NCERT Exemplar Solutions Class 10 Maths Surface Areas and Volumes — Exercise 12.1

NCERT Exemplar Problems, Class 10 Mathematics — Surface Areas and Volumes, Exercise 12.1. All 20 multiple-choice questions solved.

Twenty questions in three groups. The first seven ask you to recognise an everyday object as a combination of standard solids. The middle group asks you to recast one solid into another. The last group applies the frustum formulae.

The formulae you need, and there is no way round learning them:

$$\text{sphere: } \ V = \frac43\pi r^3, \ \ S = 4\pi r^2 \qquad \text{cone: } \ V = \frac13\pi r^2 h, \ \ \text{CSA} = \pi r l$$

$$\text{cylinder: } \ V = \pi r^2 h, \ \ \text{CSA} = 2\pi r h \qquad \text{hemisphere: } \ V = \frac23 \pi r^3, \ \ \text{CSA} = 2\pi r^2$$

$$\text{frustum: } \ V = \frac{\pi h}{3}\left(r_1^2 + r_2^2 + r_1 r_2\right), \ \ \text{CSA} = \pi l (r_1 + r_2), \ \ l = \sqrt{h^2 + (r_1 – r_2)^2}$$

The one principle that carries the recasting questions is that melting changes the shape and not the amount:

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

So set the two volume expressions equal and solve. Surface area is not preserved — Question 17 is about exactly that.

A note on the current syllabus. This chapter has lost two of its four sections. The rationalised version is Introduction, Surface Area of a Combination of Solids, Volume of a Combination of Solids, Summary — and both “Frustum of a Cone” and “Conversion of Solid from One Shape to Another” are gone. The words “frustum” and “conversion” occur zero times in the chapter, and the Summary lists only two points, both about combining two basic solids.

Nine of these twenty questions — Questions 4, 6, 7, 13, 18 on frustums and 9, 10, 12, 17 on recasting — therefore test material the book no longer contains. They are all worth doing: frustums return in Class 11 and the volume-conservation idea is the most useful thing in the chapter.

Key insight. Work in exact fractions, never decimals. Question 13’s four options are $4950$, $4951$, $4952$ and $4953$, and Question 8’s are $142296$, $142396$, $142496$ and $142596$ — a rounded intermediate value will land you on the wrong one and give no hint that anything went wrong.

Choose the correct answer from the given four options:

Question 1

A cylindrical pencil sharpened at one edge is the combination of

Solution. The unsharpened body of the pencil is a cylinder — constant circular cross-section along its length. The sharpened end tapers to a single point, and a solid that tapers from a circle to a point is a cone.

So it is a cone and a cylinder.

Option (B) is the trap worth understanding. A frustum is a cone with its tip cut off, so it tapers from a larger circle to a smaller one — it ends in a flat circular face, not a point. A sharpened pencil ends in a point (that is what sharpening is for), so it is a full cone. A pencil sharpened and then snapped at the tip would be a frustum.

Question 2

A surahi is the combination of

Solution. A surahi is the traditional round-bodied water pot with a long narrow neck. The body is a full sphere — it bulges in every direction and curves back in at both top and bottom — and the neck is a cylinder rising from it.

So it is a sphere and a cylinder.

Option (B) fails because a hemisphere is flat on one face, and a surahi’s body is rounded underneath as well as on top. That is what distinguishes it from a bowl.

Question 3

A plumbline (sahul) is the combination of (see Fig. 12.2)

Solution. Read the figure from the top down. The upper portion is a dome — a curved surface meeting the join in a circle, with no vertical sides. That is a hemisphere, not a sphere, because it is flat where it joins rather than curving back in.

Below the join the solid tapers straight to a single point, so it is a cone.

The absence of any straight vertical side rules out (A), (C) and (D) at once — all three contain a cylinder, and a cylinder would show as two parallel lines in the outline. There are none.

Question 4

The shape of a glass (tumbler) (see Fig. 12.3) is usually in the form of

Solution. The sides are straight but sloping, and the two ends are circles of different sizes — wide at the open top, narrower at the base.

• Straight sloping sides rule out a sphere.
• Two circular ends of different sizes rule out a cylinder, whose ends are equal.
• Ending in a flat circle rather than a point rules out a cone.

A solid with straight sloping sides between two unequal parallel circles is a frustum of a cone — precisely a cone with its tip removed by a cut parallel to the base.

This question uses the frustum, which is not in the current Class 10 textbook.

