Two wires, one made of copper and other of steel are joined end to end (as shown in figure). The area of cross section of copper wire is twice that of steel wire. They are placed under compressive force of magnitudes F . Find the ratio of their lengths such that change in lengths of both wires are same ( Y S = 2 × 10 11 N / m 2 and Y C = 1.1 × 10 11 N / m 2 )

Step-by-step solution

Young's modulus for steel Y S = FL S A S ∆ l S Where, Y S is Young's modulus, L S is length, A S is area of cross section, ∆ l S is change in length, F is force applied for steel wire respectively. ⇒ L S = Y S A S ∆ l S F .......(i) Young's modulus for copper Y C = FL C A C ∆ l C Where, Y C is Young modulus, L C is length, A C is area of cross section, ∆ l C is change in length, F is force applied for copper wire respectively. ⇒ L C = Y C A C ∆ l C F .....