Class 9 Mathematics Notes Chapter 13 (Chapter 13) – Examplar Problem (Englisha) Book
Alright class, let's focus on Chapter 13: Surface Areas and Volumes from your NCERT Exemplar. This chapter is crucial, not just for your school exams, but also forms a strong foundation for various competitive government exams where quantitative aptitude is tested. The Exemplar problems often require a deeper understanding than the textbook, so pay close attention.
Chapter 13: Surface Areas and Volumes - Detailed Notes
1. Introduction:
This chapter deals with calculating the surface area (the total area of the surface of a three-dimensional object) and volume (the space occupied by a three-dimensional object) of various solid shapes. Mastery of the formulas and their application is key.
2. Solid Shapes and Formulas:
-
Cuboid: A solid bounded by six rectangular faces. Let length = l, breadth = b, height = h.
- Lateral Surface Area (LSA) or Area of four walls = 2(l + b)h
- Total Surface Area (TSA) = 2(lb + bh + hl)
- Volume = l × b × h
- Length of diagonal = √(l² + b² + h²)
-
Cube: A special cuboid where all edges are equal. Let edge = a.
- Lateral Surface Area (LSA) = 4a²
- Total Surface Area (TSA) = 6a²
- Volume = a³
- Length of diagonal = a√3
-
Right Circular Cylinder: A solid generated by revolving a rectangle about one of its sides. Let radius of the base = r, height = h.
- Curved Surface Area (CSA) = 2πrh
- Total Surface Area (TSA) = CSA + Area of two circular bases = 2πrh + 2πr² = 2πr(h + r)
- Volume = Area of base × height = πr²h
- Hollow Cylinder: If R is the external radius and r is the internal radius,
- Thickness = R - r
- Area of cross-section = π(R² - r²)
- External CSA = 2πRh
- Internal CSA = 2πrh
- TSA = (External CSA) + (Internal CSA) + (Area of two rings) = 2πRh + 2πrh + 2π(R² - r²)
- Volume of material = π(R² - r²)h
-
Right Circular Cone: A solid generated by revolving a right-angled triangle about one of its sides containing the right angle. Let radius = r, height = h, slant height = l.
- Slant height, l = √(h² + r²)
- Curved Surface Area (CSA) = πrl
- Total Surface Area (TSA) = CSA + Area of circular base = πrl + πr² = πr(l + r)
- Volume = (1/3) × Volume of cylinder with same base radius and height = (1/3)πr²h
-
Sphere: A perfectly round geometrical object in three-dimensional space. Let radius = r.
- Surface Area (SA) = 4πr² (Note: For a sphere, CSA and TSA are the same)
- Volume = (4/3)πr³
-
Hemisphere: Exactly half of a sphere. Let radius = r.
- Curved Surface Area (CSA) = 2πr²
- Total Surface Area (TSA) = CSA + Area of circular base = 2πr² + πr² = 3πr²
- Volume = (1/2) × Volume of sphere = (2/3)πr³
- Spherical Shell (Hollow Sphere): If R is the external radius and r is the internal radius,
- Thickness = R - r
- Volume of material = (4/3)π(R³ - r³)
3. Key Concepts & Problem Types (Emphasis from Exemplar):
- Unit Consistency: Always ensure all dimensions are in the same unit before applying formulas. If areas or volumes are given, ensure units are consistent (e.g., cm², m², cm³, m³). Pay attention to conversions (1 m = 100 cm, 1 m² = 10000 cm², 1 m³ = 1000000 cm³, 1 Litre = 1000 cm³).
- Surface Area vs. Volume: Read the question carefully to determine whether you need to calculate surface area (for painting, covering, wrapping) or volume (for capacity, amount of material, space occupied, melting and recasting).
- Melting and Recasting: When one solid shape is melted and recast into another shape (or multiple smaller shapes), the volume remains constant.
- Example: A metallic sphere is melted and recast into small cones. Volume of Sphere = n × Volume of one small cone (where n is the number of cones).
- Flow Rate Problems: Problems involving water flowing through pipes often relate volume to time.
- Volume of water flowing out per unit time = Area of cross-section of the pipe × Speed of water flow.
- Ensure units of speed and dimensions are consistent (e.g., if speed is in m/s, dimensions should be in meters, and time in seconds).
- Embankments/Platforms: When earth is dug out (forming a pit, usually cuboidal or cylindrical) and used to form an embankment (often a hollow cylindrical shape) or a platform (cuboidal), the volume of earth dug out equals the volume of earth used in the embankment/platform.
- Ratio Problems: Questions might involve ratios of dimensions, surface areas, or volumes.
- Example: If the radii of two spheres are in the ratio 2:3, their surface areas will be in the ratio 2²:3² = 4:9, and their volumes will be in the ratio 2³:3³ = 8:27.
