A hospital room is to accommodate 56 patients. It should be done in such a way that every patient gets 2.2 m ² of floor and 8.8 ³ of space. If the length of the room is 14 m, then breadth and height of the room are respectively?

Volume of water in the tank = 100 x 500 x 2 = 30000 m³
Speed of water = 20 x 1000/60 = 1000/3 m/min
water flow per minute = (15/10) x (125/100) x (1000/3) = 625 m³
Time taken = 30000/625 = 48 min

Volume of the earth dugout as a tunnel = ?r²h = (22/7) x 2 x 2 x 56 = 704 m³
Volume of the ditch = (48 x 33)/(2 x 4) = 24 x 33 x 4 = 3168 m³
? Part required = 704/3168 = 2/9

r = Radius of the cone = 14/2 = 7 cm
h = Height of the cone = 12 x 2 = 24 cm
? Slant height (l) = ?r² + h²
= ?(7)² + (24)²
= ?49 + 576
= ?625
= 25 cm
Area of the sheet = Total surface area
= (?rl + ?r²)
= ?r(l + r)
= (22/7) x 7 x (25 + 7) = 704 sq cm

Let length of the wire be h According to the question,
Volume of sphere = Volume of wire
(4/3) x ? x 18 x 18 x 18 = ? x (2/10) x (2/10) x h
? h = (100 x 1944) cm
= (100 x 1944)/100 m = 1944 m [? 1 cm = 1/100 m]

Given, diameter = 150 cm
? r = 150/2 cm
According to the question,
2/3? (150/2)³ = 120 ?r² x 15
? (2/3) x 150 x 150 x 150 / 8 = 120 x 15 x r²
? r² = (150 x 150 x 150) / (12 x 120 x 5)
? r² = 625/4
? r = ?625/4 = 25/2
? Diameter = 2r = 2 x 25/2 = 25 cm

Given that, the height and radius of a right circular ,etal cone (solid) are 8 cm and 2 cm, respectively.
i.e., h = 8 cm and r = 2 cm
Let the radius of the sphere is R
Then, by condition,
1/3 ? r² h = 4/3 ? R³ h
? 4 x 8 = 4 R³
? R³ = (2)³ = ? R = 2
? Radius odf the sphere = 2 cm

Given that, Height = 1 m
Diameter = 140 cm or 140/100 m = 1.40 m
? r = 1.40/2 = 0.7 m

Now,
Required sheet = Total surface area = 2?r( h + r)
= (22/7) x 1.4 x (1 + 0.7) = 7.48 sq m
? Sheet required = 7.48 sq m

Height of the cylinder = 10 x Diameter of each ball
= 10 x 10 = 100 cm
? Required empty space
= ? (5)² (100) - 4/3 ? (5)³ x 10
= ? (5)² x 10 [10 - 20/3]
= ? (5)² x 10 [30 - 20/3] = 2500/3 ? cm³

Let h be the height,
Volume of the cone = V
and curved surface area of the cone = c
? 1/3 ?r²h = V, ?rl = c
i.e., ?r?r² + h² = c
?²r²(r² + h²) = c²
Consider 3?Vh³ - c²h² + 9V²
= 3? x 1/3?r²h x h³ - ?²r²(r² + h²)h + 9 x 1/9 x ?²r⁴ h²
= ?²r²h⁴ - ?²r⁴h² - ?²r²h⁴ + ?²r⁴ = 0

Let the side of the cube be 3k, 4k and 5k, respectively.
So, volumes of these cubes are 27 k³, 64k³, 125 k³ respectively.
? Volume of the new bigger cube = 27 k³ + 64k³ + 125 k³
= 216k³
So, side of the new cube = 6k
Since, diagonal of the cube = ?(6k)² + (6k)² + (6k)²
= 12?3
? 108k² = 432
?k = 2
So, sides of the three cubes were 6 cm, 8 cm and 10 cm respectively.