Exercise 12.2 — Surface Areas and Volumes

1. 13 marksNCERT Cl-10 Maths, Ex 12.2, Q1

   A solid is a cone standing on a hemisphere, both with radius 1 cm, and the cone's height equals its radius. Find the volume of the solid in terms of π.

   Hint. Unlike surface area, volumes of a combination simply add — no faces disappear.

   Step 1 — Note the given values. radius = 1 cm for both pieces, and the cone's height also equals 1 cm.

   Step 2 — Volume is always additive for combinations — nothing is hidden the way surface area is. volume = (1/3)πr²h + (2/3)πr³

   Step 3 — Substitute r = 1, h = 1. = (1/3)π(1)(1) + (2/3)π(1) = π/3 + 2π/3

   Step 4 — Add the fractions, since they already share a denominator. = 3π/3 = π

   ✦ Answer: π cm³ (exactly)

   Where students slip. Worrying about which surfaces are 'hidden' the way you would for a surface-area question. Volume has no such subtlety — every combination's volume is simply the sum of its parts.

   Another way. With r = h = 1, both terms simplify before adding: the cone contributes exactly a third of π, the hemisphere exactly two-thirds, and together they make a whole π — a tidy result that comes from the specific numbers chosen.

2. 24 marksNCERT Cl-10 Maths, Ex 12.2, Q2

   Rachel made a model shaped like a cylinder with two cones attached at its ends, using thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. Each cone has height 2 cm. Find the volume of air contained in the model.

   Hint. Subtract both cones' heights from the total length to get the cylinder's own length.

   Step 1 — Find the radius and the cylinder's length. radius = 3/2 = 1.5 cm Each cone contributes its height to the total length, so the cylinder's own length is 12 − 2(2) = 8 cm

   Step 2 — Volume is the sum of the three pieces: one cylinder, two cones. volume = πr²h_cyl + 2 × (1/3)πr²h_cone = πr²[h_cyl + (2/3)h_cone]

   Step 3 — Substitute r = 1.5, h_cyl = 8, h_cone = 2. = π(1.5)²[8 + (2/3)(2)] = π(2.25)[8 + 4/3]

   Step 4 — Simplify the bracket: 8 + 4/3 = 28/3. = π × 2.25 × 28/3

   Step 5 — Compute. 2.25 × 28 = 63, and 63/3 = 21. = 21π = 21 × (22/7) = 66

   ✦ Answer: 66 cm³

   Where students slip. Using the full 12 cm as the cylinder's length and adding the cones on top of that, which double-counts the space the cones actually occupy at the ends.

   Another way. Since the model is thin aluminium sheet (hollow), the volume of *air* inside is simply the volume enclosed by the shape — which is exactly what was computed. No separate adjustment for the sheet's thickness is needed, since the question treats inner and outer dimensions as the same.

3. 35 marksNCERT Cl-10 Maths, Ex 12.2, Q3 (Fig. 12.15)

   A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends, of length 5 cm and diameter 2.8 cm.

   Hint. This is the same shape as the capsule and the model — find the cylinder's own length first, then add the volumes.

   Step 1 — Recover the cylinder's own length from the total. radius = 2.8/2 = 1.4 cm Each hemispherical end contributes its radius to the total length, so the cylinder's length is 5 − 2(1.4) = 5 − 2.8 = 2.2 cm

   Step 2 — Volume of one gulab jamun: cylinder + two hemispheres (= one sphere). volume = πr²h + (4/3)πr³

   Step 3 — Substitute r = 1.4, h = 2.2. πr²h = (22/7)(1.4)²(2.2) = (22/7)(1.96)(2.2) ≈ 13.552 (4/3)πr³ = (4/3)(22/7)(1.4)³ = (4/3)(22/7)(2.744) ≈ 11.499

   Step 4 — Add for one jamun. one jamun ≈ 13.552 + 11.499 = 25.051 cm³

   Step 5 — Scale up to 45 jamuns. 45 × 25.051 ≈ 1127.28 cm³

   Step 6 — Take 30% of that as the syrup. 0.30 × 1127.28 ≈ 338.18

   ✦ Answer: ≈ 338.19 cm³ of syrup (approximately)

   Where students slip. Applying the 30% to only one jamun and forgetting to scale by 45, or the reverse — scaling by 45 but forgetting the 30% altogether. Both steps are needed, in either order.

   Another way. Multiply the 30% into the single-jamun volume first: 0.3 × 25.051 ≈ 7.515 cm³ of syrup per jamun, then × 45 ≈ 338.2 cm³ — the same answer, reached by scaling in the opposite order.

