Chapter 12: Surface Areas and Volumes

NCERT Solutions for CBSE Class 10 Mathematics — 20 solved questions with detailed explanations.

Questions

Topics

Important Formulas

Standard Solids

Cylinder: V = πr²h, CSA = 2πrh, TSA = 2πr(r+h)
Cone: V = (1/3)πr²h, CSA = πrl, TSA = πr(r+l), l = √(r²+h²)
Sphere: V = (4/3)πr³, SA = 4πr²
Hemisphere: V = (2/3)πr³, CSA = 2πr², TSA = 3πr²
Frustum: V = (πh/3)(R²+r²+Rr), CSA = π(R+r)l

Solved Questions

Q1. A solid is in the shape of a cone mounted on a hemisphere. The radius is 7 cm and height of the cone is 24 cm. Find the TSA.

Difficulty: Easy-Medium · Topic: Surface Area of Combination of Solids

Solution

l = √(7²+24²) = √(625) = 25. CSA cone = 22/7×7×25 = 550.

CSA hemisphere = 2×22/7×49 = 308. TSA = 550+308 = 858 cm².

Q2. A gulab jamun is made of a cylinder of length 5 cm and diameter 2.8 cm with two hemispheres at each end. Find its volume.

Difficulty: Easy-Medium · Topic: Volume of Combination of Solids

Solution

r = 1.4 cm. Cylinder length (excluding hemispheres) = 5−2(1.4) = 2.2 cm.

V = πr²h + (4/3)πr³ = 22/7×1.96×2.2 + (4/3)×22/7×2.744

= 22/7(4.312+3.659) = 22/7×7.971 = 25.05 cm³.

Q3. A metallic sphere of radius 4.2 cm is melted and recast into a cylinder of radius 6 cm. Find the height of the cylinder.

Difficulty: Easy-Medium · Topic: Conversion of Solids

Solution

(4/3)π(4.2)³ = π(6)²h → h = 4×74.088/(3×36) = 296.352/108 = 2.744 cm.

Q4. How many silver coins, 1.75 cm in diameter and 2 mm thick, can be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Difficulty: Easy-Medium · Topic: Conversion of Solids

Solution

Volume of cuboid = 5.5×10×3.5 = 192.5 cm³.

Volume of coin = π(0.875)²(0.2) = 22/7×0.765625×0.2 = 0.48125 cm³.

Number = 192.5/0.48125 = 400.

Q5. The slant height of a frustum with radii R, r and height h is:

Difficulty: Easy-Medium · Topic: Frustum of a Cone

Solution

Slant height l = √[h² + (R−r)²].

Q6. A solid ball is exactly fitted inside a cube of side a. The volume of the ball is:

Difficulty: Easy-Medium · Topic: Volume of Combination of Solids

Solution

Diameter = a → r = a/2. V = (4/3)π(a/2)³ = (4/3)πa³/8 = πa³/6.

Q7. A bucket in the shape of a frustum has top radius 28 cm, bottom radius 21 cm and height 45 cm. Find its capacity in litres.

Difficulty: Medium · Topic: Frustum of a Cone

Solution

V = (π×45/3)(28²+21²+28×21) = 15π(784+441+588) = 15π(1813) = 15×22/7×1813

= 15×22×259 = 85470 cm³. Wait: 1813/7 = 259. V = 15×22×259 = 85470 cm³ = 85.47 litres.

Let me recompute: V = (22/7 × 45/3)(784+441+588) = (22×45)/(7×3) × 1813 = (990/21)×1813 = 47.143 × 1813 = 85471 cm³ ≈ 85.47 litres.

Q8. A fez cap is in the shape of a frustum with radii 10 cm (top, smaller) and 20 cm (bottom, larger), height 15 cm. Find the CSA and total cost at Rs 2 per cm².

Difficulty: Medium · Topic: Frustum of a Cone

Solution

l = √(15²+10²) = √(225+100) = √325 = 5√13 ≈ 18.03 cm.

CSA = π(20+10)×5√13 = 150π√13/... = 30×5√13×π = 150√13×π ≈ 150×3.606×3.14 ≈ 1699 cm². Cost ≈ Rs 3398.

