Question

The diameter of brass road is $4mm$. Young's modulus of brass is $9\times 10^{10}N/m^2$. The force required to stretch $0.1\%$ of its length is

• A. $360\pi N$
• B. $36N$
• C. $36\pi \times 10^{5}N$
• D. $144\pi \times 10^{3}N$

Answer 1

Answer: (A)

The ratio of stress to strain is known as Young's modulus of wire
$Y=\frac{F/A}{l/L}$
$\Rightarrow F=\frac{YAl}{L}$
Given, $Y=9\times 10^{10}N/m^2,$
$A=\pi r^2=\pi {\left(2\times 10^{-3}\right)}^2,$
$l=0.1\%L$
$\therefore F=\frac{9\times 10^{10}\times \pi \times {\left(2\times 10^{-3}\right)}^2\times 0.1}{100}$
$\Rightarrow F=360\pi N$