4. 44 marksNCERT Cl-10 Maths, Ex 12.2, Q4 (Fig. 12.16)

   A pen stand made of wood is a cuboid 15 cm × 10 cm × 3.5 cm, with four conical depressions to hold pens. Each depression has radius 0.5 cm and depth 1.4 cm. Find the volume of wood in the entire stand.

   Hint. The depressions remove material — this is a subtraction, not an addition.

   Step 1 — Volume of the plain cuboid. cuboid = 15 × 10 × 3.5 = 525 cm³

   Step 2 — Volume of one conical depression. cone = (1/3)πr²h = (1/3)(22/7)(0.5)²(1.4) = (1/3)(22/7)(0.25)(1.4)

   0.25 × 1.4 = 0.35, and (22/7)(0.35) = 1.1, so cone = (1/3)(1.1) ≈ 0.3667 cm³

   Step 3 — Four depressions. 4 × 0.3667 ≈ 1.4667 cm³

   Step 4 — Subtract from the cuboid, since wood is what's left after removing the depressions. wood = 525 − 1.4667 ≈ 523.53

   ✦ Answer: ≈ 523.53 cm³

   Where students slip. Adding the cones' volume to the cuboid instead of subtracting it. The depressions are holes carved *into* the wood, so their volume must come out of the total, not be added to it.

   Another way. Factor the four cones together before dividing by 3: 4 × (1/3)πr²h = (4/3)(22/7)(0.25)(1.4) = (4/3)(1.1) ≈ 1.4667 — the same number, computed in one combined step rather than one-then-times-four.

5. 55 marksNCERT Cl-10 Maths, Ex 12.2, Q5

   A vessel is an inverted cone of height 8 cm and top radius 5 cm, filled with water to the brim. Lead shots, each a sphere of radius 0.5 cm, are dropped in until one-fourth of the water flows out. Find the number of lead shots dropped in.

   Hint. The water that overflows has the same volume as the shots that were dropped in — the shots displace exactly that much water.

   Step 1 — Volume of water in the full cone. cone = (1/3)πr²h = (1/3)(22/7)(5²)(8) = (1/3)(22/7)(200)

   (22/7)(200) = 4400/7 ≈ 628.57, so cone ≈ 209.52 cm³

   Step 2 — The volume that overflows is one-fourth of this. overflow = (1/4) × 209.52 ≈ 52.38 cm³

   Step 3 — This overflow equals the total volume of the lead shots dropped in, since each shot displaces its own volume of water. volume of one shot = (4/3)πr³ = (4/3)(22/7)(0.5)³ = (4/3)(22/7)(0.125)

   (22/7)(0.125) = 2.75/7 ≈ 0.3929, and (4/3)(0.3929) ≈ 0.5238 cm³

   Step 4 — Divide the overflow by one shot's volume. number of shots = 52.38 / 0.5238 ≈ 100

   ✦ Answer: 100 lead shots

   Where students slip. Computing the volume of water that *remains* (three-quarters) rather than the volume that *overflowed* (one-quarter). It is the overflow that equals the total volume of the shots — that displaced water had nowhere else to go.

   Another way. Work with exact fractions to avoid rounding: cone volume = (1/3)(22/7)(200) = 4400/21; overflow = 1100/21; one shot = (4/3)(22/7)(1/8) = 11/21. Number of shots = (1100/21)/(11/21) = 1100/11 = 100 exactly.

6. 64 marksNCERT Cl-10 Maths, Ex 12.2, Q6

   A solid iron pole is a cylinder of height 220 cm and base diameter 24 cm, 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 mass approximately 8 g. (Use π = 3.14)

   Hint. Two cylinders, straightforward addition of volumes — then convert volume to mass using the given density.

   Step 1 — Volume of the lower cylinder. radius = 24/2 = 12 cm, height = 220 cm volume₁ = πr²h = 3.14 × 12² × 220 = 3.14 × 144 × 220

   144 × 220 = 31,680, so volume₁ = 3.14 × 31,680 = 99,475.2 cm³

   Step 2 — Volume of the upper cylinder. radius = 8 cm, height = 60 cm volume₂ = 3.14 × 8² × 60 = 3.14 × 64 × 60

   64 × 60 = 3840, so volume₂ = 3.14 × 3840 = 12,057.6 cm³

   Step 3 — Total volume. total = 99,475.2 + 12,057.6 = 111,532.8 cm³

   Step 4 — Convert to mass using 8 g per cm³. mass = 111,532.8 × 8 = 892,262.4 g

   Step 5 — Convert to kilograms. 892,262.4 / 1000 = 892.2624

   ✦ Answer: ≈ 892.26 kg

   Where students slip. Using the diameter (24 cm) directly as the radius in the first cylinder's formula, which would inflate that volume by a factor of 4.

