There is an error of pm 0.04 text{ cm} in the | vedclass

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(D) Given, error in diameter $\Delta D = \pm 0.04 	ext{ cm}$.Since $D = 2r$, the error in radius is $\Delta r = \frac{\Delta D}{2} = \pm 0.02 	ext{

Vedclass · Accepted Answer

$\pm 0.6\%$

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(D) Given, error in diameter $\Delta D = \pm 0.04 	ext{ cm}$.Since $D = 2r$, the error in radius is $\Delta r = \frac{\Delta D}{2} = \pm 0.02 	ext{ cm}$.The volume of a sphere is $V = \frac{4}{3} \pi r^3$.Differentiating with respect to $r$, we get $dV = 4 \pi r^2 dr$.The relative error in volume is $\frac{dV}{V} = \frac{4 \pi r^2 dr}{\frac{4}{3} \pi r^3} = 3 \frac{dr}{r}$.Percentage error in volume $= \frac{dV}{V} 	imes 100 = 3 	imes \frac{\Delta r}{r} 	imes 100$.Substituting the values $r = 10 	ext{ cm}$ and $\Delta r = \pm 0.02 	ext{ cm}$:Percentage error $= 3 	imes \frac{\pm 0.02}{10} 	imes 100 = 3 	imes (\pm 0.002) 	imes 100 = \pm 0.6\%$.

There is an error of $\pm 0.04 \text{ cm}$ in the measurement of the diameter of a sphere. When the radius is $10 \text{ cm}$, the percentage error in the volume of the sphere is

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