The resistance of a metal wire is `10 Omega `. A current of 30 mA is flowing in it at `20^@ `C. If p.d. across its ends is constant, then its temperature is increased to `120^@ C`, then the current flowing in the wire will be in mA`( alpha = 5 xx 10^(-3//0) C)`

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the potential difference (V) across the wire at 20°C.
According to Ohm's law, the current (I) flowing through a resistor (R) is given by:
\[ I = \frac{V}{R} \]
Given:
- Resistance, \( R = 10 \, \Omega \)
- Current, \( I = 30 \, \text{mA} = 0.03 \, \text{A} \)

Rearranging the formula to find \( V \):
\[ V = I \times R \]
Substituting the values:
\[ V = 0.03 \, \text{A} \times 10 \, \Omega = 0.3 \, \text{V} \]

### Step 2: Determine the new resistance (R') at 120°C.
The resistance of a metal wire changes with temperature according to the formula:
\[ R' = R (1 + \alpha \Delta T) \]
Where:
- \( \alpha = 5 \times 10^{-3} \, \text{°C}^{-1} \)
- \( \Delta T = T_f - T_i = 120°C - 20°C = 100°C \)

Substituting the values:
\[ R' = 10 \, \Omega \times \left(1 + 5 \times 10^{-3} \times 100\right) \]
\[ R' = 10 \, \Omega \times (1 + 0.5) = 10 \, \Omega \times 1.5 = 15 \, \Omega \]

### Step 3: Calculate the new current (I') at 120°C.
Using Ohm's law again:
\[ I' = \frac{V}{R'} \]
Substituting the values:
\[ I' = \frac{0.3 \, \text{V}}{15 \, \Omega} \]
\[ I' = 0.02 \, \text{A} \]

### Step 4: Convert the current from Amperes to milliamperes.
Since \( 1 \, \text{A} = 1000 \, \text{mA} \):
\[ I' = 0.02 \, \text{A} \times 1000 = 20 \, \text{mA} \]

### Final Answer:
The current flowing in the wire at 120°C will be **20 mA**.

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