Question 5

The shape of a gilli, in the gilli-danda game (see Fig. 12.4), is a combination of

Solution. The middle section has parallel sides — a cylinder. Each end tapers to a point — a cone.

There are two such ends, so the gilli is two cones and a cylinder.

Option (B) is what you get by counting only one of the two tapered ends; the figure is symmetric and both are there. Counting is the whole question, so trace the outline from one tip to the other and name each section as you pass it: cone, cylinder, cone.

Question 6

A shuttle cock used for playing badminton has the shape of the combination of

Solution. A shuttlecock has two parts: a rounded cork base, and a skirt of feathers flaring out from it.

The cork is a rounded dome sitting on a flat circular face — a hemisphere.

The feather skirt flares outward from the narrow cork end to a wider open rim. Its sides are straight and sloping, and both ends are circles of different sizes, with the wide end open. That is a frustum of a cone.

Option (C) is the tempting one, because “cone” describes the flare loosely. But a cone ends in a point, and the shuttlecock’s skirt ends in a wide open circle — it is the frustum, the cone with its tip cut away.

This question uses the frustum, which is not in the current Class 10 textbook.

Question 7

A cone is cut through a plane parallel to its base and then the cone that is formed on one side of that plane is removed. The new part that is left over on the other side of the plane is called

Solution. This is the definition, stated in words.

A cut parallel to the base produces a circular cross-section. Above the cut sits a small cone, similar to the original; below it sits the original base together with the new circular face. Removing the small cone leaves the lower part — bounded by two parallel circles of different sizes and a sloping curved surface.

That solid is a frustum of a cone. The word comes from the Latin for “piece” or “morsel”.

Note what the cut must be parallel for. A cut at any other angle gives an ellipse rather than a circle, and the piece left is not a frustum of a right circular cone. Question 19 of this exercise is the companion fact: the cross-section itself is a circle.

This question is the definition of the frustum, which is not in the current Class 10 textbook.

Question 8

A hollow cube of internal edge $22$ cm is filled with spherical marbles of diameter $0.5$ cm and it is assumed that $\frac18$ space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is

Solution. The four options differ by $100$ each, so the arithmetic has to be exact throughout. Keep everything as fractions.

Volume available. The cube holds $22^3 = 10648$ cm³, of which $\frac18$ is unfilled, so $\frac78$ is occupied:

$$10648 \times \frac78 = 1331 \times 7 = 9317 \text{ cm}^3$$

using $10648 \div 8 = 1331$.

Volume of one marble. Diameter $0.5$ cm gives $r = \frac14$ cm:

$$\frac43 \pi r^3 = \frac43 \times \frac{22}{7} \times \frac{1}{64} = \frac{88}{21 \times 64} = \frac{11}{168} \text{ cm}^3$$

The count:

$$\frac{9317}{\frac{11}{168}} = 9317 \times \frac{168}{11} = 847 \times 168 = 142\,296$$

using $9317 \div 11 = 847$.

Every division here came out whole, which is the sign that the exact route is the intended one. Working in decimals gives a marble volume of $0.0654761\ldots$ cm³, and rounding that to six places shifts the answer by hundreds — straight onto one of the wrong options.

Question 9

A metallic spherical shell of internal and external diameters $4$ cm and $8$ cm, respectively is melted and recast into the form a cone of base diameter $8$ cm. The height of the cone is

Solution. Melting preserves volume, so set the shell’s volume equal to the cone’s.

Halve every diameter first. Internal radius $r = 2$ cm, external radius $R = 4$ cm, cone base radius $= 4$ cm.

The shell is the material between two spheres:

$$V = \frac43\pi\left(R^3 – r^3\right) = \frac43\pi\left(64 – 8\right) = \frac{224\pi}{3}$$

The cone:

$$V = \frac13\pi(4)^2 h = \frac{16\pi h}{3}$$

Equate:

$$\frac{16\pi h}{3} = \frac{224\pi}{3} \implies 16h = 224 \implies h = 14 \text{ cm}$$

The $\pi$ and the $\frac13$ cancel on both sides, which is worth noticing before substituting a value for $\pi$ — none is needed.

The commonest slip is using the diameters directly: $\frac43\pi(8^3 – 4^3)$ would be eight times too large, and the resulting height of $112$ cm is not on the list, which at least warns you.

This question is a conversion of a solid from one shape to another, which is not in the current Class 10 textbook.

