Question

The diameter of a brass rod is 4 mm and Young’s modulus of brass is $9\times {10}^{10}\frac{N}{m^2}$. The force required to stretch by 0.1% of its length is

• A.
• B.
• C. 144\pi\times10^{3}N
• D. 36\pi\times10^{5}N

Answer 1

The correct option is A

$r=2\times {10}^{-3}m,Y=9\times {10}^{10}\frac{N}{m^2},l=0.1\%L\Rightarrow \frac{l}{L}=0.001$
As $Y=\frac{F}{A}\frac{L}{l}$ $\therefore F=YA\frac{l}{L}=9\times {10}^{10}\times \pi {(2\times {10}^{-3})}^2\times 0.001=360\pi N$