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) $

Upper radius of the reservoir $r_1 = 100\ m$

Lower radius of the reservoir $r_2 = 50\ m$

Capacity of water in the reservoir $= 44 \times 10^7$ litres

Therefore,

Volume $=\frac{44 \times 10^{7}}{1000} \mathrm{~m}^{3}$

$=\frac{44 \times 10^{7}}{10^{3}}$

$=44 \times 10^{7-3}$

$=44 \times 10^{4} \mathrm{~m}^{3}$

$=440000 \mathrm{~m}^{3}$

Let $h$ be the depth of the reservoir.

This implies,
Volume $=\frac{\pi}{3}(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}) h$

$\Rightarrow 440000=\frac{22}{7 \times 3}[(100)^{2}+100 \times 50+(50)^{2}] h$

$\Rightarrow 440000=\frac{22}{21}[10000+5000+2500] h$

$\Rightarrow 440000=\frac{22}{21} \times 17500 \times h$

$\Rightarrow h=\frac{440000 \times 21}{22 \times 17500}$

$=\frac{200 \times 3}{1 \times 25}$

$=8 \times 3$

$=24 \mathrm{~m}$

Depth of the reservoir $=24 \mathrm{~m}$

Slant depth of the reservoir $l=\sqrt{(h)^{2}+(r_{1}-r_{2})^{2}}$

$=\sqrt{(24)^{2}+(100-50)^{2}}$

$=\sqrt{(24)^{2}+(50)^{2}}$

$=\sqrt{576+2500}$

$=\sqrt{3076}$

$=55.46 \mathrm{~m}$

Lateral surface area of the reservoir $=\pi(r_{1}+r_{2}) l$

$=\frac{22}{7}(100+50) \times 55.46$

$=\frac{22}{7} \times 150 \times 55.46$

$=26145.43 \mathrm{~m}^{2}$

The depth of water is $55.46\ m$ and the lateral surface area of the reservoir is $26145.43\ m^2$.