Question

Calculate the percentage increase in length of a wire of diameter $2mm$ stretched by force of $1kg$. Youngs modulus of the material of wire $15\times {10}^{10}N{m}^{-1}$.

Answer 1

Given that,

Young’s modulus of the material $Y=15\times {10}^{10}N/m$

Diameter $d=2mm$

Radius $r=0.02m$

Force $F=1kg$

Now,

$Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}$

$\frac{\Delta l}{l}=\frac{F}{A\times Y}$

$\frac{\Delta l}{l}\times 100=\left[\frac{9.8}{3.14\times {\left(0.002\right)}^2\times 15\times {10}^{10}}\right]\times 100$

$\frac{\Delta l}{l}\times 100=\left[\frac{9.8}{3.14\times 4\times {10}^{-6}\times 15\times {10}^{10}}\right]\times 100$

$\frac{\Delta l}{l}\times 100=[0.052017\times {10}^{-4}]\times 100$

$\frac{\Delta l}{l}\times 100=0.00052$%

Hence, the percentage increase in length of a wire is $0.00052$ %