Question

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

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

Answer 1

Answer: (A)

$360\pi N$

Let $F$ be force

$Y$ be Young's modulus

$\Delta$ be change in length

$l$ be length

$F=\frac{YA\Delta l}{l}$

Given:

$\frac{\Delta l}{l}=0.001$

$Y=9\times {10}^{10}N/m^2$

$r=\frac{d}{2}=2mm$

$A=\pi r^2=\pi \times (2\times {10}^{-3}{)}^2{m}^2$

$F=\frac{YA\Delta l}{l}=9\times {10}^{10}\times \pi \times 4\times (10{)}^{-6}\times 0.001=36\pi$

Hence the force is $360\pi N$