A cylindrical closed tank contains 36π cubic metres of water, and is filled to half of its capacity. When the cylindrical tank is placed upright on its circular base on level ground, the height of the water in the tank is 4 metres But when this tank is placed on its side on level ground, what will be the height of the surface of the water above the ground ?

Correct — C, (c) 3 metres.

THE INSIGHT IS SYMMETRY, NOT CALCULUS. A cylinder lying on its side has a circular cross-section, and a circle is symmetric about its horizontal diameter. So when a horizontal cylinder is exactly HALF FULL, the water surface must sit exactly on the axis — any higher and more than half the circle is covered, any lower and less than half is. The depth of the water is therefore the RADIUS, and the whole problem reduces to finding the radius.

The upright position supplies it:

upright, water occupies a cylinder of height 4 m πr² × 4 = 36π r² = 9, so r = 3 metres

Lay the tank on its side and the water surface stands at the axis, 3 metres above the ground. That is option (c).

WHY NO INTEGRATION IS NEEDED. Finding the depth of liquid in a horizontal cylinder filled to some arbitrary fraction is genuinely hard — it requires the area of a circular segment, which brings in an inverse trigonometric function. At exactly one half, symmetry does the work instead and the answer is exact and immediate. That is why the question specifies half of capacity rather than any other fraction, and recognising the special case is the skill being tested.

WHAT IS NOT NEEDED. The height of the tank never enters the answer. It can be found — the full capacity is twice 36π, so π × 9 × H = 72π gives H = 8 metres — but a horizontal cylinder has the same circular cross-section along its entire length, so the water stands at the same depth whether the tank is 8 metres long or 80. Recognising which given quantity is inert is part of reading a problem well.

A SANITY CHECK THAT SETTLES MOST OF THE OPTION SET. Lying on its side, the tank's own total height above the ground is its diameter, 6 metres. The water surface cannot be higher than that, and being half full it cannot be at the very top or near the bottom either. Only one offered value is consistent with 'half'.

THE STEM CARRIES A PRINTING DEFECT, and it is in the English column only: there is no full stop after '4 metres', so the sentence runs on as '… is 4 metres But when this tank …' with the next sentence beginning mid-line. The Hindi column prints its sentence-ending mark correctly, so this is a one-column slip and not an ambiguity in the problem. The volume is printed as 36π with the Greek letter set immediately after the digits.

Why the others are wrong

• (a)9 metres — The value of r SQUARED, reported as though it were the radius. From πr² × 4 = 36π the working gives r² = 9, and a candidate who stops at that line has 9 sitting in front of them and an option that matches it. Taking the square root is the step that is skipped, and the option exists to catch exactly that. It is also impossible on physical grounds, which is the quicker way to see it: the tank's radius is r, so lying on its side its total height above the ground is 2r. A water surface at 9 metres would need the tank to be more than 9 metres tall lying down, while the upright water column alone was only 4 metres deep. Any answer larger than the tank cannot be right.
• (b)6 metres — The DIAMETER, 2r = 6 metres — which is the height of the very top of the tank when it lies on its side, and therefore the answer to a different question: how high above the ground is the highest point of the tank, or how high would the water stand if the tank were completely full. The question says the tank is filled to HALF of its capacity, and by the symmetry of the circular cross-section half fills the tank exactly to the axis, which is half the diameter. This is the most reasonable of the wrong answers because it involves no arithmetic error at all, only answering the wrong question, and it is worth noting that the correct value is exactly half of it.
• (d)1 metre — The one option below the answer, and nothing in the data produces it. It fails the same physical check that disposes of option (a), applied in the other direction: a tank of radius 3 metres lying on its side stands 6 metres high, and a water surface at 1 metre would leave it barely a sixth of the way up its own height — nowhere near half full. Its purpose on the page is to complete a descending set, 9, 6, 3 and 1, in which each value except the answer corresponds to some other quantity or to none. Before computing anything on a problem of this shape it is worth fixing the bounds — the surface must lie between the ground and the top of the tank, and 'half full' pins it to the middle — because those bounds alone leave only one candidate standing.

THE VOLUME OF A CYLINDER is πr²h — the area of the circular cross-section multiplied by the length of the cylinder. What changes when the cylinder is reoriented is not the volume of water but the SHAPE of the region the water occupies, and that is the whole content of this problem.

