These are the Exercise 12.2 Class 10 Maths NCERT Solutions for Surface Areas and Volumes, following the rationalised NCERT edition. This is the chapter’s volume exercise: all eight questions join two or three basic solids — cone, cylinder, hemisphere, sphere — and every answer obeys one rule: the volumes of the joined parts always add.
Nothing is lost at the join when you measure volume, even though joined faces vanish when you measure surface area. That single contrast between volume and surface area decides the setup of every question below.
Every question is reproduced word for word from the official NCERT Class 10 Mathematics textbook, then solved with the reasoning, the formula, the worked arithmetic and the unit. You can open the textbook PDF on ncert.nic.in and verify any step page by page.
Exercise 12.2 Class 10 Maths NCERT Solutions
The idea behind the whole exercise is stated in section 12.3 of the textbook: the volume of a solid formed by joining two basic solids is the sum of the volumes of the constituents (NCERT, p. 168). This is the opposite of the surface-area rule from section 12.2, where the faces at the join disappear and are not counted.
So each question is a two-step reading problem:
- Name the basic solids in the object — cone, cylinder, hemisphere, sphere or cuboid.
- Add their volumes. Subtract only when material is removed (the pen stand’s depressions in Q4) or water is displaced (Q5 and Q7).
This page solves exactly the eight questions grounded in the exercise; there are no intext questions. The surface-area problems of the chapter live in Exercise 12.1, and Exercise 12.2 is the chapter’s last exercise — section 12.4 is only the summary.
Formulas You Need: Volumes of the Combined Solids
The table collects every volume formula the eight solutions call on, in the form the textbook prints it.
| Solid | Volume formula | Used in |
|---|---|---|
| Cuboid (length \( l \), breadth \( b \), height \( h \)) | \( l \times b \times h \) | Q4 — the pen stand |
| Cylinder (radius \( r \), height \( h \)) | \( \pi r^2 h \) | Q2, Q3, Q6, Q7, Q8 |
| Cone (radius \( r \), height \( h \)) | \( \frac{1}{3}\pi r^2 h \) | Q1, Q2, Q4, Q5, Q7 |
| Hemisphere (radius \( r \)) | \( \frac{2}{3}\pi r^3 \) | Q1, Q3, Q7 |
| Sphere (radius \( r \)) | \( \frac{4}{3}\pi r^3 \) | Q5 — each lead shot; Q8 — the glass vessel |
The line that decides every setup: total volume = sum of the volumes of the parts. Subtract only for removed or displaced volume — the four conical depressions of Q4, and the displaced water of Q5 and Q7.
Unless a question says otherwise, take \( \pi = \frac{22}{7} \) (NCERT, p. 170). Q6 and Q8 switch to \( \pi = 3.14 \). Q1 asks for the volume ‘in terms of \( \pi \)’, so leave the symbol in the answer.
Section 12.2 builds the slant height \( l = \sqrt{r^2 + h^2} \) for curved surface areas (NCERT, p. 163), but no volume formula uses \( l \). If you catch yourself computing it in this exercise, you have slipped into a surface-area problem. The \( \pi r^2 \) area you practised in Areas Related to Circles is the factor that becomes \( \pi r^2 h \) here.
When density is given, the flow is fixed: mass = volume × density. Q6 uses 8 g for every 1 cm³ of iron.
Question 1: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
Answer: A cone standing on a hemisphere is two solids joined at a circular face. In a volume calculation that face is internal, so the combined volume is simply cone + hemisphere. The phrase ‘height of the cone is equal to its radius’ fixes \( h = 1 \) cm directly — it is the cone’s own height, not the total height of the solid.
Step 1: Note the given values: radius \( r = 1 \) cm for both solids and cone height \( h = 1 \) cm.
Step 2: Volume of the cone:
\[ \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (1)^2 (1) = \frac{\pi}{3} \text{ cm}^3 \]
Step 3: Volume of the hemisphere:
\[ \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (1)^3 = \frac{2\pi}{3} \text{ cm}^3 \]
Step 4: Add the two volumes:
\[ \frac{\pi}{3} + \frac{2\pi}{3} = \frac{3\pi}{3} = \pi \text{ cm}^3 \]
Final answer: The volume of the solid is \( \pi \) cm³, about \( 3.14 \) cm³.
