CUET UG 2025 — Mathematics Calculus

The rate of change of volume with respect to surface area is $\frac{dV}{dS}$dSdV​ when radius is 4 cm.

For a sphere with radius $r$r:

Volume: $V = \frac{4}{3}\pi r^3$V=34​πr3

Surface Area: $S = 4\pi r^2$S=4πr2

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Using the chain rule:

$$\frac{dV}{dS} = \frac{dV/dr}{dS/dr}$$dSdV​=dS/drdV/dr​

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$$V = \frac{4}{3}\pi r^3$$V=34​πr3

$$\frac{dV}{dr} = \frac{4}{3}\pi \times 3r^2$$drdV​=34​π×3r2

$$\frac{dV}{dr} = 4\pi r^2$$drdV​=4πr2

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$$S = 4\pi r^2$$S=4πr2

$$\frac{dS}{dr} = 4\pi \times 2r$$drdS​=4π×2r

$$\frac{dS}{dr} = 8\pi r$$drdS​=8πr

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$$\frac{dV}{dS} = \frac{dV/dr}{dS/dr}$$dSdV​=dS/drdV/dr​

$$\frac{dV}{dS} = \frac{4\pi r^2}{8\pi r}$$dSdV​=8πr4πr2​

$$\frac{dV}{dS} = \frac{r}{2}$$dSdV​=2r​

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At $r = 4$r=4 cm:

$$\frac{dV}{dS} = \frac{4}{2}$$dSdV​=24​

$\frac{dV}{dS} = 2$dSdV​=2 cm

Therefore, the rate of change of volume with respect to surface area when radius is 4 cm equals 2 cm.