Class 10 Mathematics Notes Chapter 13 (Chapter 13) – Examplar Problems (English) Book

Alright class, let's focus on Chapter 13: Surface Areas and Volumes from the NCERT Exemplar. This is a crucial chapter, not just for your board exams but also frequently tested in various government exams due to its practical application and reliance on formulas and spatial reasoning. Pay close attention to the formulas and how they are applied, especially when dealing with combined or converted solids.

Chapter 13: Surface Areas and Volumes - Detailed Notes for Government Exams

1. Basic Solids: Formulas are Key!

You must memorize these formulas accurately. There's no way around it.

• Cuboid: (Dimensions: length 'l', breadth 'b', height 'h')

  • Volume = l × b × h
  • Lateral Surface Area (LSA) = 2h(l + b) (Area of four walls)
  • Total Surface Area (TSA) = 2(lb + bh + hl)
  • Length of Diagonal = √(l² + b² + h²)

• Cube: (Side 'a')

  • Volume = a³
  • Lateral Surface Area (LSA) = 4a²
  • Total Surface Area (TSA) = 6a²
  • Length of Diagonal = a√3

• Right Circular Cylinder: (Radius 'r', height 'h')

  • Volume = πr²h
  • Curved Surface Area (CSA) = 2πrh
  • Total Surface Area (TSA) = 2πrh + 2πr² = 2πr(h + r)
  • Hollow Cylinder: (Outer radius 'R', inner radius 'r', height 'h')
    • Volume = π(R² - r²)h

• Right Circular Cone: (Radius 'r', height 'h', slant height 'l')

  • Slant Height, l = √(r² + h²)
  • Volume = (1/3)πr²h
  • Curved Surface Area (CSA) = πrl
  • Total Surface Area (TSA) = πrl + πr² = πr(l + r)

• Sphere: (Radius 'r')

  • Volume = (4/3)πr³
  • Surface Area (SA) = 4πr² (Note: CSA and TSA are the same)

• Hemisphere: (Radius 'r')

  • Volume = (2/3)πr³
  • Curved Surface Area (CSA) = 2πr²
  • Total Surface Area (TSA) = 2πr² + πr² = 3πr² (CSA + Area of circular base)

2. Combination of Solids

This is where application skills are tested. Visualize the combined shape.

• Volume: The easiest part. Simply add the volumes of the individual solids that make up the combined object.
  • Example: A toy rocket (Cone + Cylinder). Volume = Volume of Cone + Volume of Cylinder.
• Surface Area: Be careful! Do not just add the TSAs of individual solids.
  • Identify the surfaces that are visible in the combined solid.
  • Add the relevant surface areas (CSA or base areas) of the individual parts.
  • Crucially, subtract the areas that become hidden or overlap when the solids are joined.
  • Example: A capsule (Cylinder + 2 Hemispheres). TSA = CSA of Cylinder + CSA of Hemisphere 1 + CSA of Hemisphere 2. (The circular bases of the cylinder and hemispheres are hidden).
  • Example: A tent (Cylinder + Cone). TSA = CSA of Cylinder + CSA of Cone + Area of Cylindrical Base (if it rests on the ground).

3. Conversion of Solids

When a solid is melted and recast into a different shape (or shapes), the key principle is:
Volume remains constant.

• Process: Calculate the volume of the original solid. Equate it to the volume of the new solid (or the sum of volumes if recast into multiple smaller objects). Solve for the unknown dimension (radius, height, number of objects, etc.).
• Example: A metallic sphere is melted and drawn into a wire (cylinder).
  • Volume of Sphere = Volume of Cylindrical Wire
  • (4/3)πR³ = πr²h (where R is sphere radius, r is wire radius, h is wire length).
• Example: How many small spherical balls can be made from a large metallic cone?
  • Volume of Large Cone = n × Volume of one Small Sphere (where 'n' is the number of spheres).
  • (1/3)πR²H = n × (4/3)πr³

4. Frustum of a Cone

This is the shape you get when you slice a cone horizontally. It looks like a bucket. (Let R be the radius of the larger circular end, r be the radius of the smaller circular end, h be the vertical height, and l be the slant height).

