A voltmeter reads the potential difference across the terminals of an old battery as `1.40 V`, while a potentiometer reads its voltage to be `1.55 V`. The voltmeter resistance is `280 Omega`.

Options

A. The emf of the battery is `1.4 V`.
B. The emf of the battery is `1.55 V`.
C. The internal resistance `r` of the battery is `30 Omega`
D. The internal resistance `r` of the battery is `5 Omega`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the given values
- The voltmeter reading (V) = 1.40 V
- The potentiometer reading (E) = 1.55 V
- The resistance of the voltmeter (R_v) = 280 Ω

### Step 2: Understand the relationship between EMF, terminal voltage, and internal resistance
The relationship can be expressed as:
\[ V = E - I \cdot r \]
Where:
- \( V \) = terminal voltage (voltmeter reading)
- \( E \) = EMF of the battery (potentiometer reading)
- \( I \) = current flowing through the circuit
- \( r \) = internal resistance of the battery

### Step 3: Calculate the current (I) using the voltmeter reading
Using Ohm's law:
\[ I = \frac{V}{R_v} \]
Substituting the values:
\[ I = \frac{1.40 \, \text{V}}{280 \, \Omega} \]
\[ I = 0.005 \, \text{A} \]

### Step 4: Use the relationship to find the internal resistance (r)
Rearranging the equation from Step 2 gives:
\[ r = \frac{E - V}{I} \]
Substituting the known values:
\[ r = \frac{1.55 \, \text{V} - 1.40 \, \text{V}}{0.005 \, \text{A}} \]
\[ r = \frac{0.15 \, \text{V}}{0.005 \, \text{A}} \]
\[ r = 30 \, \Omega \]

### Step 5: Conclusion
The internal resistance of the battery is \( 30 \, \Omega \) and the EMF of the battery is \( 1.55 \, V \).

### Final Results
- EMF of the battery, \( E = 1.55 \, V \)
- Internal resistance, \( r = 30 \, \Omega \)