Question

The slant height of a cone is fixed at $7cm$. If the rate of increase of its height is $0.3cm/sec$, then the rate of increase of its volume when its height is $4$ cm is

• A. $\frac{\pi }{2}cc/sec$
• B. $\pi cc/sec$
• C. $\frac{\pi }{5}cc/sec$
• D. $\frac{\pi }{10}cc/sec$

Answer 1

Answer: (D)

We have,
Slant height of cone $(l)$ = 7 cm
$\Rightarrow l^2=h^2+r^2$ ....(i)
$\Rightarrow r^2=7^2-4^2$ (when $h$ = 4 cm)
$\Rightarrow r^2=33$
$\Rightarrow r=\sqrt{33}$ cm
Now, differentiating equation (i) w.r.t 't'
$\Rightarrow 0=2h\frac{dh}{dt}+2r\frac{dr}{dt}$
$\Rightarrow \frac{dr}{dt}=-\frac{h}{r}\frac{dh}{dt}$ ...(ii)
Volume of the cone, $V=\frac{1}{3}\pi r^{2}h$
$\Rightarrow \frac{dV}{dt}=\frac{1}{3}\pi [2rh\frac{dr}{dt}+r^2\frac{dh}{dt}]$
$\Rightarrow \frac{dV}{dt}=\frac{1}{3}\pi \left[2rh\right(\frac{-h}{r}\frac{dh}{dt})+r^2\frac{dh}{dt}]$
$\Rightarrow \frac{dV}{dt}=\frac{1}{3}\pi [-2h^2\frac{dh}{dt}+r^2\frac{dh}{dt}]$
$\Rightarrow \frac{dV}{dt}=\frac{1}{3}\pi \frac{dh}{dt}[-2h^2+r^2]$
$\Rightarrow \left[\frac{dV}{dt}\right]=\frac{1}{3}\pi (0.3)[-2\times 4^2+(\sqrt{33}{)}^2]$
$=\frac{\pi }{10}[1]=\frac{\pi }{10}cc/sec$