$A$ container is made by closing one end of a cylinder with radius $35 \,cm$ and height $52 \,cm$ by a cone with height $12 \,cm$. How many litres of water can it store? Find the total surface area of this closed container.

(N/A) $1$. Volume of the container: The container consists of a cylinder and a cone. The radius $r = 35 \,cm$,height of cylinder $h_1 = 52 \,cm$,and height of cone $h_2 = 12 \,cm$.
Volume of cylinder $V_1 = \pi r^2 h_1 = \frac{22}{7} \times 35 \times 35 \times 52 = 200,200 \,cm^3$.
Volume of cone $V_2 = \frac{1}{3} \pi r^2 h_2 = \frac{1}{3} \times \frac{22}{7} \times 35 \times 35 \times 12 = 15,400 \,cm^3$.
Total volume $V = V_1 + V_2 = 200,200 + 15,400 = 215,600 \,cm^3$.
Since $1,000 \,cm^3 = 1 \,litre$,the volume is $215.6 \,litres$.
$2$. Total Surface Area: The surface area includes the base of the cylinder,the curved surface area of the cylinder,and the curved surface area of the cone.
Slant height of cone $l = \sqrt{r^2 + h_2^2} = \sqrt{35^2 + 12^2} = \sqrt{1225 + 144} = \sqrt{1369} = 37 \,cm$.
Base area of cylinder $= \pi r^2 = \frac{22}{7} \times 35 \times 35 = 3,850 \,cm^2$.
Curved surface area of cylinder $= 2 \pi r h_1 = 2 \times \frac{22}{7} \times 35 \times 52 = 11,440 \,cm^2$.
Curved surface area of cone $= \pi r l = \frac{22}{7} \times 35 \times 37 = 4,070 \,cm^2$.
Total surface area $= 3,850 + 11,440 + 4,070 = 19,360 \,cm^2$.