Question

The diameter of brass road is $4\, mm$. 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. $36\, N$
• 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
$Y =\frac{F / A}{l / L}$
$\Rightarrow F=\frac{Y A l}{L}$
Given, $Y=9 \times 10^{10} N / m ^{2},$
$A=\pi r^{2}=\pi\left(2 \times 10^{-3}\right)^{2},$
$l=0.1 \% L$
$\therefore F=\frac{9 \times 10^{10} \times \pi \times\left(2 \times 10^{-3}\right)^{2} \times 0.1}{100}$
$\Rightarrow F=360\, \pi\, N$