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(B) Let the radius of the sphere be $r = 4 	ext{ cm}$.The volume $V$ of a sphere is given by $V = \frac{4}{3} \pi r^3$.The surface area $S$ of a sphe

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$2 	ext{ cm}^3 	ext{ cm}^{-2}$

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(B) Let the radius of the sphere be $r = 4 	ext{ cm}$.The volume $V$ of a sphere is given by $V = \frac{4}{3} \pi r^3$.The surface area $S$ of a sphere is given by $S = 4 \pi r^2$.Differentiating $V$ and $S$ with respect to $r$:$\frac{dV}{dr} = 4 \pi r^2$$\frac{dS}{dr} = 8 \pi r$Using the chain rule,the rate of change of volume with respect to surface area is:$\frac{dV}{dS} = \frac{dV/dr}{dS/dr} = \frac{4 \pi r^2}{8 \pi r} = \frac{r}{2}$.At $r = 4 	ext{ cm}$,we have:$\frac{dV}{dS} = \frac{4}{2} = 2 	ext{ cm}^3 	ext{ cm}^{-2}$.

The rate of change of volume of a sphere with respect to its surface area when the radius is $4 \text{ cm}$ is

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