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If the radius of a spherical balloon increases by $0.1 \%$, then its volume increases approximately by
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The correct answer is:
$0.3 \%$
Let $V$ be the volume of spherical ballon of
radius $r$
Then, $V=\frac{4}{3} \pi r^{3}$
$\Rightarrow \quad \log V=\log \left(\frac{4 \pi}{3}\right)+3 \log r \Rightarrow \frac{1}{V} \frac{d V}{d r}=0+\frac{3}{r}$
$\Rightarrow \quad \frac{1}{V} \frac{d V}{d r}=\frac{3}{r} \Rightarrow \frac{1}{V} \frac{\Delta V}{\Delta r}=\frac{3}{r}$
$\Rightarrow \quad \frac{\Delta V}{V}=3 \frac{\Delta r}{r}$
$\Rightarrow \frac{\Delta V}{V} \times 100=3 \frac{\Delta r}{r} \times 100$
$\Rightarrow \frac{\Delta V}{V} \times 100=3 \times 01=0.3$
$\therefore$ Percentage increase in volume is $0.3 \%$
radius $r$
Then, $V=\frac{4}{3} \pi r^{3}$
$\Rightarrow \quad \log V=\log \left(\frac{4 \pi}{3}\right)+3 \log r \Rightarrow \frac{1}{V} \frac{d V}{d r}=0+\frac{3}{r}$
$\Rightarrow \quad \frac{1}{V} \frac{d V}{d r}=\frac{3}{r} \Rightarrow \frac{1}{V} \frac{\Delta V}{\Delta r}=\frac{3}{r}$
$\Rightarrow \quad \frac{\Delta V}{V}=3 \frac{\Delta r}{r}$
$\Rightarrow \frac{\Delta V}{V} \times 100=3 \frac{\Delta r}{r} \times 100$
$\Rightarrow \frac{\Delta V}{V} \times 100=3 \times 01=0.3$
$\therefore$ Percentage increase in volume is $0.3 \%$
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