The volume of a spherical ball is increasing at a rate of $4 \pi \mathrm{cm}^3 \mathrm{~s}^{-1}$. The rate…
The volume of a spherical ball is increasing at a rate of $4 \pi \mathrm{cm}^3 \mathrm{~s}^{-1}$. The rate at which its radius increases, when its volume is $288 \pi \mathrm{cm}^3$, is.......$\mathrm{cm} \mathrm{s}^{-1}$
$\frac{1}{6}$
$\frac{1}{36}$
$\frac{1}{9}$
$\frac{1}{24}$
Solution
Given, $\frac{d v}{d t}=4 \pi \mathrm{cm}^3 / \mathrm{s}$ ...(i)
To find, $r / d t=$ ?
When, $V=288 \pi \mathrm{cm}^3$
Let $r$ be the radius of spherical ball
$\therefore$ Volume $(V)=\frac{4}{3} \pi r^3$
On differentiating w.r.t. $t$,
$\frac{d V}{d t}=\frac{4 \pi}{3} \cdot 3 r^2 \frac{d r}{d t} \Rightarrow 4 \pi=\frac{12}{3} \pi r^2 \frac{d r}{d t}$
$\frac{1}{r^2}=\frac{d r}{d t}$ ...(ii)
When, $V=288 \pi$
$\therefore 288 \pi=\frac{4}{3} \pi r^3=r^3=72 \times 3=216$
$\Rightarrow \quad r=\sqrt[3]{216}=6 \mathrm{~cm}$
From Eq. (ii), we get
$\frac{d r}{d t}=\frac{1}{r^2}=\frac{1}{6^2}=\frac{1}{36}$