CBSE Class 9 Maths: Surface Area and Volume Exercise 13.9

Welcome to Surface Area and Volume Exercise 13.9 for Class 9, an exciting optional exercise designed to deepen your understanding of 3D shapes. While not always directly asked in board exams, this exercise is crucial for developing advanced problem-solving skills that will serve you well in higher classes and competitive exams. Here, you'll move beyond basic calculations and learn to tackle problems involving combinations of different solids, or situations where one solid is melted and recast into another. This chapter will equip you with the ability to analyse complex geometrical situations, apply appropriate formulas, and think critically about how surfaces and volumes interact in multi-part objects. By the end of this page, you'll be confident in approaching challenging problems, meticulously calculating surface areas and volumes for composite figures, and understanding the principle of conservation of volume during transformations.

Understanding Complex Problems in Surface Area and Volume

Exercise 13.9 often presents scenarios that go beyond simple, individual shapes. You'll encounter two main types of complex problems:

  1. Combined Solids: Here, two or more basic solids (like a cylinder and a cone, or a cuboid with a hemisphere scooped out) are joined or integrated. The key challenge is to correctly identify the individual surfaces that contribute to the total surface area and to sum the volumes of the constituent parts. Remember, when calculating surface area, you only consider the exposed surfaces. For instance, if a hemisphere is placed on a cylinder, the base of the hemisphere and the top circular area of the cylinder are no longer exposed. For volume, it's usually the sum or difference of individual volumes, depending on whether parts are added or removed.
  1. Conversion of Solids: These problems involve melting one solid shape (or a combination of shapes) and recasting it into another. The fundamental principle here is the conservation of volume. This means the total volume of the material before conversion is equal to the total volume of the material after conversion. This principle allows you to equate the volume formulas of the initial and final shapes to find an unknown dimension. Careful attention to units is paramount in all these problems.

Key Terms for Exercise 13.9

Combined Solid
A three-dimensional object formed by joining two or more basic geometric solids, such as a cylinder topped with a cone, or a cuboid with a hemispherical depression.
Conservation of Volume
A principle stating that when a solid is melted and recast into another shape, the total volume of the material remains constant, assuming no material is lost or added during the process.
Exposed Surface Area
For a combined solid, it refers to the total area of all surfaces that are visible and can be touched from the outside. Internal joining surfaces are not counted.

Worked Example: Conversion of Solids

  1. Problem Statement — A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
  2. Step 1: Identify Given Information and Goal — Given: Radius of sphere (r_sphere) = 4.2 cm Radius of cylinder (r_cylinder) = 6 cm To find: Height of cylinder (h_cylinder)
  3. Step 2: Apply Conservation of Volume Principle — When the sphere is melted and recast into a cylinder, their volumes are equal. Volume of sphere = Volume of cylinder
  4. Step 3: Write Down Formulas — Volume of sphere = (4/3) π (r_sphere)^3 Volume of cylinder = π (r_cylinder)^2 h_cylinder
  5. Step 4: Equate Volumes and Substitute Values — (4/3) π (4.2)^3 = π (6)^2 h_cylinder Notice that π cancels out from both sides, simplifying the calculation.
  6. Step 5: Solve for the Unknown (Height of Cylinder) — (4/3) (4.2 4.2 4.2) = 36 h_cylinder (4/3) 74.088 = 36 h_cylinder 98.784 = 36 * h_cylinder h_cylinder = 98.784 / 36 h_cylinder = 2.744 cm
  7. Final Answer — The height of the cylinder is 2.744 cm.

Exam Tip: Common Mistakes to Avoid

  1. Incorrect Surface Area for Combined Solids: Students often calculate the total surface area by simply adding the individual surface areas of the constituent solids, forgetting that some surfaces are 'hidden' when combined. Always visualise the exposed surfaces.
  2. Units Inconsistency: Ensure all dimensions are in the same unit (e.g., all in cm or all in m) before performing any calculations. A mix of units will lead to incorrect answers.
  3. Calculation Errors: Be meticulous with arithmetic, especially when dealing with squares, cubes, and values of π. Use the value of π (22/7 or 3.14) as specified in the question or choose the one that simplifies calculations.
  4. Misapplying Conservation Principle: Remember that conversion of solids is about volume conservation, not surface area conservation. Surface area changes during conversion.

