Question

A reservoir in the form of the frustum of a right circular cone contains $44\times {10}^7$ litres of water which fills it completely. The radii of the bottom and top of the reservoir are 50 metres and 100 metres respectively. Find the depth of water and the lateral surface area of the reservoir. (Take : $\pi$ = 22/7)

Answer 1

Here,
Volume of frustum of cone → 440000000 liters
⇒ $\frac{440000000}{1000}=440000m^3$
Also,
r = 50 m
R = 100 m
Now,

$=>\frac{1}{3}\pi (r_1^2+r_2^2+r_1{r}_2)h=440000m^3$
$=>\frac{1}{3}\times \frac{22}{7}\times ({50}^2+{100}^2+50\times 100)\times h=440000m^3$

$=>\frac{1}{3}\times \frac{22}{7}\times \left(17500\right)\times h=440000m^3$

$=>h=\frac{440000}{17500}=24m$

Now, curved surface area of the frustum of the cone
$=>\pi l(r+R)$
Where,

$L=\sqrt{(r_1-r_2{)}^2+h^2}$

$=>\sqrt{{50}^2+576}=55.4671cm$

So,
$=>\frac{22}{7}\times 55.461\times 150=26146.2m^2$