A closed cylindrical container 10 feet in height and 4 feet in diameter contains water with depth of 3 feet and 5 inches. What would be the level of the water when the tank is lying in horizontal position?

Well, if the water level in the cylindrical container is 3 feet and 5 inches when it's in an upright position, and we tilt it to a horizontal position, there won't be any water left in the container! It's like magic, my friend. The water will spill out and find its own level outside the container. So, to answer your question, the level of the water when the tank is lying in a horizontal position would be, drumroll please...on the floor!

To determine the new water level when the cylindrical tank is lying in a horizontal position, we need to understand the change in volume caused by the change in orientation.

Step 1: Calculate the current volume of water in the tank:
Given that the depth of the water is 3 feet and 5 inches, we need to convert the depth to feet. There are 12 inches in 1 foot, so 5 inches is equal to 5/12 feet.
The current depth of water = 3 feet + 5/12 feet = 3.4167 feet.

To calculate the volume of water, we use the formula for the volume of a cylinder, V = πr²h, where r is the radius and h is the height.
Given:
Height (h) = 10 feet
Diameter (d) = 4 feet

The radius (r) = d/2 = 4/2 = 2 feet.

Volume of water = π(2 feet)²(3.4167 feet)
Volume of water = 43.03 cubic feet (approximately)

Step 2: Calculate the new water level when the tank is lying in a horizontal position:
When the tank is lying horizontally, the water will spread out evenly along the floor of the tank, forming a new circular shape.

The area of the circle formed by the water when the tank is horizontal will be equal to the current volume of water.

Given:
Radius (r) = 2 feet
Area (A) = 43.03 cubic feet

To find the radius (R) when the tank is lying horizontally, we can use the formula for the area of a circle, A = πR².

43.03 = πR²
R² = 43.03/π
R = √(43.03/π)

Now we can calculate the new water level by finding the height (H) of the circular disk formed when the tank is horizontal.

Given:
Height (H) = ?

The new volume of water = Area (A) × Height (H)
43.03 = πR²H

Substituting the value of R obtained previously:
43.03 = π(√(43.03/π))²H
43.03 = 43.03H
H = 1 foot (approximately)

Therefore, the new water level when the tank is lying in a horizontal position will be approximately 1 foot.

To find the level of water when the tank is lying in a horizontal position, we can use the concept of the volume of water remaining constant.

First, let's convert the given measurements to a consistent unit – feet.

Height of the container = 10 feet
Diameter of the container = 4 feet
Depth of the water = 3 feet and 5 inches = 3 + (5/12) feet = 3.4167 feet (rounded to four decimal places)

Now, let's calculate the volume of water in the vertical position of the container:
Volume = π × (radius)^2 × depth

The radius of the container = (1/2) × diameter = (1/2) × 4 feet = 2 feet

Volume = π × (2 feet)^2 × 3.4167 feet
Volume ≈ 13.6346 cubic feet (rounded to four decimal places)

Now, when the container is in a horizontal position, the shape of the water column changes from a vertical cylinder to a horizontal cylinder. However, the volume of water remains the same.

To find the new depth, we can use the formula for the volume of a horizontal cylinder, which is given by:
Volume = π × (radius)^2 × length

Let's assume the length of the water column when the container is horizontal is L (in feet).

Using the volume formula and the volume of water we found earlier (13.6346 cubic feet), we can set up the equation:
π × (2 feet)^2 × L = 13.6346 cubic feet

Simplifying this equation and solving for L, we have:
L = (13.6346 cubic feet) / (π × (2 feet)^2)

L ≈ 1.0823 feet (rounded to four decimal places)

Therefore, when the container is lying in a horizontal position, the level of the water would be approximately 1.0823 feet.

Am I overthinking this ????

volume of water in cylinder being upright
= π(2^2)(41/12) inches^3 , (arrhhhg -- non-metric units)
= (41/3)π in^3

Now lay it on its side, and make a diagram of the cross section
Shade in the water which is a segment of the circle, label the endpoints as A and B. Mark the centre as C
we know the volume = segment x 10
(41/3)π = 10segment
segment area = (41/30)π in^2

We can see that
segment = area of sector - area of triangle.
(this is becoming harder than I anticipated)
let the central angle of all this mess be θ radians
area of triangle = (1/2)(2)(2)sinθ
= 2sinθ

area of whole circle = 4π
our sector = θ/2π of that
or (θ/2π)(2π) = θ in^2

so the area of our triangle is θ - (41/30)π
But we know area of triangle is 2sinθ

This is getting really messy!!!!

2sinθ = θ - (41/30)π
I used Wolfram to find θ=3.53213 radians, but that makes no sense!!!!

If this had been correct, I could now find the height of the triangle and then easily find the depth of the water.