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# KCC's Quizzes AQQ262 about conical bottle height

1. Quote of the week: "If each day is a "gift", I'd like to know where I can return Monday" - Unknown

2. New quiz AQQ262 about a conical bottle height

Good luck!

Kuo-Chang

P.S. Don't hesitate to share those weekly quizzes in EZ to colleagues or friends!

• 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)*r2(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  ! 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  ! You found the solution with crystal clear development; congratulations!

• Thanks  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  , 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)