Melt a Solid Into a New Shape and Only One Quantity Survives Unchanged

Melt a metal sphere and pour it into a cone-shaped mould. The surface area changes completely. The shape changes completely. The volume does not change at all.

That single sentence solves most of the hard questions in this chapter, and it is the reason volumes behave so differently from surface areas. Material is conserved, so however you reshape it:

$$\text{volume before}=\text{volume after}$$

And it applies to far more situations than melting.

- Recasting a solid into a different solid, or into several smaller ones
- Emptying a bucket of sand into a heap on the ground
- Digging a well and spreading the earth to make a platform
- Dropping an object into water, where the water pushed up equals the object's volume
- Filling cones from a cylindrical tub of ice cream

In every one of those the method is identical: write the volume of what you started with, write the volume of what you ended with, and set them equal. The unknown is whatever the question asks for.

Joined and carved solids work by the same arithmetic as before, but with one important difference from Part 1. When two solids are glued, their volumes simply add — no correction is needed, because the interface has no volume. Volumes are easier than surface areas, and that is worth knowing: there is no equivalent here of the disappearing glued faces.

$$\text{combined volume}={\text{volume}}_1+{\text{volume}}_2$$
$$\text{carved solid}=\text{volume of the whole}-\text{volume removed}$$

**Take $\pi =\frac{22}{7}$ where the numbers suit it**, and leave $\pi$ in the answer when a question says "in terms of $\pi$" — which it often does, precisely because the $\pi$ cancels in a conservation problem.

This page covers the second part of the CBSE Class 10 Maths chapter on surface areas and volumes: the volume formulas, combined and carved solids, conservation of volume, and capacity and cost problems.

Add the two volumes. Nothing is lost at the join, because a surface has no volume.

Worked example 1 — a cone on a hemisphere. A solid is in the shape of a cone standing on a hemisphere, both of radius $1$ cm, with the cone's height equal to its radius. Find the volume in terms of $\pi$.

$$\text{cone}=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi \times 1\times 1=\frac{\pi }{3}{\text{ cm}}^3$$
$$\text{hemisphere}=\frac{2}{3}\pi r^3=\frac{2\pi }{3}{\text{ cm}}^3$$
$$\text{total}=\frac{\pi }{3}+\frac{2\pi }{3}=\pi {\text{ cm}}^3$$

**An answer of exactly $\pi$**, which is why the question asks for it in terms of $\pi$ — substituting $\frac{22}{7}$ would hide how clean the result is.

And notice the contrast with Part 1. For the surface area of this solid you had to leave out two circles; for the volume you simply add. That is the one respect in which volume questions are kinder than surface-area questions.

Worked example 2 — an ice-cream cone with a domed top. A cone of height $12$ cm and diameter $6$ cm is topped by a hemisphere of the same radius. Find its volume in terms of $\pi$.

The radius is $3$ cm, so

$$\text{cone}=\frac{1}{3}\pi \times 9\times 12=36\pi {\text{ cm}}^3$$
$$\text{hemisphere}=\frac{2}{3}\pi \times 27=18\pi {\text{ cm}}^3$$
$$\text{total}=54\pi {\text{ cm}}^3$$

**Keeping $\pi$ unevaluated was deliberate**, because the next section divides this into a cylinder's volume and the $\pi$ cancels.

Worked example 3 — a capsule. A capsule is a cylinder of length $9$ mm with a hemisphere of radius $2.5$ mm on each end. Find its volume. Take $\pi =\frac{22}{7}$.

$$\text{cylinder}=\pi r^{2}h=\frac{22}{7}\times 6.25\times 9=\frac{1237.5}{7}{\text{ mm}}^3$$

The two hemispheres make one sphere:

$$\frac{4}{3}\pi r^3=\frac{4}{3}\times \frac{22}{7}\times 15.625=\frac{1375}{21}{\text{ mm}}^3$$

$$\text{total}=\frac{3712.5}{21}+\frac{1375}{21}=\frac{5087.5}{21}\approx 242.3{\text{ mm}}^3$$

Two hemispheres making a sphere is the same shortcut as in Part 1, and it works for volumes too: $2\times \frac{2}{3}\pi r^3=\frac{4}{3}\pi r^3$.

Worked example 4 — a solid iron pole. A solid iron pole consists of a cylinder of height $220$ cm and base diameter $24$ cm, surmounted by another cylinder of height $60$ cm and radius $8$ cm. Find its volume. Take $\pi =\frac{22}{7}$.

Lower cylinder, radius $12$ cm:

$$\frac{22}{7}\times 144\times 220=\frac{22\times 31680}{7}=\frac{696960}{7}\approx 99565.7{\text{ cm}}^3$$

Upper cylinder, radius $8$ cm:

$$\frac{22}{7}\times 64\times 60=\frac{22\times 3840}{7}=\frac{84480}{7}\approx 12068.6{\text{ cm}}^3$$

$$\text{total}\approx 99565.7+12068.6=111634.3{\text{ cm}}^3$$

The two radii were different, which is the point of this example — you cannot factorise $\pi r^2$ out of the two terms, so each must be computed separately. A question with a shared radius lets you factorise; one without it does not, and noticing which you have saves or costs a line.

