If there is an error of pm 0.04 text{ cm} in | vedclass

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

Vedclass · Accepted Answer

$\pm 0.6$

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

If there is an error of $\pm 0.04 \text{ cm}$ in the measurement of the diameter of a sphere, then the approximate percentage error in its volume, when the radius is $10 \text{ cm}$, is

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