Math, asked by SushilKesarwani8693, 1 year ago

A closed cylindrical tank contains 36 cubic feet of water and its filled to half its capacity. When the tank is placed upright on its circular base on level ground, the height of water in the tank is 4 feet. When the tank is placed on its side on level ground, what is the height, in feet, of the surface of the water above the ground?

Answers

Answered by knjroopa
3

Answer:

3 ft

Step-by-step explanation:

Given A closed cylindrical tank contains 36 π cubic feet of water and its filled to half its capacity. When the tank is placed upright on its circular base on level ground, the height of water in the tank is 4 feet. When the tank is placed on its side on level ground, what is the height, in feet, of the surface of the water above the ground?

We know that volume of a cylinder = π r²h

                      36 π = π x r² x 4

                      4  r² = 36

                        r² = 36 / 4

                       r² = 9

                         r = 3 ft

So radius is 3 ft and diameter is 2 r = 2 x 3 = 6 ft

So water comes to the level of radius = 3 ft.

                     

Answered by prashilpa
1

Answer:

2.39 feet.

Step-by-step explanation:

Water level = 36 cubic feet. (This is half of the tank)

=> Cylinder capacity = 72 Cubic Feet.

Height of Cylinder = 4 feet.

Volume of Cylinder =  π r²h

72 = π r²* 4

r² = 72/(π * 4) = 18/π

r = √(18/π) = 2.39 feet.

When we tilt the tank, the water in cylinder is still half.

That mean, it is filled up to the height of radius.

=> Surface of the water above ground = 2.39 feet.

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