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

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

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

Solution:

Let $S$ be the surface area and $V$ be the volume of

sphere of radius $r$ . Then,

$\frac{d V}{d t} = \frac{d r}{d t}$ [Given]

$\begin{array}{l} \Rightarrow r = \frac{1}{2 \sqrt{π}} [S e e i l l u s t r a t i o n 2] \\ ∴ S = 4 π r^{2} \Rightarrow S = 1 \end{array}$

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