The habit to build. Write the two volumes as separate lines with their own formulas before adding, and **keep $\pi$ symbolic until the end.** Combination questions are long enough that a decimal introduced early grows into a rounding error by the last line.

Subtract for carving; equate for displacement. In both cases the key sentence is that volume is conserved.

Worked example 1 — a cone scooped from a cylinder. A solid cylinder of radius $7$ cm and height $10$ cm has a cone of the same radius and height $6$ cm scooped out of one end. Find the volume of the remaining solid. Take $\pi =\frac{22}{7}$.

$$\text{cylinder}=\frac{22}{7}\times 49\times 10=22\times 70=1540{\text{ cm}}^3$$
$$\text{cone}=\frac{1}{3}\times \frac{22}{7}\times 49\times 6=\frac{1}{3}\times 22\times 42=308{\text{ cm}}^3$$
$$\text{remaining}=1540-308=1232{\text{ cm}}^3$$

Check the proportion. The cone occupies $\frac{308}{1540}=\frac{1}{5}$ of the cylinder, which is right: it has $\frac{6}{10}$ of the height and $\frac{1}{3}$ of the volume for that height, and $\frac{3}{5}\times \frac{1}{3}=\frac{1}{5}$. Correct.

Worked example 2 — a sphere dropped into water. A cylindrical vessel of internal radius $10$ cm contains water. A solid sphere of radius $5$ cm is dropped in and is completely submerged. Find the rise in the water level.

The water pushed up occupies exactly the sphere's volume, and it forms a cylindrical layer of radius $10$ cm and height equal to the rise $x$:

$$\pi \times 100\times x=\frac{4}{3}\pi \times 125$$

**The $\pi$ cancels**, which is why the answer is exact:

$$100x=\frac{500}{3}⟹x=\frac{5}{3}\approx 1.67\text{ cm}$$

Two conditions this depends on, and both should be stated. The sphere must be fully submerged, or only part of its volume displaces water; and the vessel must not overflow, or the displaced water is lost rather than raising the level. A question that says "completely submerged" is telling you the first condition holds.

Worked example 3 — water flowing out. A vessel is an inverted cone of height $8$ cm with a top radius of $5$ cm, filled with water to the brim. Lead shots, each of radius $0.5$ cm, are dropped in until one fourth of the water flows out. Find the number of shots dropped.

The cone's volume:

$$\frac{1}{3}\pi \times 25\times 8=\frac{200\pi }{3}{\text{ cm}}^3$$

The water that flowed out is one fourth of that, and it equals the total volume of the shots:

$$\frac{1}{4}\times \frac{200\pi }{3}=\frac{50\pi }{3}{\text{ cm}}^3$$

One shot:

$$\frac{4}{3}\pi \times 0.5^3=\frac{4}{3}\pi \times 0.125=\frac{\pi }{6}{\text{ cm}}^3$$

The number of shots:

$$\frac{50\pi /3}{\pi /6}=\frac{50}{3}\times 6=100$$

**The $\pi$ cancelled again, and the answer is a whole number — as it must be, since you cannot drop a fraction of a lead shot. A non-integer answer here means an arithmetic slip, and that is the most useful check available in this whole family of questions.

Worked example 4 — a bucket of sand into a heap.** A cylindrical bucket of height $32$ cm and radius $18$ cm is full of sand. The sand is emptied onto the ground and forms a conical heap of height $24$ cm. Find the radius and the slant height of the heap.

Volume is conserved:

$$\frac{1}{3}\pi r^2\times 24=\pi \times 324\times 32$$
$$8r^2=10368⟹r^2=1296⟹r=36\text{ cm}$$

Then the slant height, which needs the Pythagoras theorem:

$$l=\sqrt{36^2+24^2}=\sqrt{1296+576}=\sqrt{1872}=12\sqrt{13}\approx 43.27\text{ cm}$$

Notice that the heap is wider than the bucket. Its radius is $36$ cm against the bucket's $18$ cm, which makes sense: a cone holds only a third as much as a cylinder of the same base and height, so to hold the same sand at a smaller height it must spread much wider. That plausibility check is worth a moment — an answer of $r=9$ cm would have been impossible.

Worked example 5 — one sphere into several. A solid metallic sphere of radius $3$ cm is melted and recast into three smaller spheres. Two of them have radii $1.5$ cm and $2$ cm. Find the radius of the third.

**Volume is conserved, and the $\frac{4}{3}\pi$ cancels from every term**, leaving just the cubes:

$$3^3=1.5^3+2^3+r^3$$
$$27=3.375+8+r^3$$
$$r^3=27-11.375=15.625⟹r=2.5\text{ cm}$$

Check: $2.5^3=15.625$, and $3.375+8+15.625=27$. Correct.

**Cancelling the $\frac{4}{3}\pi$ first is what makes this a one-line problem. Whenever every solid in a conservation equation is the same type, the common factor cancels and you are left with a relation between the linear dimensions alone. Spotting that saves most of the arithmetic.

