A composite wire is formed by connecting a steel wire of length 2.0 m and an aluminium wire of length 2.1 m end to end. Both wires have the same uniform radius. The Young's modulus of steel is 2.0 10¹¹ N m ⁻² and that of aluminium is 7.0 10¹⁰ N m ⁻² . When the composite wire is stretched by a force of 100 N , the total net elongation produced is 1.0 mm . The common radius of the wires, in mm, is:

Step-by-step solution

When the wires are connected in series, the tension in both wires is equal to the applied force F = 100 N . The total elongation L is the sum of the elongations of the individual wires: L = L_S + L_A Using Hooke's law, L = F L A Y , we can write: L = F r^2 ( L_S Y_S + L_A Y_A ) Substitute the given values: 1.0 10⁻³ = 100 r^2 ( 2.0 2.0 10¹¹ + 2.1 7.0 10¹⁰ ) 1.0 10⁻³ = 100 r^2 ( 1.0 10⁻¹¹ + 3.0 10⁻¹¹ ) 1.0 10⁻³ = 100 r^2 ( 4.0 10⁻¹¹ ) r^2 = 4.0 10⁻⁹ 1.0 10⁻³ = 4.0 10⁻⁶ m ^2 r = 2.0 10⁻³ m = 2 mm Answer: 2