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. `400 Omega`
B. `500 Omega`
C. `510 Omega`
D. `490 Omega`

Explanation

To solve the problem step by step, we can follow these instructions:

### 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 resistance per cm of the potentiometer wire
The resistance of the potentiometer wire can be calculated using the formula:
\[ R_{\text{wire}} = \frac{r_{\text{total}}}{L} \]
Where \( r_{\text{total}} \) is the resistance corresponding to the initial balancing length.

The total resistance for the initial balancing length (500 cm) can be calculated as:
\[ r_{\text{total}} = \frac{r}{l1} \]
\[ r_{\text{total}} = \frac{10 \, \Omega}{500 \, \text{cm}} = \frac{1}{50} \, \Omega/\text{cm} \]

### Step 3: Calculate the total resistance for the new balancing length
Now, we calculate the resistance corresponding to the new balancing length (490 cm):
\[ r_{\text{new}} = \frac{1}{50} \times 490 = \frac{490}{50} = 9.8 \, \Omega \]

### Step 4: Set up the equation for the equivalent resistance
When the voltmeter is connected across the cell, the equivalent resistance \( R_{eq} \) of the voltmeter \( R_v \) and the internal resistance \( r \) of the cell is in parallel:
\[ R_{eq} = \frac{R_v \cdot r}{R_v + r} \]

This equivalent resistance should equal the resistance corresponding to the new balancing length:
\[ R_{eq} = 9.8 \, \Omega \]

### Step 5: Substitute and solve for \( R_v \)
Substituting the values into the equation:
\[ \frac{R_v \cdot 10}{R_v + 10} = 9.8 \]

Cross-multiplying gives:
\[ R_v \cdot 10 = 9.8(R_v + 10) \]
\[ 10R_v = 9.8R_v + 98 \]

Rearranging the equation:
\[ 10R_v - 9.8R_v = 98 \]
\[ 0.2R_v = 98 \]

Now, solving for \( R_v \):
\[ R_v = \frac{98}{0.2} = 490 \, \Omega \]

### Final Answer
The resistance of the voltmeter is \( R_v = 490 \, \Omega \).

---