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.
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
? 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.
(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
? 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
Let h and H be the height of water level before and after the dropping of the sphere.
Then, [? x (30)2 x H ] - [? x (30)2 x h ]
= 4/3 ? x (30)3
? ? x 900 x (H -h) = 4/3 ? x 27000
? (H- h) = 40 cm
Given, diagonal = 2?3
? a?3 = 2?3 [a = edge of the cube]
? a = 2
? Required surface area = 6a2
= 6 x 22 = 24 sq cm
Total surface area of a cube = 6a2
? 6 = 6a2
? a2 = 1
? a = 1
Now, volume of the cube = a3 = 13 = 1 cu unit
Volume of cube = (Side)3
? 729 = a3
? a = 9 cm
Diagonal of cube = side x ?3 = 9 x ?3
= 9?3 cm
Given that, l = 10 m , b = 5 m, h = 3 m
lb = 10 x 5 = 50, bh = 5 x 3 = 15,
lh = 10 x 3 = 30
? Surface area of a cuboid
= 2(lb + bh + lh)
= 2(50 + 15 + 30)
= 2 x 95 = 190 sq m
According to the question,
6a2 = 726 [a = edge of the cube]
? a2 = 726/6 = 121
? a = ?121 = 11 cm
? Required volume = a3 = 113 = 1331 cm3
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