1. Let the common terminal voltage across the parallel combination and the load be $V$.

   Since both batteries are connected in parallel across the same load resistor, the voltage across each branch and across the load is the same, equal to $V$.

2. Write current supplied by each battery.

   For a battery of emf $E$ and internal resistance $r$, the current delivered at terminal voltage $V$ is $I = \frac{E - V}{r}.$

   So for the two batteries:

   • First battery: $E_1=12\,\text{V},\ r_1=1\,\Omega$ $I_1 = \frac{12 - V}{1} = 12 - V$
   • Second battery: $E_2=13\,\text{V},\ r_2=2\,\Omega$ $I_2 = \frac{13 - V}{2}$

3. Write current through the load resistor.

   Load resistance is $R=10\,\Omega$, so $I_L = \frac{V}{10}.$

4. Apply Kirchhoff's current law.

   Total current supplied by the two batteries equals the load current: $I_1 + I_2 = I_L.$

   Therefore, $\left(12 - V\right) + \frac{13 - V}{2} = \frac{V}{10}.$

5. Solve the equation.

   Multiply throughout by $10$: $10(12 - V) + 5(13 - V) = V.$

   $120 - 10V + 65 - 5V = V$ $185 - 15V = V$ $185 = 16V$ $V = \frac{185}{16} = 11.5625\,\text{V}.$

6. Identify the correct interval.

   $11.5 < 11.5625 < 11.6$

   So the voltage across the load lies between $11.5\,\text{V}$ and $11.6\,\text{V}$.

7. Check options.

   • A: $11.7$ to $11.8$ ✗
   • B: $11.6$ to $11.7$ ✗
   • C: $11.5$ to $11.6$ ✓
   • D: $11.4$ to $11.5$ ✗

Therefore, the correct option is C.