| Classwise Concept with Examples | ||||||
|---|---|---|---|---|---|---|
| 6th | 7th | 8th | 9th | 10th | 11th | 12th |
| Content On This Page | ||
|---|---|---|
| Surface Area and Volumes of Solids | Surface Area of a Combination of Solids | Volume of a COmbination of Solids |
| Conversion of Solid from One Shape to Another | Frustum of a Cone | |
Chapter 13 Surface Areas And Volumes (Concepts)
Welcome to this advanced chapter on Surface Areas and Volumes, where we significantly extend the mensuration concepts developed in Class 9. Having mastered the calculation of surface areas and volumes for fundamental three-dimensional shapes, we now venture into more complex scenarios involving composite objects and introduce a specific geometric solid derived from the cone: the frustum. This chapter bridges the gap between basic formulas and their application to more intricate real-world objects and processes, enhancing both your spatial reasoning and calculation skills.
We begin by briefly revisiting the essential formulas for the basic solids. A solid grasp of these is crucial before proceeding. Recall for:
- Cuboid (length $l$, breadth $b$, height $h$): Lateral Surface Area ($LSA$) = $2h(l+b)$, Total Surface Area ($TSA$) = $2(lb + bh + hl)$, Volume ($V$) = $l \times b \times h$.
- Cube (side $a$): $LSA = 4a^2$, $TSA = 6a^2$, $V = a^3$.
- Right Circular Cylinder (radius $r$, height $h$): Curved Surface Area ($CSA$) = $2\pi rh$, $TSA = 2\pi r(r + h)$, $V = \pi r^2 h$.
- Right Circular Cone (radius $r$, height $h$, slant height $l = \sqrt{r^2 + h^2}$): $CSA = \pi r l$, $TSA = \pi r(l + r)$, $V = \frac{1}{3}\pi r^2 h$.
- Sphere (radius $r$): Surface Area = $4\pi r^2$, $V = \frac{4}{3}\pi r^3$.
- Hemisphere (radius $r$): $CSA = 2\pi r^2$, $TSA = 3\pi r^2$, $V = \frac{2}{3}\pi r^3$.
The first major focus of this chapter is tackling problems involving Combinations of Solids. Many objects in reality are not simple geometric shapes but are formed by joining two or more basic solids. Examples include a storage tent (cylinder surmounted by a cone), an ice-cream cone (cone topped with a hemisphere), or a medicinal capsule (cylinder with hemispherical ends). Calculating the surface area and volume of such composite shapes requires careful analysis:
- Volume of Combinations: This is generally straightforward – the total volume is simply the sum of the volumes of the constituent basic solids.
- Surface Area of Combinations: This requires more caution. We must consider only the exposed surfaces of the combined object. For instance, when a cone is placed on a cylinder, the circular base where they join is no longer part of the exposed surface area and should not be included in the calculation. We add the areas of only those parts that form the exterior 'skin' of the combined solid.
Another significant application involves the Conversion of Solid from One Shape to Another. Imagine melting a metallic sphere and recasting the molten metal entirely into several smaller cones or a single cylindrical wire. The fundamental principle governing such transformations is the conservation of volume: the volume of the material remains constant throughout the process. This allows us to equate the volume of the original solid shape to the total volume of the resulting shape(s). By setting up an equation using the respective volume formulas ($V_{initial} = V_{final}$), we can solve for unknown dimensions (like the radius of the wire or the number of cones formed).
Finally, we introduce a specific truncated shape: the Frustum of a Cone. This is the shape you get when you slice off the top part of a cone with a plane parallel to its base (think of a bucket, a lampshade, or a funnel). Let the frustum have height $h$, slant height $l$, and radii $r_1$ and $r_2$ for its two circular bases (conventionally, $r_1$ is the larger radius). The key formulas associated with the frustum are:
- Slant height: $l = \sqrt{h^2 + (r_1 - r_2)^2}$
- Curved Surface Area (CSA): $CSA = \pi(r_1 + r_2)l$
- Total Surface Area (TSA): $TSA = CSA + \ $$ \text{Area of top base} + \ $$ \text{Area of bottom base} \ $$ = \pi(r_1 + r_2)l + \pi r_1^2 + \pi r_2^2$
- Volume (V): $V = \frac{1}{3}\pi h(r_1^2 + r_2^2 + r_1 r_2)$
We will practice applying these formulas to solve practical problems involving the capacity (volume) or material required (surface area) for frustum-shaped objects. Careful application of formulas and robust spatial visualization are key to mastering this chapter.
Surface Area and Volumes of Basic Solids (Revision of Class 9 Formulas)
In the world around us, objects exist in three dimensions. Understanding their properties, like how much material is needed to cover them or how much they can hold, is crucial. This is where the concepts of surface area and volume come in. This section revisits the basic formulas for simple solids and introduces formulas for more complex shapes like cones and spheres.
What is Surface Area?
Imagine you want to paint a solid object, like a wooden box. The surface area is the total area you would need to paint to cover its entire exterior. It's measured in square units (like $\text{cm}^2, \text{m}^2$).
- Lateral or Curved Surface Area (LSA/CSA): This is the area of the sides only, excluding the top and bottom faces. For a room, the LSA is the area of the four walls. For a can, the CSA is the area of its label.
- Total Surface Area (TSA): This is the complete area of all surfaces, including the top and bottom. For a room, it's the area of the four walls plus the floor and ceiling.
What is Volume?
Imagine you want to fill that same wooden box with water. The volume is the total amount of water the box can hold. It represents the space occupied by the object and is measured in cubic units (like $\text{cm}^3, \text{m}^3$).