• Slant Height, l = √[h² + (R - r)²]
• Volume = (1/3)πh(R² + r² + Rr)
• Curved Surface Area (CSA) = πl(R + r)
• Total Surface Area (TSA) = CSA + Area of Top Circle + Area of Bottom Circle
  • TSA = πl(R + r) + πR² + πr²

Important Tips for Government Exams:

• Units: Always ensure consistency in units. If dimensions are given in cm and m, convert them to a single unit before calculation. Volume units will be cubic (cm³, m³), and area units will be square (cm², m²).
• π Value: Use π = 22/7 unless specified otherwise (e.g., 3.14). Sometimes, π might cancel out, simplifying calculations. Look for this possibility.
• Visualization: Try to draw a rough sketch of the solid(s) described. This helps immensely in identifying the correct formulas and surfaces for combined solids.
• Calculation Speed & Accuracy: Practice calculations. These questions can be calculation-intensive. Avoid silly errors.
• Read Carefully: Understand exactly what is being asked – Volume, CSA, TSA, number of items, ratio, etc.

Multiple Choice Questions (MCQs)

Here are 10 MCQs to test your understanding:

1. If the radius of a sphere is doubled, its volume becomes:
   (a) Double
   (b) Four times
   (c) Six times
   (d) Eight times

2. A solid cylinder of radius 'r' and height 'h' is placed over a solid hemisphere of the same radius 'r'. The total surface area of the combined solid is:
   (a) 2πr(h + r) + 2πr²
   (b) 2πrh + 3πr²
   (c) πr²h + (2/3)πr³
   (d) 2πrh + 4πr²

3. A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively, is melted and recast into the form of a cone of base diameter 8 cm. The height of the cone is:
   (a) 12 cm
   (b) 14 cm
   (c) 15 cm
   (d) 18 cm

4. The ratio of the volumes of a cube to that of a sphere which will exactly fit inside the cube is:
   (a) 6 : π
   (b) π : 6
   (c) 3 : π
   (d) π : 3

5. A cone and a hemisphere have equal bases and equal volumes. The ratio of their heights is:
   (a) 1 : 2
   (b) 2 : 1
   (c) 1 : 3
   (d) 3 : 1

6. A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. The volume of the toy is: (Use π = 3.14)
   (a) 25.12 cm³
   (b) 50.24 cm³
   (c) 12.56 cm³
   (d) 37.68 cm³

7. The curved surface area of a frustum of a cone is given by: (R, r are radii, l is slant height)
   (a) πl(R - r)
   (b) πl(R + r)
   (c) πh(R² + r² + Rr)
   (d) 2πl(R + r)

8. If the surface area of a sphere is 144π cm², then its radius is:
   (a) 6 cm
   (b) 8 cm
   (c) 12 cm
   (d) 7 cm

9. How many bags of grain can be stored in a cuboidal granary 12 m × 6 m × 5 m, if each bag occupies a space of 0.48 m³?
   (a) 750
   (b) 75
   (c) 1500
   (d) 375

10. The radii of the ends of a bucket 30 cm high are 21 cm and 7 cm. Its capacity in litres is: (Use π = 22/7)
    (a) 19.84 L
    (b) 20.02 L
    (c) 21.56 L
    (d) 22.72 L

Answers to MCQs:

1. (d) Eight times (Volume ∝ r³)
2. (b) 2πrh + 3πr² (CSA Cylinder + CSA Hemisphere + Area of Cylindrical Base = 2πrh + 2πr² + πr²)
3. (b) 14 cm (Volume of shell = Volume of cone => (4/3)π(4³ - 2³) = (1/3)π(4²)h => h=14)
4. (a) 6 : π (Volume Cube = a³; Sphere radius = a/2, Volume Sphere = (4/3)π(a/2)³ = πa³/6. Ratio = a³ / (πa³/6) = 6/π)
5. (b) 2 : 1 (Volume Cone = (1/3)πr²h_c; Volume Hemisphere = (2/3)πr³. Equating gives h_c = 2r. Height of hemisphere is its radius 'r'. Ratio h_c : r = 2r : r = 2:1)
6. (a) 25.12 cm³ (Radius r = 2 cm. Vol = Vol Cone + Vol Hemisphere = (1/3)πr²h + (2/3)πr³ = (1/3)π(2²)(2) + (2/3)π(2³) = (8/3)π + (16/3)π = (24/3)π = 8π = 8 * 3.14 = 25.12)
7. (b) πl(R + r) (Direct formula)
8. (a) 6 cm (Surface Area = 4πr² = 144π => r² = 36 => r = 6)
9. (b) 750 (Volume of granary = 1265 = 360 m³. Number of bags = Total Volume / Volume per bag = 360 / 0.48 = 750)
10. (b) 20.02 L (Volume of Frustum (Bucket) = (1/3)πh(R² + r² + Rr) = (1/3)(22/7)30(21² + 7² + 217) = (1022/7)(441 + 49 + 147) = (220/7)*(637) = 220 * 91 = 20020 cm³. Since 1000 cm³ = 1 L, Volume = 20.02 L)

Make sure you practice a variety of problems from the Exemplar book, focusing on combinations and conversions. Good luck!