   Another way. Keep the volumes in cm³ and multiply by 8 g/cm³ at the very end, rather than converting each cylinder's volume to a mass separately — one conversion instead of two reduces the chance of a slip.

7. 75 marksNCERT Cl-10 Maths, Ex 12.2, Q7

   A solid cone of height 120 cm and radius 60 cm stands on a hemisphere of radius 60 cm, placed upright in a right circular cylinder full of water, touching the bottom. Find the volume of water left in the cylinder if the cylinder has radius 60 cm and height 180 cm.

   Hint. The solid displaces water equal to its own volume. What's left is the cylinder's volume minus the solid's.

   Step 1 — Volume of the cone. cone = (1/3)πr²h = (1/3)π(60²)(120) = (1/3)π(3600)(120)

   3600 × 120 = 432,000, and 432,000/3 = 144,000, so cone = 144,000π

   Step 2 — Volume of the hemisphere. hemisphere = (2/3)πr³ = (2/3)π(60³) = (2/3)π(216,000)

   (2/3)(216,000) = 144,000, so hemisphere = 144,000π

   Step 3 — Total volume of the solid (cone + hemisphere). solid = 144,000π + 144,000π = 288,000π

   Step 4 — Volume of the full cylinder. cylinder = πr²h = π(60²)(180) = π(3600)(180) = 648,000π

   Step 5 — Water left = cylinder's volume − solid's volume, since the solid displaces exactly its own volume of water. water left = 648,000π − 288,000π = 360,000π

   Step 6 — Substitute π = 22/7. = 360,000 × (22/7) = 7,920,000/7

   ✦ Answer: ≈ 1,131,428.57 cm³

   Where students slip. Adding the solid's volume to the cylinder's, as though the solid were extra water rather than something displacing water that was already there. The solid takes up space the water would otherwise occupy — hence subtraction.

   Another way. Notice the cone and hemisphere happen to have equal volumes here (both 144,000π), a coincidence of the specific radius and height chosen — a useful check that both computations were done correctly, since two independent formulas landing on the same number is not an accident to dismiss.

8. 85 marksNCERT Cl-10 Maths, Ex 12.2, Q8

   A spherical glass vessel has a cylindrical neck 8 cm long and 2 cm in diameter, and a spherical part of diameter 8.5 cm. A child measures its capacity as 345 cm³. Check whether she is correct, using these as the inside measurements and π = 3.14.

   Hint. Compute the actual volume from the given dimensions and compare it with the claimed 345 cm³.

   Step 1 — Volume of the cylindrical neck. radius = 2/2 = 1 cm, height = 8 cm cylinder = πr²h = 3.14 × 1² × 8 = 25.12 cm³

   Step 2 — Volume of the spherical part. radius = 8.5/2 = 4.25 cm sphere = (4/3)πr³ = (4/3)(3.14)(4.25)³

   4.25³ = 76.765625 (4/3)(3.14) ≈ 4.1867 sphere ≈ 4.1867 × 76.765625 ≈ 321.39

   Step 3 — Add the two volumes. total ≈ 25.12 + 321.39 = 346.51

   Step 4 — Compare with the child's claim. 346.51 vs. 345 — the two are close but not equal, a difference of about 1.5 cm³.

   ✦ Answer: the actual capacity is ≈ 346.51 cm³, so the child's measurement of 345 cm³ is not quite correct — though close enough that it may reflect the natural imprecision of physically measuring how much water the vessel holds, rather than a mistake in reasoning.

   Where students slip. Rounding 346.51 down to 345 and declaring the child correct just because the numbers are near each other. The question asks you to *check*, which means computing the true value precisely and comparing — not eyeballing whether the two look similar.

   Another way. Compute (4.25)³ as 4.25 × 4.25 × 4.25 in two steps to catch errors early: 4.25 × 4.25 = 18.0625, then 18.0625 × 4.25 = 76.765625 — breaking the cube into two multiplications makes it easier to verify by hand.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Mathematics textbook, Reprint 2026-27 (this chapter now has two exercises, 12.1 with 9 questions and 12.2 with 8; the frustum of a cone and conversion of solids, with the old third, fourth and fifth exercises, are no longer part of this chapter). Questions are referenced from the NCERT textbook for identification.