A

To find the length AJ such that the galvanometer shows no deflection, we will first set the potential differences across the potentiometer wire and the cell C equal to each other as they balance each other.

1. **Given Information**:
- E.M.F. of cell D = ε
- Internal resistance of cell D = r
- E.M.F. of cell C = ε/2
- Internal resistance of cell C = 2r
- Resistance of potentiometer wire AB = 9r
- Total length of potentiometer wire = L

2. **Potential Difference Across Potentiometer Wire**:
The potential difference across the potentiometer wire of length AJ is given by:
\[ V = I \times R = I \times \frac{9r \cdot AJ}{L} \] Where I is the current through the potentiometer wire, \( AJ = \text{length between point A and J} \).

3. **Total Current Through Cell D**:
The total current supplied by cell D is given by:
\[ I = \frac{E_{cell}}{R + r} = \frac{\epsilon}{9r + r} = \frac{\epsilon}{10r} \]
4. **Potential Drop Across Potentiometer Wire**:
Therefore, the potential drop in length AJ is:
\[ V_{AJ} = I \times \frac{9r \cdot AJ}{L} = \frac{\epsilon}{10r} \times \frac{9r \cdot AJ}{L} = \frac{9\epsilon\,AJ}{10L} \]

5. **Potential Drop Across Cell C**:
The potential drop across C is given by:
\[ V_C = I_C imes R_C = \frac{\epsilon/2}{2r} = \frac{\epsilon}{4r} \]
6. **Setting Equal for No Deflection**:
We set these two equations equal for the state of no deflection:
\[ \frac{9\epsilon AJ}{10L} = \frac{\epsilon}{4r} \]
7. **Solving for AJ**:
Cancel ε from both sides:
\[ \frac{9AJ}{10L} = \frac{1}{4} \]
Rearranging gives:
\[ AJ = \frac{10L}{9} \cdot \frac{1}{4} = \frac{10L}{36} = \frac{5L}{18} \]

Therefore, the length AJ is \( \frac{5L}{18} \).

Hence, we have found the length AJ where the galvanometer shows no deflection.