Common error: reading ‘height of the cone is equal to its radius’ as ‘the whole solid is 2 cm tall’ and substituting \( h = 2 \). Also, ‘in terms of \( \pi \)’ is an instruction: do not multiply by \( \frac{22}{7} \). The exact answer stays \( \pi \) cm³.
Question 2: Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Answer: The 12 cm length spans the whole model — tip of one cone to tip of the other. Recover the cylinder’s own height first by subtracting the two cone heights: \( 12 – 2 – 2 = 8 \) cm.
The diameter 3 cm is shared by the cylinder and both cone bases, so every radius is \( 1.5 \) cm. The air the model holds is the sum of the cylinder and the two cones; the thin sheet’s own thickness is ignored by the assumption the question gives.
Step 1: Radius from the diameter: \( r = \frac{3}{2} = 1.5 \) cm.
Step 2: Cylinder height: \( 12 – 2 – 2 = 8 \) cm.
Step 3: Volume of the cylinder:
\[ \pi r^2 h = \pi (1.5)^2 (8) = 18\pi \text{ cm}^3 \]
Step 4: Volume of one cone:
\[ \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (1.5)^2 (2) = 1.5\pi \text{ cm}^3 \]
Step 5: Add the cylinder and both cones:
\[ 18\pi + 3\pi = 21\pi = 21 \times \frac{22}{7} = 66 \text{ cm}^3 \]
Final answer: The model contains \( 66 \) cm³ of air.
Common error: treating all 12 cm as the cylinder’s height. Sketch the model and label the two cone heights (2 cm each) before writing any formula; the cylinder’s 8 cm then falls out of \( 12 – 4 \).
Question 3: A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig. 12.15).
Answer: The figure below shows that the 5 cm length is the whole jamun — it already includes the two hemispherical ends. Each end reaches 1.4 cm (its own radius) into that length, so the cylinder between them is only \( 5 – 2(1.4) = 2.2 \) cm long.
Two hemispheres of the same radius always combine into exactly one sphere, so one jamun is one cylinder plus one sphere. Only 30% of that total volume is syrup, and there are 45 jamuns to account for.
Step 1: Radius \( r = \frac{2.8}{2} = 1.4 \) cm; cylinder height \( h = 5 – 2(1.4) = 2.2 \) cm.
Step 2: Cylindrical part of one jamun:
\[ \pi (1.4)^2 (2.2) = 4.312\pi \approx 13.55 \text{ cm}^3 \]
Step 3: The two hemispherical ends together form one sphere:
\[ \frac{4}{3}\pi (1.4)^3 = \frac{4}{3}\pi (2.744) \approx 3.66\pi \approx 11.50 \text{ cm}^3 \]
Step 4: Volume of one jamun \( \approx 7.97\pi \approx 25.05 \) cm³.
Step 5: Syrup in one jamun \( = 0.30 \times 25.05 \approx 7.52 \) cm³.
In 45 jamuns:
\[ 7.52 \times 45 \approx 338.2 \text{ cm}^3 \]
Final answer: About \( 338.2 \) cm³ of syrup is found in the 45 gulab jamuns.
Common error: using the diameter 2.8 cm as the radius, or taking the whole 5 cm as the cylinder height. Label the figure before calculating — radius 1.4, cylinder 2.2, sphere 1.4 — and ‘30% of the volume’ becomes one multiplication by 0.3.
Question 4: A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig. 12.16).
Answer: The stand starts as a solid cuboid of wood, and the four conical holes are drilled out of it, as Fig. 12.16 shows. So the wood that remains equals the cuboid’s volume minus four cone volumes. The word ‘depression’ signals removal — the subtract case of this exercise, the opposite of Q1 where stacked solids add.
Step 1: Volume of the cuboid:
\[ 15 \times 10 \times 3.5 = 525 \text{ cm}^3 \]
Step 2: Volume of one conical depression (radius 0.5 cm, depth 1.4 cm):
\[ \frac{1}{3}\pi (0.5)^2 (1.4) = \frac{1}{3} \times \frac{22}{7} \times 0.25 \times 1.4 \approx 0.37 \text{ cm}^3 \]
Step 3: Four depressions \( \approx 4 \times 0.37 = 1.47 \) cm³.
Step 4: Wood remaining:
\[ 525 – 1.47 = 523.53 \text{ cm}^3 \]
Final answer: The volume of wood in the entire stand is about \( 523.53 \) cm³.