Q9. A cone of height 24 cm and radius 6 cm is divided into 3 parts by two planes parallel to the base at heights 8 cm and 16 cm. Find the ratio of volumes of the three parts.

Difficulty: Medium · Topic: Conversion of Solids

Solution

By similar triangles, at h=8: r₁=2, at h=16: r₂=4, at h=24: r₃=6.

V₁ (top cone, h=8) = (1/3)π(4)(8) = 32π/3.

V₂ (frustum 8-16) = (π×8/3)(4+16+8) = (8π/3)(28) = 224π/3.

V₃ (frustum 16-24) = (π×8/3)(16+36+24) = (8π/3)(76) = 608π/3.

Ratio = 32:224:608 = 1:7:19.

Q10. A toy is in the shape of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height is 15.5 cm. Find the TSA.

Difficulty: Medium · Topic: Volume of Combination of Solids

Solution

Hemisphere radius = 3.5, cone height = 15.5−3.5 = 12 cm.

l = √(12²+3.5²) = √(144+12.25) = √156.25 = 12.5 cm.

TSA = πrl + 2πr² = 22/7×3.5×12.5 + 2×22/7×12.25 = 137.5+77 = 214.5 cm².

Q11. A medicine capsule is a cylinder of diameter 5 mm and length 2 mm, with two hemispheres at each end. Find its surface area.

Difficulty: Medium · Topic: Surface Area of Combination of Solids

Solution

r = 2.5 mm = 0.25 cm. Cylinder height = 2 mm = 0.2 cm. CSA cyl = 2π(0.25)(0.2) = 0.1π.

SA of 2 hemispheres = 4π(0.0625) = 0.25π. Total = 0.35π ≈ 1.1 cm².

Q12. Water flows at 7 m/min through a cylindrical pipe of diameter 1.4 cm. How long will it take to fill a conical vessel of radius 35 cm and depth 42 cm?

Difficulty: Medium · Topic: Conversion of Solids

Solution

Volume of cone = (1/3)π(35)²(42) = (1/3)×22/7×1225×42 = 22×1225×2 = 53900 cm³.

Flow rate: r = 0.7 cm. Cross-section = π(0.49). Speed = 700 cm/min.

Volume/min = 22/7×0.49×700 = 22×0.49×100 = 1078 cm³/min.

Time = 53900/1078 = 50 min. Hmm, let me recheck: 22/7×0.49 = 22×0.07 = 1.54. ×700 = 1078.

53900/1078 ≈ 50 min. But if the answer given in NCERT is 15 min, let me recheck the flow: 7 m/min = 700 cm/min. Volume/min = π(0.7)²×700 = 22/7 × 0.49 × 700 = 1078 cm³.

Cone volume: (1/3)×(22/7)×1225×42 = (22/7)×(1225×42/3) = (22/7)×17150 = 22×2450 = 53900.

Time = 53900/1078 = 50 min.

Q13. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, surmounted by a cone of height 60 cm. Find the mass if 1 cm³ of iron has 8 g mass.

Difficulty: Medium · Topic: Volume of Combination of Solids

Solution

r = 12 cm. Cyl volume = π(144)(220) = 31680π. Cone volume = (1/3)π(144)(60) = 2880π.

Total = 34560π ≈ 108573 cm³. Mass = 108573×8 = 868586 g ≈ 892.26 kg (with π = 22/7: 34560×22/7 = 108617. ×8 = 868937 g ≈ 868.9 kg).

Q14. A well of diameter 3 m is dug 14 m deep. The earth taken out is spread evenly all around to form a 4 m wide embankment. Find the height of the embankment.

Difficulty: Medium · Topic: Conversion of Solids

Solution

Well volume = π(1.5)²(14) = 31.5π.

Embankment is a ring: outer radius = 1.5+4 = 5.5. Area = π(5.5²−1.5²) = π(30.25−2.25) = 28π.

Height = 31.5π/(28π) = 31.5/28 = 1.125 m.

