1. Key idea: condition for same null point
   In a potentiometer/meter bridge wire, the null point depends on the potential gradient along the wire.
   If the null point remains the same for two different cells, then the potential gradient along the wire must remain the same in both cases.

2. Resistance of potentiometer wire
   The resistance of wire $AB$ is given as $R_{AB}=4\,\Omega$ The rheostat is in series with this wire.

So total series resistance is: $R_{\text{total}}=R_{AB}+R_h$

3. Current in first case
   For the first cell, $\varepsilon_1=0.5\,\text{V}, \qquad R_h=2\,\Omega$ Hence total resistance is $R_1=4+2=6\,\Omega$ So current through the potentiometer wire is $I_1=\frac{\varepsilon_1}{R_1}=\frac{0.5}{6}$

4. Current in second case
   For the second cell, $\varepsilon_2=? , \qquad R_h=6\,\Omega$ Hence total resistance is $R_2=4+6=10\,\Omega$ So current is $I_2=\frac{\varepsilon_2}{10}$

5. Same null point implies same potential gradient
   Potential gradient along the wire is proportional to current through the wire.
   Since the same null point is obtained, $I_1=I_2$ Therefore, $\frac{0.5}{6}=\frac{\varepsilon_2}{10}$

6. Solve for $\varepsilon_2$ $\varepsilon_2=10\cdot \frac{0.5}{6}=\frac{5}{6}\approx 0.833\,\text{V}$

Thus, $\boxed{\varepsilon_2\approx 0.83\,\text{V}}$

7. Compare with options
   Given options are:

• A: $0.3$ V
• B: $0.6$ V
• C: $0.5$ V
• D: $0.4$ V

None of these matches $0.83$ V.

So the given stored answer $0.3$ V is not consistent with the stated data.

8. Likely issue
   Either the figure contains additional resistances not described in the text, or there is a typo in the problem statement/options. Based on the information provided, the correct value should be $\boxed{0.83\,\text{V}}$