The surface area of a sphere when. its volume is increasing at the same rate as its radius , is

A
1
B
$\frac{1}{2 \sqrt{π}}$
C
$4 π$
D
$\frac{4 π}{3}$

Solution:

At any time $t$ , let $V$ , Sand $r$ denote respectively

the volume, surface area and radius of the sphere. Then,

$\begin{array}{l} V = \frac{4}{3} π r^{3} and S = 4 π r^{2} \\ \frac{d V}{d t} = \frac{d r}{d t} \Rightarrow 4 π r^{2} \frac{d r}{d t} = \frac{d r}{d t} \Rightarrow 4 π r^{2} = 1 \Rightarrow S = 1 \end{array}$

Solution:

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