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What is the surface area of a sphere when the volume is increasing at the same rate as its radius?

\[\begin{array}{1 1}(1)1&(2)\frac{1}{2\pi}\\(3)4\pi&(4)\frac{4\pi}{3}\end{array}\]

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Let S be the surface area, V be the volume, r be the radius of a sphere at time t
Given $\large\frac{dv}{dt} =\frac{dr}{dt}$
$V= \large\frac{4}{3}$$\pi r^3$
Surface area $S= 4 \pi r^2$
$V= \large\frac{4}{3} $$\pi r^3$
$\large\frac{dV}{dt} =\frac{4}{3} $$3 \pi ^2 \large\frac{dr}{dt}$
$\large\frac{dV}{dt}$$ =4 \pi ^2 \large\frac{dV}{dt}$
$1=S\qquad => S=1$
Surface area= 1
Hence 1 is the correct answer.
answered May 19, 2014 by meena.p

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