The ratio of stress to tensile strain is called Young's modulus. 

 

              Consider a long bar of length L0 and cross-sectional area A. Let an external force F equal to the weight w stretches it such that the stretched length becomes L. According to Hooke's law, the ratio of this stress to tensile strain is constant within the elastic limit of the body.

 

Young’s modulus        Y      =  $\frac{Stress}{(Tensile strain)}$     

Let $\triangle L$ be the change in length of the rod, then

 

$\triangle L$   =      L – L0  

Since                           Stress      = \frac{Force}{Area}     =    \frac{F}{A}

And                             Tensile strain  =    $(L - L_{o})/( L_{o} )$   =   $\frac{(\triangle L )}{L_{o}}$ 

As                               Y                  = $\frac{\triangle L}{(Tensile strain)}$ 

 

                                 = $\frac{F}{A} \times \frac{(L_{o})}{\triangle L}$

                          Y = $\frac{(〖FL〗_{o})}{A\triangleL}$  

 

Unit:  SI unit of Young's modulus is newton per square metre $(Nm ^{-2})$.