Question

The diameter of a brass rod is $4mm$ 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 $k\pi$ newton. Find $k$

Answer 1

Formula used: $Y=\frac{FL}{Al}$
$\text{given,}d=4mm$
$Y=9\times {10}^{10}N/m^2$
$\text{As we know,}\frac{F}{A}=Y\frac{\Delta l}{l}$

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

$=\pi {(2\times {10}^{-3})}^2\times 9\times {10}^{10}$

$\times \frac{0.1}{100}(\because r=\frac{d}{2})$
$=\pi \times 4\times {10}^{-6}\times 9\times {10}^7$

$=360\pi N$

Final Answer: (360)