3080 cm³ of water is required to fill a cylindrical vessel completely and 2310 cm³ of water is required to fill it upto 5 cm below the top. Find :

(i) radius of the vessel.

(ii) height of the vessel.

(iii) wetted surface area of the vessel when it is half-filled with water.

Let radius of vessel be r cm and height be h cm.

Given,

Volume of cylindrical vessel = 3080 cm³

∴ πr²h = 3080 ………..(1)

Given,

It takes 2310 cm³ of water to fill cylinder upto 5 cm below the top.

∴ πr²(h - 5) = 2310 ………..(2)

Dividing (1) by (2) we get,

$\Rightarrow \dfrac{πr^2h}{πr^2(h - 5)} = \dfrac{3080}{2310} \\[1em] \Rightarrow \dfrac{h}{h - 5} = \dfrac{308}{231} \\[1em] \Rightarrow 231h = 308(h - 5) \\[1em] \Rightarrow 231h = 308h - 1540 \\[1em] \Rightarrow 308h - 231h = 1540 \\[1em] \Rightarrow 77h = 1540 \\[1em] \Rightarrow h = \dfrac{1540}{77} \\[1em] \Rightarrow h = 20 \text{ cm}$

(i) Substituting value of h in equation (1), we get :

$\Rightarrow πr^2 \times 20 = 3080 \\[1em] \Rightarrow \dfrac{22}{7} \times r^2 \times 20 = 3080 \\[1em] \Rightarrow r^2 = \dfrac{3080 \times 7}{20 \times 22} \\[1em] \Rightarrow r^2 = 49 \\[1em] \Rightarrow r = 7 \text{ cm}.$

Hence, radius of vessel = 7 cm.

(ii) From above,

h = 20 cm.

Hence, height of vessel = 20 cm.

(iii) When vessel is half-filled, water will be filled upto $\dfrac{20}{2}$ = 10 cm.

Wetted surface area = 2πrh + πr²

= πr(2h + r)

= $\dfrac{22}{7} \times 7 \times (2 \times 10 + 7)$

= $22 \times 27$

= 594 cm².

Hence, wetted surface area = 594 cm².