The resistance of a wire of iron is `10 ohm` and temperature coefficient of resistivity is `5 xx 10^-3//.^@C`, At `20^@C` it carries `30 mA` of current. Keeping constant potential difference between its ends. The temperature of the wire is raised to `120^@C`. The current in `mA` that flows in the wire now is.

Options

A. 20
B. 15
C. 10
D. 40

Answer

To find the current flowing through the iron wire when its temperature is raised from \(20^\circ C\) to \(120^\circ C\), we can follow these steps:

### Step 1: Identify the given values
- Initial resistance of the wire, \(R_0 = 10 \, \Omega\)
- Temperature coefficient of resistivity, \(\alpha = 5 \times 10^{-3} \, \text{°C}^{-1}\)
- Initial temperature, \(T_0 = 20 \, \text{°C}\)
- Final temperature, \(T_1 = 120 \, \text{°C}\)
- Initial current, \(I_0 = 30 \, \text{mA} = 0.03 \, \text{A}\)

### Step 2: Calculate the change in temperature
\[ \Delta T = T_1 - T_0 = 120 \, \text{°C} - 20 \, \text{°C} = 100 \, \text{°C} \]

### Step 3: Calculate the new resistance \(R_1\) at \(120^\circ C\)
Using the formula for resistance change with temperature:
\[ R_1 = R_0 \left(1 + \alpha \Delta T\right) \]
Substituting the values:
\[ R_1 = 10 \left(1 + 5 \times 10^{-3} \times 100\right) \]
\[ R_1 = 10 \left(1 + 0.5\right) = 10 \times 1.5 = 15 \, \Omega \]

### Step 4: Use Ohm's Law to find the new current \(I_1\)
Since the potential difference \(V\) across the wire remains constant, we can use the relationship:
\[ V = I_0 R_0 = I_1 R_1 \]
From this, we can express \(I_1\):
\[ I_1 = \frac{I_0 R_0}{R_1} \]
Substituting the known values:
\[ I_1 = \frac{0.03 \, \text{A} \times 10 \, \Omega}{15 \, \Omega} \]
\[ I_1 = \frac{0.3}{15} = 0.02 \, \text{A} = 20 \, \text{mA} \]

### Final Answer
The current flowing through the wire at \(120^\circ C\) is \(20 \, \text{mA}\).

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