Surface Areas and Volumes Class 10 Maths Worksheet with Answers Chapter 12

This worksheet is based on the chapter Surface Areas and Volumes, that is chapter 12. You will practise questions on curved/total surface area, volume/capacity, and real-life applications like painting and water filled/left. π is going to be your best friend, so give it lot of respect.

Surface Areas and Volumes (Combination of Solids) | Practice Worksheet

Note: Take π = 22/7. Use proper units in every final answer.

A. Surface Area of Combination of Solids (7 Questions)

1. A capsule is made of a cylinder with two hemispherical ends.
   The total length is 18 cm and diameter is 6 cm.
   Find the total surface area of the capsule.
2. A toy is a cone mounted on a hemisphere (same radius).
   Radius = 3.5 cm, total height = 15.5 cm.
   Find the area to be painted (only the outer surface).
3. A cylinder of radius 4 cm and height 10 cm has a hemisphere of the same radius attached on one end.
   Find the total outer surface area of the solid.
4. A cubical block of side 8 cm is surmounted by a hemisphere.
   (a) What is the greatest possible diameter of the hemisphere?
   (b) Find the total surface area of the new solid.
5. A solid cylinder (radius 5 cm, height 14 cm) has a conical cavity (same radius and same height) hollowed out from the top.
   Find the total surface area of the remaining solid.
6. A wooden article is made by scooping out a hemisphere from each end of a solid cylinder.
   Cylinder radius = 3.5 cm, height = 12 cm.
   Find the total surface area of the article.
7. A tent is shaped like a cylinder with a conical top.
   Cylindrical part: height 2.5 m, radius 2 m.
   Conical top: slant height 3 m.
   Find the area of canvas needed (base not covered).

B. Volume of Combination of Solids (7 Questions)

8. A solid is made by mounting a cone on a hemisphere (same radius).
   Radius = 3 cm, height of cone = 4 cm.
   Find the volume of the solid (in terms of π).
9. A model is shaped like a cylinder with two identical cones at both ends.
   Total length = 18 cm. Cylinder part length = 12 cm.
   Diameter (everywhere) = 6 cm.
   Find the volume of air in the model.
10. A gulab jamun is shaped like a cylinder with two hemispherical ends.
    Total length = 5 cm, diameter = 2.8 cm.
    If syrup is about 30% of its volume, find syrup in 40 such jamuns.
11. A solid iron pole consists of:
    
    • Cylinder 1: height 220 cm, diameter 24 cm
    
    • Cylinder 2 on top: height 60 cm, radius 8 cm
    If 1 cm³ of iron = 8 g, find the mass of the pole (in kg). (Use π = 3.14)
12. A right circular cylinder is full of water.
    Radius = 7 cm, height = 20 cm.
    A solid hemisphere of radius 7 cm is dipped completely into it (water overflows).
    Find the volume of water overflowed.
13. A spherical vessel has an inner radius 4.2 cm.
    Find its capacity in cm³ and in litres. (Use π = 22/7, 1 litre = 1000 cm³)
14. A glass is a cylinder of inner radius 3 cm and height 12 cm, but it has a hemispherical raised bottom of radius 3 cm.
    Find:
    
    (a) apparent capacity, (b) actual capacity.

C. Mixed Applications: Painting, Cost, Capacity (7 Questions)

15. A toy rocket is a cone on a cylinder.
    Total height = 30 cm; cone height = 6 cm.
    Cone base diameter = 6 cm; cylinder base diameter = 4 cm.
    If cone is painted red and cylinder yellow, find the area painted red and area painted yellow. (Use π = 3.14)
16. A bird-bath is a cylinder with a hemispherical depression on top (same radius).
    Radius = 28 cm, cylinder height = 70 cm.
    Find the total surface area exposed.
17. A tank is shaped like a cylinder with two hemispherical ends.
    Radius = 1.4 m, length of cylindrical part = 4 m.
    Find the capacity of the tank in m³.
18. The tank in Q17 is to be painted from outside at ₹120 per m².
    Find the cost of painting.
19. A conical vessel (open at top) has radius 7 cm and height 24 cm.
    It is full of water. Some identical lead shots (spheres) of radius 0.5 cm are dropped in and 1/4 of the water overflows.
    Find the number of shots.
20. A cuboid of dimensions 15 cm × 10 cm × 4 cm has four conical depressions.
    Each depression: radius 0.7 cm, depth 2.1 cm.
    Find the volume of wood left.
21. A shopkeeper uses a cylindrical jar of radius 10 cm and height 35 cm.
    He fills it with oil up to 80% of its capacity.
    Find the volume of oil in litres.

