A 100 m long wire having cross-sectional area $6.25 × 10^{−4} m^2$ and Young's modulus is $10^{10}Nm^{−2}$ is subjected to a load of 250 N, then the elongation in the wire will be

Answer (4)

Given,

Length of wire, $l=100m$

Cross sectional area, $A=6.25\times10^{-4}m^2$

Young's modulus, $Y=10^{10} Nm^{-2}$

Load, $F=250N$

Elongation in the wire, $\Delta l$ = ?

We know that, Stress=Y$\times$ Strain

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

$\displaystyle \begin{aligned}\therefore \Delta l&=\frac{F\ l}{AY}\\&=\frac{250\times100}{6.25\times10^{-4}\times10^{10}}\\&=4\times10^{-3}m \end{aligned}$