Worked example 6 — filling cones from a tub.** A cylindrical container of diameter $12$ cm and height $15$ cm is full of ice cream, which is to be filled into cones of height $12$ cm and diameter $6$ cm, each with a hemispherical top. How many such cones can be filled?

The cylinder:

$$\pi \times 36\times 15=540\pi {\text{ cm}}^3$$

Each cone with its dome, from worked example 2 of the previous section: $54\pi$ cm³.

$$\text{number}=\frac{540\pi }{54\pi }=10$$

**The $\pi$ cancelled and the answer is a whole number.** Both of those are signs the working is right, and both are reasons to keep $\pi$ symbolic rather than substituting $\frac{22}{7}$ at the start.

Compute the volume in cubic metres or cubic centimetres, convert with the litre relation, and then apply the rate. As in Part 1, the marks are in the conversion, not the formula.

Worked example 1 — a water tank. A cylindrical water tank has diameter $2.8$ m and height $2.5$ m. Find its capacity in litres, and the cost of filling it at ₹$20$ per $1000$ litres. Take $\pi =\frac{22}{7}$.

The radius is $1.4$ m, so

$$V=\frac{22}{7}\times 1.96\times 2.5=22\times 0.28\times 2.5=15.4{\text{ m}}^3$$

Convert, using $1$ m³ $=1000$ litres:

$$15.4\times 1000=15400\text{ litres}$$

**The rate is per $1000$ litres**, so the number of units is $15.4$, and

$$\text{cost}=15.4\times 20=₹308$$

Notice the shortcut. Because the rate is per $1000$ litres and $1$ m³ is $1000$ litres, the cost is the volume in cubic metres times the rate. Reading the rate before converting saved a step.

Worked example 2 — a well and a platform. A well of diameter $7$ m is dug $20$ m deep, and the earth taken out is spread evenly to form a rectangular platform $22$ m by $14$ m. Find the height of the platform. Take $\pi =\frac{22}{7}$.

The earth dug out:

$$V=\frac{22}{7}\times 3.5^2\times 20=\frac{22}{7}\times 12.25\times 20=22\times 1.75\times 20=770{\text{ m}}^3$$

**That same earth forms a cuboid of base $22\times 14$:**

$$22\times 14\times h=770⟹308h=770⟹h=2.5\text{ m}$$

Check: $308\times 2.5=770$ m³. Correct.

Worked example 3 — a well and a circular embankment. A well of diameter $3$ m is dug $14$ m deep, and the earth is spread evenly all around it to form a circular embankment of width $4$ m. Find the height of the embankment.

The earth dug out, with radius $1.5$ m:

$$V=\pi \times 2.25\times 14=31.5\pi {\text{ m}}^3$$

The embankment is a ring, with outer radius $1.5+4=5.5$ m and inner radius $1.5$ m, so its base area is

$$\pi (5.5^2-1.5^2)=\pi (30.25-2.25)=28\pi {\text{ m}}^2$$

Setting the volumes equal:

$$28\pi h=31.5\pi ⟹h=\frac{31.5}{28}=1.125\text{ m}$$

**The $\pi$ cancelled once more. And the shape of the embankment is the step that decides the question: it is a ring, not a disc, because the well itself is still there in the middle. Using $\pi \times 5.5^2$ as the base area would have given $1.04$ m, and the error would be invisible in the working.

Worked example 4 — how many items.** How many spherical lead shots of diameter $4$ cm can be made from a solid cube of lead whose edge is $44$ cm? Take $\pi =\frac{22}{7}$.

The cube:

$$44^3=85184{\text{ cm}}^3$$

One shot, of radius $2$ cm:

$$\frac{4}{3}\times \frac{22}{7}\times 8=\frac{704}{21}{\text{ cm}}^3$$

The number:

$$85184÷\frac{704}{21}=85184\times \frac{21}{704}=121\times 21=2541$$

Check: $\frac{85184}{704}=121$ exactly, and $121\times 21=2541$. A whole number, as it must be.

Worked example 5 — filling bottles from a bowl. A hemispherical bowl of internal radius $9$ cm is full of liquid, which is to be filled into cylindrical bottles of diameter $3$ cm and height $4$ cm. How many bottles are needed?

$$\text{bowl}=\frac{2}{3}\pi \times 729=486\pi {\text{ cm}}^3$$
$$\text{one bottle}=\pi \times 2.25\times 4=9\pi {\text{ cm}}^3$$
$$\text{number}=\frac{486\pi }{9\pi }=54$$

The question asks how many bottles are needed, not how many can be completely filled — and here the division is exact so the two answers agree. When such a division is not exact, read the wording carefully: "how many can be filled" wants the whole number below, while "how many are needed" wants the next one up, because a part-filled bottle is still a bottle.

The layout that protects these marks. Write the starting volume, write the volume of one piece, then divide — three lines, each markable. **And keep $\pi$ symbolic**, because in every conservation question it cancels, and cancelling a symbol is safer than dividing two decimals.