A lead bullet strikes against a steel plate with a velocity `200ms^-1`. If the impact is perfectly inelastic and the heat produced is equally shared between the bullet and the target, thent he rise in temperature of the bullet is (specific heat capacity of lead `=125Jkg^-1K^-1`)

Options

A. `80^@C`
B. `60^@C`
C. `160^@C`
D. `40^@C`

Explanation

To solve the problem of the lead bullet striking a steel plate and determining the rise in temperature of the bullet, we can follow these steps:

### Step 1: Understand the scenario
We have a lead bullet with an initial velocity of \( v = 200 \, \text{m/s} \) striking a steel plate. The collision is perfectly inelastic, meaning the bullet comes to rest after the collision. We need to find the rise in temperature of the bullet after the collision, given that the specific heat capacity of lead is \( c = 125 \, \text{J/kg/K} \).

### Step 2: Calculate the initial kinetic energy of the bullet
The kinetic energy (KE) of the bullet before the collision can be calculated using the formula:
\[ \text{KE} = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the bullet and \( v \) is its velocity.

### Step 3: Determine the heat produced in the collision
Since the bullet comes to rest after the collision, all of its initial kinetic energy is converted into heat energy. Therefore, the heat produced \( Q \) is equal to the initial kinetic energy:
\[ Q = \frac{1}{2} mv^2 \]

### Step 4: Share the heat energy between the bullet and the target
According to the problem, the heat produced is equally shared between the bullet and the target. Thus, the heat absorbed by the bullet is:
\[ Q_{\text{bullet}} = \frac{Q}{2} = \frac{1}{4} mv^2 \]

### Step 5: Relate heat absorbed to temperature rise
The heat absorbed by the bullet can also be expressed in terms of its specific heat capacity and the rise in temperature:
\[ Q_{\text{bullet}} = mc\Delta T \]
where \( \Delta T \) is the rise in temperature.

### Step 6: Set the two expressions for heat equal
Now we can set the two expressions for heat equal to each other:
\[ \frac{1}{4} mv^2 = mc\Delta T \]

### Step 7: Cancel mass \( m \) from both sides
Since mass \( m \) appears on both sides of the equation, we can cancel it out:
\[ \frac{1}{4} v^2 = c\Delta T \]

### Step 8: Solve for \( \Delta T \)
Rearranging the equation gives us:
\[ \Delta T = \frac{1}{4c} v^2 \]

### Step 9: Substitute the known values
Substituting \( v = 200 \, \text{m/s} \) and \( c = 125 \, \text{J/kg/K} \):
\[ \Delta T = \frac{1}{4 \times 125} (200)^2 \]
Calculating \( (200)^2 = 40000 \):
\[ \Delta T = \frac{40000}{500} = 80 \, \text{K} \]

### Conclusion
The rise in temperature of the bullet is \( \Delta T = 80 \, \text{K} \) or \( 80 \, \text{°C} \).

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