Chapter 12: Surface Areas and Volumes Class 10 Maths NCERT Solutions

★ 4.7 (223) (2.2K)
Padhai.co
Padhai.co
View Profile

Contents

Class 10 Maths NCERT Solutions - Chapter wise

1

Chapter 1: Real Numbers

2

Chapter 2: Polynomials

3

Chapter 3 : Pair of Linear Equations in Two Variables

4

Chapter 4 : Quadratic Equations

5

Chapter 5 : Arithmetic Progressions

6

Chapter 6 : Triangles

7

Chapter 7 : Coordinate Geometry

8

Chapter 8: Introduction to Trigonometry

9

Chapter 9: Some Applications of Trigonometry

10

Chapter 10 : Circles

11

Chapter 11 : Areas Related to Circles

12

Chapter 12: Surface Areas and Volumes

13

Chapter 13: Statistics

14

Chapter 14 : Probability

Chapter Notes

Chapter 12: Surface Areas and Volumes — Exercise 12.1

Welcome to the world of 3D shapes! This chapter is all about “Combination of Solids”—think of it like LEGO math, where we join cubes, cones, and hemispheres to calculate the total space they take up or the area we can touch on the outside.

Q1. 2 cubes each of volume 64 cm^3 are joined end to end. Find the surface area of the resulting cuboid.

  • The Logic: When you join two cubes side-by-side, they don’t stay cubes anymore; they turn into a Cuboid. The height and width stay the same, but the length doubles. First, we need to find the side of the original cube using the volume.

  • The Steps:

    • Volume of cube = s^3 = 64. So, side (s) = 4 cm.

    • When joined, for the new cuboid:

      • Length (l) = 4 + 4 = 8 cm

      • Breadth (b) = 4 cm

      • Height (h) = 4 cm

    • Surface Area formula = 2(lb + bh + hl)

    • Area = 2(84 + 44 + 4*8) = 2(32 + 16 + 32) = 2(80)

  • The Result: The surface area of the resulting cuboid is 160 cm^2.

Q2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

  • The Logic: Imagine a bowl with a pipe standing on top of it. To find the “touchable” area inside, we add the Curved Surface Area (CSA) of the Cylinder and the CSA of the Hemisphere. Human Tip: The radius of the cylinder is the same as the hemisphere (7 cm). To find the cylinder’s height, subtract the radius from the total height!

     

  • The Steps:

    • Radius (r) = 14 / 2 = 7 cm.

    • Height of cylinder (h) = Total height – Radius = 13 – 7 = 6 cm.

    • Inner Surface Area = CSA of Cylinder + CSA of Hemisphere

    • Formula = 2pirh + 2pir^2 = 2pi*r(h + r)

    • Area = 2 * (22/7) * 7 * (6 + 7)

    • Area = 44 * 13

  • The Result: The inner surface area of the vessel is 572 cm^2.

Q3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

  • The Logic: This is like an ice-cream cone with a single scoop. The total surface area is the CSA of the Cone + the CSA of the Hemisphere. Pro-Tip: For the cone, you need the “Slant Height” (l), not just the vertical height.

  • The Steps:

    • Radius (r) = 3.5 cm.

    • Vertical height of cone (h) = 15.5 – 3.5 = 12 cm.

    • Slant height (l) = root(r^2 + h^2) = root(3.5^2 + 12^2) = root(12.25 + 144) = root(156.25) = 12.5 cm.

    • Total Surface Area = CSA of Cone + CSA of Hemisphere

    • Formula = pirl + 2pir^2 = pi*r(l + 2r)

    • Area = (22/7) * 3.5 * (12.5 + 2*3.5) = 11 * (12.5 + 7) = 11 * 19.5

  • The Result: The total surface area of the toy is 214.5 cm^2.

Q4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

  • The Logic: The hemisphere sits on top of the cube. The “greatest diameter” it can have is exactly equal to the side of the cube (7 cm). When calculating area, remember that the base of the hemisphere covers a circular part of the cube’s top face—so we subtract that circle but add the curved top of the hemisphere.

