A steel rod has a radius of 20mm and a length of 2 . 0m . A force...

Young's modulus is given as

$Y =\dfrac{Stress}{Strain}= \dfrac{\dfrac{F}{A}}{\dfrac{∆ L}{L}}$

The given data is~

Stress$= 62 . 8kN$

$r = 20 \times {10}^{- 3}m$

$Y = 2 \times {10}^{11}Nm^{- 2}$

Thus,

Strain = \dfrac{F}{A \times Y} ⇒ \dfrac{∆ L}{L} = \dfrac{62 . 8 \times {10}^{3}N}{\pi \times \left(20 \times {10}^{- 3}m\right)2 \times \left(2 \times {10}^{11}Nm^{- 2}\right)} = 25 \times {10}^{- 5}N