Question 1106544
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A close cylindrical tank 10ft in height and 4ft in diameter contains water with depth of 3ft and 5 inches. 
What would be the height of the water when the tank is lying in a horrizontal  position?
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<pre>
First, convert dimensions from feet to inches:  H = 10 ft = 120 in,  h = 3 feet 5 inches = 12*3+5 = 41 in,  r = 2 ft = 24 inches.

The volume of the tank is  V = {{{pi*r^2*H}}} = {{{pi*r^2*120}}} = {{{120pi*r^2}}}.

The filled volume of the tank is  F = {{{pi*r^2*h}}} = {{{41pi*r^2}}}.

The ratio of the filled part to the total volume is  {{{F/V}}} = {{{41/120}}}.

In any position (vertical/horizontal) the filled part is  {{{41/120}}} of the total volume.

Hence, we need to find the central angle of the circle which subtends the segment of the circle whose area is {{{41/120}}} of the circle area.

The area of the segment of a circle is  A = {{{(1/2)*r^2*alpha - (1/2)*r^2*sin(alpha)}}},  where {{{alpha}}} is the central angle in radians.

Hence, we need to solve the equation

{{{(1/2)*r^2*alpha - (1/2)*r^2*sin(alpha)}}} = {{{(41/120)*pi*r^2}}},

or, canceling r^2 in both sides

{{{(1/2)*alpha - (1/2)*sin(alpha)}}} = {{{(41*pi/120)}}} = 1.07283.


I solved this non-linear equation using Excel function  "Goal Seek" of  the section "What-if"  in my computer.

The answer is {{{alpha}}} = 2.633 radians.

Since the problem asks about the depth, it is  {{{r-r*cos(alpha/2)}}} = {{{24-24*c0s(2.633/2)}}} = 17.96 inches.     (24 = 24 inches = r = 2 ft is the radius)
</pre>

<U>Answer</U>.  The depth under the question is 17.96 inches.