=\frac{F / A}{A \Delta l}$ Given, $\frac{ F }{ A }=$ stress $=3 \cdot 18 \times 10^{8} Nm ^{-2}$ $l=1\, m,\, Y =2 \times 10^{11} N m ^{-2}$ $\Delta l=\frac{l\left(\frac{F}{A}\right)}{Y}=\frac{1 \times 3.18 \times 10^{8}}{2 \times 10^{11}}$ $=1.59 \times 10^{-3} m =1.59\, mm$