Step 1 (Formula / Governing Law & Given Data): [1/4 Mark]

The volume of a sphere $V$ in terms of its radius $r$ is given by:

$$V = \frac{4}{3}\pi r^3$$

Since the diameter $D = 2r$, we have $r = \frac{D}{2}$. Substituting this into the volume formula:

$$V = \frac{4}{3}\pi \left( \frac{D}{2} \right)^3 = \frac{4}{3}\pi \left( \frac{D^3}{8} \right) = \frac{\pi D^3}{6}$$

Given: Radius $r = 5 \text{ cm}$, which implies Diameter $D = 2r = 10 \text{ cm}$.

Step 2 (Step-by-Step Mathematical Working): [1/2 Mark]

We need to find the rate of change of volume with respect to the diameter, which is $\frac{dV}{dD}$. Differentiating $V = \frac{\pi D^3}{6}$ with respect to $D$:

$$\frac{dV}{dD} = \frac{d}{dD} \left( \frac{\pi D^3}{6} \right)$$ $$\frac{dV}{dD} = \frac{\pi}{6} \cdot (3D^2) = \frac{\pi D^2}{2}$$

Now, substitute $D = 10 \text{ cm}$ into the derivative:

$$\left. \frac{dV}{dD} \right|_{D=10} = \frac{\pi (10)^2}{2} = \frac{100\pi}{2} = 50\pi$$

Step 3 (Final Concluded Answer): [1/4 Mark]

The rate of change of volume with respect to the diameter is $50\pi \text{ cm}^3/\text{cm}$.

Correct Option: (C) $50\pi \text{ cm}^3/\text{cm}$

💡 EL10 Concept Insight: Think of it like this: if you grow a balloon, its volume increases based on how much you stretch its diameter; by turning the volume formula into a "diameter-only" version first, we can easily calculate exactly how much "space" is added for every tiny bit we increase the width.