  • The Steps:

    • Greatest diameter = 7 cm (Radius = 3.5 cm).

    • Total Surface Area = Surface Area of Cube + CSA of Hemisphere – Area of Hemisphere Base

    • Formula = 6s^2 + 2pir^2 – pir^2 = 6s^2 + pir^2

    • Area = 6*(7^2) + (22/7) * 3.5 * 3.5

    • Area = 6*49 + 38.5 = 294 + 38.5

  • The Result: The greatest diameter is 7 cm and the total surface area is 332.5 cm^2.

Q5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

  • The Logic: Whether you put a hemisphere on a cube or dig a hole into it, the surface area added is the same! You lose the flat circle on top but gain the curved “bowl” inside.

  • The Steps:

    • Side of cube = l. Radius of hemisphere = l/2.

    • Surface Area = Area of 6 faces of cube – Area of circle + CSA of hemisphere

    • Formula = 6l^2 – pi(l/2)^2 + 2pi(l/2)^2 = 6l^2 + pi(l/2)^2

    • Area = 6l^2 + (pil^2)/4 = (1/4)*l^2 * (24 + pi)

  • The Result: The surface area is (1/4)l^2(24 + pi) square units.

Q6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

  • The Logic: Think of the capsule as three separate parts joined together: a cylinder in the middle and two “caps” (hemispheres) on the ends. Since the hemispheres are “stuck” to the cylinder, we only care about their Curved Surface Areas (CSA).

  • The Steps:

    1. Radius (r): Diameter is 5 mm, so radius = 5/2 = 2.5 mm.

    2. Height of Cylinder (h): The total length is 14 mm. We must subtract the radius of both hemispheres (2.5 + 2.5 = 5 mm). So, h = 14 – 5 = 9 mm.

    3. Total Surface Area: CSA of Cylinder + 2 * CSA of Hemisphere.

    4. Formula: 2pirh + 2 * (2pir^2) = 2pi*r(h + 2r).

    5. Calculation: 2 * (22/7) * 2.5 * (9 + 5) = 2 * (22/7) * 2.5 * 14.

    6. Calculation: 2 * 22 * 2.5 * 2 = 220.

  • The Result: The surface area of the capsule is 220 mm^2.

Q7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m^2. (Note that the base of the tent will not be covered with canvas.)

  • The Logic: To find the canvas needed, we calculate the “outer skin” of the tent. This includes the CSA of the Cylinder (the walls) and the CSA of the Cone (the roof). We don’t include the floor!

  • The Steps:

    1. Dimensions: Radius (r) = 2 m, Cylinder Height (h) = 2.1 m, Slant Height (l) = 2.8 m.

    2. Total Area: CSA of Cone + CSA of Cylinder.

    3. Formula: pirl + 2pirh = pir(l + 2h).

    4. Calculation: (22/7) * 2 * (2.8 + 2 * 2.1) = (22/7) * 2 * (2.8 + 4.2) = (22/7) * 2 * 7.

    5. Calculation: 22 * 2 = 44.

    6. Cost: 44 m^2 * 500 = 22,000.

  • The Result: The area of canvas is 44 m^2 and the total cost is Rs 22,000.

Q8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm^2.

  • The Logic: When you “hollow out” a cone from a cylinder, you actually increase the surface area you can touch. You now have the outside walls of the cylinder, the flat base at the bottom, and the new “sloping” wall inside where the cone used to be.

  • The Steps:

    1. Dimensions: Radius (r) = 0.7 cm, Height (h) = 2.4 cm.

    2. Slant Height (l): root(0.7^2 + 2.4^2) = root(0.49 + 5.76) = root(6.25) = 2.5 cm.

    3. Total Area: CSA of Cylinder + CSA of Cone + Area of Cylinder Base.

    4. Formula: 2pirh + pirl + pir^2 = pi*r(2h + l + r).

    5. Calculation: (22/7) * 0.7 * (2 * 2.4 + 2.5 + 0.7) = 2.2 * (4.8 + 2.5 + 0.7) = 2.2 * 8 = 17.6.

  • The Result: The total surface area is 17.6 cm^2, which is 18 cm^2 to the nearest cm^2.

