In an experiment a student measures terminal potential difference (TPD) of a cell. When the same cell is connected to an external resistance of `5 Omega` then the TPD is 2 V and when the same is repeated for an external resistance of `10 Omega` then the TPD is 2.4 V. What is the emf and internal resistance of cell?

Options

A. `2.5V, 3 Omega`
B. `2.5V, 3.5 Omega`
C. `3V, 2.5 Omega`
D. `3V, 3.5 Omega`

Explanation

To solve the problem, we need to find the electromotive force (emf) of the cell and its internal resistance using the given terminal potential differences (TPD) for two different external resistances.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have two scenarios:
- When the external resistance \( R_1 = 5 \, \Omega \), the TPD \( V_1 = 2 \, V \).
- When the external resistance \( R_2 = 10 \, \Omega \), the TPD \( V_2 = 2.4 \, V \).
- We need to find the emf \( E \) and the internal resistance \( r \) of the cell.

2. **Using the Formula for Terminal Potential Difference:**
- The terminal potential difference \( V \) can be expressed as:
\[ V = \frac{E \cdot R}{R + r} \]
- Where:
- \( E \) = emf of the cell
- \( R \) = external resistance
- \( r \) = internal resistance of the cell

3. **Setting Up the Equations:**
- For the first case (with \( R_1 = 5 \, \Omega \)):
\[ V_1 = \frac{E \cdot 5}{5 + r} = 2 \]
Rearranging gives:
\[ 5E = 2(5 + r) \quad \Rightarrow \quad 5E = 10 + 2r \quad \text{(Equation 1)} \]

- For the second case (with \( R_2 = 10 \, \Omega \)):
\[ V_2 = \frac{E \cdot 10}{10 + r} = 2.4 \]
Rearranging gives:
\[ 10E = 2.4(10 + r) \quad \Rightarrow \quad 10E = 24 + 2.4r \quad \text{(Equation 2)} \]

4. **Solving the Equations:**
- We have two equations:
- \( 5E = 10 + 2r \) (1)
- \( 10E = 24 + 2.4r \) (2)

- From Equation (1), we can express \( r \):
\[ r = \frac{5E - 10}{2} \]

- Substitute \( r \) into Equation (2):
\[ 10E = 24 + 2.4\left(\frac{5E - 10}{2}\right) \]
Simplifying:
\[ 10E = 24 + 1.2(5E - 10) \]
\[ 10E = 24 + 6E - 12 \]
\[ 10E - 6E = 12 \quad \Rightarrow \quad 4E = 12 \quad \Rightarrow \quad E = 3 \, V \]

5. **Finding the Internal Resistance \( r \):**
- Substitute \( E = 3 \, V \) back into Equation (1):
\[ 5(3) = 10 + 2r \quad \Rightarrow \quad 15 = 10 + 2r \quad \Rightarrow \quad 2r = 5 \quad \Rightarrow \quad r = 2.5 \, \Omega \]

### Final Answer:
- The emf \( E \) of the cell is \( 3 \, V \).
- The internal resistance \( r \) of the cell is \( 2.5 \, \Omega \).