Question

A water is poured into an inverted cone whose semi-vertical angle is ${45}^0$ so that the level of the water increases at the rate of 1 cm/sec., then the rate at which the volume of water is increasing when height of water is 2 cm is

• A. $4\pi c{m}^3/sec$
• B. $48\pi c{m}^3/sec$
• C. $16\pi c{m}^3/sec$
• D. $64\pi c{m}^3/sec$

Answer 1

The correct option is A $4\pi c{m}^3/sec$

Let us consider the cane drawn beside. Let the height of the conies radius = r

& $\theta =$ semivertical angle $={45}^0$

According to the question $\frac{dh}{dt}=1cm/sec$

& $\tan \theta =\frac{r}{h}\Rightarrow \tan {45}^0=\frac{r}{h}$

$\therefore r=h$ ------eq.1

Now for a cane, $v=\frac{1}{3}\pi r^{2}h$

differentiating w.r.t 't' we get,

$\frac{dv}{dt}=\frac{\pi }{3}\frac{d}{dt}\left(r^{2}h\right)=\frac{\pi }{3}\frac{d}{dt}(h^2.h)$

$\frac{dv}{dt}=\frac{\pi }{3}.3h^2\frac{dh}{dt}$

$\therefore \frac{dv}{dt}=\pi h^2[\because \frac{dh}{dt}=1]$

when h = 2 cm $\Rightarrow \frac{dv}{dt}=\pi \times 2^2=4\pi c{m}^2/sec$

$\therefore$ option A is correct.