Common error: adding the cone volumes instead of subtracting them, or forgetting that there are four depressions. Check the sign against common sense: the remaining wood cannot exceed the 525 cm³ cuboid it came from.
Question 5: A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Answer: Dropping solids into a brim-full vessel pushes out water equal in volume to the solids themselves. ‘One-fourth of the water flows out’ therefore says: total volume of all shots = one-fourth of the cone’s capacity.
Find that displaced volume, then divide it by the volume of a single lead shot. The cm³ units cancel, leaving a pure count — no rounding is needed because \( \pi \) cancels completely.
Step 1: Capacity of the conical vessel:
\[ \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (5)^2 (8) = \frac{200\pi}{3} \text{ cm}^3 \]
Step 2: Displaced water, one-fourth of it:
\[ \frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3} \text{ cm}^3 \]
Step 3: Volume of one lead shot (a sphere of radius 0.5 cm):
\[ \frac{4}{3}\pi (0.5)^3 = \frac{4}{3}\pi (0.125) = \frac{\pi}{6} \text{ cm}^3 \]
Step 4: Number of shots:
\[ \frac{50\pi}{3} \div \frac{\pi}{6} = \frac{50\pi}{3} \times \frac{6}{\pi} = 100 \]
Final answer: 100 lead shots were dropped into the vessel.
Common error: the radius 0.5 cm is already a radius — do not halve it again — and some students divide the full cone volume instead of one-fourth of it. Notice \( \pi \) cancels in the division; if your count still carries \( \pi \), you divided the wrong pair of volumes.
Question 6: A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8g mass. (Use π = 3.14)
Answer: The pole is two cylinders stacked, so its total volume is the sum of their volumes. Mass is a different quantity: the question gives the density as 8 g per cm³, so mass = volume × density.
One number is a trap — the 24 cm is a diameter, so the taller cylinder’s radius is 12 cm, while the 8 cm is already a radius and needs no change. This question also switches to \( \pi = 3.14 \).
Step 1: Taller cylinder: radius \( r = \frac{24}{2} = 12 \) cm, height 220 cm.
\[ V_1 = 3.14 \times 12^2 \times 220 = 3.14 \times 144 \times 220 = 99,475.2 \text{ cm}^3 \]
Step 2: Upper cylinder: radius 8 cm, height 60 cm.
\[ V_2 = 3.14 \times 8^2 \times 60 = 3.14 \times 64 \times 60 = 12,057.6 \text{ cm}^3 \]
Step 3: Total volume \( = 99,475.2 + 12,057.6 = 111,532.8 \) cm³.
Step 4: Mass \( = 111,532.8 \times 8 = 892,262.4 \) g.
Convert to kilograms: \( 892,262.4 \div 1000 = 892.2624 \) kg.
Final answer: The mass of the pole is approximately \( 892.26 \) kg.
Common error: using 24 cm as the taller cylinder’s radius, and leaving the answer in grams. The conversion \( 892,262.4 \) g \( = 892.26 \) kg is part of the answer — a pole’s mass is reported in kilograms.
Question 7: A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Answer: The solid rests on the cylinder’s bottom and is completely submerged, so it displaces its whole volume — no part sticks out of the water. Water left = volume of the cylinder − volume of the solid.
The solid is cone + hemisphere, and all three shapes share the same radius 60 cm, which keeps the arithmetic clean: the cone and hemisphere even come out equal at 144,000π each. Keep \( \pi \) as a symbol until the last line.
Step 1: Volume of the cylinder:
\[ \pi r^2 h = \pi (60)^2 (180) = 648,000\pi \text{ cm}^3 \]
Step 2: Volume of the cone:
\[ \frac{1}{3}\pi (60)^2 (120) = 144,000\pi \text{ cm}^3 \]
Step 3: Volume of the hemisphere:
\[ \frac{2}{3}\pi (60)^3 = \frac{2}{3}\pi (216,000) = 144,000\pi \text{ cm}^3 \]
Step 4: Volume of the solid \( = 144,000\pi + 144,000\pi = 288,000\pi \) cm³.
Step 5: Water left in the cylinder:
\[ 648,000\pi – 288,000\pi = 360,000\pi \approx 360,000 \times \frac{22}{7} = 1,131,428.6 \text{ cm}^3 \]
Final answer: The volume of water left is \( 360,000\pi \) cm³, approximately \( 1,131,428.6 \) cm³.
