A `1^(@) C` rise in temperature is observed in a conductor by passing a certain current . If the current is doubled , then the rise in temperature is approximately

Options

A. `2.5 ^(@) C`
B. `4 ^(@) C`
C. `2 ^(@) C`
D. `1 ^(@) C`

Explanation

To solve the problem, we need to understand the relationship between the current flowing through a conductor and the rise in temperature it produces.

### Step-by-Step Solution:

1. **Understand the Heating Effect of Current**:
The heating effect of current in a conductor can be expressed by the formula:
\[ Q = I^2 R t \]
where \( Q \) is the heat produced, \( I \) is the current, \( R \) is the resistance, and \( t \) is the time for which the current flows.

2. **Relate Heat to Temperature Rise**:
The heat produced in the conductor also causes a rise in temperature, which can be expressed as:
\[ Q = m s \Delta T \]
where \( m \) is the mass of the conductor, \( s \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.

3. **Establish the Proportionality**:
Since both expressions for heat \( Q \) are equal, we can set them equal to each other:
\[ I^2 R t = m s \Delta T \]
From this, we can express the temperature rise \( \Delta T \) in terms of current \( I \):
\[ \Delta T \propto I^2 \]
This means that the temperature rise is proportional to the square of the current.

4. **Set Up the Ratios**:
Let’s denote the initial temperature rise as \( \Delta T_1 \) when the current is \( I_1 \) and the new temperature rise as \( \Delta T_2 \) when the current is \( I_2 \). We can write:
\[ \frac{\Delta T_1}{\Delta T_2} = \frac{I_1^2}{I_2^2} \]

5. **Substituting Known Values**:
From the problem, we know:
- \( \Delta T_1 = 1^\circ C \)
- \( I_2 = 2I_1 \) (since the current is doubled)

Substituting these values into the ratio gives:
\[ \frac{1}{\Delta T_2} = \frac{I_1^2}{(2I_1)^2} \]
Simplifying the right side:
\[ \frac{1}{\Delta T_2} = \frac{I_1^2}{4I_1^2} = \frac{1}{4} \]

6. **Cross-Multiplying to Find \( \Delta T_2 \)**:
Cross-multiplying gives:
\[ 1 \cdot 4 = \Delta T_2 \]
Therefore,
\[ \Delta T_2 = 4^\circ C \]

### Final Answer:
The rise in temperature when the current is doubled is approximately \( 4^\circ C \).