An ideal battery of emf 2 V and a series resistance R are connected in the primary circuit of a potentiometer of length 1 m and resistance `5Omega` The value of R, to give a potential difference of 5 m V across 10 cm of potentiometer wire is

Options

A. `180Omega`
B. `190Omega`
C. `195Omega`
D. `200Omega`

Answer

To solve the problem, we need to find the value of the series resistance \( R \) that will give a potential difference of 5 mV across 10 cm of the potentiometer wire. Here are the steps to find the solution:

### Step 1: Understand the given values
- EMF of the battery, \( E = 2 \, \text{V} \)
- Total resistance of the potentiometer wire (1 m), \( R_{wire} = 5 \, \Omega \)
- Length of the potentiometer wire, \( L = 1 \, \text{m} \)
- Potential difference across 10 cm of wire, \( V = 5 \, \text{mV} = 5 \times 10^{-3} \, \text{V} \)
- Length of the wire segment for which we want the potential difference, \( l = 10 \, \text{cm} = 0.1 \, \text{m} \)

### Step 2: Calculate the resistance of the 10 cm segment
The resistance of the 10 cm segment of the potentiometer wire can be calculated using the formula:
\[ R_{segment} = R_{wire} \times \frac{l}{L} \]
Substituting the values:
\[ R_{segment} = 5 \, \Omega \times \frac{0.1 \, \text{m}}{1 \, \text{m}} = 0.5 \, \Omega \]

### Step 3: Use Ohm's Law to find the current
Using Ohm's Law \( V = I \times R \), we can express the current \( I \) flowing through the circuit:
\[ I = \frac{V}{R_{segment}} = \frac{5 \times 10^{-3} \, \text{V}}{0.5 \, \Omega} = 0.01 \, \text{A} = 10 \, \text{mA} \]

### Step 4: Relate the current to the total resistance in the circuit
The total resistance in the circuit is the sum of the series resistance \( R \) and the resistance of the potentiometer wire:
\[ R_{total} = R + R_{wire} = R + 5 \, \Omega \]
Using the EMF of the battery, we can express the current as:
\[ I = \frac{E}{R_{total}} = \frac{2 \, \text{V}}{R + 5} \]

### Step 5: Set the two expressions for current equal to each other
Equating the two expressions for current:
\[ \frac{2}{R + 5} = 0.01 \]

### Step 6: Solve for \( R \)
Cross-multiplying gives:
\[ 2 = 0.01(R + 5) \]
Expanding this we get:
\[ 2 = 0.01R + 0.05 \]
Subtracting 0.05 from both sides:
\[ 1.95 = 0.01R \]
Now, solving for \( R \):
\[ R = \frac{1.95}{0.01} = 195 \, \Omega \]

### Final Answer
The value of the series resistance \( R \) is \( 195 \, \Omega \).

---