Question 10

A solid piece of iron in the form of a cuboid of dimensions $49$ cm $\times$ $33$ cm $\times$ $24$ cm, is moulded to form a solid sphere. The radius of the sphere is

Solution. Volume is conserved:

$$\frac43\pi r^3 = 49 \times 33 \times 24$$

Rather than multiplying out, keep the factors and cancel:

$$r^3 = \frac{3}{4\pi} \times 49 \times 33 \times 24 = \frac{3 \times 7}{4 \times 22} \times 49 \times 33 \times 24$$

Now $\frac{33}{22} = \frac32$ and $\frac{24}{4} = 6$, so

$$r^3 = 3 \times 7 \times 49 \times \frac32 \times 6 = 21 \times 49 \times 9 = 9261$$

And $9261 = 21^3$, so

$$r = 21 \text{ cm}$$

The number $9261$ is worth recognising: $21^3$. If yours is not a perfect cube, the cancellation has gone wrong somewhere, because these questions are always built to come out exactly.

A rough check: the cuboid holds $38\,808$ cm³, and a sphere of radius $21$ holds $\frac43 \times \frac{22}{7} \times 9261 = 38\,808$ cm³ ✓.

This question is a conversion of a solid from one shape to another, which is not in the current Class 10 textbook.

Question 11

A mason constructs a wall of dimensions $270$ cm $\times$ $300$ cm $\times$ $350$ cm with the bricks each of size $22.5$ cm $\times$ $11.25$ cm $\times$ $8.75$ cm and it is assumed that $\frac18$ space is covered by the mortar. Then the number of bricks used to construct the wall is

Solution. The same structure as Question 8: total volume, reduce by the fraction taken up by something else, divide by one unit.

The wall:

$$270 \times 300 \times 350 = 28\,350\,000 \text{ cm}^3$$

Space for bricks, mortar taking $\frac18$:

$$28\,350\,000 \times \frac78 = 3\,543\,750 \times 7 = 24\,806\,250 \text{ cm}^3$$

One brick:

$$22.5 \times 11.25 \times 8.75 = 2214.84375 \text{ cm}^3$$

The count:

$$\frac{24\,806\,250}{2214.84375} = 11\,200$$

The decimals here are exact — $22.5$, $11.25$ and $8.75$ are $\frac{45}{2}$, $\frac{45}{4}$ and $\frac{35}{4}$ — so the brick volume is $\frac{45 \times 45 \times 35}{32} = \frac{70875}{32}$ exactly, and the division comes out whole.

Note the direction of the mortar correction. The mortar occupies an eighth, so the bricks get only seven-eighths; dividing by $\frac78$ instead of multiplying would give $12\,800$ and is the error to watch for.

Question 12

Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter $2$ cm and height $16$ cm. The diameter of each sphere is

Solution. The cylinder’s base diameter is $2$ cm, so its radius is $1$ cm.

The cylinder:

$$V = \pi r^2 h = \pi (1)^2 (16) = 16\pi \text{ cm}^3$$

Each sphere gets a twelfth of that:

$$\frac43 \pi R^3 = \frac{16\pi}{12} = \frac{4\pi}{3}$$

Cancelling $\frac{4\pi}{3}$ from both sides:

$$R^3 = 1 \implies R = 1 \text{ cm}$$

The question asks for the diameter:

$$2R = 2 \text{ cm}$$

Option (A), $4$ cm, is what you get by reporting $2R$ after mistakenly finding $R = 2$ — or by using the cylinder’s diameter as its radius, which inflates the volume fourfold. Option (B) is the radius doubled after a different slip.

The final halving-or-doubling step catches people in nearly every question of this kind: find $R$, then read the question again to see whether $R$ or $2R$ was wanted.

This question is a conversion of a solid from one shape to another, which is not in the current Class 10 textbook.

Question 13

The radii of the top and bottom of a bucket of slant height $45$ cm are $28$ cm and $7$ cm, respectively. The curved surface area of the bucket is

Solution. The bucket is a frustum, and the slant height is given directly — so no Pythagoras is needed, which is the small kindness in this question.

$$\text{CSA} = \pi l \left(r_1 + r_2\right) = \frac{22}{7} \times 45 \times (28 + 7) = \frac{22}{7} \times 45 \times 35$$

Cancel the $7$ into the $35$:

$$= 22 \times 45 \times 5 = 22 \times 225 = 4950 \text{ cm}^2$$

The options are $4950$, $4951$, $4952$, $4953$. That is deliberate: they are one apart, so any rounding of $\frac{22}{7}$ to $3.14$ lands you elsewhere. With $3.14$ the answer would be $3.14 \times 45 \times 35 = 4945.5$, which is not on the list at all — a useful sign that the exact fraction was intended.

