Question

A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa. If a 610N load( about that exerted by the mass of an average human) is applied to the other end, what will be the elongation of the brass rod in mm?

Answer 1

elongation of the brass rod is 0.01956 mm

Step-by-step 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 =
`(PL)/(AE)` ................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 =
`(PL)/(AE)`

δ =
`(610*50)/((\pi)/(4) * 4.50^2 * 98*10^3)`

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm