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` calories /`gm^@C`, latent jeat of fusion of lead=`6`calories`/gm`,`J=4.2 `joules /calorie.

Explanation

To solve the problem, we need to calculate the velocity of the lead bullet that just melts upon hitting an obstacle, considering that 25% of the heat is absorbed by the obstacle. Here's the step-by-step solution:

### Step 1: Understand the heat transfer
When the bullet hits the obstacle, it will lose some of its kinetic energy, which will be converted into heat. The heat will be used to raise the temperature of the bullet to its melting point and to melt it.

### Step 2: Calculate the temperature change
The initial temperature of the bullet is \(27^\circ C\) and the melting point of lead is \(327^\circ C\). Therefore, the change in temperature (\(\Delta T\)) is:
\[ \Delta T = 327^\circ C - 27^\circ C = 300^\circ C \]

### Step 3: Calculate the heat required to raise the temperature and melt the bullet
The heat required to raise the temperature of the bullet to its melting point is given by:
\[ Q_1 = m \cdot c \cdot \Delta T \]
where:
- \(m\) = mass of the bullet (in grams)
- \(c\) = specific heat of lead = \(0.03 \, \text{cal/g}^\circ C\)
- \(\Delta T = 300^\circ C\)

The heat required to melt the bullet is given by:
\[ Q_2 = m \cdot L \]
where:
- \(L\) = latent heat of fusion of lead = \(6 \, \text{cal/g}\)

### Step 4: Total heat required
The total heat (\(Q\)) required to raise the temperature to the melting point and to melt the bullet is:
\[ Q = Q_1 + Q_2 = m \cdot c \cdot \Delta T + m \cdot L \]
Substituting the values:
\[ Q = m \cdot (0.03 \cdot 300 + 6) = m \cdot (9 + 6) = m \cdot 15 \, \text{cal} \]

### Step 5: Kinetic energy of the bullet
The kinetic energy (\(KE\)) of the bullet before hitting the obstacle is given by:
\[ KE = \frac{1}{2} m v^2 \]
Since 25% of the heat is absorbed by the obstacle, 75% of the heat is used to raise the temperature and melt the bullet. Therefore:
\[ 0.75 \cdot KE = Q \]
Substituting for \(KE\):
\[ 0.75 \cdot \frac{1}{2} m v^2 = m \cdot 15 \]

### Step 6: Simplifying the equation
We can cancel \(m\) from both sides (assuming \(m \neq 0\)):
\[ 0.75 \cdot \frac{1}{2} v^2 = 15 \]
\[ 0.375 v^2 = 15 \]
\[ v^2 = \frac{15}{0.375} = 40 \]
\[ v = \sqrt{40} \approx 6.32 \, \text{m/s} \]

### Final Answer
The velocity of the bullet is approximately \(6.32 \, \text{m/s}\).

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