Practice Questions with Solutions

  • Q: 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 π. A: Step 1: Identify shapes and dimensions. Cone: radius (r) = 1 cm, height (h) = 1 cm (given height = radius). Hemisphere: radius (r) = 1 cm. Step 2: Write down formulas for individual volumes. Volume of cone = (1/3)πr²h Volume of hemisphere = (2/3)πr³ Step 3: Calculate individual volumes. Volume of cone = (1/3)π(1)²(1) = (1/3)π cm³ Volume of hemisphere = (2/3)π(1)³ = (2/3)π cm³ Step 4: Add volumes to find the total volume of the solid. Total Volume = Volume of cone + Volume of hemisphere = (1/3)π + (2/3)π = (3/3)π = π cm³ Final answer: The volume of the solid is π cm³.
  • Q: 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 radius is 3.5 cm, find the total surface area of the article. A: Step 1: Identify surfaces for total surface area. The article consists of the curved surface area (CSA) of the cylinder and the curved surface areas of two hemispheres. The base areas of the hemispheres and the cylinder's top/bottom circles are not exposed. Given: Height of cylinder (h) = 10 cm, Radius (r) = 3.5 cm. Step 2: Write down formulas for relevant surface areas. CSA of cylinder = 2πrh CSA of hemisphere = 2πr² Step 3: Calculate individual curved surface areas. CSA of cylinder = 2 (22/7) 3.5 10 = 2 22 0.5 10 = 220 cm² CSA of one hemisphere = 2 (22/7) (3.5)² = 2 (22/7) 12.25 = 2 22 1.75 = 77 cm² Step 4: Calculate the total surface area. Total Surface Area = CSA of cylinder + 2 (CSA of one hemisphere) Total Surface Area = 220 + 2 77 = 220 + 154 = 374 cm² Final answer: The total surface area of the article is 374 cm².
  • Q: How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm? (Use π = 22/7) A: Step 1: Convert all units to the same measurement (cm). Diameter of coin = 1.75 cm, so radius (r) = 1.75 / 2 = 0.875 cm. Thickness of coin (h) = 2 mm = 0.2 cm. Cuboid dimensions: length (l) = 5.5 cm, breadth (b) = 10 cm, height (H) = 3.5 cm. Step 2: Calculate the volume of one silver coin (which is cylindrical in shape). Volume of one coin = πr²h = (22/7) (0.875)² 0.2 = (22/7) 0.765625 0.2 = 0.48125 cm³ (approx). Step 3: Calculate the volume of the cuboid. Volume of cuboid = l b H = 5.5 10 3.5 = 192.5 cm³. Step 4: Determine the number of coins by dividing the cuboid's volume by the volume of one coin. Number of coins = Volume of cuboid / Volume of one coin = 192.5 / 0.48125 = 400. Final answer: 400 silver coins must be melted.
  • Q: 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 cm³ of iron has approximately 8 g mass. (Use π = 3.14) A: Step 1: Identify dimensions of both cylinders. Lower cylinder: Height (H1) = 220 cm, Diameter = 24 cm, so Radius (R1) = 12 cm. Upper cylinder: Height (H2) = 60 cm, Radius (R2) = 8 cm. Step 2: Calculate the volume of each cylinder. Volume of lower cylinder (V1) = πR1²H1 = 3.14 (12)² 220 = 3.14 144 220 = 99475.2 cm³. Volume of upper cylinder (V2) = πR2²H2 = 3.14 (8)² 60 = 3.14 64 60 = 12057.6 cm³. Step 3: Calculate the total volume of the iron pole. Total Volume (V_total) = V1 + V2 = 99475.2 + 12057.6 = 111532.8 cm³. Step 4: Calculate the mass of the pole. Mass = Total Volume Mass per cm³ = 111532.8 cm³ 8 g/cm³ = 892262.4 g. Convert mass to kg (optional, but good practice for large values): 892262.4 g = 892.2624 kg. Final answer: The mass of the pole is approximately 892262.4 g or 892.26 kg.

Frequently Asked Questions

What is special about Exercise 13.9 in Class 9 Maths?

Exercise 13.9 is an 'Optional Exercise' in the NCERT textbook, meaning its problems are often more challenging than regular exercises. It focuses on developing advanced problem-solving skills, particularly for combined solids and conversion of solids, which are essential for deeper understanding of geometry.

How do I calculate the surface area of combined solids?

To calculate the surface area of combined solids, you must consider only the *exposed* surfaces. Identify the individual shapes, calculate their curved surface areas, and then sum these. Do not include the areas of surfaces that are joined together and therefore not visible.

What is the 'Conservation of Volume' principle?

The Conservation of Volume principle states that when a solid material is melted and recast into a new shape, its total volume remains unchanged. This principle is fundamental for solving problems where one solid is transformed into another, allowing you to equate their volumes to find unknown dimensions.