An iron rocket fragment initially at `-100^@C` enters the earth's atmosphere almost horizontally and quickly fuses completely in atmospheric friction. Specific heat of iron is `0.11 kcal//kg^@C`. Its melting point is `1535^@C` and the latent heat of fusion is `3 kcals//kg`. The minimum velocity with which the fragmento must have entered the atmosphere is

Options

A. `0.45km//s`
B. `1.32km//s`
C. `2.32km//s`
D. zero

Explanation

To find the minimum velocity with which the iron rocket fragment must have entered the atmosphere, we will use the concepts of heat transfer and kinetic energy. Here’s the step-by-step solution:

### Step 1: Calculate the heat required to raise the temperature of the iron fragment to its melting point.
The heat required (Q) to raise the temperature can be calculated using the formula:
\[ Q = m \cdot s \cdot \Delta T \]
Where:
- \( m \) = mass of the fragment (in kg)
- \( s \) = specific heat of iron = 0.11 kcal/kg°C
- \( \Delta T \) = change in temperature = (melting point - initial temperature)

Given:
- Initial temperature = -100°C
- Melting point = 1535°C
- \( \Delta T = 1535 - (-100) = 1635°C \)

Thus,
\[ Q = m \cdot 0.11 \cdot 1635 \]

### Step 2: Calculate the heat required for melting the iron fragment.
The heat required for melting (Q_melt) can be calculated using the formula:
\[ Q_{melt} = m \cdot L_f \]
Where:
- \( L_f \) = latent heat of fusion = 3 kcal/kg

Thus,
\[ Q_{melt} = m \cdot 3 \]

### Step 3: Total heat required for the process.
The total heat (Q_total) required to raise the temperature to the melting point and then melt the iron is:
\[ Q_{total} = Q + Q_{melt} \]
Substituting the values from Steps 1 and 2:
\[ Q_{total} = m \cdot 0.11 \cdot 1635 + m \cdot 3 \]
\[ Q_{total} = m \cdot (0.11 \cdot 1635 + 3) \]

### Step 4: Set the total heat equal to the kinetic energy.
The kinetic energy (KE) of the fragment can be expressed as:
\[ KE = \frac{1}{2} m v^2 \]
Setting the total heat equal to the kinetic energy gives:
\[ m \cdot (0.11 \cdot 1635 + 3) = \frac{1}{2} m v^2 \]

### Step 5: Cancel mass and solve for velocity.
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 0.11 \cdot 1635 + 3 = \frac{1}{2} v^2 \]
Now, calculate \( 0.11 \cdot 1635 \):
\[ 0.11 \cdot 1635 = 179.85 \]
Thus,
\[ 179.85 + 3 = \frac{1}{2} v^2 \]
\[ 182.85 = \frac{1}{2} v^2 \]

### Step 6: Solve for \( v^2 \) and then \( v \).
Multiply both sides by 2:
\[ 365.7 = v^2 \]
Now, take the square root:
\[ v = \sqrt{365.7} \]
Calculating the square root:
\[ v \approx 19.1 \text{ m/s} \]

### Step 7: Convert to km/s.
To convert from m/s to km/s, divide by 1000:
\[ v \approx 0.0191 \text{ km/s} \]

### Final Result
The minimum velocity with which the fragment must have entered the atmosphere is approximately **0.0191 km/s**.