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A cube has total surface area 150 cm². Find its side and volume.
6a² = 150, a² = 25, a = 5 cm.
V = 5³ = 125 cm³.
Answer: a = 5 cm, V = 125 cm³
Find the TSA and volume of a cuboid 8 cm × 5 cm × 3 cm.
TSA = 2(8·5 + 5·3 + 3·8) = 2(40 + 15 + 24) = 158 cm².
V = 8 × 5 × 3 = 120 cm³.
Answer: 158 cm², 120 cm³
A tank is 2 m × 1.5 m × 1 m. How many litres does it hold?
V = 2 × 1.5 × 1 = 3 m³ = 3 × 10⁶ cm³.
1000 cm³ = 1 litre, so 3000 litres.
Answer: 3000 litres
A cube of side 4 cm is painted and cut into 1 cm cubes. Count cubes with 3, 2, 1, 0 painted faces.
Corners: 8 cubes with 3 faces.
Edges: 12 edges × (4 − 2) = 24 with 2 faces.
Faces: 6 × (2 × 2) = 24 with 1 face.
Inside: 2 × 2 × 2 = 8 with none. Total 8 + 24 + 24 + 8 = 64.
Answer: 8, 24, 24, 8
A cylinder has r = 7 cm and h = 10 cm (π = 22/7). Find CSA, TSA and volume.
CSA = 2 × 22/7 × 7 × 10 = 440 cm².
2πr² = 2 × 22/7 × 49 = 308, so TSA = 440 + 308 = 748 cm².
V = 22/7 × 49 × 10 = 1540 cm³.
Answer: 440, 748 cm²; 1540 cm³
A tumbler is open at the top with r = 3.5 cm and h = 10 cm. Find the metal sheet needed.
Area = πr(2h + r) = 22/7 × 3.5 × (20 + 3.5) = 11 × 23.5.
= 258.5 cm².
Answer: 258.5 cm²
A cylindrical tank has radius 3.5 m and height 4 m. How many litres?
V = 22/7 × 3.5² × 4 = 154 m³.
1 m³ = 1000 litres, so 154,000 litres.
Answer: 154000 litres
A solid cylinder r = 3 cm, h = 8 cm is melted into spheres of radius 1 cm. How many spheres?
Cylinder volume = π × 9 × 8 = 72π.
Sphere volume = 4π/3.
Number = 72π ÷ (4π/3) = 54.
Answer: 54 spheres
A cone has r = 6 cm and h = 8 cm. Find l, CSA and TSA (π = 22/7).
l = √(36 + 64) = 10 cm.
CSA = πrl = 60π ≈ 188.57 cm².
TSA = πr(l + r) = 96π ≈ 301.71 cm².
Answer: l = 10; 188.57; 301.71 cm²
Find the volume of a cone with r = 7 cm and h = 24 cm.
V = ⅓ × 22/7 × 49 × 24 = 22 × 7 × 8 = 1232 cm³.
Answer: 1232 cm³
A cylinder has volume 96 cm³. What is the volume of a cone of the same base and height?
Cone volume is one third of the cylinder: 96 ÷ 3 = 32 cm³.
Answer: 32 cm³
A cone has r = 7 cm and CSA 550 cm². Find l and h.
πrl = 550: 22/7 × 7 × l = 550, so l = 25 cm.
h = √(25² − 7²) = √576 = 24 cm.
Answer: l = 25 cm, h = 24 cm
A square pyramid has base side 6 cm and height 4 cm. Find its volume.
Base area = 36 cm².
V = ⅓ × 36 × 4 = 48 cm³.
Answer: 48 cm³
A pyramid has base area 15 cm² and height 9 cm. Find its volume.
V = ⅓ × 15 × 9 = 45 cm³.
Answer: 45 cm³
Find the surface area and volume of a sphere of radius 7 cm (π = 22/7).
A = 4 × 22/7 × 49 = 616 cm².
V = (4/3) × 22/7 × 343 = 4312/3 ≈ 1437.33 cm³.
Answer: 616 cm²; ≈ 1437.33 cm³
Find the TSA and volume of a hemisphere of radius 14 cm.
