A lead bullet penetrates into a solid object and melts. Assuming that 40 % of its kinetic energy is used to heat it, the initial speed of bullet is (Given, initial temperature of the bullet = 127 ° C , Melting point of the bullet = 327 ° C , Latent heat of fusion of lead = 2 . 5 × 10 4 J kg - 1 , Specific heat capacity of lead = 125 J kg K - 1 )

Step-by-step solution

Loss of kinetic energy of the lead bullet = Heat generated to raise its temperature from 127   ° C to 327   ° C . 40   % loss of the kinetic energy of the lead bullet = Heat gained by the lead bullet to raise its temperature. 0 . 4 1 2 m v 2 = m × S Pb × ∆ T + m × L Pb 0 . 4 × 1 2 m v 2 = m S Pb T 2 - t 1 + m L Pb 0 . 2 m v 2 = m × 125 × 327 - 127 + 2 . 5 × 10 4 v 2 = 5 × 125 × 200 + 2 . 5 × 10 4 v 2 = 5 × 25000 + 25000 v = 500 &#