Question

two wires of length l,radius r and length 2l and radius 2r respectively having same youngs modulus are hung with a weight mg.net elongation is

Answer 1

Given :
Let us say the Young's Modulus of two wires are E.
Area of first wire=A=πr²
Area of second wire =π(2r)²=4πr²=4A
If elongations are Δ1 and Δ2

ThenΔ 1=mgxL/AreaxE
=mgL/πr² xE

Δ 2=mgx2L/4xAreaxE
=mgL/2πr² x E

Δ1: Δ2=mgL/πr² xE :mgL/2πr² x E=2:1