The correct answer is C because:

As all of the springs have a negligable mass, the force which is causing spring A and B to extend is F N
Consider replacing springs A and B with a single spring which produces the same extension

ma27_bulk-properties-of-solids_aqa-as-physics-mcq

As springs A and B are connected in parallel they will both produce the same extension, ΔL
When a load of F N is applied to the system, both springs will be under tension
- Spring A will have a tension of T_A and spring B will have a tension of T_B
As the system is in equilibrium
- F = T_A + T_B
We can use Hooke’s law, F = kΔL, to calculate the tension in each spring, T_A and T_B
- T_A= k_AΔL
- T_B = k_BΔL
Hence, F = k_AΔL + k_BΔL
- F = (k_A + k_B)ΔL
Therefore if we replaced springs A and B with a single spring with a spring constant of k_A + k_B it would produce the same effect as the two springs in parallel
Now consider combining all three springs:

ma27-2_bulk-properties-of-solids_aqa-as-physics-mcq

As the springs all have a negligible mass, all of the springs will experience a force of F N acting on them
If the single spring is to produce the same effect as the two springs in parallel then the total extension of the two springs must equal the total extension of the single spring
Let ΔL_AB be the extension of springs A and B combined, ΔL_C be the extension of spring C and ΔL_T be the extension of the single spring
- ΔL_T= ΔL_AB + ΔL_C
We can rearrange the Hooke’s law equation to calculate extension of each spring
- ΔL = $\frac{F}{k}$
- Therefore, ΔL_AB = $\frac{F}{k_{A} + k_{B}}$ and ΔL_C = $\frac{F}{k_{C}}$
- Hence ΔL_T = $\frac{F}{k_{A} + k_{B}}$ + $\frac{F}{k_{C}}$
- ΔL_T = F $(\frac{1}{k_{A} + k_{B}} + \frac{1}{k_{C}})$
The common demominator for adding the fractions will be k_C(k_A + k_B)
- ΔL_T = F $(\frac{k_{c}}{k_{C} (k_{A} + k_{B})} + \frac{k_{A} + k_{B}}{k_{C} (k_{A} + k_{B})})$
- ΔL_T= F $(\frac{k_{A} + k_{B} + k_{c}}{k_{C} (k_{A} + k_{B})})$
Rearranging this equation so that it takes the same form as the Hooke’s law equation:
- F = ΔL_{T $\div$ $(\frac{k_{A} + k_{B} + k_{c}}{k_{C} (k_{A} + k_{B})})$}
- F = $(\frac{k_{C} (k_{A} + k_{B})}{k_{A} + k_{B} + k_{c}})$ x ΔL_T
Hence the combined spring constant for all three springs is $\frac{k_{C} (k_{A} + k_{B})}{k_{A} + k_{B} + k_{c}}$

Bulk Properties of Solids (AQA AS Physics)

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