A 2 volt battery, a 15 `Omega` resistor and a potentiometer of 100 cm length, all are connected in series. If the resistance of potentiometer wire is 5 `Omega`, then the potential gradient of the potentiometer wire is:

Options

A. `0.005 V//cm`
B. `0.05V//cm`
C. `0.02 V//cm`
D. `0.2 V//cm`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the total resistance in the circuit.
The total resistance in the circuit is the sum of the resistance of the 15 Ω resistor and the resistance of the potentiometer wire (5 Ω).

\[ R_{\text{total}} = R_{\text{resistor}} + R_{\text{potentiometer}} = 15 \, \Omega + 5 \, \Omega = 20 \, \Omega \]

### Step 2: Calculate the current flowing through the circuit.
Using Ohm's law, the current (I) can be calculated using the formula:

\[ I = \frac{V}{R_{\text{total}}} \]

Where:
- \( V = 2 \, \text{volts} \)
- \( R_{\text{total}} = 20 \, \Omega \)

Substituting the values:

\[ I = \frac{2 \, \text{V}}{20 \, \Omega} = 0.1 \, \text{A} \, \text{(or 1/10 A)} \]

### Step 3: Calculate the potential difference across the potentiometer wire.
The potential difference (V) across the potentiometer wire can be calculated using Ohm's law again:

\[ V = I \times R_{\text{potentiometer}} \]

Where:
- \( I = 0.1 \, \text{A} \)
- \( R_{\text{potentiometer}} = 5 \, \Omega \)

Substituting the values:

\[ V = 0.1 \, \text{A} \times 5 \, \Omega = 0.5 \, \text{V} \]

### Step 4: Calculate the potential gradient of the potentiometer wire.
The potential gradient (k) is defined as the potential difference per unit length of the wire:

\[ k = \frac{V}{L} \]

Where:
- \( V = 0.5 \, \text{V} \)
- \( L = 100 \, \text{cm} \)

Substituting the values:

\[ k = \frac{0.5 \, \text{V}}{100 \, \text{cm}} = 0.005 \, \text{V/cm} \]

### Final Answer:
The potential gradient of the potentiometer wire is \( 0.005 \, \text{V/cm} \).

---