Common error: subtracting only part of the solid because it ‘touches the bottom’ — it displaces its full volume either way. Spotting the equal volumes (144,000π each) turns the final subtraction into \( 648 – 288 = 360 \) thousand π.
Question 8: A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Answer: The vessel is two parts: a large sphere with a narrow cylindrical neck on top. Its true capacity is the sum of the two volumes. ‘Inside measurements’ means the dimensions given are the ones that hold water, so we calculate directly with \( \pi = 3.14 \).
The comparison at the end needs a sentence, not just a number: state the calculated value and say whether 345 cm³ matches it.
Step 1: Radii from the diameters: sphere \( r = \frac{8.5}{2} = 4.25 \) cm; neck \( r = \frac{2}{2} = 1 \) cm, height 8 cm.
Step 2: Volume of the spherical part:
\[ \frac{4}{3} \times 3.14 \times (4.25)^3 \approx 321.4 \text{ cm}^3 \]
Step 3: Volume of the cylindrical neck:
\[ 3.14 \times (1)^2 \times 8 = 25.12 \text{ cm}^3 \]
Step 4: True capacity \( \approx 321.4 + 25.12 = 346.5 \) cm³.
Step 5: Compare: the child’s 345 cm³ is about 1.5 cm³ short of 346.5 cm³.
Final answer: She is not exactly correct — the vessel actually holds about \( 346.5 \) cm³, not 345 cm³.
Common error: using 8.5 cm directly as the sphere’s radius, which gives a sphere of over 2,500 cm³ — far too large for a vessel the child measured at 345 cm³. Always halve both diameters before substituting.
How to Check Your Answers Before Submitting
These five checks are not general advice; each one catches a specific slip this exercise is built around.
- The add-or-subtract decision. Decide before any arithmetic. Stacked solids add (Q1, Q2, Q3, Q6, Q7, Q8); scooped-out depressions subtract (Q4); displaced water subtracts from the container (Q5 and Q7). A leftover amount can never exceed the full container.
- The radius-or-diameter scan. Circle every ‘diameter’ and halve it: 2.8 → 1.4 (Q3), 24 → 12 (Q6), 8.5 → 4.25 and 2 → 1 (Q8). A radius given as a radius stays as it is — 0.5 in Q5, 8 in Q6.
- The \( \pi \) instruction. The exercise opens with ‘Unless stated otherwise, take \( \pi = \frac{22}{7} \)’ (NCERT, p. 170). Q6 and Q8 say use \( \pi = 3.14 \). Q1 says ‘in terms of \( \pi \)’ — substitute nothing at all.
- The unit chain. Carry cm³ through every volume step. In Q6, volume × density gives grams, and grams ÷ 1000 gives kilograms — state that conversion in the final line.
- The magnitude sanity check. Ask whether the size is believable before submitting. The Q8 sphere alone is about 321 cm³, so the child’s 345 cm³ is plausible but short; had you used the diameter as the radius, that sphere alone would pass 2,500 cm³, which no small vessel would hold.
Mistakes this exercise catches
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Using the diameter as the radius (2.8 in Q3, 24 in Q6, 8.5 and 2 in Q8) | Always halve a ‘diameter’; a radius given as a radius stays unchanged | Circle every ‘diameter’ in the question and write \( r \) beside it before any formula |
| Taking the whole given length as the cylinder’s height (5 cm in Q3, 12 cm in Q2) | Subtract the ends: \( 5 – 2(1.4) = 2.2 \) cm; \( 12 – 2 – 2 = 8 \) cm | Sketch the object and label the ends before substituting |
| Adding the cone volumes to the cuboid in Q4 | Depressions are removed, so subtract: \( 525 – 4(0.37) \approx 523.53 \) cm³ | The remaining wood must be less than the plain cuboid’s 525 cm³ |
| Substituting \( \frac{22}{7} \) into Q1 | ‘In terms of \( \pi \)’ bans substitution | The answer should be a pure multiple of \( \pi \), not a decimal |
| Forgetting the gram-to-kilogram step in Q6 | Mass = volume × density, then divide by 1000 | Ask what unit the question expects — a pole’s mass is reported in kg |
Worked example with new numbers: the add rule and the slant-height trap
Try this fresh setup before the test: a cone of height 9 cm and base radius 3 cm is mounted on a cylinder of height 7 cm and the same radius 3 cm. Find the total volume.