Q15. A container made of a metal sheet is in the shape of a frustum with radii 8 cm (bottom) and 20 cm (top) and height 16 cm. Find the cost of the metal sheet at Rs 8 per 100 cm². (Use π = 3.14)

Difficulty: Medium · Topic: Frustum of a Cone

Solution

l = √(16²+(20−8)²) = √(256+144) = √400 = 20 cm.

TSA = π(20+8)(20) + π(8²) = 560π + 64π = 624π = 624×3.14 = 1959.36 cm².

Cost = 1959.36×8/100 = Rs 156.75.

Q16. A decorative block is a cube of edge 5 cm with a hemisphere of largest possible diameter on the top face. Find the TSA of the block.

Difficulty: Medium · Topic: Surface Area of Combination of Solids

Solution

Hemisphere radius = 2.5 cm. TSA = 5 faces of cube + top face minus circle + CSA hemisphere.

= 5(25) + 25 − π(6.25) + 2π(6.25) = 125 + 25 + π(6.25) = 150 + 6.25π ≈ 150 + 19.64 = 169.64 cm².

Wait: TSA = 5×25 (five full faces) + (25 − π×2.5²) (top face minus circle) + 2π(2.5²) (hemisphere CSA).

= 125 + 25 − 6.25π + 12.5π = 150 + 6.25π ≈ 150 + 19.64 = 169.64 cm².

Q17. A wooden article is made by scooping out a hemisphere from each end of a solid cylinder. If the cylinder height is 10 cm and radius 3.5 cm, find the TSA.

Difficulty: Medium-Hard · Topic: Volume of Combination of Solids

Solution

TSA = CSA cylinder + 2×CSA hemisphere = 2πrh + 2×2πr² = 2πr(h+2r).

= 2×22/7×3.5×(10+7) = 2×11×17 = 374 cm².

Q18. A solid metallic sphere of diameter 4.2 cm is melted and recast in the form of a cone of diameter 2.8 cm. Find the height of the cone.

Difficulty: Medium-Hard · Topic: Conversion of Solids

Solution

(4/3)π(2.1)³ = (1/3)π(1.4)²h → 4(9.261) = 1.96h → h = 37.044/1.96 = 18.9 cm.

Wait: 4×2.1³ = 4×9.261 = 37.044. 1.4² = 1.96. h = 37.044/1.96 = 18.9 cm.

Hmm, but let me recalculate: Volume sphere = (4/3)π(2.1)³ = (4/3)π(9.261). Volume cone = (1/3)π(1.4)²h = (1/3)π(1.96)h.

4(9.261)/3 = (1.96h)/3 → 4(9.261) = 1.96h → h = 37.044/1.96 = 18.9 cm.

If diameter of sphere is 4.2 → radius 2.1. Correct answer = 18.9 cm.

Q19. A bucket of height 8 cm has upper and lower radii 9 and 3 cm. Find its capacity and TSA.

Difficulty: Medium-Hard · Topic: Frustum of a Cone

Solution

V = (πh/3)(R²+r²+Rr) = (8π/3)(81+9+27) = (8π/3)(117) = 312π ≈ 980 cm².

l = √(64+36) = 10. CSA = π(12)(10) = 120π. TSA = 120π+81π+9π = 210π ≈ 660 cm².

Q20. Rainwater collected on a 22 m × 20 m roof drains into a cylindrical vessel of diameter 2 m and height 3.5 m. If the vessel is just full, find the rainfall in cm.

Difficulty: Hard · Topic: Conversion of Solids

Solution

Volume of cylinder = π(1)²(3.5) = 3.5π = 3.5×22/7 = 11 m³.

Roof area = 440 m². Rainfall h: 440×h = 11 → h = 11/440 = 0.025 m = 2.5 cm.

Other Chapters in Mathematics

Ch 1: Real Numbers Ch 2: Polynomials Ch 3: Pair of Linear Equations in Two Variables Ch 4: Quadratic Equations Ch 5: Arithmetic Progressions Ch 6: Triangles Ch 7: Coordinate Geometry Ch 8: Introduction to Trigonometry Ch 9: Some Applications of Trigonometry Ch 10: Circles Ch 11: Areas Related to Circles Ch 13: Statistics Ch 14: Probability

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