The emf of a cell is 6 V and internal resistance is `0.5 kOmega`. What will be the reading (in volt) of a voltmeter having an internal resistance of `2.5kOmega`?

Answer

To solve the problem, we need to determine the reading of a voltmeter connected across a cell with a given EMF and internal resistance. Here's a step-by-step solution:

### Step-by-Step Solution:

1. **Identify Given Values:**
- EMF of the cell (E) = 6 V
- Internal resistance of the cell (r) = 0.5 kΩ = 500 Ω
- Internal resistance of the voltmeter (R_v) = 2.5 kΩ = 2500 Ω

2. **Calculate Total Resistance in the Circuit:**
The voltmeter is connected in parallel with the internal resistance of the cell. The total resistance (R_total) can be calculated using the formula for resistors in parallel:
\[ \frac{1}{R_{total}} = \frac{1}{r} + \frac{1}{R_v} \]
Substituting the values:
\[ \frac{1}{R_{total}} = \frac{1}{500} + \frac{1}{2500} \]
Finding a common denominator (2500):
\[ \frac{1}{R_{total}} = \frac{5}{2500} + \frac{1}{2500} = \frac{6}{2500} \]
Therefore,
\[ R_{total} = \frac{2500}{6} \approx 416.67 \, \Omega \]

3. **Calculate the Current Flowing in the Circuit:**
Using Ohm's Law, the total current (I) flowing through the circuit can be calculated as:
\[ I = \frac{E}{R_{total} + r} \]
Substituting the values:
\[ I = \frac{6}{416.67 + 500} = \frac{6}{916.67} \approx 0.00654 \, A \, \text{(or 6.54 mA)} \]

4. **Calculate the Voltage Across the Voltmeter:**
The voltage (V) across the voltmeter can be calculated using Ohm's Law:
\[ V = I \times R_v \]
Substituting the values:
\[ V = 0.00654 \times 2500 \approx 16.35 \, V \]

5. **Adjust for the Voltage Drop Across the Internal Resistance of the Cell:**
The voltage drop across the internal resistance (r) is given by:
\[ V_{drop} = I \times r = 0.00654 \times 500 \approx 3.27 \, V \]
Therefore, the voltage reading on the voltmeter will be:
\[ V_{reading} = E - V_{drop} = 6 - 3.27 \approx 2.73 \, V \]

### Final Answer:
The reading of the voltmeter will be approximately **2.73 V**.