In an experiment for calibration of voltmeter, a standard cell of emf 1.5V is balanced at 300cm length of potentiometer wire. The P.D across a resistance in the circuit is balancedat 1.25m. If a voltmeter is connected across the same resistance. It reads 0.65V. The errorin the volt meter is

Options

A. 0.05V
B. 0.025V
C. 0.5V
D. 0.25V

Answer

To solve the problem step by step, we will follow the outlined procedure to find the error in the voltmeter reading.

### Step 1: Understand the relationship between emf and balancing length
In a potentiometer, the emf (E) is directly proportional to the balancing length (L) of the wire. Therefore, we can express this relationship as:
\[ \frac{E_2}{E_1} = \frac{L_2}{L_1} \]
where:
- \(E_1\) is the emf of the standard cell (1.5 V),
- \(L_1\) is the balancing length for the standard cell (300 cm or 3 m),
- \(E_2\) is the emf corresponding to the balancing length \(L_2\) (1.25 m).

### Step 2: Calculate the value of \(E_2\)
Using the relationship established in Step 1, we can rearrange the formula to solve for \(E_2\):
\[ E_2 = E_1 \times \frac{L_2}{L_1} \]

Substituting the known values:
- \(E_1 = 1.5 \, \text{V}\)
- \(L_2 = 1.25 \, \text{m}\)
- \(L_1 = 3 \, \text{m}\)

We find:
\[ E_2 = 1.5 \times \frac{1.25}{3} \]
Calculating this gives:
\[ E_2 = 1.5 \times 0.4167 = 0.625 \, \text{V} \]

### Step 3: Find the error in the voltmeter reading
The voltmeter reads \(0.65 \, \text{V}\). The error (\(\Delta E\)) can be calculated as:
\[ \Delta E = E_{\text{measured}} - E_2 \]
Substituting the values:
\[ \Delta E = 0.65 - 0.625 \]
Calculating this gives:
\[ \Delta E = 0.025 \, \text{V} \]

### Conclusion
The error in the voltmeter is \(0.025 \, \text{V}\).

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