Step 1: Cone: \( \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3)^2 (9) = 27\pi \) cm³.
Step 2: Cylinder: \( \pi r^2 h = \pi (3)^2 (7) = 63\pi \) cm³.
Step 3: Total: \( 27\pi + 63\pi = 90\pi \approx 90 \times \frac{22}{7} \approx 282.9 \) cm³.
Final answer: The combined volume is \( 90\pi \) cm³, about \( 282.9 \) cm³.
Now the trap: the cone’s slant height is \( l = \sqrt{3^2 + 9^2} = \sqrt{90} \approx 9.5 \) cm. That number is needed for the curved surface area \( \pi r l \), but it plays no role in volume, which uses only perpendicular height. Computing \( l \) inside Exercise 12.2 usually means you slipped into a surface-area method.
Which solid combinations the chapter repeats
| Combination of solids | Where it appears | Volume setup |
|---|---|---|
| Cone on a hemisphere | Q1, Q7; Example 7 of section 12.3 | \( \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 \) |
| Cylinder with two cones | Q2 | \( \pi r^2 h + 2\left( \frac{1}{3}\pi r^2 h_c \right) \) |
| Cylinder with two hemispherical ends | Q3; the truck container in the chapter’s introduction | \( \pi r^2 h + \frac{4}{3}\pi r^3 \) (two hemispheres = one sphere) |
| Cuboid with conical depressions | Q4 | \( lbh – 4\left( \frac{1}{3}\pi r^2 h \right) \) |
| Solid dropped into a brim-full vessel | Q5 (shots into a cone), Q7 (solid into a cylinder) | water left = container − solid |
Exercise 12.2 is the last exercise in this chapter; section 12.4 is only the summary. The same habit — read the question, name the solids, add or subtract the volumes — carries into every later chapter of Class 10 Maths, and our Class 10 Maths notes collect the formulas chapter by chapter. When you reach a data-heavy chapter, the Statistics solutions show the same step-by-step discipline.
Frequently Asked Questions About Exercise 12.2
How many questions are there in Exercise 12.2 of Class 10 Maths Surface Areas and Volumes?
Eight questions in total, all on the volume of combined solids: cone on a hemisphere (Q1), cylinder with two cones (Q2), gulab jamuns (Q3), pen stand (Q4), lead shots in a cone (Q5), iron pole (Q6), displacement in a cylinder (Q7), and a spherical vessel with a cylindrical neck (Q8). No intext questions appear in this exercise.
What shape is a gulab jamun in Question 3, and how do we find its volume?
It is a cylinder with two hemispherical ends (Fig. 12.15). The two hemispheres together equal one sphere of radius 1.4 cm, and the cylinder between them is \( 5 – 2(1.4) = 2.2 \) cm long. So one jamun is \( \pi (1.4)^2 (2.2) + \frac{4}{3}\pi (1.4)^3 \approx 25.05 \) cm³, and 30% of that is syrup.
Should I use π as 22/7 or 3.14 in Exercise 12.2?
The exercise states ‘Unless stated otherwise, take \( \pi = \frac{22}{7} \)’ (NCERT, p. 170). Use \( \frac{22}{7} \) for Q1–Q5 and Q7. Q6 and Q8 explicitly switch to \( \pi = 3.14 \). Q1 is special: ‘in terms of \( \pi \)’ means keep the symbol and substitute no value.
Why do we add volumes in some questions but subtract in others?
The volume of a combined solid is always the sum of the volumes of its constituents (NCERT, p. 168), which is why stacked objects add.
Subtraction appears only when material is removed, like the four conical depressions of the pen stand (Q4), or when an immersed solid displaces water, so the water left is the container minus the solid (Q5 and Q7).
What does the ‘one-fourth of the water flows out’ statement in Question 5 test?
It tests the displacement idea: overflowing water has exactly the volume of the objects dropped in. So the total shot volume is one-fourth of the cone’s capacity, \( \frac{50\pi}{3} \) cm³, and dividing by one shot’s volume \( \frac{\pi}{6} \) cm³ gives 100 shots. The same idea returns in Q7 with a cylinder of water.
Revision pages for every chapter and subject start from the CBSE notes home page, and the Class 10 section keeps everything for this grade in one place.
Reference: NCERT Class 10 Mathematics textbook, chapter Surface Areas and Volumes.
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