A steel wire of length 3.2 m (Yₛ = 2.0 × 10¹¹ N m⁻²) and a copper wire of length 4.4 m (Y_c = 1.1 × 10¹¹ 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)

Question

Accepted Answer

154. A = (22/7)(1.4e-3)^2 = 6.16e-6 m^2. L1/Y1 + L2/Y2 = 3.2/2.0e11 + 4.4/1.1e11 = 1.6e-11 + 4.0e-11 = 5.6e-11. F = elongation*A/sum = (1.4e-3)(6.16e-6)/(5.6e-11) = 154 N.

A steel wire of length 3.2 m (Yₛ = 2.0 × 10¹¹ N m⁻²) and a copper wire of length 4.4 m (Y_c = 1.1 × 10¹¹ 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)

Solution

Related JEE Main Physics questions