A steel wire of length 3 . 2 m Y S = 2 . 0 × 10 11 N m - 2 and a copper wire of length 4 . 4 m Y C = 1 . 1 × 10 11 N m - 2 , both of radius 1 . 4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1 . 4 mm . The load applied, in Newton, will be: (Given π = 22 7 )

Step-by-step solution

When two wires are connected end to end, the tension in them will be same. Change in length using Hooke's law can be found as, Y = F l A ∆ l ⇒ Δ l = F l A Y Total change in length can be written as, Δ l = Δ l 1 + Δ l 2 ⇒ Δ l = F l 1   A 1 Y 1 + F l 2   A 2 Y 2 ⇒ F = Δ l l 1   A 1 Y 1 + l 2   A 2 Y 2 = 1 . 4 × 10 - 3 3 . 2 π 1 . 4 × 10 - 3 2 × 2 . 0 × 10 11 + 4 . 4 π 1 . 4 × 10 - 3 2 × 1 . 1 × 1