Two metallic wires P and Q have same volume and are made up of same material. If their area of cross sections are in the ratio 4 : 1 and force F 1 is applied to P , an extension of ∆ l is produced. The force which is required to produce same extension in Q is F 2 . The value of F 1 F 2 is ______.

Step-by-step solution

The formula for Young's modulus is given by Y = F A ∆ l l = F l A ∆ l From above equation, it follows that Δ l = F l A Y . . . 1 Again, volume of the wire is given by V = A l ⇒ l = V A Hence, from equation (1), ∆ l = F V A 2 Y . . . 2 As, Y and V is the same for both the wires, ∆ l ∝ F A 2 Hence, for two wires, it can be written that ∆ l 1 ∆ l 2 = F 1 A 1 2 × A 2 2 F 2 . . . 3 As ∆ l 1 = ∆ l 2 , from equation (3), it follows that F 1 A 2 2 = F 2 A 1 2 ⇒ F 1 F 2 = A 1 2 A 2 2 = 4 1 2 = 16