If an error of 0.02 text{ cm}^2 is found in t | vedclass

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(D) The surface area of a sphere is given by $S = 4 \pi r^2$. Taking the derivative with respect to $r$, we get the differential $\Delta S \approx dS

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$0.1$

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(D) The surface area of a sphere is given by $S = 4 \pi r^2$. Taking the derivative with respect to $r$, we get the differential $\Delta S \approx dS = 8 \pi r \Delta r$. Given $\Delta S = 0.02 	ext{ cm}^2$ and $r = 10 	ext{ cm}$, we substitute these values: $0.02 = 8 \pi (10) \Delta r$. Solving for $\Delta r$, we get $\Delta r = \frac{0.02}{80 \pi} = \frac{0.001}{4 \pi} 	ext{ cm}$. The volume of a sphere is $V = \frac{4}{3} \pi r^3$. The approximate error in volume is $\Delta V \approx dV = 4 \pi r^2 \Delta r$. Substituting $r = 10$ and $\Delta r = \frac{0.001}{4 \pi}$: $\Delta V = 4 \pi (10)^2 	imes \frac{0.001}{4 \pi} = 100 	imes 0.001 = 0.1 	ext{ cm}^3$.

If an error of $0.02 \text{ cm}^2$ is found in the surface area of a sphere when its radius is measured as $10 \text{ cm}$, then the approximate error that occurs in the volume of the sphere, in cubic centimetres, is

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