If the radius of a sphere is measured as 9 te | vedclass

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(A) Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius. Given $r = 9 	ext{ m}$ and $\Delta r = 0.03 	ext{ m}$.T

Vedclass · Accepted Answer

$2.16$

Vedclass · Answer

(A) Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius. Given $r = 9 	ext{ m}$ and $\Delta r = 0.03 	ext{ m}$.The surface area $S$ of a sphere is given by $S = 4 \pi r^2$.Differentiating $S$ with respect to $r$,we get $\frac{dS}{dr} = 8 \pi r$.The approximate error in the surface area is given by $dS = \left( \frac{dS}{dr} ight) \Delta r$.Substituting the values,$dS = (8 \pi 	imes 9) 	imes 0.03$.$dS = 72 \pi 	imes 0.03 = 2.16 \pi 	ext{ m}^2$.Thus,the approximate error in calculating the surface area is $2.16 \pi 	ext{ m}^2$.

If the radius of a sphere is measured as $9 \text{ m}$ with an error of $0.03 \text{ m}$,then find the approximate error in calculating its surface area. (in $\pi \text{ m}^2$)

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