Two cells having an internal resistance of `0.2 Omega` and `0.4 Omega` are connected in parallel, the voltage across the battery is 1.5 V. If the emf of one cell is 1.2 V, then the emf of second cell is

Options

A. 2.1 V
B. 2.7 V
C. 3 V
D. 4.2 V

Explanation

To solve the problem, we need to find the electromotive force (emf) of the second cell when two cells with internal resistances of \(0.2 \, \Omega\) and \(0.4 \, \Omega\) are connected in parallel, with a voltage across the battery of \(1.5 \, V\) and the emf of the first cell being \(1.2 \, V\).

### Step-by-Step Solution:

1. **Identify the known values:**
- Internal resistance of the first cell, \(R_1 = 0.2 \, \Omega\)
- Internal resistance of the second cell, \(R_2 = 0.4 \, \Omega\)
- Voltage across the battery, \(V = 1.5 \, V\)
- Emf of the first cell, \(E_1 = 1.2 \, V\)
- Emf of the second cell, \(E_2 = ?\)

2. **Use the formula for voltage in parallel circuits:**
The total voltage across the battery in a parallel circuit can be expressed as:
\[ V = \frac{E_1 R_2 + E_2 R_1}{R_1 + R_2} \]

3. **Substitute the known values into the equation:**
\[ 1.5 = \frac{1.2 \times 0.4 + E_2 \times 0.2}{0.2 + 0.4} \]

4. **Calculate the denominator:**
\[ 0.2 + 0.4 = 0.6 \]

5. **Rewrite the equation:**
\[ 1.5 = \frac{0.48 + 0.2 E_2}{0.6} \]

6. **Multiply both sides by \(0.6\) to eliminate the denominator:**
\[ 1.5 \times 0.6 = 0.48 + 0.2 E_2 \]
\[ 0.9 = 0.48 + 0.2 E_2 \]

7. **Rearrange the equation to isolate \(E_2\):**
\[ 0.2 E_2 = 0.9 - 0.48 \]
\[ 0.2 E_2 = 0.42 \]

8. **Solve for \(E_2\):**
\[ E_2 = \frac{0.42}{0.2} = 2.1 \, V \]

### Final Answer:
The emf of the second cell is \(E_2 = 2.1 \, V\).