Two wires are made of the same material and have the same volume. However, wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by Δx on applying force F, how much force is needed to stretch wire 2 by the same amount?

Let L1 and L2 be the lengths of wire 1 and wire 2 respectively.
Let A1 and A2 be the cross-sectional areas of wire 1 and wire 2 respectively.
Given that A1 = A and A2 = 3A.
The volume of both wires is the same, so V1 = V2.
Since V = AL, we have A1L1 = A2L2.
Therefore, AL1 = 3AL2, which implies L1 = 3L2.
Young's modulus (Y) is given by the formula:
Y = (F/A) / (ΔL/L)
For wire 1:
Y = (F/A) / (Δx/L1) (i)
For wire 2:
Y = (F'/3A) / (Δx/L2) (ii)
Since both wires are made of the same material, their Young's modulus is the same.
Equating (i) and (ii):
(F/A) / (Δx/L1) = (F'/3A) / (Δx/L2)
F/A * L1/Δx = F'/3A * L2/Δx
F L1 = F' L2 /3
Since L1 = 3L2:
3F L2 = F' L2 /3
F' = 9F