TSA = 3πr² = 3 × 22/7 × 196 = 1848 cm².
V = (2/3) × 22/7 × 2744 = 17248/3 ≈ 5749.33 cm³.
Answer: 1848 cm²; ≈ 5749.33 cm³
The radius of a sphere is doubled. What happens to its surface area and volume?
A ∝ r², so the area becomes 2² = 4 times.
V ∝ r³, so the volume becomes 2³ = 8 times.
Answer: ×4 and ×8
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Each thin layer has base area lw and a tiny thickness. Stacking layers adds the thicknesses to the height h, so the total volume is (lw) × h.
With l = w = h = a: TSA = 2(a² + a² + a²) = 6a²; V = a·a·a = a³.
V(A) = 8000 ft³ ≈ 2.26 lakh litres; V(B) = 9000 ft³ ≈ 2.55 lakh litres. A is deeper (8 ft is unsafe for children and costs more to excavate); B is shallower and longer, better for lap swimming and young learners. Considering depth safety, length for swimming and cost, B seems more suitable for a school; the choice depends on the parameters one weighs.
Both volumes are 216 cm³; SA(A) = 216 cm², SA(B) = 228 cm². The cuboid has the larger surface area.
CSA = 2πrh = 2 × 22/7 × 3.5 × 25 = 550 cm².
Cone with r = 5, h = 12: V = ⅓π × 25 × 12 = 100π cm³.
V = ⅔ × 22/7 × 3.5³ = 89.8 cm³ (about 90 mL).
Divide the surface into tiny regions and join them to the centre: thin cones of height r. V = ⅓ × (sum of base areas) × r = ⅓ × 4πr² × r = (4/3)πr³, i.e. ⅓ × surface area × radius.
Golgappas: 5000π cm³ of pani divided by 8.5 cm³ (60% of 14.14) gives about 1850; with 80% of 22.43 ≈ 18 cm³ gives about 870. Tennis balls: upper bound 3.78 × 10⁸ ÷ 113 ≈ 33.5 lakh; square packing 150 × 150 × 75 = 16,87,500. For the other two (closer) packings, with density about 74% instead of 52%, the estimate rises to roughly 2.5 million.
a³ = 64, a = 4; TSA = 6 × 16 = 96 cm².
2 m = 200 cm; 200 ÷ 20 = 10 along each edge; 10³ = 1000 cubes.
40 × 25 × 10 = 10000 m³; box = 2.5 m³; number = 4000.
Side 5 cm; the cuboid is 10 × 5 × 5. SA = 2(50 + 50 + 25) = 250 cm².
Original: 6 × 16 = 96 cm². There are 64 small cubes, each 6 cm²: 384 cm². Ratio = 96 : 384 = 1 : 4.
lw = 6, wh = 15, lh = 10. Multiply: (lwh)² = 900, so V = 30 cm³ (edges 2, 3, 5).
Need lw + wh + hl = 50 with l ≤ w ≤ h. For l = 1: (w + 1)(h + 1) = 51 = 3 × 17 gives (1, 2, 16). For l = 2: (w + 2)(h + 2) = 54 = 6 × 9 gives (2, 4, 7). For l = 3: (w + 3)(h + 3) = 59 (prime), none. l ≥ 5 is impossible since 3l² ≤ 50; l = 4: (w + 4)(h + 4) = 66 has no factor pair with both ≥ 8. So (i) yes; (ii),(iii) exactly two: 1 × 2 × 16 and 2 × 4 × 7.
CSA(A) = 2πrh; CSA(B) = 2π(2r)(h/2) = 2πrh. Ratio 1 : 1. V(A) = πr²h; V(B) = π(4r²)(h/2) = 2πr²h. Ratio 1 : 2.
(a) V ∝ r²h: 4 × 3 : 9 × 2 = 12 : 18 = 2 : 3. (b) CSA ∝ rh: 2 × 3 : 3 × 2 = 1 : 1.
Its radius is r/2 and height r: V = π(r/2)² × r = πr³/4 cm³.
(1.1)² × (1 − x/100) = 1, so 1 − x/100 = 1/1.21 and x = 100 × 0.21/1.21 = 2100/121 ≈ 17.36%.
