Question

Young's Modulus of Aluminum is $7\times {10}^{10}Pa$. The force needed to stretch an aluminum wire with $2mm$ diameter and $800mm$ length by $1mm$ is

Answer 1

Given,

$Y=7\times {10}^{10}Pa$

$D=2mm$

Area, $A=\pi \frac{D^2}{4}=\pi \frac{2mm\times 2mm}{4}$

$A=3.14\times {10}^{-6}{m}^2$

$L=800mm$

$\Delta L=1mm$

Young's modulus of aluminium,

$Y=\frac{F/A}{\Delta L/L}$

$\frac{F}{A}=Y\frac{1mm}{800mm}=Y\frac{1}{800}$

$F=Y\frac{A}{800}=Y\frac{3.14\times {10}^{-6}}{800}$

$F=7\times {10}^{10}\times 3.925\times {10}^{-9}N$

$F=274.75N$