Volume and Surface Area (Mensuration of solids) is one of the most reliable scoring areas in CDS and OTA Maths. The questions are formula-driven, so once you memorise a small table of formulas and practise a few traps, you can finish each sum in under a minute. This Cavalier guide gives you every 3D formula, the units logic, and exam-style solved questions.
Why this topic is a guaranteed scorer
In almost every CDS Elementary Mathematics paper, 3 to 6 marks come directly from solids — volume, total surface area (TSA), curved/lateral surface area (CSA), and recasting problems. Unlike geometry proofs that demand reasoning and construction, these questions are pure substitution: identify the solid, pick the correct formula, plug in the numbers, and you are done. That predictability is exactly why disciplined students treat this chapter as guaranteed marks rather than a gamble.
The OTA paper (for short-service commission) leans even more heavily on quick mensuration, because it rewards speed above all else. With negative marking in play, a candidate who can clear a mensuration sum in forty seconds gets both the mark and the saved time to attempt harder algebra or trigonometry questions. If you lock these formulas into memory and drill ten questions a day for a fortnight, you will bank easy marks while others fumble over which surface to count.
The chapter draws directly from the NCERT Class 8, 9 and 10 surface-area-and-volume chapters, so nothing here is beyond your school syllabus. What changes at the CDS level is the speed and the traps, not the depth of the mathematics.
Volume is a 3D quantity, measured in cubic units (cm3, m3). Surface area is a 2D quantity wrapped around a solid, measured in square units (cm2, m2). Mixing the two is the No.1 silly mistake.
Core terms: volume, TSA and CSA
Before the formulas, fix three ideas in your head.
- Volume — the space a solid occupies, or how much liquid it can hold (capacity).
- Total Surface Area (TSA) — the area of every outer face, including flat tops and bottoms.
- Curved / Lateral Surface Area (CSA/LSA) — only the curved or side faces, excluding the flat top and base.
For a closed tin, you want TSA. For the label wrapped around a can, or the paint on a pillar's side, you want CSA/LSA. For the area of cloth needed to make a conical tent, you again want only the curved surface, because the floor is open ground. Reading the question carefully decides which formula wins, and a single misread word can turn a correct calculation into a wrong answer.
A helpful mental model: imagine peeling the solid like an orange. The total peel is the TSA. If you throw away the flat circular lids and keep only the side strip, that strip is the CSA. Holding this picture stops you from blindly writing TSA every time.
Underline the words "open", "closed", "curved", "lateral" and "hollow" in the question. They tell you exactly which surface to include or drop.
Cube and cuboid formulas
A cuboid has length l, breadth b and height h. A cube is a special cuboid where l = b = h = a.
Cuboid: Volume = l × b × h; TSA = 2(lb + bh + hl); LSA = 2h(l + b).
Cube: Volume = a3; TSA = 6a2; LSA = 4a2; Diagonal = a√3.
Cuboid diagonal = √(l2 + b2 + h2).
The diagonal formula is a frequent CDS favourite, because it links mensuration with the Pythagoras idea in three dimensions. A common question gives you the diagonal and one or two edges and asks for the missing edge or the volume, so be ready to work the formula both forwards and backwards.
Another classic asks how many small cubes of edge 1 cm fit inside a larger cuboid. The answer is simply the larger volume divided by the small volume, because every cube fills perfectly with no gaps. Surface area, by contrast, does not divide so cleanly, which is exactly the kind of distinction examiners test.
Cylinder: solid and hollow
A right circular cylinder has base radius r and height h. Throughout, take π = 22/7 unless the question says otherwise.
Solid cylinder: Volume = πr2h; CSA = 2πrh; TSA = 2πr(r + h).
Hollow cylinder/pipe (outer R, inner r): Volume of material = πh(R2 − r2); TSA = 2π(R + r)(R − r + h).
For an open pipe or a tube, do not add the two circular ends. Tube questions usually want only the curved surfaces, so include the top and bottom rings (annulus area) only if the ends are explicitly described as solid rims.
Cone and slant height
A cone has base radius r, vertical height h, and slant height l. These three are tied together by Pythagoras.
Slant height: l = √(r2 + h2).
Volume = (1/3)πr2h; CSA = πrl; TSA = πr(l + r).
Notice the volume of a cone is exactly one-third that of a cylinder with the same base and height. Examiners love comparing the two: a typical question fills a conical vessel with water and pours it into a cylindrical one of the same base, and the water level rises to only one-third of the cone's height. Recognising this 1 : 3 relationship lets you answer such items almost without calculation. The same logic explains why three identical cones can be melted to fill exactly one cylinder of matching base and height.
If a question gives h but the surface-area formula needs l, immediately compute l = √(r2 + h2) before doing anything else. Many cone errors come from using h where l was required.
Sphere and hemisphere
For a sphere of radius r, surface area and volume depend only on r.
Sphere: Volume = (4/3)πr3; Surface area = 4πr2.
Solid hemisphere: Volume = (2/3)πr3; CSA = 2πr2; TSA = 3πr2.
Hollow sphere (outer R, inner r): Volume of material = (4/3)π(R3 − r3).