UPRIGHT, on its circular base: the water forms a shorter cylinder of the same radius. Its depth and its volume are proportional, so half the capacity means half the height. The geometry is easy. ON ITS SIDE, axis horizontal: the water forms a horizontal prism whose cross-section is a CIRCULAR SEGMENT — the region of a circle cut off by a horizontal chord. Depth and volume are no longer proportional, because a circle is wide in the middle and narrow at top and bottom. Raising the surface by a metre near the axis adds far more water than raising it by a metre near the top.

THE HALF-FULL CASE IS THE SPECIAL ONE. The horizontal diameter divides a circle into two congruent halves, so a half-filled horizontal cylinder has its surface exactly on the axis and its depth is exactly the radius. No other fill fraction is that simple: for a general fraction the depth must be found from the area of a circular segment,

segment area = r²(θ − sin θ)/2, with θ the angle the chord subtends at the centre,

which cannot be solved for θ in elementary terms. Examinations therefore ask the half-full case, and recognising it as the symmetric one is the intended skill.

ORIENTATION AND WHAT IT DOES NOT CHANGE. The tank's length is irrelevant once it is horizontal, because the cross-section is identical everywhere along it. Doubling the length doubles both the capacity and the water, and the depth is unmoved. Spotting an inert quantity is as useful as spotting a needed one.

THE FAMILY OF ERRORS THIS SHAPE PRODUCES. Reporting r² instead of r after solving for the square is the commonest by far, and it appears in almost every mensuration item that requires a square root. Next is answering the diameter where the radius was wanted, or the reverse. Third is assuming the upright depth carries over to the horizontal position — it does not; here the water is 4 metres deep standing and 3 metres deep lying down, with the same volume in the tank both times.

Key facts

• The volume of a cylinder is πr²h; reorienting the tank changes the shape of the water, not its volume.
• Standing upright, water of volume 36π at a depth of 4 metres gives πr² × 4 = 36π, so r² = 9 and r = 3 metres.
• A horizontal cylinder that is exactly half full has its water surface on the axis, by the symmetry of the circular cross-section, so the depth equals the RADIUS — here 3 metres.
• The length of the cylinder does not affect the depth once it lies on its side, because the cross-section is the same along its entire length.
• The tank's full capacity is twice 36π, giving a height of 8 metres, but that figure is not needed for the answer.
• For any fill fraction other than one half, the depth in a horizontal cylinder requires the area of a circular segment, r²(θ − sin θ)/2, which has no elementary closed-form solution for the depth.
• Lying on its side the tank stands one diameter — 6 metres — above the ground, which bounds every admissible answer.
• Depth and volume are proportional in an upright cylinder but NOT in a horizontal one, because a circle is widest at its middle.

Common traps

• Reporting r² instead of r. Solving πr² × 4 = 36π gives 9, and taking the square root is the step most often skipped.
• Answering with the diameter where the radius was wanted. Here the diameter, 6 metres, is the height of the top of the tank, not of the water.
• Assuming the upright depth carries over to the horizontal position. The same water stands 4 metres deep upright and 3 metres deep on its side.
• Trying to compute a circular segment when the fill is exactly one half, where symmetry gives the answer at sight.
• Using the tank's height, which is recoverable but plays no part in the answer once the tank is horizontal.
• Failing to bound the answer before choosing: the surface must lie between the ground and the top of the tank, which is one diameter up.

• practice — not a real PYQ

  A closed cylindrical tank of radius 2 metres and length 10 metres is exactly half full of water. It is laid on its side on level ground. What is the depth of the water in the tank ?

  • (a)1 metre
  • (b)2 metres
  • (c)4 metres
  • (d)5 metres

  Answer(b) 2 metres — a horizontal cylinder that is half full has its water surface exactly on the axis, by the symmetry of the circular cross-section, so the depth is the radius. The length of 10 metres is irrelevant, because the cross-section is identical along the whole tank; 4 metres is the diameter, which is the height of the top of the tank rather than of the water.

• practice — not a real PYQ

  A cylindrical vessel standing upright on its circular base contains 45π cubic metres of water, and the water stands 5 metres deep. What is the radius of the vessel ?

  • (a)3 metres
  • (b)4·5 metres
  • (c)9 metres
  • (d)45 metres

  Answer(a) 3 metres — from πr² × 5 = 45π we get r² = 9 and therefore r = 3. The value 9 is r squared and is offered to catch a candidate who stops before taking the square root, which is the single commonest slip in mensuration problems that need one.