Question

A tensile force of $1000N$ applied to a $10m$ metal beam having a cross sectional area $4\times {10}^{-4}\left(m^2\right)$. Then the corresponding strain in $\left(\mu m/m\right)$ is $[$Elasticity of given metal is $10\times {10}^{10}\left(N/m^2\right)]$

• A. 25

Answer 1

The correct option is A 25
$\text{modulus elasticity}\left(E\right)=\frac{F/A}{\Delta l/l}$

$Strain=\frac{\Delta l}{l}=\frac{F}{AE}$

$=\frac{1000}{4\times {10}^{-8}\times 10\times {10}^{10}}$

$=25\left(\frac{\mu m}{m}\right)$