Combination Solids Word Problems MCQs Quiz | Class 10
This quiz is designed for **Class X** students, focusing on **Mathematics (Code 041)** from **Unit VI: Mensuration**. Test your understanding of **Surface Area and Volume in real-life contexts** related to combination solids. Answer all 10 multiple-choice questions and then submit to see your score. You can also download a PDF of the answer sheet for revision.
Understanding Combination Solids: Surface Area and Volume
Combination solids are formed when two or more basic solid shapes are joined together. In Class X Mathematics, particularly in Unit VI: Mensuration, understanding how to calculate the surface area and volume of these composite figures is crucial for solving real-life problems.
Key Concepts for Combination Solids:
- Identifying Component Shapes: The first step is to break down the combination solid into its constituent basic shapes (e.g., cylinder, cone, hemisphere, cuboid, sphere).
- Surface Area Calculation: When finding the surface area of a combination solid, you add the areas of all the *exposed* surfaces. The areas of the surfaces where the solids are joined together are NOT included, as they are hidden internally. For example, if a cone is placed on a cylinder, you would calculate the curved surface area of the cone and the curved surface area of the cylinder, plus the area of the cylinder’s base (if it’s exposed).
- Volume Calculation: Calculating the volume of a combination solid is generally simpler than surface area. Volumes are additive. You simply find the volume of each individual component shape and add them together (or subtract, if a portion is hollowed out).
Important Formulas for Basic 3D Shapes:
| Shape | Curved/Lateral Surface Area (CSA/LSA) | Total Surface Area (TSA) | Volume (V) |
|---|---|---|---|
| Cylinder | 2 x pi x r x h | 2 x pi x r x (r + h) | pi x r^2 x h |
| Cone | pi x r x l | pi x r x (r + l) | (1/3) x pi x r^2 x h |
| Sphere | 4 x pi x r^2 | 4 x pi x r^2 | (4/3) x pi x r^3 |
| Hemisphere | 2 x pi x r^2 | 3 x pi x r^2 | (2/3) x pi x r^3 |
| Cuboid | 2 x h x (l + b) | 2 x (l x b + b x h + h x l) | l x b x h |
(Where r = radius, h = height, l = slant height, b = breadth, l = length)
Real-Life Contexts:
These concepts are widely applicable in various real-world scenarios:
- Tents and Buildings: Calculating the canvas needed for a tent (cone on a cylinder) or the amount of paint for a building involves surface area.
- Storage and Capacity: Determining the volume of grains a silo can hold (cylinder with a conical top) or the amount of liquid a vessel contains (hemisphere with a cylindrical neck).
- Manufacturing and Design: Engineers and designers use these calculations to optimize material usage, understand structural stability, and predict costs for products like medicine capsules, toys, or decorative items.
- Melting and Recasting: Problems involving melting one solid shape and recasting it into another demonstrate the principle of volume conservation.
Quick Revision Tips:
- Draw a clear diagram of the combination solid.
- Label all given dimensions accurately.
- Carefully identify which surfaces are exposed for surface area calculations.
- For volume, simply sum or subtract the volumes of the component parts.
- Always include the correct units in your final answer.
Practice Questions:
- A decorative block is made of two solids – a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and a hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. (Use pi = 22/7)
- 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. Determine the volume of the toy. (Use pi = 22/7)
- A golf ball has a diameter of 4.1 cm. Its surface has 150 dimples each of radius 2 mm. Calculate the total surface area of the dimples. (Assume dimples are hemispheres, use pi = 22/7)
- How many spherical bullets each of 5 mm diameter can be obtained from a cuboid of lead with dimensions 11 cm x 10 cm x 5 cm?
- 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 cubic cm of iron has approximately 8 g mass. (Use pi = 3.14)
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