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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

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  • 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%

Reply
  • 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%

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