A hemisphere's TSA is 3πr2 because it adds the flat circular base (πr2) to the curved 2πr2. Forgetting the base is a classic slip.
Frustum of a cone
A frustum is what remains when the top of a cone is sliced off parallel to the base. It appears as buckets, glasses and lampshades in CDS word problems. Let the two radii be R (bottom) and r (top), height h, and slant height l.
Slant height: l = √(h2 + (R − r)2).
Volume = (1/3)πh(R2 + r2 + Rr).
CSA = πl(R + r); TSA = πl(R + r) + πR2 + πr2.
Bucket questions often ask only for CSA plus the bottom (since a bucket is open at the top), so read carefully before adding both circular ends.
Units and conversion logic
Many marks are lost not on formulas but on unit conversion. Keep these relations ready.
- 1 m = 100 cm, so 1 m2 = 10,000 cm2 and 1 m3 = 10,00,000 cm3.
- 1 litre = 1000 cm3; 1 m3 = 1000 litres.
- For capacity, always convert the final volume to litres if the question asks how much water it holds.
When you scale a length by k, area scales by k2 and volume by k3. So doubling a sphere's radius makes its volume 8 times bigger, not twice. Many candidates answer 2× and lose the mark.
Melting, recasting and combination solids
A whole sub-class of questions melts one solid and reshapes it into another. The trick is simple: volume is conserved during melting, even though surface area changes.
When a solid is melted and recast, set Volume(old) = Volume(new) and solve for the unknown dimension. Surface area is irrelevant to the volume equation.
For combination solids (a cone on a hemisphere, a capsule, a tent = cylinder + cone), add the volumes and add only the exposed surfaces. Do not count the joining face, because it is hidden inside the body and contributes nothing to paint, cloth or polish.
Take a circus tent shaped like a cylinder topped by a cone. Its volume is the cylinder's volume plus the cone's volume. But the canvas needed is the cylinder's curved surface plus the cone's curved surface only — not the shared circle where they meet, and not the open floor. A toy made of a cone mounted on a hemisphere works the same way: total height is cone height plus radius, and only the cone's slant surface and the hemisphere's curved surface are visible. Train your eye to spot which faces vanish inside the join, and combination questions become straightforward bookkeeping.
Worked example: cone from a cylinder
A solid metallic cylinder of radius 6 cm and height 15 cm is melted and recast into solid cones, each of radius 2 cm and height 3 cm. How many cones are formed?
So 135 cones are obtained. Notice π cancels neatly — a sign you set up the equation correctly. Whenever π survives to the final answer in a melting problem, recheck your working, because the question almost always engineers it to cancel. This single example captures the whole melting-and-recasting family: write the conserved volume on each side, cancel common factors, and divide.
More traps examiners set
Beyond units, watch for these recurring tricks in CDS papers.
- Radius vs diameter: if the question gives diameter, halve it before substituting. The single most common error in the topic.
- Open vs closed: an open box or cup drops one face; an open cylindrical tank drops the top circle.
- Hemisphere TSA: include the flat base (3πr2), but for a bowl's inner curved surface use only 2πr2.
- Same volume, different shape: equal volumes do not mean equal surface areas. A sphere has the least surface area for a given volume.
Sketch the solid in the margin and label r, h and l. A 5-second diagram prevents 90% of formula mix-ups.
Previous-year style question
Q. The radii of two spheres are in the ratio 2 : 3. What is the ratio of their volumes?
Answer: Volume scales as the cube of the radius. So the ratio of volumes = 23 : 33 = 8 : 27. The radius ratio 2 : 3 maps to a volume ratio of 8 : 27.
- Volume is cubic units; surface area is square units — never mix them.
- Cube: V = a3, TSA = 6a2; Cuboid diagonal = √(l2+b2+h2).
- Cylinder V = πr2h, CSA = 2πrh; Cone V = (1/3)πr2h, CSA = πrl.
- Sphere V = (4/3)πr3, SA = 4πr2; Hemisphere TSA = 3πr2.
- Melting conserves volume; scaling length by k scales volume by k3.
- Check radius vs diameter, and open vs closed, before substituting.
Frequently asked questions
Which formulas must I memorise first for CDS mensuration?
Start with cube, cuboid, cylinder, cone, sphere and hemisphere — their volume, CSA and TSA. These six solids cover the large majority of CDS and OTA questions on the topic.
What value of pi should I use in the exam?
Use 22/7 unless the question specifies otherwise, because most CDS numbers are chosen to cancel cleanly with 22/7. If π cancels out entirely, the exact value does not matter.
How do I handle melting and recasting questions?
Equate the volume of the original solid to the total volume of the new solids, since melting conserves volume. Surface area changes and should be ignored in that equation.
Why is the hemisphere's total surface area 3 pi r squared?
A hemisphere has a curved surface of 2πr2 plus a flat circular base of πr2. Adding them gives 3πr2. For an open bowl's inner surface, use only the curved 2πr2.
If a sphere's radius doubles, how does volume change?
Volume depends on r3, so doubling the radius multiplies the volume by 23 = 8 times. Surface area, depending on r2, becomes 4 times larger.
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