Answer:
3.5 cm
Explanation:
mass, m = 50 kg
diameter = 1 mm
radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m
L = 11.2 m
Y = 2 x 10^11 Pa
Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3
= 7.85 x 10^-7 m^2
Let the wire is stretch by ΔL.
The formula for Young's modulus is given by
[tex]Y =\frac{mgL}{A\Delta L}[/tex]
[tex]\Delta L =\frac{mgL}{A\times Y}[/tex]
ΔL = 0.035 m = 3.5 cm
Thus, the length of the wire stretch by 3.5 cm.
