Q 11-10-084JEE MainJEE Main 2022 (27 Jun, Shift 2)Medium
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 $= 127^\circ$C, Melting point of the bullet $= 327^\circ$C, Latent heat of fusion of lead $= 2.5\times10^4\ \text{J kg}^{-1}$, Specific heat capacity of lead $= 125\ \text{J kg}^{-1}\text{K}^{-1}$)
Answer: (B) $500\ \text{m s}^{-1}$
Per kilogram, heat needed to reach the melting point and melt:
$$c\Delta T + L = 125\times200 + 2.5\times10^4 = 5\times10^4\ \text{J kg}^{-1}$$
$$0.4\times\frac12 v^2 = 5\times10^4 \Rightarrow v^2 = 2.5\times10^5 \Rightarrow v = 500\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics