A tank 3 m long, 2 m wide and 1.5 m deep is dug in a field 22 m long and 14 m wide. If the earth dug out is evenly spread out over the field, the rose in level of the field will be?

Given that,
Volume of sphere = Volume of cone
? (4/3)?r³ = (1/3)?r²h
? 4r = h

Total surface area of the pyramid
= [1/2(perimeter of the base) (Slant height) ] + Area of the base
= 1/2 (4) (4) (8) + 4² = 80 cm²

volume = (2/3)?r³
= (2/3) x ? x 3³
= (2/3) x ? x 27
= 18? cm³

Required surface area = 4?r²
= 4 x ? x 4² = 664? sq cm

Given, side of cube, a = 9
? Diagonal of cube = a?3 = 9?3

Let the volume be x³ and 27x³
? Their edges are x and 3x

Now Ratio of their surface area = 6x² : 54x² = 1 : 9

Let original length of cube = L
Then, its surface area = 6L²
New edge = (150 x L)/10 = 3L/2
New surface area = 6 x (3L/2)² = (6 x 9 x L²)/4 = (27/2) x L²
Increase in surface area = (27/2 - 6) x L² = 15 x L²/2
? Increase percent = (15 x L²/2 ) / (6L²) x 100 % = 125 %

Let the edges of the two cubes be x and y metres
Then, x³ - y³ = 152
and x² - y² = 20
Also, x + y = 10
So, x - y = (x² - y²) / (x + y) = 20/10 = 2
Now, (x³ - y³) / (x - y) = 152/2
? x² + y² + xy = 76
? (x + y)² - xy = 76
? xy = (x + y)² - 76 = (10)² - 76 = 24

? l + b + h = 19
and l² + b² + h² = (5?5)² = 125
? (l + b + h)² = (19)²
? (l² + b²+ h²) + 2(lb + bh + lh) = 361
? 2 (lb + bh + lh) = (361 - 125) = 236
? Surface area = 236 cm²

Volume of the Earth taken out = 30 x 20 x 12
= 7200 m³

Area of the remaining portion (leaving the area of dug out portion) = 470 x 30 + 30 x 10
= 14100 + 300 = 14400 m³

Let h be height to which the field is raised by spreading the Earth dug out.
Then, 14400 x h = 7200
? h = 7200/14400 = 0.5 m