D. Higher Order Thinking (HOTS) (7 Questions)

22. A capsule has total length 22 cm and radius 3 cm.
    Find the length of the cylindrical part.
23. A cylinder and a cone have the same base radius and same height.
    If the volume of the cylinder is 330 cm³, what is the volume of the cone?
24. A solid is formed by joining two cubes of side 5 cm end-to-end to form a cuboid.
    Find the surface area of the new cuboid.
25. A sphere of radius r is cut into two equal halves.
    Compare:
    
    (a) TSA of the sphere and (b) total surface area of two hemispheres together.
    Which is greater and by how much (in terms of r and π)?
26. A cylindrical container has radius 7 cm and water height 16 cm.
    A solid sphere of radius 3.5 cm is dropped in completely.
    By how much does the water level rise?
27. A cone (radius 7 cm, height 24 cm) is mounted on a hemisphere (radius 7 cm).
    Find the total surface area (outer surface only).
    (Do not include the common circular face.)
28. A student wrote: “TSA of a capsule = TSA of cylinder + TSA of two hemispheres.”
    Explain why this is wrong in one sentence, and write the correct expression.

Answer Key with Hints

A. Surface Area of Combination of Solids

1. Answer: TSA = 2πrh + 4πr²
   Here r = 3 cm, total length = 18 cm ⇒ h = 18 − 2r = 12 cm
   TSA = 2π×3×12 + 4π×3² = 72π + 36π = 108π = 339.43 cm² (approx.)
   
   Hint: Capsule = cylinder (curved only) + sphere (because 2 hemispheres).
2. Answer: r = 3.5 cm, total height = 15.5 cm ⇒ cone height h = 15.5 − 3.5 = 12 cm
   Slant height l = √(r² + h²) = √(3.5² + 12²) = √(12.25 + 144) = √156.25 = 12.5 cm
   TSA = 2πr² + πrl = 2π(3.5²) + π(3.5)(12.5)
   = 2π(12.25) + π(43.75) = (24.5π + 43.75π) = 68.25π = 214.5 cm²
   
   Hint: Do not add the circular base of the cone because it is joined to the hemisphere.
3. Answer: TSA = (CSA of cylinder) + (CSA of hemisphere) + (area of one circular base)
   = 2πrh + 2πr² + πr² = 2πrh + 3πr²
   = 2π×4×10 + 3π×16 = 80π + 48π = 128π = 402.29 cm²
   
   Hint: One end is covered by hemisphere, the other end is a flat circular base.
4. Answer: (a) Greatest diameter = 8 cm
   (b) TSA = TSA of cube − area of circle covered + CSA of hemisphere
   = 6a² − πr² + 2πr² = 6a² + πr²
   a = 8 cm, r = 4 cm ⇒ TSA = 6×64 + π×16 = 384 + 16π = 384 + 50.29 = 434.29 cm²
   
   Hint: The circle where hemisphere touches cube is not exposed.
5. Answer: TSA (remaining) = CSA of cylinder + base area (bottom) + CSA of conical cavity
   = 2πrh + πr² + πrl
   r = 5, h = 14, l = √(5²+14²)=√221 = 14.87
   TSA = 2π×5×14 + π×25 + π×5×14.87
   = 140π + 25π + 74.35π = 239.35π = 752.2 cm²
   
   Hint: The top circular face is removed (it becomes the opening).
6. Answer: TSA = CSA of cylinder + CSA of two hemispherical hollows
   = 2πrh + 2(2πr²) = 2πrh + 4πr²
   r = 3.5, h = 12 ⇒ TSA = 2π×3.5×12 + 4π×(3.5²)
   = 84π + 4π×12.25 = 84π + 49π = 133π = 418 cm²
   
   Hint: The flat circular ends are not present (they are scooped).
7. Answer: Canvas area = CSA of cylinder + CSA of cone
   = 2πrh + πrl
   r = 2, h = 2.5, l = 3 ⇒ area = 2π×2×2.5 + π×2×3 = 10π + 6π = 16π = 50.29 m²
   
   Hint: “Base not covered” means do not add circular base area.

