`A` lead bullet of `10g` travelling at `300m//s` strikes against a block of wood and comes to rest. Assuming `50%` heat is absorbed by the bullet, the increase in its temperature is (sp-heat of lead is `150J//Kg-K`)

Options

A. `100^(@)C`
B. `125^(@)C`
C. `150^(@)C`
D. `200^(@)C`

Explanation

To solve the problem, we need to find the increase in temperature of a lead bullet after it strikes a block of wood and comes to rest, given that 50% of its kinetic energy is converted into heat absorbed by the bullet.

### Step-by-Step Solution:

1. **Calculate the initial kinetic energy (KE) of the bullet:**
The formula for kinetic energy is:
\[ KE = \frac{1}{2} mv^2 \]
where:
- \( m = 10 \, \text{g} = 0.01 \, \text{kg} \) (since 1 g = 0.001 kg)
- \( v = 300 \, \text{m/s} \)

Substituting the values:
\[ KE = \frac{1}{2} \times 0.01 \, \text{kg} \times (300 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \times 0.01 \times 90000 \]
\[ KE = 0.005 \times 90000 = 450 \, \text{J} \]

2. **Determine the heat energy absorbed by the bullet:**
Since 50% of the kinetic energy is absorbed as heat:
\[ Q = 0.5 \times KE = 0.5 \times 450 \, \text{J} = 225 \, \text{J} \]

3. **Use the heat energy to find the increase in temperature:**
The heat absorbed by the bullet can also be expressed using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \( Q = 225 \, \text{J} \)
- \( m = 0.01 \, \text{kg} \)
- \( c = 150 \, \text{J/kg-K} \)
- \( \Delta T \) is the increase in temperature.

Rearranging the formula to solve for \( \Delta T \):
\[ \Delta T = \frac{Q}{m \cdot c} \]
Substituting the values:
\[ \Delta T = \frac{225 \, \text{J}}{0.01 \, \text{kg} \times 150 \, \text{J/kg-K}} \]
\[ \Delta T = \frac{225}{1.5} = 150 \, \text{K} \]

4. **Final result:**
The increase in temperature of the lead bullet is:
\[ \Delta T = 150 \, \text{K} \, \text{or} \, 150 \, \text{°C} \]