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If the radius of a sphere is measured as $9 \mathrm{~m}$ with an error of $0.03 \mathrm{~m}$, then find the approximate error in calculating its surface area.
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Radius of the sphere $=9 \mathrm{~m}: \Delta \mathrm{r}=0.03 \mathrm{~m}$
Surface area of sphere $S=4 \pi r^2$
$\Delta \mathrm{s}=\frac{\mathrm{ds}}{\mathrm{dr}} \times \Delta \mathrm{r}=8 \pi \mathrm{r} \times \Delta \mathrm{r}=8 \pi \times 9 \times 0 \cdot 03=2 \cdot 16 \pi \mathrm{m}^2$
Surface area of sphere $S=4 \pi r^2$
$\Delta \mathrm{s}=\frac{\mathrm{ds}}{\mathrm{dr}} \times \Delta \mathrm{r}=8 \pi \mathrm{r} \times \Delta \mathrm{r}=8 \pi \times 9 \times 0 \cdot 03=2 \cdot 16 \pi \mathrm{m}^2$
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