(i) 12³ = 1728 cm³. (ii) π × 4 × 12 = 48π ≈ 150.86 cm³ (22/7). (iii) 1728 ÷ 150.86 ≈ 11.45, so 11 complete rods.
r = 12. TSA = πr(l + r) = 22/7 × 12 × 33 = 8712/7 ≈ 1244.57 m².
CSA = πrl = 22/7 × 7 × 10 = 220 cm².
l = √(256 + 144) = 20. CSA = π × 12 × 20 = 240π ≈ 754.29 cm². TSA = π × 12 × 32 = 384π ≈ 1206.86 cm².
⅓πr² × 15 = 1570, 5πr² = 1570, r² = 314/π = 100 (π = 3.14). r = 10 cm.
(i) πr × 14 = 308: r = 308 ÷ 44 = 7 cm. (ii) TSA = πr(l + r) = 22/7 × 7 × 21 = 462 cm².
l = 25. CSA = 22/7 × 7 × 25 = 550 cm². For 10 caps: 5500 cm².
l = 10. CSA = π × 6 × 10 = 60π ≈ 188.57 m². Length = 188.57 ÷ 3 ≈ 62.86 m; add 0.2 m: ≈ 63.06 m.
r = 6, h = 8, l = 10. V = ⅓π × 36 × 8 = 96π ≈ 301.71 cm³. CSA = π × 6 × 10 = 60π ≈ 188.57 cm².
The water forms a smaller cone similar to the cup with ratio ½, so its volume is (½)³ = 1/8 of the cup.
A = 4πr² = 4 × 22/7 × 0.49 = 6.16 cm².
Same density, so weights are in the ratio of volumes: 5920/740 = 8 = (R/2.5)³, so R/2.5 = 2 and R = 5 cm.
Areas are in the ratio of squares of radii: moon : earth = (1/4)² = 1 : 16.
(i) CSA = 2πr² = 2 × 22/7 × 441 = 2772 cm². (ii) TSA = 3πr² = 4158 cm².
Box = 1024 cm³. Spheres: 16 × (4/3)π × 8 ≈ 536.4 cm³. Liquid ≈ 1024 − 536.4 ≈ 488 cm³.
New volume factor = (1.1)³ = 1.331, an increase of 33.1%.
(1 + x/100)³ = 1.728 = 1.2³, so 1 + x/100 = 1.2 and x = 20.
2πr = 35.2, so r = 35.2 × 7/44 = 5.6 m. CSA = 2πr² = 2 × 22/7 × 31.36 = 197.12 m² = 19,71,200 cm². Cost = 19,71,200 ÷ 100 × 10 = ₹1,97,120.
Container = 3000 cm³. A scoop of radius about 3 cm is ≈ 113 cm³; upper bound ≈ 26. Allowing for the air gaps and a flat top, roughly 20–25 scoops. (Estimates vary with the scoop size assumed.)
(a + 1)³ − a³ = 3a² + 3a + 1.
Drinking about 2 litres a day for 70 years ≈ 51,000 litres ≈ 51 m³. A 30 × 30 × 15 ft classroom ≈ 13,500 ft³ ≈ 380 m³. So no: a lifetime of drinking fills only about one seventh of it.
Assume a 6 m × 5 m × 3 m room: wall area ≈ 2 × 11 × 3 = 66 m²; less doors and windows ≈ 60 m². A 20 cm thick wall: 12 m³. A brick 20 × 10 × 10 cm = 0.002 m³ gives about 6000 bricks (about 5000 with the mortar).
A: 50³ = 1,25,000 mm³. B: 100 × 10³ = 1,00,000 mm³. Option A gives more chocolate.
About 8 × 10⁹ people at 0.25 m² each (0.5 m × 0.5 m) ≈ 2 × 10⁹ m² = 2000 km², i.e. a square about 45 km on a side (a large city).
One glass: 22/7 × 3.5² × 12 = 462 cm³. For 1600: 7,39,200 cm³ = 739.2 litres.
Sphere area = 100π, so cone CSA = 20π = π × 4 × l and l = 5 cm. h = √(25 − 16) = 3 cm. V = ⅓π × 16 × 3 = 16π ≈ 50.27 cm³.
