Enforcement Officer / Accounts Officer (EO/AO) 2017 recruitment test. Transcribed from the Commission's own booklet (series A, GRS-A-TNS); answer as per the official UPSC final answer key. This question is in Part B. Answer last reviewed 24 August 2026.

If the radius of the new spherical container is double the radius of the old spherical container, then the ratio of the volume of the new container and the volume of the old container is

(a)2 : 1
(b)4 : 1
(c)8 : 1
(d)2π : 1

Answer

Why

Correct — C, (c) 8 : 1. Volume scales with the cube of a linear dimension, and that single law decides the item without any arithmetic worth the name. A sphere of radius r has volume four-thirds π r³. Doubling the radius replaces r by 2r, and (2r)³ = 8r³, so the volume is eight times what it was. The ratio of the new container to the old is 8 : 1, which is option (c).

The step worth noticing is what happens to the constant. Both containers are spheres, so both volumes carry the same factor of four-thirds π, and in a ratio that factor appears above and below the line and cancels. What survives is r³ against (2r)³, that is 1 against 8. This is why a scaling question never requires the formula to be remembered in full: as long as the two solids have the same shape, only the power of the linear dimension matters — the first power for any length, the second for any area, the third for any volume.

A concrete instance settles any doubt. A sphere of radius 1 cm has volume four-thirds π, about 4·19 cm³. A sphere of radius 2 cm has volume four-thirds π × 8, about 33·5 cm³. The second is eight times the first, and the ratio 33·5 : 4·19 reduces to 8 : 1 with π nowhere in sight. The same law is worth holding alongside its companion for area: doubling the radius multiplies the surface area by four, since surface area goes with the square, while it multiplies the volume by eight. A container built to twice the dimensions holds eight times as much but needs only four times as much sheet metal, which is the reason large tanks are cheaper per litre than small ones.

Why the others are wrong

(a)2 : 1 — Two to one is the ratio of the radii themselves, offered to a candidate who carries the linear ratio straight across to the volumes. It would be right if the question asked for the ratio of the diameters, the circumferences of the great circles, or any other length on the two spheres, because every length doubles when the shape is scaled up by a factor of two. Volume does not. A solid grows in three independent directions at once, so scaling by two lengthens, widens and deepens it, and the product of those three doublings is a factor of eight. The clearest way to see the difference is to think of unit cubes: a cube of side 2 contains eight cubes of side 1, not two. Whenever a ratio of volumes is wanted, the linear ratio must be cubed before it is reported.
(b)4 : 1 — Four to one is the square of the linear ratio, and it is the correct answer to a different question — the ratio of the surface areas of the two containers. Surface area goes as the square of the linear dimension, so doubling the radius multiplies the area of the sphere, 4πr², by four. The option is therefore not nonsense but a substitution of one scaling law for another, and it is the single most instructive wrong answer in the set, because the two laws differ in exactly the way that makes large vessels efficient: eight times the capacity for four times the material. A candidate who chooses it has usually recalled the exponent from the wrong formula. The safeguard is to name the quantity before choosing the power — length takes the first power, area the second, volume the third.
(d)2π : 1 — This option carries π into the answer, and π cannot appear in a ratio of two volumes of the same shape. Both spheres have volumes of the form four-thirds π times a cube, so the constant divides out exactly and what remains is a pure number. A ratio compares two quantities of the same kind, and the units and constants they share always vanish; if π survived, the ratio would change according to whether the volumes were measured in cubic centimetres or cubic metres, which is impossible. The option is aimed at a candidate who writes out both volume formulas in full, cancels carelessly, and leaves one π stranded — a mechanical slip rather than a conceptual error, and one avoided entirely by working with the scaling law instead of the formula.

Concept

Scaling is one of the most useful ideas in mensuration and one of the least often stated explicitly. If two solids have the same shape and every linear dimension of the second is k times the corresponding dimension of the first, then every length on the second is k times the length on the first, every area is k² times, and every volume is k³ times. The rule holds for any shape at all — spheres, cubes, cylinders, cones, irregular castings — because it follows from the dimensions of the quantities rather than from any particular formula. Three consequences are worth carrying. First, the constants in the formulas, such as the four-thirds π of a sphere or the one-third π of a cone, never appear in a ratio of two similar solids, because they are common to both. Second, the ratio of surface area to volume behaves as 1 ÷ k, so larger bodies have proportionately less surface — which is why a large vessel loses heat more slowly than a small one holding the same substance, and why small organisms lose water faster than large ones. Third, the rule runs backwards: if the volumes of two similar solids are in the ratio 27 : 64, their linear dimensions are in the ratio 3 : 4 and their surface areas in the ratio 9 : 16.

This item and question 111 are the mensuration pair of the paper, and they test opposite instincts. Question 111 punishes a candidate who reaches for the volume formula when a counting argument is wanted; this one rewards a candidate who realises that the formula's constant is irrelevant. Both belong to the same practical skill, which is knowing what a question actually constrains before computing anything. The stem here is also written in a way that repays close reading: it mentions a new container and an old container without ever giving a size, which is the clearest possible signal that the answer must be a ratio derived from a scaling law rather than a pair of numbers. When no dimension is supplied, none is needed, and a candidate who invents one and computes two volumes will reach the same 8 : 1 by a much longer road.

