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`

Answer

To solve the problem, we need to determine the resistance that must be connected in series with a potentiometer wire to achieve a specific potential gradient. Let's break down the solution step by step.

### Step 1: Understand the Problem
We have a potentiometer wire of length \( L = 4 \, \text{m} \) (which is \( 400 \, \text{cm} \)) and resistance \( R_w = 8 \, \Omega \). We need to achieve a potential gradient of \( 1 \, \text{mV/cm} \) using an accumulator with an EMF of \( 2 \, \text{V} \).

### Step 2: Calculate the Total Voltage Drop Required
The potential gradient is given as \( 1 \, \text{mV/cm} \). To find the total voltage drop (\( \Delta V \)) across the entire length of the wire, we multiply the potential gradient by the length of the wire in cm:

\[ \Delta V = \text{Potential Gradient} \times \text{Length} = 1 \, \text{mV/cm} \times 400 \, \text{cm} = 400 \, \text{mV} = 0.4 \, \text{V} \]

### Step 3: Apply Ohm's Law
Using Ohm's Law, we know that the voltage drop across a resistor is given by:

\[ \Delta V = I \times R_{\text{total}} \]

Where \( R_{\text{total}} \) is the total resistance in the circuit, which is the sum of the resistance of the wire and the series resistance \( R_s \):

\[ R_{\text{total}} = R_w + R_s = 8 \, \Omega + R_s \]

### Step 4: Calculate the Current
The current \( I \) can be expressed in terms of the EMF (\( V \)) and the total resistance:

\[ I = \frac{V}{R_{\text{total}}} = \frac{2 \, \text{V}}{8 + R_s} \]

### Step 5: Substitute Current into Voltage Equation
Substituting the expression for current into the voltage equation gives:

\[ 0.4 = \left(\frac{2}{8 + R_s}\right) \times (8) \]

### Step 6: Solve for \( R_s \)
Now we can rearrange the equation to solve for \( R_s \):

\[ 0.4 = \frac{16}{8 + R_s} \]

Cross-multiplying gives:

\[ 0.4(8 + R_s) = 16 \]

Expanding this:

\[ 3.2 + 0.4R_s = 16 \]

Subtracting \( 3.2 \) from both sides:

\[ 0.4R_s = 12.8 \]

Dividing by \( 0.4 \):

\[ R_s = \frac{12.8}{0.4} = 32 \, \Omega \]

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

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