One end of a horizontal thick copper wire of length 2 L and radius 2 R is welded to an end of another horizontal thin copper wire of length L and radius R . When the arrangement stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is

Step-by-step solution

As the two rods are in series, the force at the connecting surface will be same. F 1 = F 2 ⇒ Y × ∆ l 1 l 1 × A 1 = Y × ∆ l 2 l 2 × A 2 Where Y is Young's modulus, ∆ l 1 and ∆ l 2 is increase in length, l 1 and l 2 are lengths, A 1 and A 2 are areas, 2 R and R are radii of thick and thin wires respectively. ∵ F = Y × strain × Area ⇒ ∆ l 1 l 1 × A 1 = ∆ l 2 l 2 × A 2 Here, A 1 = π 2 R 2 , l 1 = 2 L A 2 = πR 2 , l 2 = L ⇒ ∆ l 1 2 L × π 2 R 2 = ∆ l 2 L × πR 2 ⇒ ∆ l 2 ∆ l 1 = 2