A potentiometer wire has length `4 m` and resistance `8 Omega`. The resistance that must be connected in series with the wire and an accumulator of e.m.f. `2 V`, so as the get a potential gradient `1 mV` per cm` on the wire is

Options

A. `32 Omega`
B. `40 Omega`
C. `44 Omega`
D. `48 Omega`

Explanation

To solve the problem step by step, we need to find the resistance that must be connected in series with the potentiometer wire and an accumulator to achieve a potential gradient of 1 mV per cm.

### Step 1: Understand the given data
- Length of the potentiometer wire, \( L = 4 \, \text{m} \)
- Resistance of the potentiometer wire, \( R_w = 8 \, \Omega \)
- E.M.F of the accumulator, \( E = 2 \, \text{V} \)
- Desired potential gradient, \( V_g = 1 \, \text{mV/cm} = 0.001 \, \text{V/cm} \)

### Step 2: Calculate the total potential difference across the wire
The total potential difference across the potentiometer wire can be calculated using the desired potential gradient and the length of the wire.

\[ \text{Total potential difference} = V_g \times L = 0.001 \, \text{V/cm} \times 400 \, \text{cm} = 0.4 \, \text{V} \]

### Step 3: Set up the circuit equation
The total potential difference across the circuit (including the resistance \( R \) that we need to find) is equal to the E.M.F of the battery:

\[ E = I(R + R_w) \]

Where:
- \( I \) is the current flowing through the circuit.
- \( R \) is the resistance we need to find.

### Step 4: Calculate the current \( I \)
Using Ohm's law, the current \( I \) can be expressed as:

\[ I = \frac{V}{R + R_w} \]

From the previous step, we know that the potential difference across the wire is \( 0.4 \, \text{V} \):

\[ 0.4 = I \cdot R_w \implies I = \frac{0.4}{R_w} = \frac{0.4}{8} = 0.05 \, \text{A} \]

### Step 5: Substitute \( I \) back into the circuit equation
Now substituting \( I \) back into the circuit equation:

\[ 2 = 0.05(R + 8) \]

### Step 6: Solve for \( R \)
Rearranging the equation:

\[ 2 = 0.05R + 0.4 \]
\[ 2 - 0.4 = 0.05R \]
\[ 1.6 = 0.05R \]
\[ R = \frac{1.6}{0.05} = 32 \, \Omega \]

### Conclusion
The resistance that must be connected in series with the wire is \( R = 32 \, \Omega \).

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