When a current is passed in a conductor, `3^(@)C` rise in temperature is observed. If the strength of current is made thrice, then rise in temperature will approximately be

Options

A. `36^(@)C`
B. `27^(@)C`
C. `186^(@)C`
D. `9^(@)C`

Explanation

To solve the problem, we can use the relationship between the heat generated in a conductor and the rise in temperature due to the current flowing through it. The heat generated (Q) in a conductor is given by the formula:

\[ Q = I^2 R t \]

Where:
- \( Q \) is the heat generated,
- \( I \) is the current,
- \( R \) is the resistance,
- \( t \) is the time for which the current flows.

Since the problem states that the rise in temperature is directly proportional to the heat generated, we can express this relationship as:

\[ \Delta T \propto Q \]

This means that if we know the initial rise in temperature (\( \Delta T_1 \)) for a certain current (\( I_1 \)), we can find the rise in temperature for a different current (\( I_2 \)) using the following relationship:

\[ \frac{\Delta T_1}{\Delta T_2} = \frac{I_1^2}{I_2^2} \]

### Step-by-step Solution:

1. **Identify the initial conditions**:
- Given that the initial rise in temperature (\( \Delta T_1 \)) is \( 3^\circ C \) when the current is \( I_1 \).

2. **Determine the new current**:
- The new current (\( I_2 \)) is thrice the initial current, so \( I_2 = 3I_1 \).

3. **Set up the ratio of temperature rises**:
- Using the relationship derived above, we have:
\[ \frac{\Delta T_1}{\Delta T_2} = \frac{I_1^2}{(3I_1)^2} \]

4. **Simplify the equation**:
- This simplifies to:
\[ \frac{\Delta T_1}{\Delta T_2} = \frac{I_1^2}{9I_1^2} = \frac{1}{9} \]

5. **Express \( \Delta T_2 \) in terms of \( \Delta T_1 \)**:
- Rearranging gives:
\[ \Delta T_2 = 9 \Delta T_1 \]

6. **Substitute the known value**:
- Now substitute \( \Delta T_1 = 3^\circ C \):
\[ \Delta T_2 = 9 \times 3^\circ C = 27^\circ C \]

### Final Answer:
The rise in temperature when the current is made thrice will be approximately \( 27^\circ C \).