A lead bullet just melts when stopped by an obstacle. Assuming that 25 per cent of the heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is `27^@C`. (Melting point of lead=`327^@C`, specific heat of lead =`0.03 cal //gm-^@C`, latent heat of fusion of lead=`6 cal//gm,J=4.2 J //cal`).

Explanation

To find the velocity of the lead bullet that just melts upon being stopped by an obstacle, we can follow these steps:

### Step 1: Determine the heat required to melt the lead bullet
The heat required to melt the lead bullet can be calculated using the formula:

\[ Q = m \cdot C \cdot \Delta T + m \cdot L \]

Where:
- \(m\) = mass of the bullet (in grams)
- \(C\) = specific heat of lead = 0.03 cal/g°C
- \(\Delta T\) = change in temperature = \(327°C - 27°C = 300°C\)
- \(L\) = latent heat of fusion of lead = 6 cal/g

Substituting the values:

\[ Q = m \cdot (0.03 \cdot 300 + 6) \]

Calculating \(0.03 \cdot 300\):

\[ 0.03 \cdot 300 = 9 \text{ cal} \]

So,

\[ Q = m \cdot (9 + 6) = m \cdot 15 \text{ cal} \]

### Step 2: Convert the heat from calories to joules
Since 1 joule = 4.2 calories, we can convert \(Q\) to joules:

\[ Q = m \cdot 15 \text{ cal} \cdot 4.2 \text{ J/cal} = m \cdot 63 \text{ J} \]

### Step 3: Relate the heat to the kinetic energy of the bullet
The kinetic energy (KE) of the bullet is given by:

\[ KE = \frac{1}{2} m v^2 \]

Since 75% of the kinetic energy is used for melting the bullet, we have:

\[ 0.75 \cdot KE = Q \]

Substituting for KE:

\[ 0.75 \cdot \frac{1}{2} m v^2 = m \cdot 63 \]

### Step 4: Simplify the equation
We can cancel \(m\) from both sides (assuming \(m \neq 0\)):

\[ 0.75 \cdot \frac{1}{2} v^2 = 63 \]

### Step 5: Solve for \(v^2\)
Rearranging gives:

\[ \frac{0.75}{2} v^2 = 63 \]

\[ 0.375 v^2 = 63 \]

Dividing both sides by 0.375:

\[ v^2 = \frac{63}{0.375} = 168 \]

### Step 6: Calculate \(v\)
Taking the square root:

\[ v = \sqrt{168} \approx 12.96 \text{ m/s} \]

### Final Answer
The velocity of the lead bullet is approximately **12.96 m/s**.

---