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 $=127^{\circ} \mathrm{C}$, Melting point of the bullet $=327^{\circ} \mathrm{C}$, Latent heat of fusion of lead $=2.5 \times 10^{4} \mathrm{~J} \mathrm{Kg}^{-1}$, Specific heat capacity of lead $=125 \mathrm{~J} / \mathrm{kg} \mathrm{K}$ )

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Heat & Thermodynamics

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

= 12 7^{\circ} C

, Melting point of the bullet

= 32 7^{\circ} C

,

Latent heat of fusion of lead

= 2.5 \times 1 0^{4} J Kg^{- 1}

,
Specific heat capacity of lead

= 125 J / kg K

)

Answer: (B)

Heat required to raise temperature from 127°C to 327°C and melt:

Q = m c Δ T + m L = m (125 \times 200 + 2.5 \times 1 0^{4}) = m (25000 + 25000) = 50000 m

J. This equals 40% of kinetic energy:

0.4 \times \frac{1}{2} m v^{2} = 50000 m

. Solving:

0.2 v^{2} = 50000

, so

v^{2} = 250000

and

v = 500

m/s.

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Single correct option