EPFO APFC 2023 General Studies · Question 103 of 120

Correct — B, (b) 2·25. Melting conserves volume and nothing else. The metal in the old disc is exactly the metal in the new one, so the whole question is the single equation 'volume before = volume after', and every other quantity in the stem is decoration. A solid circular disc is a cylinder, whose volume is π r² h with h the thickness. Writing the old disc first, π × 12² × 4 = π × 144 × 4 = 576π cubic inches. Writing the new disc with the unknown thickness t, π × 16² × t = 256π t. Setting them equal, 256π t = 576π; the π cancels because it multiplies both sides, and t = 576 ÷ 256 = 2·25 inches. That is option (b). It is worth doing the same calculation a second way, in ratios, because the ratio form is faster and it exposes the trap the option set is built around. The radius grows from 12 to 16, a factor of 16/12 = 4/3. The face area of a circle grows as the square of the radius, so the face area grows by (4/3)² = 16/9. Since the volume is face area multiplied by thickness and the volume is fixed, the thickness must fall by exactly the reciprocal of that factor, 9/16. So t = 4 × 9/16 = 36/16 = 9/4 = 2·25 inches. The number that matters is the square: a candidate who scales the thickness by the radius ratio itself rather than by its square gets 4 × 3/4 = 3 inches, which is not offered at all, and the absence of 3 from the option list is a quiet signal that squaring is the point of the item. Two further observations make the answer robust. The result does not depend on π, so no approximation of it is needed and no value of it can be got wrong; nor does it depend on the metal, since density cancels along with everything else, which is why the stem can say 'metallic' without saying which metal. And the answer must be smaller than 4 inches, because the disc has been made wider without adding material, so any option above the original thickness could have been discarded on sight — a check that is free here because all four options happen to respect it.

• (a)2 — This is the only whole number in a set of quarters, and it is where a candidate lands who has not carried out the division at all. Two competing rough arguments both end here. The first is a vague sense that a much wider disc must be much thinner, so the thickness is halved from 4 to 2. The second is a completed calculation abandoned at the last step: 576 ÷ 256 is a little over 2, and a candidate who stops at 'a bit more than two' and reaches for the printed 2 has thrown away the only digits the question is testing. The exact quotient is available and it is not close to a whole number: 256 × 2 = 512, leaving 64, and 64 is exactly a quarter of 256, so the quotient is 2 and a quarter, not 2. Put in ratio form, the thickness must be multiplied by 9/16, and 4 × 9/16 = 2·25. Nothing in the problem halves anything.
• (c)2·50 — This sits a quarter of an inch above the true answer, in the middle of the four printed values, and it is the option a candidate chooses who has decided the answer lies between 2 and 3 and picks the tidiest figure in that gap. It is also where a slip in the division lands: 576 ÷ 256 is sometimes misread as 640 ÷ 256, which is 2·5, and 640 is what 256 × 2·5 comes to — so the option is exactly self-consistent with a wrong numerator. The way to be certain is to keep the numbers as fractions instead of decimals. The volume ratio is 12² : 16², that is 144 : 256, which cancels to 9 : 16 — both sides divide by 16. The new thickness is therefore 4 × 9/16 = 9/4, and 9/4 is 2·25 exactly. A thickness of 2·5 inches would require the new radius to satisfy r² = 576/2·5 = 230·4, which no whole number does.
• (d)2·75 — This is the largest of the four and the furthest from the answer, and it is placed to catch a candidate who has reasoned that the change in radius is small — 12 to 16 is only four inches — and that the thickness should therefore change only a little from 4. That intuition fails because area grows with the square of the radius, so a one-third increase in the radius is close to a four-ninths increase in the face, and the thickness has to fall by more than a third to compensate. The exact factor is 9/16, a fall of nearly forty-four per cent, which takes 4 inches down to 2·25 and nowhere near 2·75. A thickness of 2·75 inches would hold π × 256 × 2·75 = 704π cubic inches of metal, against the 576π the original disc contained, so this option quietly creates 128π cubic inches of metal out of nothing.

Recasting problems are conservation problems, and the quantity conserved is volume. When a solid is melted and reshaped, no material is added or lost, so the volume of the new shape equals the volume of the old one; surface area, height, radius and shape may all change freely, and only the volume is fixed. That single sentence solves every question of the type, whatever the two shapes are. The formulae needed are few and should be held exactly: a cylinder, which is what a solid circular disc is, has volume π r² h and curved surface area 2π r h; a sphere has volume (4/3)π r³ and surface area 4π r²; a cone has volume (1/3)π r² h and slant height l = √(r² + h²); a cube of side a has volume a³. Because both sides of a recasting equation carry the same π, it always cancels, and answers are exact rational numbers rather than decimals needing approximation — if a recasting answer seems to require a value of π, something has gone wrong. The second idea in the topic is scaling. If a linear dimension of a shape is multiplied by k while the shape is preserved, every area scales by k² and every volume by k³. That is why a disc whose radius grows by a factor of 4/3 gains face area by (4/3)², and why its thickness must shrink by the reciprocal of that square if the volume is to stay put. Getting the exponent right — one, two or three — is what separates a correct answer from a plausible one in almost every mensuration item of this kind.

