To the potentiometer wire of L and `10Omega` resistance, a battery of emf 2.5 volts and a resistance R are connected in series. If a potential difference of 1 volt is balanced across L/2 length, the value of R is `Omega` will be

Options

A. `30 m//s`
B. `20 m//s`
C. `10 m//s`
D. `5 m//s`

Explanation

To solve the problem, we need to analyze the given information and apply the principles of a potentiometer circuit.

### Step 1: Understand the Circuit
We have a potentiometer wire of length \( L \) and a resistance of \( 10 \, \Omega \). A battery with an emf of \( 2.5 \, \text{V} \) is connected in series with a resistance \( R \).

### Step 2: Determine the Total Resistance
The total resistance in the circuit is the sum of the resistance of the potentiometer wire (which we can consider negligible for the purpose of this calculation) and the fixed resistor:
\[ R_{\text{total}} = R + 10 \, \Omega \]

### Step 3: Calculate the Current in the Circuit
Using Ohm's Law, the current \( I \) flowing through the circuit can be calculated using the formula:
\[ I = \frac{V}{R_{\text{total}}} \]
Substituting the values, we have:
\[ I = \frac{2.5 \, \text{V}}{R + 10} \]

### Step 4: Analyze the Potentiometer Wire
The potential difference (voltage) across a length \( L/2 \) of the potentiometer wire is balanced at \( 1 \, \text{V} \). The potential difference across the entire length \( L \) can be expressed as:
\[ V_L = I \cdot L \]
Since \( V_L \) is the total voltage across the potentiometer wire, we can express the voltage across \( L/2 \) as:
\[ V_{L/2} = I \cdot \frac{L}{2} \]
Setting \( V_{L/2} = 1 \, \text{V} \):
\[ 1 = I \cdot \frac{L}{2} \]

### Step 5: Substitute the Expression for Current
Substituting the expression for \( I \) from Step 3 into the equation from Step 4:
\[ 1 = \left(\frac{2.5}{R + 10}\right) \cdot \frac{L}{2} \]

### Step 6: Rearranging the Equation
Rearranging the equation to solve for \( R \):
\[ 1 = \frac{2.5L}{2(R + 10)} \]
\[ 2(R + 10) = 2.5L \]
\[ R + 10 = 1.25L \]
\[ R = 1.25L - 10 \]

### Step 7: Conclusion
The value of \( R \) can be expressed in terms of \( L \):
\[ R = 1.25L - 10 \, \Omega \]