A cylindrical vessel 60 cm in diameter is partially filled with water. A sphere 60 cm in diameter is gently dropped into the vessel. To what further height will water rise in the cylinder?

? 1/3 x (22/7) x r² x 24 = 1232
? r² = 1232 x (7/22) x (3/24) = 49
? r = 7 cm
Now, h = 24
So, l = ?7² + (24)²
= ?625
= 25 cm
? Curved surface area = ?rl
= (22/7) x 7 x 25 cm²
= 550 cm²

Since the diameters are in the ratio 4 : 5. It follows that their radii are in the ratio 4 : 5.
Let them be 4r and 5r. Let the height be h and H .
? Ratio of volumes = [(1/3)?(4r)²h / (1/3)?(5r)²H
= (16h)/(25H)
? h/H = (1/4) x (25/16) = 25/64 = 25: 64.

? 1/3 ? x (2)² x h = ? x (2)² x 6
? h = 18 cm

? (1/3)?r² x h = ?r² x 5
? h = 15 cm

Total volume of cuboid = (10 x 5 x 2) cm³
= 100 cm³
Volume curved = 1/3 x (22/7) x 3 x 3 x 7 cm³
= 66 cm³
% of wood wasted = (100 - 66)% = 34 %

? 22/7 x r² = 154
? r² = 154 x (7/22) = 49
? r = 7 cm and h = 14
So, l = ?(7)² + (14)² = ?245 = 7?5 cm

? Area of curved surface = ?rl
= (22/7) x 7 x 7?5 cm²
= 154?5 cm²

(2/3) x (22/7) x r³ = 19404
? r³ = 19404 x (7/22) x (3/2) = 9261 = (21)³
? r = 21 cm
Total Surface area = 3?r²
= 3 x (22/7) x 21 x 21 cm²
= 4158 cm²

? 4/3?r³ = 4/3? x [(3/2)³ - {(3/4)³ + 1³ } ]
? r³ = 125/64 = (5/4)³
? r = 5/4
? Diameter = (5/4) x 2 cm = 2.5 cm.

Area of 4 walls of the room = [2 (l + b ) x h ]m²
Area of 4 walls of new room = [2 (3l + 3b) x 3h]m²
= 9 x [2(l + b) x h ]m²
? Cost of painting the 4 walls of the new room = Rs. (9 x 350)
= Rs. 3150

Let the height of the cylinder be H and its radius = r
Then, (?r² H) + (1/3) x (?r²h) = 3 x (1/3)?r²h
? ?r²H = (2/3)?r²h
? H = 2h/3.