A one-equation mensuration item like this is placed in an EPFO paper to be finished quickly, and its discriminating power lies entirely in the exponent. Almost every candidate knows that the volume is conserved; the paper is testing whether the volume of a cylinder is remembered as π r² h rather than π r h, and whether the radius ratio is squared rather than used raw. The option set is built to make that the only decision that matters: the four values 2, 2·25, 2·50 and 2·75 step in quarters, so no estimate can be matched to a printed figure, and the value 3 inches, which is what the un-squared ratio produces, is deliberately absent, denying the candidate who makes that mistake any confirmation at all. The measurements are in inches rather than centimetres, which is unusual in an Indian paper and can distract for a moment; it changes nothing, since the units cancel on both sides of the equation and the answer is asked for in inches too. The decimals are printed with a raised middle dot — 2·25, not 2.25 — which is this booklet's convention throughout and appears in eight of its questions. Read it as an ordinary decimal point. Time management is the practical lesson: this item should take under a minute, and the minute saved is what pays for the data-interpretation set that follows immediately after it.

• Melting and recasting conserves volume and nothing else. Set the volume of the original solid equal to the volume of the new solid and solve for the unknown dimension. Surface area is not conserved, mass per unit area is not conserved, and the material is irrelevant because density cancels from both sides of the equation.
• A solid circular disc is a cylinder, of volume π r² h. Here π × 12² × 4 = 576π cubic inches must equal π × 16² × t = 256π t, so t = 576/256 = 2·25 inches. The π cancels, which is why recasting answers are exact rational numbers and never require an approximation of π.
• In ratio form the same result is quicker: the radius grows by 16/12 = 4/3, so the face area grows by (4/3)² = 16/9, and the thickness must fall by the reciprocal 9/16. Thus t = 4 × 9/16 = 9/4 = 2·25. Scaling a linear dimension by k scales areas by k² and volumes by k³.
• The volumes worth holding exactly for this topic: cylinder π r² h, sphere (4/3)π r³, hemisphere (2/3)π r³, cone (1/3)π r² h with slant height √(r² + h²), and cube a³. A cone has exactly one third the volume of the cylinder that shares its base and height, which is the fact most recasting questions between the two shapes turn on.
• Two free checks on any recasting answer. The new dimension must move in the direction the geometry demands — a wider disc made from the same metal must be thinner than 4 inches — and scaling the radius without squaring it gives 4 × 3/4 = 3 inches, a figure this option set deliberately does not offer.

• Scaling the thickness by the radius ratio instead of by its square. That gives 4 × 12/16 = 3 inches, a value the paper pointedly does not offer, so a candidate who makes this error finds no printed answer and should treat that as a signal to square the ratio rather than to guess.
• Misremembering the volume of a cylinder. It is π r² h; using π r h, or using the curved surface area 2π r h, changes the exponent on the radius and therefore changes the answer entirely. The radius is squared and the thickness is not.
• Rounding to the tidiest option. The exact quotient 576 ÷ 256 is 2·25 and no rounding of it produces either 2 or 2·5; when four options step in quarters, the arithmetic has to be finished rather than estimated.
• Assuming the material or the units matter. Density cancels, so 'metallic' tells the candidate nothing beyond that the solid can be melted, and the inches cancel on both sides of the equation, so no conversion is needed anywhere.

Mensuration in EPFO papers is almost always a single-equation item, and the recasting form is the most frequent of them because it can be posed between any two shapes and still be finished in a minute. The shapes rotate — a sphere into a cylinder, a cylinder into a wire, several small spheres from one large one, a cone into a hemisphere — while the method does not: equate the volumes, cancel π, solve. A second common form asks for a ratio rather than a value, giving two similar solids and asking how their surface areas or volumes compare, which tests the exponents one, two and three directly. A third gives a cost — of painting, plating or filling — and tests whether the candidate reaches for an area formula or a volume formula. In every case the preparation is the same short list of formulae held exactly, plus the discipline of asking which quantity the physical process actually preserves before writing anything down.

No directly related past PYQ was found.

• practice — not a real PYQ

  A solid metallic cylinder of radius 6 cm and height 24 cm is melted and recast into a solid cylinder of height 6 cm. What is the radius of the new cylinder?

  • (a)8 cm
  • (b)10 cm
  • (c)12 cm
  • (d)14 cm

  Answer(c) 12 cm — the volume is conserved, so π × 6² × 24 = π × r² × 6. That gives 864 = 6 r², so r² = 144 and r = 12 cm. The height has been divided by four, so the face area must be multiplied by four, and multiplying an area by four multiplies the radius by two — which is the scaling argument that gives the same answer without any arithmetic.

• practice — not a real PYQ

  A solid metallic sphere of radius 6 cm is melted and recast into a solid circular disc of radius 12 cm. What is the thickness of the disc?

  • (a)1 cm
  • (b)1.5 cm
  • (c)2 cm
  • (d)3 cm

  Answer(c) 2 cm — the sphere's volume is (4/3)π × 6³ = 288π cubic centimetres, and the disc is a cylinder of volume π × 12² × t = 144π t. Equating them gives 144 t = 288, so t = 2 cm. The point of the item is that the two shapes use different formulae but share the one quantity that melting preserves, and that the π cancels here exactly as it does when both shapes are cylinders.