Key facts

The volume of a sphere is four-thirds π r³, so volume varies as the cube of the radius.
Doubling the radius multiplies the volume by 2³ = 8, giving a ratio of 8 : 1.
The factor four-thirds π is common to both containers and cancels in the ratio, so π cannot appear in the answer.
For similar solids, scaling every linear dimension by k multiplies lengths by k, areas by k² and volumes by k³.
Doubling the radius therefore multiplies the surface area of a sphere by 4 while multiplying its volume by 8.
The ratio of surface area to volume falls as 1 ÷ k, which is why larger vessels are more economical in material and lose heat more slowly.
The rule works backwards: volumes in the ratio 27 : 64 imply linear dimensions in the ratio 3 : 4.
A ratio of two quantities of the same kind is a pure number, with shared units and constants divided out.

Length goes as k, area as k², volume as k³

(2r)³ = 8r³, so the ratio is 8 : 1→Volume varies as the cube of a linear dimension, and that single law decides the item without any arithmetic worth the name. A sphere of radius r has volume four-thirds π r³, and doubling the radius replaces r by 2r.

The constant cancels→Both containers are spheres, so both volumes carry the same factor of four-thirds π, and in a ratio that factor appears above and below the line and divides out. What survives is r³ against (2r)³, that is 1 against 8 — so the formula never has to be remembered in full.

(a) 2 : 1 — the linear ratio, uncubed→Right for the diameters, the circumferences of the great circles, or any other length on the two spheres, because every length doubles when the shape is scaled up by two. But a solid grows in three independent directions at once: a cube of side 2 contains eight cubes of side 1, not two.

(b) 4 : 1 — the area ratio→The correct answer to a different question. Surface area goes as the square, so doubling the radius multiplies 4πr² by four. The most instructive wrong answer in the set, because the two laws differ in exactly the way that makes large vessels efficient: eight times the capacity for four times the material.

(d) 2π : 1 — π cannot survive a ratio→A ratio compares two quantities of the same kind, and the units and constants they share always vanish. If π survived, the ratio would change according to whether the volumes were measured in cubic centimetres or cubic metres, which is impossible. A mechanical slip from writing both formulas out and cancelling carelessly.

The rule, and what follows from it→For similar solids, scaling every linear dimension by k multiplies lengths by k, areas by k² and volumes by k³, whatever the shape. The ratio of surface area to volume therefore falls as 1 ÷ k, which is why a large vessel loses heat more slowly than a small one. And the rule runs backwards: volumes in the ratio 27 : 64 imply linear dimensions in the ratio 3 : 4.

Study next

Scaling laws for similar solids, and the corresponding rules for similar plane figuresSurface area and volume formulas for the sphere, cylinder, cone and hemisphereProblems in which a solid is melted and recast into smaller solids, where volume is conservedSurface-area-to-volume ratio and its consequences in physics and biologyRatio and proportion generally, including how to move between a ratio of areas and a ratio of volumes

Common traps

Reporting the linear ratio as the volume ratio, which answers a question about lengths rather than about capacity.
Using the square instead of the cube, which gives the ratio of surface areas rather than of volumes.
Leaving π in the answer, when a ratio of two similar volumes is always a pure number.
Assuming a size for the old container and computing two volumes, which wastes time and invites arithmetic slips.
Forgetting that the rule applies only to solids of the same shape; two dissimilar solids have no such simple relation.

Scaling questions are a fixture in these papers and appear in four guises. The first states the change in a linear dimension and asks for the effect on area or volume, as here. The second gives the change in area or volume and asks for the change in the linear dimension, which needs the square or cube root. The third disguises the idea in a physical story — a tank enlarged, a ball melted and recast, an ice cube divided into smaller ones — where the candidate must first identify what is conserved. The fourth asks for a percentage change rather than a ratio, so that a doubling becomes an increase of seven hundred per cent, which catches candidates who report the multiplier instead of the increase. All four are handled by the same three exponents, and it is worth being able to write them down without hesitation: one for length, two for area, three for volume.

Related PYQs

EPFO_APFC_2016_Q113

If the radius of a circle is reduced by 50%, its area will be reduced by

(a) 30%
(b) 50%
(c) 60%
(d) 75%

Answer(d) 75%

The same law one dimension lower — by how much the area of a circle falls when its radius is reduced by half.

EPFO_EOAO_2023_Q55

An ice cube with 10 cm side is divided into eight smaller cubes, each with same side. Which one of the following statements is correct in this context ?

(a) Total volume will increase and total surface area will decrease.
(b) Total volume will decrease and total surface area will increase.
(c) Total volume will remain the same and total surface area will increase.
(d) Total volume will increase and total surface area will remain the same.

Answer(c) Total volume will remain the same and total surface area will increase.

Scaling inside a physical story: a ten-centimetre ice cube cut into eight smaller cubes, with the effect on total volume and total surface area to be identified.

Practice

practice — not a real PYQ
If the radius of a spherical ball is halved, the ratio of the volume of the smaller ball to that of the original ball is
- (a)1 : 2
- (b)1 : 4
- (c)1 : 6
- (d)1 : 8
Answer(d) 1 : 8
practice — not a real PYQ
The surface areas of two spheres are in the ratio 9 : 16. What is the ratio of their volumes?
- (a)3 : 4
- (b)9 : 16
- (c)27 : 64
- (d)81 : 256
Answer(c) 27 : 64

Why