If the radius of a spherical balloon increase | vedclass

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(B) Let $V$ be the volume of a spherical balloon of radius $r$. Then,$V = \frac{4}{3} \pi r^3$. Differentiating both sides with respect to $r$,we get

Vedclass · Accepted Answer

$0.3$

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(B) Let $V$ be the volume of a spherical balloon of radius $r$. Then,$V = \frac{4}{3} \pi r^3$. Differentiating both sides with respect to $r$,we get $\frac{dV}{dr} = 4 \pi r^2$. We know that the approximate change in volume $\Delta V$ is given by $\Delta V \approx \frac{dV}{dr} 	imes \Delta r$. So,$\Delta V \approx (4 \pi r^2) 	imes \Delta r$. The percentage increase in volume is given by $\frac{\Delta V}{V} 	imes 100$. Substituting the values,we get $\frac{\Delta V}{V} 	imes 100 \approx \frac{4 \pi r^2 	imes \Delta r}{\frac{4}{3} \pi r^3} 	imes 100 = 3 	imes \frac{\Delta r}{r} 	imes 100$. Given that the radius increases by $0.1 \%$,we have $\frac{\Delta r}{r} 	imes 100 = 0.1$. Therefore,the percentage increase in volume is $3 	imes 0.1 \% = 0.3 \%$.

If the radius of a spherical balloon increases by $0.1 \%$,then its volume increases approximately by (in $\%$)

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