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 \times 104 J kg - 1, Specific heat capacity of lead = 125 J/kg K)

Heat and Thermodynamics PYQ — JEE Main 2022 · 27 Jun · Shift 2 · Q53 with Solution

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Correct answer: B

The bullet is initially at $127^\circ\text{C}$ and must be heated to the melting point $327^\circ\text{C}$ .

So the rise in temperature is

\Delta T = 327 - 127 = 200\,\text{K}

Heat required per unit mass to raise temperature to melting point:

q_1 = c\Delta T = 125 \times 200 = 25000\,\text{J kg}^{-1}

Latent heat required per unit mass to melt it:

q_2 = L = 2.5\times 10^4\,\text{J kg}^{-1}

Hence total heat required per unit mass is

q = q_1 + q_2 = 25000 + 25000 = 50000\,\text{J kg}^{-1}

Only $40\%$ of the initial kinetic energy is used to heat the bullet.

Initial kinetic energy per unit mass is

\frac{1}{2}v^2

Given that $40\%$ of this becomes heat:

0.4\left(\frac{1}{2}v^2\right) = 50000

0.2v^2 = 50000

v^2 = \frac{50000}{0.2} = 250000

v = 500\,\text{m s}^{-1}

Thus the correct option is B.