A potentiometer wire has a length of `10 m` and resistance `4 Omega m^(-1)`. An accumulator of emf `2 V` and a resistance box are connected in series with it. Culculate the resistance to be introduced in the box so as to get a potential gradient of
(a) `0.1 V//m` and (b) `0.1 mVm^(-1)`.

Explanation

To solve the problem step by step, we will calculate the resistance to be introduced in the resistance box for two different potential gradients: (a) 0.1 V/m and (b) 0.1 mV/m.

### Given Data:
- Length of the potentiometer wire, \( L = 10 \, m \)
- Resistance per unit length, \( R' = 4 \, \Omega/m \)
- EMF of the accumulator, \( E = 2 \, V \)

### Step 1: Calculate the total resistance of the potentiometer wire
The total resistance \( R_1 \) of the wire can be calculated using the formula:
\[ R_1 = R' \times L \]
Substituting the values:
\[ R_1 = 4 \, \Omega/m \times 10 \, m = 40 \, \Omega \]

### Step 2: Calculate the current in the circuit
The total resistance in the circuit is the sum of the resistance of the wire and the resistance in the box \( R \):
\[ R_{total} = R + R_1 \]
Using Ohm's law, the current \( I \) in the circuit can be expressed as:
\[ I = \frac{E}{R_{total}} = \frac{E}{R + R_1} \]
Substituting \( E = 2 \, V \) and \( R_1 = 40 \, \Omega \):
\[ I = \frac{2}{R + 40} \]

### Step 3: Calculate the potential gradient
The potential gradient \( K \) is defined as the potential difference per unit length:
\[ K = \frac{V}{L} = \frac{I \cdot R_1}{L} \]
Substituting \( L = 10 \, m \):
\[ K = \frac{I \cdot 40}{10} = 4I \]

### Step 4: Set up the equation for the first case (a) \( K = 0.1 \, V/m \)
Substituting \( K = 0.1 \, V/m \):
\[ 0.1 = 4I \]
Thus,
\[ I = \frac{0.1}{4} = 0.025 \, A \]

### Step 5: Substitute \( I \) back to find \( R \)
Now substituting \( I \) back into the equation for current:
\[ 0.025 = \frac{2}{R + 40} \]
Cross-multiplying gives:
\[ 0.025(R + 40) = 2 \]
Expanding and rearranging:
\[ 0.025R + 1 = 2 \implies 0.025R = 1 \implies R = \frac{1}{0.025} = 40 \, \Omega \]

### Step 6: Set up the equation for the second case (b) \( K = 0.1 \, mV/m \)
Convert \( K \) to volts:
\[ K = 0.1 \, mV/m = 0.1 \times 10^{-3} \, V/m = 0.0001 \, V/m \]
Using the same relationship:
\[ 0.0001 = 4I \implies I = \frac{0.0001}{4} = 0.000025 \, A \]

### Step 7: Substitute \( I \) back to find \( R \) for the second case
Substituting \( I \) back into the equation for current:
\[ 0.000025 = \frac{2}{R + 40} \]
Cross-multiplying gives:
\[ 0.000025(R + 40) = 2 \]
Expanding and rearranging:
\[ 0.000025R + 0.001 = 2 \implies 0.000025R = 2 - 0.001 = 1.999 \implies R = \frac{1.999}{0.000025} = 79960 \, \Omega \]

### Final Answers:
(a) The resistance to be introduced in the box for a potential gradient of \( 0.1 \, V/m \) is \( 40 \, \Omega \).

(b) The resistance to be introduced in the box for a potential gradient of \( 0.1 \, mV/m \) is \( 79960 \, \Omega \).