The diameter of a sphere is measured as 42 te | vedclass

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(C) Given diameter $d = 42 	ext{ cm}$, so radius $r = \frac{d}{2} = 21 	ext{ cm}$.The error in diameter is $\Delta d = \frac{1}{77} 	ext{ cm}$, so

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$36$

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(C) Given diameter $d = 42 	ext{ cm}$, so radius $r = \frac{d}{2} = 21 	ext{ cm}$.The error in diameter is $\Delta d = \frac{1}{77} 	ext{ cm}$, so the error in radius is $\Delta r = \frac{\Delta d}{2} = \frac{1}{154} 	ext{ cm}$.The volume of a sphere is $V = \frac{4}{3} \pi r^3$.The approximate error in volume $\Delta V$ is given by $\Delta V \approx \frac{dV}{dr} 	imes \Delta r$.$\frac{dV}{dr} = \frac{d}{dr} \left( \frac{4}{3} \pi r^3 ight) = 4 \pi r^2$.Substituting the values: $\Delta V = 4 	imes \frac{22}{7} 	imes (21)^2 	imes \frac{1}{154}$.$\Delta V = 4 	imes \frac{22}{7} 	imes 441 	imes \frac{1}{154}$.$\Delta V = 4 	imes 22 	imes 63 	imes \frac{1}{154} = \frac{5544}{154} = 36$.Thus, the error in the volume is $36 	ext{ cm}^3$.

The diameter of a sphere is measured as $42 \text{ cm}$. If there is an error of $1/77 \text{ cm}$ in measuring it, then the error involved in the volume of that sphere (in cubic centimeters) is

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