QJEEM25Phf805: JEE Main 2025 Physics - Class 11 JEEM25Phf805 Solution: A gun fires a lead bullet of temperature 300 K into a wooden…

Let's determine the mass step by step:

The bullet is heated from 300 K to 600 K, so the increase in temperature is:

$$\mathrm{\Delta }T=600\text{K}-300\text{K}=300\text{K}.$$

The heat required to raise the temperature of the bullet (using the specific heat capacity of lead) is:

$$Q_1=m\times c\times \mathrm{\Delta }T=m\times 125\text{J/kg K}\times 300\text{K}=37500m\text{J},$$

where $m$ is the mass in kilograms.

After reaching 600 K, the bullet melts. The heat required for the phase change (melting) is given by:

$$Q_2=m\times L=m\times 2.5\times {10}^4\text{J/kg}=25000m\text{J}.$$

The total heat required for the process is:

$$Q=Q_1+Q_2=37500m+25000m=62500m\text{J}.$$

We are given that the total heat is 625 J, so:

$$62500m=625.$$

Solving for $m$ :

$$m=\frac{625}{62500}=0.01\text{kg}.$$

Finally, converting the mass from kilograms to grams:

$$0.01\text{kg}=0.01\times 1000\text{g}=10\text{g}.$$

Thus, the mass of the bullet is 10 grams. This corresponds to Option D.