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 $$= 2.5 \times 10^4$$ J kg$$^{-1}$$, Specific heat capacity of lead $$= 125$$ J kg$$^{-1}$$ K$$^{-1}$$)

A lead bullet penetrates a solid object and melts. 40% of its kinetic energy is used to heat it.

Initial temperature $$T_i = 127°$$C, Melting point $$T_m = 327°$$C

Latent heat of fusion $$L_f = 2.5 \times 10^4$$ J/kg

Specific heat capacity $$c = 125$$ J/(kg·K)

Calculate the heat required to melt the bullet.

Temperature rise: $$\Delta T = 327 - 127 = 200$$ K

Heat to raise temperature to melting point:

$$Q_1 = mc\Delta T = m \times 125 \times 200 = 25000m$$ J

Heat to melt:

$$Q_2 = mL_f = m \times 2.5 \times 10^4 = 25000m$$ J

Total heat required:

$$Q = Q_1 + Q_2 = 25000m + 25000m = 50000m$$ J

Relate to kinetic energy.

40% of kinetic energy equals the heat required:

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

$$0.2v^2 = 50000$$

$$v^2 = 250000$$

$$v = 500$$ m/s

The initial speed of the bullet is $$500$$ m/s.

The correct answer is Option B.