A lead bullet penetrates into a solid object and melts . Assuming that 40 % of its kinetic energy is used to heat it , the initial speed of bullet is :
(Given ,initial temperature of the bullet =` 127^(@)C`)
Latenet heat of fusion of lead = `2.5 xx 10^(4) " J kg"^(-1)`,
Specific heat capacity of lead = 125 J/KgK)

Options

A. `125 ms^(-1)`
B. `500 ms^(-1)`
C. `250 ms^(-1)`
D. `600 ms^(-1)`

Explanation

To solve the problem, we need to determine the initial speed of a lead bullet that melts upon penetrating a solid object. We are given that 40% of its kinetic energy is used to heat the bullet, and we have the necessary physical constants for lead.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Latent heat of fusion of lead, \( L = 2.5 \times 10^4 \, \text{J/kg} \)
- Specific heat capacity of lead, \( S = 125 \, \text{J/kgK} \)
- Initial temperature of the bullet, \( T_i = 127^\circ C \)
- Final temperature of the bullet after melting, \( T_f = 300^\circ C \) (melting point of lead)

2. **Calculate the Temperature Change:**
\[ \Delta T = T_f - T_i = 300^\circ C - 127^\circ C = 173 \, \text{K} \]

3. **Write the Heat Required to Melt the Bullet:**
The total heat \( Q \) required to heat the bullet and melt it can be expressed as:
\[ Q = mL + mS\Delta T \]
where \( m \) is the mass of the bullet.

4. **Express Kinetic Energy:**
The kinetic energy \( KE \) of the bullet is given by:
\[ KE = \frac{1}{2} mv^2 \]
Since 40% of the kinetic energy is used for heating, we have:
\[ 0.4 KE = 0.4 \left(\frac{1}{2} mv^2\right) = \frac{0.2 mv^2}{1} \]

5. **Set Up the Equation:**
Equating the heat required to the kinetic energy used:
\[ mL + mS\Delta T = 0.2 mv^2 \]
Dividing through by \( m \) (assuming \( m \neq 0 \)):
\[ L + S\Delta T = 0.2 v^2 \]

6. **Substituting Values:**
Substitute the values for \( L \), \( S \), and \( \Delta T \):
\[ 2.5 \times 10^4 + 125 \times 173 = 0.2 v^2 \]
Calculate \( S\Delta T \):
\[ 125 \times 173 = 21625 \, \text{J/kg} \]
Now substitute:
\[ 2.5 \times 10^4 + 21625 = 0.2 v^2 \]
\[ 25000 + 21625 = 0.2 v^2 \]
\[ 46625 = 0.2 v^2 \]

7. **Solving for \( v^2 \):**
\[ v^2 = \frac{46625}{0.2} = 233125 \]

8. **Finding \( v \):**
\[ v = \sqrt{233125} \approx 483.06 \, \text{m/s} \]

### Final Answer:
The initial speed of the bullet is approximately \( 483.06 \, \text{m/s} \).