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

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Aicharya · Accepted Answer

Given, $\Delta r= \pm \frac{0.04}{2}=0.02$ Volume of sphere $ V=\frac{4}{3} \pi r^3 $ On differentiating w.r.t. $r$, we get $ \begin{aligned} & \frac{d U}{d r}=\frac{4}{3} \pi 	imes 3 r^2=4 \pi r^2 \ & 	herefore \quad \Delta V=\frac{d U}{d r} \Delta r=4 \pi r^2 \Delta r \ & \end{aligned} $ $	herefore$ Relative per cent error $ \begin{aligned} & \frac{\Delta V}{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 \ & =\frac{3 	imes( \pm 0.02)}{10} 	imes 100 \ & = \pm \frac{6}{10}= \pm 0.6 \end{aligned} $

If there is an error of $\pm 0.04 \mathrm{~cm}$ in the measurement of the diameter of a sphere, then the…

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