Question

A sphere increases its volume at the rate of $\pi cc/s$. The rate at which its surface area increases when the radius is $1 \,cm$ is

• A. $2\, \pi\, sq.cm/s$
• B. $ \pi \, sq. \,cm/s$
• C. $\frac{3 \pi}{2} sq.\, cm/s$
• D. $\frac {\pi} {2} sq.\, cm/s$

Answer 1

Answer: (A)

Given rate of increase volume of sphere is
$\frac{d V}{d t}=\pi$...(i)
We know that,
Volume of sphere; $V=\frac{4}{3} \pi r^{3}$
$\Rightarrow \frac{d V}{d t}=\frac{4}{3} \pi \cdot 3 r^{2} \cdot \frac{d r}{d t}=4 \pi r^{2} \frac{d r}{d t}$
$\Rightarrow \pi=4 \pi r^{2} \frac{d r}{d t} ;$ [from Eq. (i)]
$\Rightarrow \frac{d r}{d t}=\frac{1}{4 r^{2}}$...(ii)
Also, we know that, surface of sphere,
$S =4 \pi r^{2}$
$\Rightarrow \frac{d S}{d t}=8 \pi r \frac{d r}{d t}=8 \pi r \cdot \frac{1}{4 r^{2}}$ [from Eq.(ii)]
$\frac{d S}{d t}=\frac{2 \pi}{r} [\because$ given $r=1]$
$\frac{d S}{d t}=2 \pi$
So, rate of increase in surface $\equiv 2 \pi\, sq\, cm / s$