Question

A 5m long aluminium wire y =7×10 diameter 3 mm supports a 40 kg mass

Answer 1

Hey dear,
Question is incomplete.

◆ Complete question should be-
A 5 m long aluminium wire, y=7×10^10 N/m^2, diameter 3 mm supports a 40 kg mass, Find the extension produced.

◆ Answer-
x = 7.072 mm


◆ Explanation-
# Given-
l = 5 m
r = 3 mm = 3×10^-3 m
Y = 10^10 N/m^2
m = 40 kg

# Solution-
Cross sectional area of string-
A = πr^2
A = 3.142 × (3×10^-3)^2
A = 2.828×10^-5 m^2

Extension produced in the wire will be-
x = mgl / YA
x = 40 × 10 × 5 / (10^10 × 2.828×10^-5)
x = 7.07×10^-3 m
x = 7.072 mm

Therefore, extension produced will be 7.072 mm.

Hope it helps...

Answer 2

I have uploaded a pic . I think so it help you