A one metre steel wire of negligible mass and area of cross-section 0.01 ~cm ^2 is kept on a smooth horizontal table with one end fixed. A ball of mass 1 ~kg is attached to the other end. The ball and the wire are rotating with an angular velocity of . If the elongation of the wire is 2 ~mm , then is (Young's modulus of steel =2 10¹¹ Nm ⁻² )

Step-by-step solution

Given, elongation of the wire, l=2 ~mm =2 10⁻³ ~m Mass of the ball, m=1 ~kg Length of wire, l=1 ~m Area of cross-sectional of wire, A=0.01 ~cm ^2=0.01 10⁻⁴ ~m Young's modulus of steel, Y=2 10¹¹ Nm ⁻² Tension force in wire, T=m ^2 l Stress = Tension Area = m ^2 l A Strain = l l = stress Young's modulus or l= m ^2 l^2 Y A or = Y A l m l^2 Putting the given values, we get aligned & = 2 10¹¹ 0.01 10⁻⁴ 2 10⁻³ 1 (1)^2 & =20 rad / sec ⁻¹ aligned