A heater coil working on mains produces 100 cal of heat in a certain time. Now, the heater coil is cut into three equal parts and one part is only used for heating. The quantity of heat produced (in calories) in the same time is

Options

A. 300
B. 200
C. `100//3`
D. `200//3`

Answer

To solve the problem, we need to understand how the heat produced by a heater coil changes when the coil is cut into three equal parts and only one part is used.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- The heater coil produces 100 calories of heat in a certain time when it is intact.
- We denote the heat produced by the entire coil as \( H = 100 \, \text{cal} \).

2. **Heat Production Formula**:
- The heat produced by a coil can be expressed using the formula:
\[ H = \frac{V^2 \cdot t}{R} \]
- Where \( V \) is the voltage, \( t \) is the time, and \( R \) is the resistance of the coil.

3. **Cutting the Coil**:
- The coil is cut into three equal parts. Therefore, the resistance of each part will be:
\[ R' = \frac{R}{3} \]
- This is because resistance is directly proportional to the length of the wire.

4. **Using One Part of the Coil**:
- When we use only one part of the coil, the new heat produced \( H' \) can be calculated using the same formula:
\[ H' = \frac{V^2 \cdot t}{R'} \]
- Substituting \( R' \) into the equation gives:
\[ H' = \frac{V^2 \cdot t}{\frac{R}{3}} = \frac{3V^2 \cdot t}{R} \]

5. **Relating New Heat to Original Heat**:
- We know from the original condition that:
\[ H = \frac{V^2 \cdot t}{R} = 100 \, \text{cal} \]
- Therefore, substituting this into our equation for \( H' \):
\[ H' = 3 \cdot H = 3 \cdot 100 \, \text{cal} = 300 \, \text{cal} \]

6. **Final Result**:
- The quantity of heat produced in the same time using one part of the heater coil is:
\[ H' = 300 \, \text{cal} \]

### Conclusion:
The heat produced by one part of the heater coil in the same time is **300 calories**.