B. Volume of Combination of Solids

8. Answer: V = volume(hemisphere) + volume(cone)
   = (2/3)πr³ + (1/3)πr²h
   = (2/3)π(27) + (1/3)π(9)(4) = 18π + 12π = 30π cm³
   
   Hint: For volume, we add volumes (nothing “disappears” inside).
9. Answer: r = 3 cm
   Cylinder length = 12 cm ⇒ Vcyl = πr²h = π×9×12 = 108π
   Remaining length = 18−12 = 6 cm ⇒ two cones together have height 6 cm ⇒ each cone height = 3 cm
   Vcones = 2×(1/3)πr²h = 2×(1/3)π×9×3 = 18π
   Total V = 108π + 18π = 126π = 396 cm³
   
   Hint: Always split the total length correctly among parts.
10. Answer: r = 1.4 cm, cylindrical length = 5 − 2r = 5 − 2.8 = 2.2 cm
    Vone = πr²h + (4/3)πr³
    = π(1.4²)(2.2) + (4/3)π(1.4³)
    = π(1.96×2.2) + (4/3)π(2.744)
    = 4.312π + 3.6587π = 7.9707π = 25.04 cm³
    Syrup in 40 jamuns = 40×0.30×25.04 = 300.48 cm³ = 0.300 L
    
    Hint: 30% means multiply by 0.30.
11. Answer: Use π = 3.14
    Cylinder 1: r = 12 cm, h = 220 ⇒ V₁ = 3.14×12²×220 = 3.14×144×220 = 994,? cm³ = 99,475.2 cm³
    Cylinder 2: r = 8 cm, h = 60 ⇒ V₂ = 3.14×64×60 = 12,057.6 cm³
    Total V = 111,532.8 cm³
    Mass = 111,532.8×8 g = 892,262.4 g = 892.26 kg
    
    Hint: Convert grams to kg by dividing by 1000.
12. Answer: Overflow = volume of hemisphere = (2/3)πr³
    r = 7 ⇒ overflow = (2/3)π×343 = 686π/3 cm³ = 718.67 cm³
    
    Hint: When dipped fully, it displaces its own volume.
13. Answer: V = (4/3)πr³ = (4/3)π(4.2³)
    4.2³ = 74.088 ⇒ V = (4/3)π×74.088 = 98.784π
    Using π = 22/7 ⇒ V = 98.784×22/7 = 310.5 cm³
    In litres = 310.5/1000 = 0.311 L (approx.)
    
    Hint: 1 litre = 1000 cm³.
14. Answer: Apparent = πr²h = π×9×12 = 108π
    Actual = 108π − (2/3)πr³ = 108π − (2/3)π×27 = 108π − 18π = 90π cm³
    Using π = 22/7 ⇒ actual = 282.86 cm³
    
    Hint: Raised hemisphere “steals” space, so subtract its volume.

C. Mixed Applications

15. Answer: Use π = 3.14
    Cone: r = 3, h = 6 ⇒ l = √(3²+6²)=√45=6.708
    Red area = CSA of cone + (base area of cone − base area of cylinder)
    Cylinder radius = 2
    Red = πrl + πr² − π(2²) = 3.14(3×6.708) + 3.14(9 − 4)
    = 63.2 + 15.7 = 78.9 cm² (approx.)
    Yellow area = CSA of cylinder + one base area = 2πrh + πr²
    h(cyl)=30−6=24, r=2 ⇒ 2×3.14×2×24 + 3.14×4 = 301.44 + 12.56 = 314.0 cm²
    
