When a 8 ~m long wire is stretched by a load of 10 ~kg - wt , it is elongated by 1.5 ~mm . The energy stored in the wire in this process is (g=10 ~ms ⁻² )

Step-by-step solution

Given that, weight of load, w=10 ~kg - wt F=w=10 10 ~N =100 ~N Elongation in wire, l=1.5 ~mm =1.5 10⁻³ ~m Length of wire, l=8 ~m We know that, elastic potential energy stored in wire, U= 1 2 F l= 1 2 100 1.5 10⁻³=0.075 ~J