KCC's Quizzes AQQ262 about conical bottle height

Volume of a right cone is  (1/3)  pi R^2 H.   Given r = p h  at  an elevation h, ( with p = R/H)   
                        we deal with  V = K  h^3       with K = (1/3) pi  p  
Volume of the water at the left = K H^3  - K*8^3   =  volume of water at the right = K (H-2) ^3
Dividing by the constant K which is not zero, and evaluation the polynomials, we get:  4H^2 - 8H -504  = 0  ;  note:  504 =  8^3 - 2^3

We get  H  = 1 +  sqrt(127) ,  (the other root is negative)  and greater than 8,  OK  (numerially:  12.269... )

The volume of water is dependant of the radius of the circle at the bottom of the cone, but any R gives the H valeur here up and thus, R is undecided.

1. The height of the bottle is 10.22cm.
2. The volume is undefined as the base radius drops out of the equations.

1. The height of the bottle is about 10.22

2. The volume of the liquid is about half of the volume of the bottle.

To calculate the height H of the bottle, you write down the equation for the liquid volume V in the two cases:

V=(PI/3)*(H-8) * (R² + r1² +R*r1) (volume of a truncated cone of height = H-8)

V=(PI/3)*(H-2)*r2² (volume of a cone of height = H-2)

Now, for geometric reasons, you have r1= 8*R/H and r2= R* (H-2)/H, so, by using the fact that volume V is the same, you get the following equation:

(H-8) * (R² + 64*R²/H² + 8*R²/H) = (H-2)³ * R² /H²)

After some calculation, you get to:

H² - 2*H - 84 = 0

You keep just the positive root, which is about 10.22 (ok: is >8)

With H=10.22 the volume of the liquid is about half of the bottle volume. You can see that by evaluating V vs the volume of the cone bottle. 

Bingo gpiarino ! You get it, congratulations! For question 2 one can also just say the volume cannot be determined with the data given (in fact one needs to give R).

Great feedback retiredEE ! You found the solution with crystal clear development; congratulations!

Thanks vanderghast for your prompt feedback! Your development is correct but I think you made a typo somewhere before ending to the polynomial 4H^2 - 8H -504  = 0

504 seems to be incorrect...

May be you can recheck?

Got a tough time to find back my original computation (the sheet of paper is far in the recycling bin) but indeed, it seems that I tripped on the development of (H-2)^3. I now get 6H^2 -12H -504 = 0, leading to H= 10.2195... 

YES! You get it now vanderghast , congratulations!

1.
It is clear that the solution is independent of the radius R. For simplicity, I choose them R=h. Both the volume of liquid and the volume of air are the same in both figures and I did the calculation for air.
Its volume in the left picture is

PI/3*h1*R1^2 = PI/3*h1^3 = PI/3*8^3.

I put together an equation for the same volume of air for the left picture.

PI/3*h^3 = PI/3*(h-2)^3 + PI/3*8^3.

The resulting equation is h^3 - 512 = (h-2)^3.

And now a bit of a "dirty" solution - I went with this equation on chatGPT. The first attempt failed - the result showed 2 imaginary roots. I pointed out the error and then once again, but the third time it showed 2 real roots. The positive one is correct one and therefore
h=10.22

2.
The exact volume is uknown, R we don't know. But its volume in % is

100% - air = 100 - 100*(8/10.22) = 52.0%

 Precision: it is of course one same bottle with the same quantity of liquid shown in the picture (they correspond to 2 different positions of the bottle)