Answer:
elongation of the brass rod is 0.01956 mm
Explanation:
given data
length = 5 cm = 50 mm
diameter = 4.50 mm
Young's modulus = 98.0 GPa
load = 610 N
to find out
what will be the elongation of the brass rod in mm
solution
we know here change in length formula that is express as
δ = [tex]\frac{PL}{AE}[/tex] ................1
here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length
so all value in equation 1
δ = [tex]\frac{PL}{AE}[/tex]
δ = [tex]\frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}[/tex]
δ = 0.01956 mm
so elongation of the brass rod is 0.01956 mm
