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} = \frac{\pm 0.04}{2}

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$\pm 0.6$

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(D) Given,error in diameter $\Delta D = \pm 0.04 	ext{ cm}$.Since $D = 2r$,the error in radius is $\Delta r = \frac{\Delta D}{2} = \frac{\pm 0.04}{2} = \pm 0.02 	ext{ cm}$.The volume of a sphere is $V = \frac{4}{3} \pi r^3$.Differentiating with respect to $r$,we get $\frac{dV}{dr} = 4 \pi r^2$,so $dV = 4 \pi r^2 \Delta r$.The percentage error in volume is given by $\frac{dV}{V} 	imes 100$.Substituting the values: $\frac{dV}{V} 	imes 100 = \frac{4 \pi r^2 \Delta r}{\frac{4}{3} \pi r^3} 	imes 100 = \frac{3 \Delta r}{r} 	imes 100$.Given $r = 10 	ext{ cm}$ and $\Delta r = \pm 0.02 	ext{ cm}$,we have:Percentage error $= \frac{3 	imes (\pm 0.02)}{10} 	imes 100 = \frac{\pm 0.06}{10} 	imes 100 = \pm 0.6 \%$.

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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