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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!

Parents
• 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).