Latest Elitmus Aptitude Question SOLUTION: 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 t

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(#M40151584) ELITMUS Numerical Ability question geometry Keep an EYE

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?

Asked In Elitmus MR. X (9 years ago)

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subhi singh  (9 years ago)

ans is cube root of (7h8)

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siddharth prakash singh  (9 years ago)

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.

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Ranjit Sah  (9 years ago)

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

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swarup  (9 years ago)

read this.it will clear the doubt
http://www.careerbless.com/qna/discuss.php?questionid=959

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SANTU MAITY  (9 years ago)

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

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sarita kumari  (9 years ago)

5 value....................................

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  (7 years ago)

nice question

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