A potentiometer wire has length 8m and resistance `16Omega` . The resistance that must be conneted in series with the wire and an accumulator of emf 4V, so as to get a potential gradient of 1mV per cm on the wire is

Options

A. `48Omega`
B. `16Omega`
C. `32Omega`
D. `64Omega`

Explanation

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

### Step-by-Step Solution:

1. **Understand the Given Values:**
- Length of the potentiometer wire (L) = 8 m
- Resistance of the potentiometer wire (R_p) = 16 Ω
- EMF of the accumulator (E) = 4 V
- Desired potential gradient (k) = 1 mV/cm = 0.001 V/cm

2. **Convert the Potential Gradient to Standard Units:**
- Since 1 cm = 0.01 m, we convert the potential gradient to volts per meter:
\[ k = 1 \, \text{mV/cm} = 0.001 \, \text{V/cm} \times 100 \, \text{cm/m} = 0.1 \, \text{V/m} \]

3. **Calculate the Total Resistance Required:**
- The potential gradient \( k \) is defined as:
\[ k = \frac{V_{AB}}{L} \]
- Rearranging gives:
\[ V_{AB} = k \times L = 0.1 \, \text{V/m} \times 8 \, \text{m} = 0.8 \, \text{V} \]

4. **Relate Voltage Drop to Current and Resistance:**
- The voltage drop across the potentiometer wire can also be expressed using Ohm's law:
\[ V_{AB} = I \times R_p \]
- Substituting for \( V_{AB} \):
\[ 0.8 \, \text{V} = I \times 16 \, \Omega \]
- Solving for current \( I \):
\[ I = \frac{0.8 \, \text{V}}{16 \, \Omega} = 0.05 \, \text{A} \]

5. **Calculate Total Resistance in the Circuit:**
- The total resistance \( R_{total} \) in the circuit is the sum of the potentiometer resistance and the series resistance \( R \):
\[ R_{total} = R + R_p = R + 16 \, \Omega \]

6. **Apply Ohm's Law to Find the Series Resistance:**
- The total current \( I \) can also be expressed using the total EMF and total resistance:
\[ I = \frac{E}{R_{total}} = \frac{4 \, \text{V}}{R + 16 \, \Omega} \]
- Setting the two expressions for current equal gives:
\[ 0.05 = \frac{4}{R + 16} \]

7. **Solve for R:**
- Rearranging the equation:
\[ 0.05(R + 16) = 4 \]
- Expanding and solving for \( R \):
\[ 0.05R + 0.8 = 4 \]
\[ 0.05R = 4 - 0.8 = 3.2 \]
\[ R = \frac{3.2}{0.05} = 64 \, \Omega \]

### Final Answer:
The resistance that must be connected in series with the wire is **64 Ω**.

---