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shubhamsrg

  • 2 years ago

Consider a cylinder with radius = 5 cm and height = 20 cm. The cylinder is placed vertically on the ground and water is filled in to a height of 10 cm from the ground. Now the cylinder is closed from the top and it is placed on the ground horizontally. Find the new height of water from the ground .

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  1. shubhamsrg
    • 2 years ago
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    |dw:1359278725894:dw|

  2. Azteck
    • 2 years ago
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    Find the volume of the water when the cylinder is vertically upwards.

  3. shubhamsrg
    • 2 years ago
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    I have tried all. Couldn't crack it.

  4. Azteck
    • 2 years ago
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    Just do it.

  5. Azteck
    • 2 years ago
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    I will tell you what to do next. I might have got it just by looking at the question.

  6. shubhamsrg
    • 2 years ago
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    V = 250pi

  7. shubhamsrg
    • 2 years ago
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    So, what next ?

  8. frx
    • 2 years ago
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    All you need to know: http://mathworld.wolfram.com/CylindricalSegment.html

  9. Azteck
    • 2 years ago
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    Do you know how to calculate the volume of apartial cylinder?

  10. shubhamsrg
    • 2 years ago
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    Nopes I don't, am referring to the link @frx gave .

  11. sirm3d
    • 2 years ago
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    |dw:1359279395322:dw| \[\Large{A=\int_{-5}^h 2\sqrt{5^2-y^2}} \;\mathrm dy\]

  12. Azteck
    • 2 years ago
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    Found this. The first suggestion works for a *vertical* cylinder, but not for a horizontal one. Your first step will depend on the fluid level in the tank... You start the same way, . I'm assuming it's a circular cylinder, where If the cylinder is more than half full, you'll want to start with the volume of the full cylinder, V = r^2*length*pi then subtract the unfilled space at the top. The cross-section of this empty space takes the form of a circular segment, whose area is given by the formula A = 1/2 * r^2 * (x - sin(x)) where r is the circle's radius, and x is the angle (measured in radians) formed between the two radii drawn from the points on the tank wall where the fluid level ends. Multiply this area by the cylinder's length to obtain the volume of empty space. Subtract this number from the full cylinder's volume to arrive at your fluid volume. If the cylinder is less than half full, you'll do something similar...only this time you'll be finding the cross sectional area of the fluid sitting on the bottom of the tank and multiplying that by the tank's length to find the volume. You can also use this method if the tank is exactly half full...you'll see that the angle between the two radii is flat (pi radians), forming a diameter...and as sin(pi) = 0, the area formula reduces to A= (1/2) * r^2 * pi

  13. sirm3d
    • 2 years ago
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    it's \[\Large A=\int_{-5}^{5+h}2\sqrt{25-y^2}\;\mathrm dy\]|dw:1359279823727:dw|

  14. shubhamsrg
    • 2 years ago
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    @Azteck how do I find "x" ?

  15. sirm3d
    • 2 years ago
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    that's \(-5+h\) for the upper limit.

  16. shubhamsrg
    • 2 years ago
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    I am unable to follow, @sirm3d

  17. sirm3d
    • 2 years ago
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    i'm sorry. i assumed you know integral calculus.

  18. shubhamsrg
    • 2 years ago
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    I do know integral calculus .

  19. shubhamsrg
    • 2 years ago
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    Wait, am getting it, 2 mins.

  20. sirm3d
    • 2 years ago
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    it's the area under the curve.

  21. shubhamsrg
    • 2 years ago
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    How do you get 5^2 - y^2 ?

  22. shubhamsrg
    • 2 years ago
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    oh wait,

  23. shubhamsrg
    • 2 years ago
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    So you're finding the length of an element at a distance y from origin.

  24. sirm3d
    • 2 years ago
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    |dw:1359280450587:dw|

  25. shubhamsrg
    • 2 years ago
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    Ah ok so our working eqn will be A(20) = 250pi Where A =integral (2 sqrt(25 -y^2)) from y= -5 to (-5+h)

  26. sirm3d
    • 2 years ago
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    yes.

  27. sirm3d
    • 2 years ago
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    in the process, you will be deriving the area bounded by the arc and a chord. you can also derive the area as the difference between the area of the sector of a circle and the isosceles triangle.

  28. shubhamsrg
    • 2 years ago
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    =>[ sqrt(25-y^2) + 25sin^-1( y/5) ] from y=-5 to -5+h = 25pi/2 =>( sqrt( 25 - (h-5)^2) + 25sin^-1((h-5)/5) ) = 25pi am getting complex solutions, please someone confirm. http://www.wolframalpha.com/input/?i=%28+sqrt%28+25+-+%28h-5%29%5E2%29+%2B+25sin%5E-1%28%28h-5%29%2F5%29+%29++%3D+25pi

  29. shubhamsrg
    • 2 years ago
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    Oh wait, RHS will be 0 ./

  30. shubhamsrg
    • 2 years ago
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    Seems fine thanks a lot @sirm3d

  31. sauravshakya
    • 2 years ago
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    Isnt it simply 5 cm as volume of water =1/2 volume of cylinder

  32. shubhamsrg
    • 2 years ago
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    I see your point, I guess it won't be 5 since the area does not increases uniformly as we move up from ground level when it is placed horizontally. Area around center is more than at the bottom; or at top.

  33. shubhamsrg
    • 2 years ago
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    When cylinder was placed vertically, area was increasing uniformly,but not in the case of horizontal positioning.

  34. sauravshakya
    • 2 years ago
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    |dw:1359287336567:dw| If I will cut the cylinder from the diameter in two parts won't their volume be equal

  35. sauravshakya
    • 2 years ago
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    |dw:1359287729887:dw|

  36. shubhamsrg
    • 2 years ago
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    Indeed! Re-checked my calculations! It is 5 only!

  37. shubhamsrg
    • 2 years ago
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    It was real simple then, guess calculus would have been involved only if, say, radius was 4 cm. hm, thanks @sauravshakya and @sirm3d

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