Answer:
0.5052 cm per sec ( approx )
Step-by-step explanation:
∵ The volume of a cone is,
[tex]V=\frac{1}{3}\pi (r)^2h[/tex]
Where,
r = radius,
h = height,
Here, r = 2 and h = 8,
Thus, the volume of the cone is,
[tex]V=\frac{1}{3}\pi (2)^2 8=\frac{32\pi }{3}[/tex]
When the cone is half filled,
Volume would be,
[tex]V_1=\frac{V}{2}=\frac{16\pi }{3}[/tex]
Now, [tex]\frac{r}{h}=\frac{1}{4}\implies r=\frac{h}{4}[/tex]
[tex]V_1=\frac{1}{3}\pi (\frac{h}{4})^2 h=\frac{\pi h^3}{48}----(1)[/tex]
[tex]\implies \frac{\pi h^3}{48} = \frac{16\pi }{3}[/tex]
[tex]\implies h \approx 6.35[/tex]
Differentiating equation (1) with respect to t ( time )
[tex]\frac{dV_1}{dt}=\frac{\pi h^2}{16}\frac{dh}{dt}[/tex]
We have, [tex]\frac{dV_1}{dt}=4\text{ cube cm per s},h=6.35\text{ cm}[/tex]
[tex]4=\frac{\pi (6.35)^2}{16}\times \frac{dh}{dt}[/tex]
[tex]\implies \frac{dh}{dt}=\frac{64}{6.35^2 \pi}\approx 0.5052\text{ cm per sec}[/tex]