Q9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

  • The Logic: Similar to the capsule, but the hemispheres are “holes” inside. We calculate the outside wall of the cylinder and add the surface area of the two inner “bowls” created by scooping.

  • The Steps:

    1. Dimensions: Radius (r) = 3.5 cm, Height (h) = 10 cm.

    2. Total Area: CSA of Cylinder + 2 * CSA of Hemisphere.

    3. Formula: 2pirh + 2 * (2pir^2) = 2pi*r(h + 2r).

    4. Calculation: 2 * (22/7) * 3.5 * (10 + 7) = 22 * 17 = 374.

  • The Result: The total surface area of the article is 374 cm^2.

Chapter 12: Surface Areas and Volumes — Exercise 12.2

Welcome to the “Volume” edition! While surface area was about the “skin” of an object, Volume is all about the “stuff” inside—like how much water a tank holds or how much Gulab Jamun syrup is absorbed.

Q1. 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 pi.

  • The Logic: This is a simple addition problem. Since the cone is sitting on the hemisphere, the total volume is just the Volume of the Cone + Volume of the Hemisphere.

  • Human Tip: When a question asks for the answer “in terms of pi,” it means you don’t need to multiply by 22/7 or 3.14. Just leave the symbol π as it is!

  • The Steps:

    • Radius (r) = 1 cm.

    • Height of cone (h) = 1 cm.

    • Total Volume = (1/3) * pi * r^2 * h + (2/3) * pi * r^3

    • Volume = (1/3) * pi * (1)^2 * 1 + (2/3) * pi * (1)^3

    • Volume = (1/3)pi + (2/3)pi = (3/3)pi

  • The Result: The volume of the solid is pi cm^3.

Q2. 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.

  • The Logic: Think of this like a rocket. You have a central cylinder and two identical cones on the ends. To find the total air inside, we add the volumes of all three parts.

     

  • The Steps:

    1. Dimensions: Radius (r) = 3/2 = 1.5 cm.

    2. Height of Cones (h1): 2 cm each.

    3. Height of Cylinder (h2): Total length – (height of both cones) = 12 – (2 + 2) = 8 cm.

    4. Total Volume: Volume of Cylinder + 2 * (Volume of Cone)

    5. Calculation: [pi * r^2 * h2] + 2 * [(1/3) * pi * r^2 * h1]

    6. Calculation: pi * (1.5)^2 * [8 + 2/3 * 2] = (22/7) * 2.25 * [8 + 4/3]

    7. Calculation: (22/7) * 2.25 * (28/3) = 22 * 2.25 * (4/3) = 66.

  • The Result: The volume of air contained in the model is 66 cm^3.

Q3. 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.

  • The Logic: This is exactly like the capsule problem from the last exercise, but for volume!

    1. Find the volume of one gulab jamun (Cylinder + 2 Hemispheres).

    2. Multiply by 45 to get the total volume.

    3. Calculate 30% of that total to find the syrup.

  • The Steps:

    1. Dimensions: Radius (r) = 1.4 cm.

    2. Cylinder Height (h): 5 – (1.4 + 1.4) = 2.2 cm.

    3. Volume of 1 Jamun: pi * r^2 * h + (4/3) * pi * r^3 = pi * r^2 [h + 4/3r]

    4. Volume of 45 Jamuns: 45 * (22/7) * 1.4 * 1.4 * [2.2 + (4/3 * 1.4)] ≈ 1127.28 cm^3.

    5. Syrup (30%): 0.30 * 1127.28 ≈ 338.18.

  • The Result: The volume of syrup is approximately 338 cm^3.

Q4. 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.

  • The Logic: When you drill holes into wood, you are removing volume. So, we calculate the volume of the solid cuboid and subtract the volume of the 4 cones.

