Let radius = r cm
and height = h cm
\(\therefore \pi r^2 h = 3080 \) .....(1)
& \(\pi r^2 (h - 5) = 2310 \) ....(2)
Dividing (1) by (2), we get
\(\frac{\pi r^2 h}{\pi r^2 (h - 5)} = \frac{3080}{2310}\)
⇒ \(2310 h = 3080 h - 15400\)
⇒ \((3080 - 2310) h = 15400\)
⇒ \(h = \frac{15400}{770} = \frac{1400}{70 } = 20\, cm\)
From (1),
\(r^2 = \frac{3080}{\pi h} = \frac{3080 \times 7}{22\times 20}\)
\(= \frac{28 \times 7}4\)
\(= 7 \times 7 = 49 = 7^2\)
\(\therefore r = 7cm\)
(i) radius of the vessel = r = 7cm
(ii) Height of the vessel = h = 20cm
(iii) When half filled then height is \(h_1 = \frac{20}2 = 10 cm\)
Then wetted surface area of the vessel
= curved surface area of half cylindrical vessel + Area of bottom
\(= 2\pi r h_1 + \pi r^2\)
\(= 2\times \frac{22}7 \times 7 \times 10 + \frac{22}7 \times 7 \times 7\)
\(= 440 + 154\)
\(= 594 cm^2\)