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