  • The Steps:

    1. Volume of Cuboid: 15 * 10 * 3.5 = 525 cm^3.

    2. Volume of 4 Cones: 4 * [(1/3) * pi * r^2 * h]

    3. Calculation: 4 * (1/3) * (22/7) * 0.5 * 0.5 * 1.4 = 1.46 cm^3.

    4. Remaining Wood: 525 – 1.46 = 523.54 cm^3.

  • The Result: The volume of wood in the entire stand is 523.53 cm^3.

Q5. 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.

  • The Logic: This is based on Archimedes’ Principle! The volume of water that flows out is exactly equal to the total volume of the lead shots dropped in. We find the volume of the cone, take one-fourth of it, and divide that by the volume of a single lead shot.

  • The Steps:

    1. Volume of Cone: (1/3) * pi * r^2 * h = (1/3) * pi * 5 * 5 * 8 = (200/3)pi cm^3.

    2. Volume of Displaced Water: (1/4) * (200/3)pi = (50/3)pi cm^3.

    3. Volume of 1 Lead Shot (Sphere): (4/3) * pi * r^3 = (4/3) * pi * (0.5)^3 = (4/3) * pi * (1/8) = (1/6)pi cm^3.

    4. Number of Lead Shots: (Volume of Water Out) / (Volume of 1 Shot)

    5. Calculation: ((50/3)pi) / ((1/6)pi) = (50/3) * 6 = 100.

  • The Result: The number of lead shots dropped in the vessel is 100.

Q6. 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 cm^3 of iron has approximately 8g mass. (Use pi = 3.14)

  • The Logic: To find the total mass, we first need the total volume. Since one cylinder is on top of the other, we simply add their volumes: Volume of Large Cylinder + Volume of Small Cylinder.

  • The Steps:

    1. Large Cylinder (R=12, H=220): Volume = pi * 12^2 * 220 = 31680 * 3.14 = 99475.2 cm^3.

    2. Small Cylinder (r=8, h=60): Volume = pi * 8^2 * 60 = 3840 * 3.14 = 12057.6 cm^3.

    3. Total Volume: 99475.2 + 12057.6 = 111532.8 cm^3.

    4. Mass: Volume * 8g = 111532.8 * 8 = 892262.4g.

    5. Convert to kg: 892.26 kg.

  • The Result: The mass of the pole is approximately 892.26 kg.

Q7. 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.

  • The Logic: The cylinder was full of water. When you put the solid inside, it “kicks out” its own volume of water. So, Water Left = Volume of Cylinder – Volume of Solid (Cone + Hemisphere).

     

  • The Steps:

    1. Volume of Cylinder: pi * 60^2 * 180 = 648000pi cm^3.

    2. Volume of Solid:

      • Cone: (1/3) * pi * 60^2 * 120 = 144000pi cm^3.

      • Hemisphere: (2/3) * pi * 60^3 = 144000pi cm^3.

      • Total Solid = 288000pi cm^3.

    3. Water Left: 648000pi – 288000pi = 360000pi cm^3.

    4. Calculation: 360000 * (22/7) ≈ 1,131,428.57 cm^3.

    5. In m^3: ≈ 1.131 m^3.

  • The Result: The volume of water left in the cylinder is approximately 1.131 m^3.

Q8. 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 cm^3. Check whether she is correct, taking the above as inside measurements, and pi = 3.14.

  • The Logic: This vessel looks like a round-bottom flask. Total Volume = Volume of Sphere + Volume of Cylinder (Neck). We calculate it ourselves to verify the child’s claim.

  • The Steps:

    1. Cylindrical Neck (r=1, h=8): Volume = pi * 1^2 * 8 = 8 * 3.14 = 25.12 cm^3.

    2. Spherical Part (r=4.25): Volume = (4/3) * pi * (4.25)^3 = (4/3) * 3.14 * 76.765 = 321.39 cm^3.

    3. Total Volume: 25.12 + 321.39 = 346.51 cm^3.

  • The Result: No, she is not correct. The actual volume is 346.51 cm^3.

Practice like a Topper with our A+ Practice System!

Rate this Resource

Community Rating

4.7

Tap to rate

Comments (0)

No comments yet. Be the first to comment!

Please log in to comment

Log In