Question: Water is poured into an inverted conical vessel of which the radius of the base is 2 m and height 4 ...

Question

Water is poured into an inverted conical vessel of which the radius of the base is 2 m and height 4 m, at the rate of 77 litre/minute. The rate at which the water level is rising at the instant when the depth is 70 cm is (use $\pi$ = 22/7)

Answer 1

Solution

Let R be the radius of the base of the cone and H be its height.
Given R = 2 m = 200 cm and H = 4 m = 400 cm.
Let V be the volume of water in the cone at time t, h be the depth of the water, and r be the radius of the water surface at depth h.
The volume of water is given by $V = \frac{1}{3}\pi r^2 h$ .
By similar triangles, we have $\frac{r}{h} = \frac{R}{H}$ .
Substituting the values of R and H, we get $\frac{r}{h} = \frac{200}{400} = \frac{1}{2}$ , so $r = \frac{h}{2}$ .
Substitute this into the volume formula: $V = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3}\pi \frac{h^2}{4} h = \frac{\pi}{12} h^3$ .
We are given the rate at which water is poured into the vessel, which is $\frac{dV}{dt} = 77$ litre/minute.
Since 1 litre = 1000 cm³, $\frac{dV}{dt} = 77 \times 1000 = 77000$ cm³/minute.
We need to find the rate at which the water level is rising, $\frac{dh}{dt}$ , when $h = 70$ cm.
Differentiate the volume equation with respect to time t:
$\frac{dV}{dt} = \frac{d}{dt}\left(\frac{\pi}{12} h^3\right) = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$ .
Now, substitute the given values: $\frac{dV}{dt} = 77000$ cm³/minute, $h = 70$ cm, and $\pi = \frac{22}{7}$ .
$77000 = \frac{1}{4} \cdot \frac{22}{7} \cdot (70)^2 \frac{dh}{dt}$
$77000 = \frac{1}{4} \cdot \frac{22}{7} \cdot 4900 \frac{dh}{dt}$
$77000 = \frac{1}{4} \cdot 22 \cdot 700 \frac{dh}{dt}$
$77000 = \frac{1}{4} \cdot 15400 \frac{dh}{dt}$
$77000 = 3850 \frac{dh}{dt}$
Solve for $\frac{dh}{dt}$ :
$\frac{dh}{dt} = \frac{77000}{3850} = \frac{7700}{385}$ .
Dividing 7700 by 385: $\frac{7700}{385} = \frac{770 \times 10}{385} = \frac{(2 \times 385) \times 10}{385} = 2 \times 10 = 20$ .
So, $\frac{dh}{dt} = 20$ cm/minute.