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 ast `500 cm`. On connecting a voltmeter acros the cell, the balancing length is decreased by `10 cm`
The voltmeter reading will be

Options

A. `1.96V`
B. `1.8V`
C. `1.64V`
D. `0.96V`

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Understand the given data
- Length of the potentiometer wire, \( L = 600 \, \text{cm} \)
- Current through the potentiometer wire, \( I = 40 \, \text{mA} = 40 \times 10^{-3} \, \text{A} \)
- EMF of the cell, \( E = 2 \, \text{V} \)
- Internal resistance of the cell, \( r = 10 \, \Omega \)
- Initial balancing length, \( l_1 = 500 \, \text{cm} \)
- New balancing length after connecting the voltmeter, \( l_2 = 500 \, \text{cm} - 10 \, \text{cm} = 490 \, \text{cm} \)

### Step 2: Calculate the potential drop per unit length of the potentiometer
The potential drop across the length of the potentiometer wire can be calculated using the formula:
\[ V = \text{Potential drop per unit length} \times \text{Length} \]
At the initial balancing length:
\[ E = V = \left( \frac{V}{L} \right) \times l_1 \]
Where \( V \) is the potential drop across the potentiometer wire.

Rearranging gives:
\[ \text{Potential drop per unit length} = \frac{E}{l_1} = \frac{2 \, \text{V}}{500 \, \text{cm}} = \frac{2}{500} \, \text{V/cm} = 0.004 \, \text{V/cm} \]

### Step 3: Calculate the potential difference across the new balancing length
Now we can find the potential drop across the new balancing length \( l_2 = 490 \, \text{cm} \):
\[ V_{AB} = \text{Potential drop per unit length} \times l_2 = 0.004 \, \text{V/cm} \times 490 \, \text{cm} = 1.96 \, \text{V} \]

### Step 4: Determine the current in the circuit with the voltmeter connected
When the voltmeter is connected, the effective voltage across the internal resistance and the voltmeter can be calculated. The total resistance in the circuit is:
\[ R_{total} = r + R_V \]
Where \( R_V \) is the resistance of the voltmeter.

Using Ohm's law, the current \( I \) in the circuit can be expressed as:
\[ I = \frac{E}{r + R_V} \]

### Step 5: Relate the voltage reading of the voltmeter to the potential drop
The potential difference across the voltmeter can be expressed as:
\[ V_{AB} = I \times R_V \]
Substituting \( I \) gives:
\[ V_{AB} = \frac{E}{r + R_V} \times R_V \]

### Step 6: Conclusion
Since we already calculated \( V_{AB} = 1.96 \, \text{V} \), this is the reading of the voltmeter when connected across the cell.

### Final Answer
The reading of the voltmeter will be \( 1.96 \, \text{V} \).

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