A voltmeter with resistance `500 Omega` is used to measure the emf of a cell of internal resistance `4 Omega`. The percentage error in the reading of the voltmeter will be

Options

A. 0.002
B. 0.008
C. 0.014
D. 0.022

Answer

To solve the problem of finding the percentage error in the reading of a voltmeter used to measure the emf of a cell with internal resistance, we can follow these steps:

### Step 1: Identify the given values
- Resistance of the voltmeter, \( R_v = 500 \, \Omega \)
- Internal resistance of the cell, \( r = 4 \, \Omega \)

### Step 2: Calculate the total resistance in the circuit
The total resistance in the circuit when the voltmeter is connected in parallel with the internal resistance of the cell is:
\[ R_{total} = R_v + r = 500 \, \Omega + 4 \, \Omega = 504 \, \Omega \]

### Step 3: Calculate the current flowing through the circuit
Using Ohm's law, the current \( I \) flowing through the circuit can be calculated as:
\[ I = \frac{E}{R_{total}} = \frac{E}{504 \, \Omega} \]
where \( E \) is the emf of the cell.

### Step 4: Calculate the voltage measured by the voltmeter
The voltage \( V_m \) measured by the voltmeter is given by:
\[ V_m = I \cdot R_v = \left(\frac{E}{504}\right) \cdot 500 \]
Substituting the expression for \( I \):
\[ V_m = \frac{500E}{504} \]

### Step 5: Calculate the error in the measurement
The error in the measurement can be calculated as:
\[ \text{Error} = \text{Actual Value} - \text{Measured Value} = E - V_m \]
Substituting the value of \( V_m \):
\[ \text{Error} = E - \frac{500E}{504} = E\left(1 - \frac{500}{504}\right) = E\left(\frac{504 - 500}{504}\right) = E\left(\frac{4}{504}\right) \]

### Step 6: Calculate the percentage error
The percentage error can be calculated using the formula:
\[ \text{Percentage Error} = \left(\frac{\text{Error}}{\text{Actual Value}}\right) \times 100 = \left(\frac{E \cdot \frac{4}{504}}{E}\right) \times 100 = \left(\frac{4}{504}\right) \times 100 \]
Simplifying this gives:
\[ \text{Percentage Error} = \frac{400}{504} \approx 0.79365 \approx 0.8\% \]

### Final Answer
The percentage error in the reading of the voltmeter is approximately **0.8%**.

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