- Percentage Change: Problems asking for the percentage change in CSA, TSA, or Volume when dimensions are changed (e.g., radius increased by 10%). Calculate the original value, the new value, find the change, and then calculate the percentage change = (Change / Original Value) × 100.
- Hollow Shapes: Be careful with hollow cylinders and spherical shells. Distinguish between internal and external dimensions, and calculate the volume of the material used.
4. Tips for Government Exams:
- Memorize Formulas: Formulas are non-negotiable. Write them down multiple times.
- Visualize: Try to visualize the shapes and the process described (e.g., melting, filling).
- Approximation: Sometimes, using π ≈ 22/7 or 3.14 is specified. If not, choose based on whether the dimensions are multiples of 7. In competitive exams, look at the options to gauge the required precision.
- Practice Exemplar Problems: These problems test application and are closer to the level sometimes seen in competitive exams.
- Time Management: Practice solving problems within a time limit. Direct formula-based questions should be very quick.
Multiple Choice Questions (MCQs)
Here are 10 MCQs based on the concepts from Chapter 13, keeping the Exemplar level in mind:
-
If the radius of a sphere is doubled, its volume becomes:
(A) Double
(B) Four times
(C) Six times
(D) Eight times -
The total surface area of a cube is 150 cm². The length of its edge is:
(A) 4 cm
(B) 5 cm
(C) 6 cm
(D) 5.5 cm -
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? (Use π = 22/7)
(A) 38.5
(B) 77
(C) 115.5
(D) 3.85 -
The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 5:3. The ratio of their volumes is:
(A) 10:17
(B) 20:27
(C) 17:27
(D) 20:37 -
The dimensions of a cuboid are 5 cm × 4 cm × 2 cm. The number of cubes of edge 1 cm that can be cut from it is:
(A) 20
(B) 40
(C) 10
(D) 18 -
The curved surface area of a right circular cylinder of height 14 cm is 88 cm². The diameter of the base is:
(A) 1 cm
(B) 2 cm
(C) 4 cm
(D) 0.5 cm -
If the lateral surface area of a cube is 100 m², then its volume is:
(A) 1000 m³
(B) 125 m³
(C) 250 m³
(D) 500 m³ -
The total surface area of a hemisphere of radius 7 cm is: (Use π = 22/7)
(A) 308 cm²
(B) 154 cm²
(C) 462 cm²
(D) 616 cm² -
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 -
The ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube is:
(A) 6 : π
(B) 4 : π
(C) 3 : π
(D) 2 : π
Answer Key for MCQs:
- (D) Eight times (Volume is proportional to r³, so (2r)³ = 8r³)
- (B) 5 cm (6a² = 150 => a² = 25 => a = 5)
- (A) 38.5 (r = 3.5/2 = 1.75 m, h = 12 m. V = (1/3)πr²h = (1/3) * (22/7) * (1.75)² * 12 = 38.5 m³. Since 1 m³ = 1 kilolitre, capacity = 38.5 kl)
- (B) 20:27 (V₁/V₂ = (πr₁²h₁)/(πr₂²h₂) = (r₁/r₂)² * (h₁/h₂) = (2/3)² * (5/3) = (4/9) * (5/3) = 20/27)
- (B) 40 (Volume of cuboid = 542 = 40 cm³. Volume of small cube = 1³ = 1 cm³. Number of cubes = Volume of cuboid / Volume of small cube = 40/1 = 40)
- (B) 2 cm (CSA = 2πrh = 88 => 2 * (22/7) * r * 14 = 88 => 88r = 88 => r = 1 cm. Diameter = 2r = 2 cm)
- (B) 125 m³ (LSA = 4a² = 100 => a² = 25 => a = 5 m. Volume = a³ = 5³ = 125 m³)
- (C) 462 cm² (TSA = 3πr² = 3 * (22/7) * 7² = 3 * 22 * 7 = 462 cm²)
- (B) 14 cm (External radius R=4 cm, Internal radius r=2 cm. Volume of shell = (4/3)π(R³ - r³) = (4/3)π(4³ - 2³) = (4/3)π(64 - 8) = (4/3)π(56). Cone radius r_c = 8/2 = 4 cm. Volume of cone = (1/3)πr_c²h = (1/3)π(4²)h = (16/3)πh. Equating volumes: (4/3)π(56) = (16/3)πh => 4 * 56 = 16h => h = (4 * 56) / 16 = 56 / 4 = 14 cm)
- (A) 6 : π (Let edge of cube be 'a'. Then radius of sphere inside it is r = a/2. Volume of cube = a³. Volume of sphere = (4/3)πr³ = (4/3)π(a/2)³ = (4/3)π(a³/8) = πa³/6. Ratio = Volume of Cube / Volume of Sphere = a³ / (πa³/6) = 6/π or 6 : π)
Study these notes thoroughly and practice the Exemplar problems. Good luck!