Question

A cone having fixed volume has semi-vertical angle of $\frac{\pi }{4}.$ At an instant when its height is decreasing at the rate of $2m / s,$ its radius increases at a rate equal to

• A. $2m / s$
• B. $4m / s$
• C. $1m / s$
• D. $8m / s$

Answer 1

Answer: (C)

As, $V=\frac{1}{3}\pi r^{2}h$
$\Rightarrow \frac{d V}{d t}=\frac{\pi }{3}\left(r^{2} \frac{d h}{d t} + h \cdot 2 r \frac{d r}{d t}\right)=0$
$\Rightarrow r\frac{d h}{d t}=-2h\frac{d r}{d t}$

Also, $\frac{r}{h}=tan\left(\frac{\pi }{4}\right)\Rightarrow r=h$
$\therefore \frac{d h}{d t}=-2\frac{d r}{d t}\Rightarrow \frac{d r}{d t}=-\frac{1}{2}\frac{d h}{d t}$
$\Rightarrow \frac{d r}{d t}=+1$
$\Rightarrow $ radius increases at $1m / s.$