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 `30 kcals//kg`. The minimum velocity with which the fragmento must have entered the atmosphere is

Options

A. 0.45 km/s
B. 1.32 km/s
C. 2.32 km/s
D. Zero

Explanation

To find the minimum velocity with which the iron rocket fragment must have entered the atmosphere, we can follow these steps:

### Step 1: Calculate the heat required to raise the temperature of the iron fragment to its melting point.

1. **Initial Temperature (T_initial)** = -100°C
2. **Melting Point of Iron (T_melting)** = 1535°C
3. **Specific Heat of Iron (c)** = 0.11 kcal/kg°C

The temperature change (ΔT) is:
\[ \Delta T = T_{melting} - T_{initial} = 1535°C - (-100°C) = 1635°C \]

The heat required to raise the temperature to the melting point (Q1) is given by:
\[ Q_1 = m \cdot c \cdot \Delta T \]
Substituting the values:
\[ Q_1 = m \cdot 0.11 \cdot 1635 \]
\[ Q_1 = 180.85m \text{ kcal} \]

### Step 2: Calculate the heat required to melt the iron fragment.

4. **Latent Heat of Fusion (L)** = 30 kcal/kg

The heat required to melt the iron (Q2) is given by:
\[ Q_2 = m \cdot L \]
Substituting the values:
\[ Q_2 = m \cdot 30 \text{ kcal} \]

### Step 3: Calculate the total heat required.

The total heat required (Q_total) to raise the temperature and then melt the iron is:
\[ Q_{total} = Q_1 + Q_2 \]
\[ Q_{total} = 180.85m + 30m = 210.85m \text{ kcal} \]

### Step 4: Convert the total heat required to Joules.

1 kcal = 4184 Joules, therefore:
\[ Q_{total} = 210.85m \cdot 4184 \text{ J} \]
\[ Q_{total} = 882,000m \text{ J} \]

### Step 5: Set the kinetic energy equal to the heat required.

The kinetic energy (KE) of the fragment is given by:
\[ KE = \frac{1}{2} mv^2 \]
Setting the kinetic energy equal to the heat required:
\[ \frac{1}{2} mv^2 = 882,000m \]

### Step 6: Cancel out mass (m) and solve for velocity (v).

Since m is present on both sides, we can cancel it out:
\[ \frac{1}{2} v^2 = 882,000 \]
Multiplying both sides by 2:
\[ v^2 = 1,764,000 \]
Taking the square root:
\[ v = \sqrt{1,764,000} \approx 1320.5 \text{ m/s} \]

### Step 7: Convert the velocity to kilometers per second.

\[ v \approx 1.32 \text{ km/s} \]

### Final Answer:
The minimum velocity with which the iron rocket fragment must have entered the atmosphere is approximately **1.32 km/s**.

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