Note also that the CSA excludes the flat base; a bucket is open at the top, and “curved surface area” means only the sloping sides. Question 8 of Exercise 12.2 is about what to include when the base is wanted too.

This question uses the frustum formulae, which are not in the current Class 10 textbook.

Question 14

A medicine-capsule is in the shape of a cylinder of diameter $0.5$ cm with two hemispheres stuck to each of its ends. The length of entire capsule is $2$ cm. The capacity of the capsule is

Solution. The cylinder is not $2$ cm long, and that is the whole question.

The diameter is $0.5$ cm, so $r = 0.25$ cm, and each hemisphere stuck on the end projects by one radius. Two of them take up

$$0.25 + 0.25 = 0.5 \text{ cm}$$

of the total length, leaving the cylinder at

$$2 – 0.5 = 1.5 \text{ cm}$$

The cylinder:

$$\pi r^2 h = \frac{22}{7} \times 0.0625 \times 1.5 = \frac{22}{7} \times 0.09375$$

The two hemispheres together make one sphere:

$$\frac43 \pi r^3 = \frac43 \times \frac{22}{7} \times 0.015625 = \frac{22}{7} \times 0.0208\overline{3}$$

Total:

$$\frac{22}{7}\left(0.09375 + 0.0208\overline{3}\right) = \frac{22}{7} \times 0.1145833\ldots = 0.36 \text{ cm}^3$$

Taking the cylinder as the full $2$ cm gives $0.43$ cm³ — not among the options, which is the question warning you. Whenever hemispheres or a dome are “stuck on the ends”, the stated overall length always includes them.

Question 15

If two solid hemispheres of same base radius $r$ are joined together along their bases, then curved surface area of this new solid is

Solution. Two hemispheres joined along their flat faces make a sphere.

The curved surface of each hemisphere is $2\pi r^2$, and both survive on the outside of the new solid:

$$2\pi r^2 + 2\pi r^2 = 4\pi r^2$$

which is exactly the surface area of a sphere of radius $r$, as it must be.

Why option (B), $6\pi r^2$, is wrong is the point of the question. A hemisphere’s total surface area is $2\pi r^2 + \pi r^2 = 3\pi r^2$ — the dome plus its flat circular face. Adding two of those gives $6\pi r^2$. But when the two are joined, the two flat faces meet and become internal: they are no longer surface at all.

The rule is general. Whenever two solids are stuck together, the faces in contact vanish from the surface area, and you subtract them from the sum of the separate totals. Questions 1 to 3 of Exercise 12.2 are all this same point.

Question 16

A right circular cylinder of radius $r$ cm and height $h$ cm $(h > 2r)$ just encloses a sphere of diameter

Solution. A sphere inside a cylinder is limited by whichever is the tighter constraint: the cylinder’s width or its height.

• To fit across, the sphere’s diameter cannot exceed $2r$.
• To fit along, it cannot exceed $h$.

The condition $h > 2r$ tells you which one binds: the cylinder is taller than it is wide, so the width is the tighter constraint and the largest sphere has diameter

$$2r \text{ cm}$$

The sphere touches the curved wall all the way round but has room to spare above and below.

The condition in the brackets is the question. Had it read $h < 2r$ — a short wide tin — the height would bind instead and the answer would be $h$, option (C). A condition tacked onto a question in brackets is almost never decoration.

Question 17

During conversion of a solid from one shape to another, the volume of the new shape will

Solution. Melting and recasting rearranges the material without adding or removing any, so the volume is conserved:

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

That single principle is what makes Questions 9, 10 and 12 solvable at all — each sets one volume expression equal to another.

Surface area is a different matter, and is not conserved. Recasting a sphere into a long thin wire increases the surface area enormously while the volume is unchanged; melting many small marbles into one large ball decreases it. Of all the measurements of a solid, volume is the one that survives melting, and that is exactly why every recasting question is a volume equation.

This question states the principle behind the conversion of solids, which is not in the current Class 10 textbook.

