Answer:
Option A
Explanation:
Let V and r be the volume and radius of spherical ball , respectively,
Volume of spherical ball = $\frac {4}{3} \pi r^{3}$
$\Rightarrow$ $ 288 \pi$ =$\frac {4}{3} \pi r^{3}$ [ given, V=288 cm3]
$\Rightarrow$ 288 =$\frac {4}{3} r^{3}$
$\Rightarrow$ r3 =72 x 3=8 x 27
$\Rightarrow$ $ r^{3} = 2^{3} \times 3^{3}$
[taking cube roots both sides]
$\Rightarrow$ r=6
On differentiating Eq.(i) w.r.t 't' , we get
$\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$
$\therefore$ $4 \pi=4\pi r^{2}\frac{dr}{dt}$
[ given $\frac{dV}{dt}=4\pi$ cubic cm/s]
$\Rightarrow$ $1=(6)^{2}\frac{dr}{dt}$ [$\because$ r=6]
$\Rightarrow$ $\frac{dr}{dt}=\frac{1}{36}$
Now, surface area of spherical ball (s) = $4\pi r^{2}$
$\Rightarrow$ s= $ 4\pi r^{2}$
On differentiating both sides, w.r.t 't', we get
$\frac{ds}{dt}=4\times 2\pi r\frac{dr}{dt}$
$=8\times \pi \times6\times\frac{1}{36}$
$\left[\because r= 6 and \frac{dr}{dt}=\frac{1}{36}\right]$
$\Rightarrow$ $\frac{ds}{dt}=\frac{4\pi}{3}cm^{2}/s$