(i) (69900/6370)³ ≈ 10.973³ ≈ 1321. (ii) (695700/6370)³ ≈ 109.2³ ≈ 1.3 × 10⁶ (about 13 lakh).
The cylinder has radius r and height 2r: V = πr² × 2r = 2πr³. Sphere = (4/3)πr³. Ratio = (4/3) ÷ 2 = 2/3.
Circles of radius R and R + d: 2π(R + d) − 2πR = 1, so d = 1/(2π) ≈ 0.159 m ≈ 16 cm. The answer does not depend on R: the same for the Earth, Moon, Jupiter and a volleyball. It is surprising because the Earth’s circumference is huge, yet the rope rises as much as it does on a small ball.
The base square is inscribed in the circle of radius r, so its diagonal is 2r and its area is 2r². V(pyramid) = ⅓ × 2r² × h. V(cylinder) = πr²h. Ratio = 2/(3π).
New V − old V = (l + 1)wh − lwh = wh. Answer (c) wh cubic units.
π(r + 1)²h − πr²h = πh(2r + 1) = 2πrh + πh. Answer (e).
Change = (4/3)π[r³ − (r − 1)³] = (4/3)π(3r² − 3r + 1) = 4πr² − 4πr + (4/3)π (a decrease in volume).
a = V^(1/3), so S = 6a² = 6V^(2/3).
a = √(S/6), so V = a³ = (S/6)^(3/2).
Water must rise 20 − 16 = 4 cm: volume = π × 16 × 4 = 64π cm³. One marble = (4/3)π. Number = 64π ÷ (4π/3) = 48 marbles.
Measure the refill: length about 10 cm, inner diameter about 1 mm (r = 0.05 cm). V = π × 0.05² × 10 ≈ 0.08 cm³, so roughly 0.1 mL (assuming it is full). Your measurements may differ.
Ball = (4/3)π × 216 = 288π ≈ 905 cm³. A chapati of radius 10 cm and thickness 1.5 mm = π × 100 × 0.15 ≈ 47 cm³. About 905 ÷ 47 ≈ 19 chapatis.
Measure the outer radius R and the flesh thickness t; flesh of a half ≈ (2/3)π[R³ − (R − t)³]. Example R = 7, t = 1: (2/3)π(343 − 216) ≈ 266 cm³ per half, about 530 cm³ for the whole fruit.
Assume 164 pages of 20 cm × 26 cm = 520 cm² each: 85,280 cm² ≈ 8.5 m². A 30 ft × 30 ft room has ≈ 84 m². So no; they cover about a tenth.
Room = 9000 ft³ ≈ 2.55 × 10⁸ cm³. A book 21 × 27 × 1.2 cm ≈ 680 cm³. Because books are cuboids that pack almost without gaps: ≈ 2.55 × 10⁸ ÷ 680 ≈ 3.7 lakh books.
8 × 10⁹ people × about 0.07 m³ (70 kg at the density of water) = 5.6 × 10⁸ m³, giving a side ≈ 825 m. Allowing room to stand (1 m³ each) it is 8 × 10⁹ m³, side 2 km. So about 1–2 km.
(i) t = (R³ + 3V/4π)^(1/3) − R (≈ V/(4πR²) for a thin layer), with R = 6371 km and V = 1.38 × 10⁹ km³. (ii) t ≈ 1.38 × 10⁹ ÷ (4π × 6371²) ≈ 2.7 km.
Read as: find two cuboids, the second with twice the surface area and half the volume. Take 8 × 14 × 14: V = 1568, S = 840. Take 1 × 28 × 28: V = 784 (half) and S = 2(28 + 28 + 784) = 1680 (double). Other pairs exist, e.g. 8 × 10 × 24 and 1 × 24 × 40.
Method: split the house into rooms (cuboids), measure or estimate length, width and height of each, add the volumes, and subtract nothing for walls inside. Example: 4 rooms of 10 × 10 × 9 ft plus a hall 20 × 8 × 9 ft gives 5400 + 1440 ≈ 6840 ft³ ≈ 1.9 × 10⁵ litres. State your assumptions.