A cone vessel containing sulphuric acid up to height level h/2 (vertex is downwards) and the remaining portion is air volume. Now, the vessel is inverted then find the height up to which there is air volume?

• ans is cube root of (7h8)
  
• 9 years agoHelpfull: Yes(11) No(0)
• ans is sqrt(3):/2 h.
  draw a triangle ABC having a vertex C downwards and a side AB upwards. let height of this triangle =h
  Now suppose acid is fille upto a mark DE at a height of h/2(given). thus another triangle CDE will be formed with height h/2.
  triangle ABC and CDE are similar. and AB:DE =h/(h/2)= 2:1
  so the ar(triangle ABC):ar(triangle CDE) =4:1 {frm prop of similar triangles}
  let ar(triangle ABC)=4x^2 and ar(triangle CDE) =x^2
  now invert the triangle ABC such that vertex C is at the top and side ab is at the bottom.
  now the acid will come down. having area x^2. Let acid is now upto the mark GH such that GH is parallel to AB. so the ar(triangle CGH)=4x^2- x^2= 3x^2.
  triangle CGH is similar to triangle CAB. also ratio of their areas= 3x^2: 4x^2
  ratio of their sides= sqrt(3):sqrt(4)=sqrt(3):2
  ratio of their heights=sqrt(3):2
  but the height of CAB= h
  let height of CGH=H
  so H:h=sqrt(3):2
  so H=sqrt(3):2*h
  which is the required height.
  
• 9 years agoHelpfull: Yes(7) No(16)
• Let 'h' be the height and 'r; be the radius of conical vessel.
  
  Total volume of vessel = 1/3 π r2 h = V
  
  At a level of h/2 the radiusof cone will be r/2 and hence its volume will be : 1/3 π (r/2)2 (h/2) = V/8
  
  Therefore volume of sulphiric acid is V/8 and hence volume of air = V - V/8 = 7V/8
  
  When the vessel is inverted the sulphuric acid will be at the side of base and then remaining portion will be occupied by air:
  
  let H be the height , therefore the radius upto height H will be (H/h) x r
  
  Therefore the volume of air = 1/3 π (Hr/h)2 H = 1/3 π r2h x 7/8
  
  or, H3/h2 = 7h/8
  
  or, H3 = 7/8 h3
  
  Therefore , H = (7/8)1/3 h
• 9 years agoHelpfull: Yes(5) No(0)
• read this.it will clear the doubt
  http://www.careerbless.com/qna/discuss.php?questionid=959
• 9 years agoHelpfull: Yes(4) No(0)
• Let 'h' be the height and 'r; be the radius of conical vessel.
  
  Total volume of vessel = 1/3 π r^2 h = V
  
  At a level of h/2 the radiusof cone will be r/2 and hence its volume will be : 1/3 π (r/2)^2 (h/2) = V/8
  
  Therefore volume of sulphiric acid is V/8 and hence volume of air = V - V/8 = 7V/8
  
  When the vessel is inverted the sulphuric acid will be at the side of base and then remaining portion will be occupied by air:
  
  let H be the height , therefore the radius upto height H will be (H/h) x r
  
  Therefore the volume of air = 1/3 π (Hr/h)^2 H = 1/3 π r^2*h x 7/8
  
  or, H^3/h^2 = 7h/8
  
  or, H^3 = 7/8 (h^3)
  
  Therefore , H = (7/8)^1/3* h
• 9 years agoHelpfull: Yes(3) No(0)
• 5 value....................................
• 9 years agoHelpfull: Yes(0) No(0)
• nice question
• 7 years agoHelpfull: Yes(0) No(0)