Question 18

The diameters of the two circular ends of the bucket are $44$ cm and $24$ cm. The height of the bucket is $35$ cm. The capacity of the bucket is

Solution. A bucket is a frustum, and “capacity” means volume. Halve the diameters first: $r_1 = 22$ cm, $r_2 = 12$ cm.

$$V = \frac{\pi h}{3}\left(r_1^2 + r_2^2 + r_1 r_2\right) = \frac{22}{7} \times \frac{35}{3} \times \left(484 + 144 + 264\right)$$

The bracket is $892$, and $\frac{35}{7} = 5$:

$$= 22 \times 5 \times \frac{892}{3} = \frac{98\,120}{3} = 32\,706.\overline{6} \text{ cm}^3$$

Convert to litres. One litre is $1000$ cm³:

$$\frac{32\,706.67}{1000} = 32.7 \text{ litres}$$

Two places to lose it. Using the diameters instead of the radii inflates the answer fourfold. And forgetting the third term $r_1r_2$ in the bracket gives $628$ instead of $892$, hence about $23$ litres — the frustum formula has three terms, not two, and the cross term is easy to drop.

This question uses the frustum volume formula, which is not in the current Class 10 textbook.

Question 19

In a right circular cone, the cross-section made by a plane parallel to the base is a

Solution. A cross-section is the flat face exposed by the cut — a two-dimensional region, not a solid.

Every horizontal slice of a right circular cone is a circle, smaller the nearer the cut is to the apex. So the cross-section is a circle.

Option (B) is the trap, and Question 7 is where it comes from. There the same cut was described and the answer was “frustum of a cone” — but that was the name of the solid left behind, not of the face. Here the question asks for the cross-section itself.

So Questions 7 and 19 describe the identical operation and want different things:

• the piece below the cut is a frustum (a solid)
• the face made by the cut is a circle (a plane region)

Options (C) and (D) can be dismissed on dimension alone: a sphere and a hemisphere are solids, and no flat cut produces one.

Question 20

Volumes of two spheres are in the ratio $64 : 27$. The ratio of their surface areas is

Solution. Go through the radii — never directly from volumes to areas.

Volumes to radii. Volume goes as the cube of the radius:

$$\frac{r_1^3}{r_2^3} = \frac{64}{27} \implies \frac{r_1}{r_2} = \frac{\sqrt[3]{64}}{\sqrt[3]{27}} = \frac43$$

Radii to surface areas. Surface area goes as the square of the radius:

$$\frac{S_1}{S_2} = \frac{r_1^2}{r_2^2} = \left(\frac43\right)^2 = \frac{16}{9}$$

So the ratio is $16 : 9$.

The general rule, worth carrying: if lengths are in the ratio $k$, then areas are in the ratio $k^2$ and volumes in the ratio $k^3$. To get from volumes to areas, take the cube root and then square — that is, raise to the power $\frac23$:

$$\left(\frac{64}{27}\right)^{2/3} = \frac{16}{9}$$

Option (B), $4 : 3$, is the ratio of the radii, found on the way. Option (C) is $16 : 9$ inverted.

Common mistakes

• Questions 9, 10, 12 and 18 — using a diameter where a radius belongs. Every one of these gives at least one diameter, and the error inflates a volume eightfold or a surface area fourfold. Write the radii on their own line before starting.
• Questions 8, 11 and 13 — rounding early. The options in Question 13 are one apart and in Question 8 are a hundred apart. Keep $\frac{22}{7}$ as a fraction and cancel; every one of these questions is built so that the divisions come out whole.
• Question 14 — taking the stated length as the cylinder’s. Hemispheres stuck on the ends each add one radius, so the cylindrical part is $2 – 0.5 = 1.5$ cm. The same applies to any “total length” or “total height” in this chapter.
• Question 15 — adding total surface areas. When two solids are joined, the faces in contact stop being surface. Two hemispheres give $4\pi r^2$, not $6\pi r^2$ — and the check is that the result is a sphere.
• Questions 7 and 19 — confusing the solid with the cross-section. The same parallel cut produces a frustum (the piece) and a circle (the face). Read which the question wants.
• Question 18 — dropping the cross term. The frustum volume has three terms, $r_1^2 + r_2^2 + r_1r_2$. Leaving out $r_1r_2$ loses about a third of the answer.
• Question 20 — going straight from volumes to areas. Lengths scale as $k$, areas as $k^2$, volumes as $k^3$. Always route through the radii: cube root, then square.
• Question 11 — dividing by the fraction instead of multiplying. The mortar occupies an eighth, so the bricks get $\frac78$ of the wall. Multiplying and dividing differ here by a factor of $\frac{64}{49}$, about $14\%$.

Practise next

• Exercises 12.2 and 12.3 — eight reasoning statements and fourteen short-answer questions, where the combination rules of Question 15 are tested directly and the recasting principle of Question 17 is applied.
• Exercise 12.4 — twenty long-answer questions, including flow-rate and cost problems built on these same formulae.
• Exercise 11.1 — worth revisiting for the circle areas that every one of these solids is built from.