    Hint: Check which circular faces are visible and which are joined.
16. Answer: TSA = CSA(cyl) + CSA(hemisphere) + area of bottom base
    = 2πrh + 2πr² + πr² = 2πrh + 3πr²
    r=28, h=70 ⇒ TSA = 2π×28×70 + 3π×784 = 3920π + 2352π = 6272π cm²
    = 6272×22/7 = 19,712 cm² = 1.9712 m²
    
    Hint: Convert cm² to m² by dividing by 10,000.
17. Answer: V = πr²h + (4/3)πr³
    r=1.4, h=4 ⇒ V = π(1.96)(4) + (4/3)π(2.744)
    = 7.84π + 3.6587π = 11.4987π
    Using π = 22/7 ⇒ V = 36.14 m³ (since units are in m)
    
    Hint: Keep all dimensions in metres to get m³.
18. Answer: Outer area = 2πrh + 4πr² = 2π(1.4)(4) + 4π(1.4²)
    = 11.2π + 7.84π = 19.04π
    Using π=22/7 ⇒ area = 59.84 m²
    Cost = 59.84 × 120 = ₹7,180.8 = ₹7,181
    
    Hint: Painting outside means curved areas only (no ends if closed).
19. Answer: Overflow = 1/4 of cone volume
    Cone volume = (1/3)πr²h = (1/3)π×49×24 = 392π
    Overflow = 98π cm³
    One shot volume = (4/3)π(0.5³) = (4/3)π(0.125) = (1/6)π
    Number = (98π) ÷ ((1/6)π) = 98×6 = 588
    
    Hint: π cancels nicely—good day for algebra.
20. Answer: Volume of cuboid = 15×10×4 = 600 cm³
    One cone volume = (1/3)πr²h = (1/3)π(0.7²)(2.1) = (1/3)π(0.49)(2.1) = 0.343π
    Four cones = 1.372π = 1.372×22/7 = 4.312 cm³
    Wood left = 600 − 4.312 = 595.69 cm³
    
    Hint: Depressions remove volume, so subtract.
21. Answer: Capacity = πr²h = π×100×35 = 3500π cm³
    80% oil = 0.8×3500π = 2800π cm³
    Using π=22/7 ⇒ oil = 2800×22/7 = 8800 cm³ = 8.8 litres
    
    Hint: Convert cm³ to litres by dividing by 1000.

D. Higher Order Thinking

22. Answer: Cylindrical length = 22 − 2r = 22 − 6 = 16 cm
    
    Hint: Two hemispheres together contribute 2r to length.
23. Answer: Cone volume = (1/3) of cylinder volume = 110 cm³
    
    Hint: Same r and h ⇒ Vcone = (1/3)πr²h.
24. Answer: New cuboid dimensions = 10 cm × 5 cm × 5 cm
    TSA = 2(lb + bh + lh) = 2(10×5 + 5×5 + 10×5) = 2(50+25+50) = 250 cm²
    
    Hint: Joining end-to-end doubles one dimension.
25. Answer: TSA of sphere = 4πr²
    Two hemispheres together: each hemisphere TSA = 3πr² (CSA 2πr² + base πr²)
    Total = 6πr²
    Difference = 6πr² − 4πr² = 2πr² (two bases add extra area)
    
    Hint: Cutting creates new flat faces.
26. Answer: Rise = displaced volume / base area
    Sphere volume = (4/3)π(3.5³) = (4/3)π(42.875) = 57.1667π
    Base area = π(7²)=49π
    Rise = (57.1667π)/(49π) = 1.167 cm (approx.)
    
    Hint: π cancels—so don’t panic.
27. Answer: TSA = CSA(hemisphere) + CSA(cone) = 2πr² + πrl
    r=7, h=24 ⇒ l = √(7²+24²)=√625=25
    TSA = 2π×49 + π×7×25 = 98π + 175π = 273π cm²
    = 273×22/7 = 858 cm²
    
    Hint: Do not include the common circular face.
28. Answer: Wrong because the circular bases that are joined are not exposed.
    Correct: TSA(capsule) = CSA of cylinder + CSA of two hemispheres = 2πrh + 4πr².
    
    Hint: “Total surface area” of each part includes bases, but joined bases disappear.