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shubhamsrg

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 .

  • one year ago
  • one year ago

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  1. shubhamsrg
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    |dw:1359278725894:dw|

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

    • one year ago
  3. shubhamsrg
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    I have tried all. Couldn't crack it.

    • one year ago
  4. Azteck
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    Just do it.

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

    • one year ago
  6. shubhamsrg
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    V = 250pi

    • one year ago
  7. shubhamsrg
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    So, what next ?

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

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

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

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

    • one year ago
  12. Azteck
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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

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

    • one year ago
  14. shubhamsrg
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    @Azteck how do I find "x" ?

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

    • one year ago
  16. shubhamsrg
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    I am unable to follow, @sirm3d

    • one year ago
  17. sirm3d
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    i'm sorry. i assumed you know integral calculus.

    • one year ago
  18. shubhamsrg
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    I do know integral calculus .

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

    • one year ago
  20. sirm3d
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    it's the area under the curve.

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

    • one year ago
  22. shubhamsrg
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    oh wait,

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

    • one year ago
  24. sirm3d
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    |dw:1359280450587:dw|

    • one year ago
  25. shubhamsrg
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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)

    • one year ago
  26. sirm3d
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    yes.

    • one year ago
  27. sirm3d
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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.

    • one year ago
  28. shubhamsrg
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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

    • one year ago
  29. shubhamsrg
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    Oh wait, RHS will be 0 ./

    • one year ago
  30. shubhamsrg
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    Seems fine thanks a lot @sirm3d

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

    • one year ago
  32. shubhamsrg
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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.

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

    • one year ago
  34. sauravshakya
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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

    • one year ago
  35. sauravshakya
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    |dw:1359287729887:dw|

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

    • one year ago
  37. shubhamsrg
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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

    • one year ago
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