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in Surface Areas And Volumes by (20 points)
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\( 3080 cm ^{3} \) of water is required to fill a cylindrical vessel completely and \( 2310 cm ^{3} \) 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.

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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

\(= 2\pi r h_1 + \pi r^2\)

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

\(= 440 + 154\)

\(= 594 cm^2\)

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