A `5^(@)C` rise in the temperature is observed in a conductor by passing some current. When the current is doubled, then rise in temperature will be equal to

Options

A. `5^(@)C`
B. `10^(@)C`
C. `20^(@)C`
D. `40^(@)C`

Explanation

To solve the problem, we need to understand the relationship between the current passing through a conductor and the temperature rise in that conductor. The heat generated in the conductor due to the current can be described using the formula:

1. **Heat generated (Q)**:
\[ Q = I^2 R t \]
where \(I\) is the current, \(R\) is the resistance of the conductor, and \(t\) is the time for which the current flows.

2. **Heat absorbed by the conductor**:
\[ Q = m s \Delta T \]
where \(m\) is the mass of the conductor, \(s\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.

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 equation, we can see that the change in temperature \(\Delta T\) is directly proportional to the square of the current:
\[ \Delta T \propto I^2 \]

3. **Initial Condition**:
Given that when the current \(I\) produces a temperature rise of \(5^\circ C\), we can denote this as:
\[ \Delta T_1 = 5^\circ C \]

4. **Doubling the Current**:
If we double the current, the new current \(I_2\) will be:
\[ I_2 = 2I \]
The new temperature rise \(\Delta T_2\) will be:
\[ \Delta T_2 \propto (2I)^2 = 4I^2 \]

5. **Calculating the New Temperature Rise**:
Since \(\Delta T\) is proportional to \(I^2\), we can express the new temperature rise as:
\[ \Delta T_2 = 4 \Delta T_1 \]
Substituting \(\Delta T_1 = 5^\circ C\):
\[ \Delta T_2 = 4 \times 5^\circ C = 20^\circ C \]

Thus, the rise in temperature when the current is doubled will be \(20^\circ C\).

**Final Answer**: \(20^\circ C\)

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