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 determine the value of the resistance \( R \) that will give a potential difference of 5 mV across 10 cm of a potentiometer wire. Here’s how we can approach this step by step:

### Step 1: Understand the relationship between potential difference, length, and total length of the potentiometer wire.
The total length of the potentiometer wire is given as 1 m (or 100 cm). We need to find the potential difference across 10 cm of this wire, which is given as 5 mV.

### Step 2: Calculate the potential difference per unit length of the potentiometer wire.
The potential difference across the entire length of the potentiometer wire can be calculated using the formula:
\[ V = \frac{V_{0} \times l}{L} \]
where:
- \( V \) is the potential difference across the length \( l \),
- \( V_{0} \) is the total voltage across the potentiometer wire,
- \( l \) is the length of the wire for which we want to find the potential difference (10 cm),
- \( L \) is the total length of the wire (100 cm).

Given that \( V \) is 5 mV (or \( 5 \times 10^{-3} \) V), we can rearrange the equation to find \( V_{0} \):
\[ 5 \times 10^{-3} = V_{0} \times \frac{10}{100} \]
\[ V_{0} = 5 \times 10^{-3} \times 10 = 5 \times 10^{-2} \text{ V} = 0.05 \text{ V} \]

### Step 3: Use Ohm's Law to find the current in the circuit.
The total resistance in the circuit is the sum of the series resistance \( R \) and the resistance of the potentiometer wire (5 ohms). The current \( I \) through the circuit can be expressed using Ohm's Law:
\[ I = \frac{V_{battery}}{R + R_{pot}} \]
where:
- \( V_{battery} = 2 \text{ V} \),
- \( R_{pot} = 5 \text{ ohm} \).

Thus,
\[ I = \frac{2}{R + 5} \]

### Step 4: Relate the potential difference across the potentiometer wire to the current and resistance.
The potential difference across the potentiometer wire can also be expressed as:
\[ V = I \times R_{pot} \]
Substituting the expression for \( I \):
\[ 0.05 = \left(\frac{2}{R + 5}\right) \times 5 \]

### Step 5: Solve for \( R \).
Rearranging the equation gives:
\[ 0.05 = \frac{10}{R + 5} \]
Cross-multiplying yields:
\[ 0.05(R + 5) = 10 \]
Expanding and simplifying:
\[ 0.05R + 0.25 = 10 \]
\[ 0.05R = 10 - 0.25 \]
\[ 0.05R = 9.75 \]
\[ R = \frac{9.75}{0.05} = 195 \text{ ohms} \]

### Final Answer:
The value of \( R \) that will give a potential difference of 5 mV across 10 cm of the potentiometer wire is \( R = 195 \text{ ohms} \).

---