A gun fires a lead bullet of temperature 300 K into a wooden block. The bullet having melting temperature of 600 K penetrates into the block and melts down. If the total heat required for the process is 625 J , then the mass of the bullet is grams. (Latent heat of fusion of lead $$=2.5\times10^{4}JKg^{-1}$$ and specific heat capacity $$=125JKg^{-1}K^{-1}$$ of lead

We need to find the mass of the bullet that melts completely after penetrating the wooden block. Since the total heat required has two components, the first part is the heat to raise the temperature from 300 K to the melting point at 600 K, which is given by $$Q_1 = mc\Delta T$$, and the second part is the heat to melt the bullet at 600 K, given by $$Q_2 = mL$$.

Given that $$c = 125 \text{ J kg}^{-1}\text{K}^{-1}$$, $$L = 2.5 \times 10^{4} \text{ J kg}^{-1}$$, and $$\Delta T = 600 - 300 = 300 \text{ K}$$, the total heat supplied is $$Q = 625 \text{ J}$$.

Substituting these values into the expression for total heat, we have $$Q = mc\Delta T + mL$$, which becomes $$625 = m\bigl(125 \times 300 + 2.5 \times 10^{4}\bigr)$$. Therefore, $$625 = m(37500 + 25000)$$, leading to $$625 = m \times 62500$$. This gives $$m = \frac{625}{62500} = 0.01 \text{ kg} = 10 \text{ grams}$$.

Therefore, the correct answer is Option 1: 10 grams.