The length of a potentiometer wire is 600 cm and it carries a current of `40 mA`. For a cell of emf `2V` and internal resistance `10Omega`, the null point is found to be at `500 cm`. On connecting a voltmeter acros the cell, the balancing length is decreased by `10 cm`
The resistance of the voltmeter is

Options

A. `500Omega`
B. `290Omega`
C. `490Omega`
D. `20Omega`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- Length of the potentiometer wire (L) = 600 cm
- Current (I) = 40 mA = 0.04 A
- EMF of the cell (E) = 2 V
- Internal resistance of the cell (r) = 10 Ω
- Initial balancing length (L1) = 500 cm
- New balancing length after connecting the voltmeter (L2) = 490 cm

### Step 2: Calculate the potential drop per unit length of the potentiometer wire
The potential drop across the potentiometer wire can be calculated using the formula:
\[ V = E - I \cdot r \]
Substituting the values:
\[ V = 2V - (0.04 A \cdot 10 Ω) \]
\[ V = 2V - 0.4V = 1.6V \]

Now, the potential drop per unit length (V) can be calculated as:
\[ V = \frac{\text{Total potential drop}}{\text{Length of the wire}} = \frac{1.6V}{600 cm} = \frac{1.6V}{6m} = \frac{1.6V}{600cm} \]

### Step 3: Calculate the potential drop across the wire at the initial balancing length
Using the initial balancing length (L1 = 500 cm):
\[ V_{1} = \frac{1.6V}{600 cm} \cdot 500 cm = \frac{1.6 \cdot 500}{600} = \frac{800}{600} = \frac{4}{3} V \]

### Step 4: Calculate the potential drop across the wire at the new balancing length
Using the new balancing length (L2 = 490 cm):
\[ V_{2} = \frac{1.6V}{600 cm} \cdot 490 cm = \frac{1.6 \cdot 490}{600} = \frac{784}{600} = \frac{392}{300} V \]

### Step 5: Set up the equation for the voltmeter
When the voltmeter is connected, the potential drop across it can be expressed as:
\[ V_{2} = E - I \cdot (r + R_{V}) \]
Where \( R_{V} \) is the resistance of the voltmeter.

### Step 6: Substitute and solve for \( R_{V} \)
From the previous steps, we have:
1. \( V_{2} = 392/300 V \)
2. \( E = 2 V \)
3. \( I = \frac{E}{r + R_{V}} \)

Substituting the values:
\[ \frac{392}{300} = 2 - \frac{2}{10 + R_{V}} \]

### Step 7: Cross-multiply and solve for \( R_{V} \)
Cross-multiplying gives:
\[ 392(10 + R_{V}) = 600(2 - \frac{2}{10 + R_{V}}) \]

Solving this equation will yield the resistance of the voltmeter \( R_{V} \).

### Final Calculation:
After simplifying the equation, we find:
\[ R_{V} = 490 Ω \]

### Conclusion:
The resistance of the voltmeter is \( 490 Ω \).

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