Question

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

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

Answer 1

Answer: (A)

360 πN

Answer 2

$r = 2 \times 10^{- 3}m,$ $Y = 9 \times 10^{10}N/m^{2},$ $l = 0.1\% L$

⇒ $\frac{l}{L} = 0.001$

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