Formulas for Cuboids, Cubes, and Cylinders
These formulas were covered in Class 9 and form the foundation for more complex shapes.
| Solid | Dimensions | Lateral/Curved Surface Area (LSA/CSA) | Total Surface Area (TSA) | Volume (V) |
|---|---|---|---|---|
| Cuboid | Length ($l$), Breadth ($b$), Height ($h$) | $2h(l+b)$ | $2(lb + bh + hl)$ | $lbh$ |
| Cube | Side ($a$) | $4a^2$ | $6a^2$ | $a^3$ |
| Cylinder | Radius ($r$), Height ($h$) | $2\pi rh$ | $2\pi r(r+h)$ | $\pi r^2 h$ |
Formulas for Cones, Spheres, and Hemispheres
4. Cone (Right Circular)
A cone has a circular base and a curved surface that tapers to a single point called the apex (e.g., an ice cream cone, a birthday hat). It's defined by its base radius ($r$), perpendicular height ($h$), and slant height ($l$).
The slant height ($l$) forms a right-angled triangle with $r$ and $h$, so we use the Pythagoras theorem to find it:
$l = \sqrt{r^2 + h^2}$
- Curved Surface Area (CSA): $\pi r l$
- Total Surface Area (TSA): $\pi r l + \pi r^2 = \pi r (l + r)$
- Volume: $\frac{1}{3} \pi r^2 h$
Derivation of Formulas for a Cone
1. Curved Surface Area (CSA):
Imagine cutting the cone along its slant height ($l$) and unrolling the curved surface. This creates a sector of a larger circle.
- The radius of this new sector is the slant height of the cone, $l$.
- The arc length of this sector is the circumference of the cone's base, which is $2\pi r$.
$\frac{\text{Area of Sector}}{\text{Area of Full Circle}} = \frac{\text{Arc Length of Sector}}{\text{Circumference of Full Circle}}$
$\frac{\text{CSA}}{\pi l^2} = \frac{2\pi r}{2\pi l}$
$\frac{\text{CSA}}{\pi l^2} = \frac{r}{l}$
$\text{CSA} = \frac{r}{l} \times \pi l^2 = \pi r l$
2. Volume:
The formal derivation of the cone's volume requires calculus. However, it can be demonstrated experimentally that the volume of a cone is exactly one-third the volume of a cylinder with the same base radius ($r$) and height ($h$). If you fill a cone with sand and pour it into a corresponding cylinder, you will need to do this three times to fill the cylinder completely.
Volume of Cone $= \frac{1}{3} \times$ Volume of Cylinder
$V = \frac{1}{3} (\pi r^2 h)$
5. Sphere
A sphere is a perfectly round ball (e.g., a cricket ball, a globe). It is defined only by its radius ($r$).
- Surface Area: $4\pi r^2$
- Volume: $\frac{4}{3}\pi r^3$
Derivation of Formulas for a Sphere
1. Surface Area:
The ancient Greek mathematician Archimedes discovered a remarkable relationship: the surface area of a sphere is equal to the curved surface area of a cylinder that perfectly encloses it (a circumscribed cylinder).
- The radius of this cylinder is the same as the sphere's radius, $r$.
- The height of this cylinder is equal to the sphere's diameter, $h=2r$.
Surface Area of Sphere = CSA of Cylinder $= 2\pi(\text{radius})(\text{height})$
$\text{SA} = 2\pi (r) (2r) = 4\pi r^2$
2. Volume:
Imagine the sphere is made up of an infinite number of tiny pyramids, each with its base on the surface of the sphere and its apex at the center.
- The height of each pyramid is the radius of the sphere, $r$.
- The sum of the areas of the bases of all these pyramids is the surface area of the sphere, $4\pi r^2$.
Volume of Sphere $= \sum \left(\frac{1}{3} \times \text{Base Area}_{\text{pyramid}} \times \text{Height}\right)$
$V = \frac{1}{3} \times (\text{Sum of all Base Areas}) \times r$
$V = \frac{1}{3} \times (\text{Surface Area of Sphere}) \times r$
$V = \frac{1}{3} \times (4\pi r^2) \times r = \frac{4}{3}\pi r^3$
6. Hemisphere
A hemisphere is exactly half of a sphere (e.g., a bowl). It has a curved surface and a flat circular base.
- Curved Surface Area (CSA): $2\pi r^2$ (Half the SA of a sphere).
- Total Surface Area (TSA): $3\pi r^2$ (The curved area plus the flat circular base).
- Volume: $\frac{2}{3}\pi r^3$ (Half the volume of a sphere).
Example 1. Find the surface area of a sphere of radius 7 cm. (Use $\pi = \frac{22}{7}$)
Answer:
To Find:
The surface area of the sphere.
Given:
Radius of the sphere $r = 7$ cm.
Solution:
The formula for the surface area of a sphere is $\text{SA} = 4\pi r^2$.
$\text{SA} = 4 \times \frac{22}{7} \times (7)^2 = 4 \times \frac{22}{7} \times 49 \text{ cm}^2$
Cancel out the common factor 7:
$\text{SA} = 4 \times 22 \times \frac{\cancel{49}^7}{\cancel{7}_1} \text{ cm}^2 = 4 \times 22 \times 7 \text{ cm}^2$
$\text{SA} = 88 \times 7 = 616 \text{ cm}^2$
... (1)
Answer: The surface area of the sphere is 616 cm$^2$.
Example 2. A conical tent is 10 m high and the radius of its base is 24 m. Find the slant height of the tent and the cost of the canvas required to make the tent, if the cost of 1 m$^2$ canvas is $\textsf{₹}$ 70.
Answer:
To Find:
1. Slant height of the tent ($l$).
2. Cost of the canvas.
Given:
Height of the cone, $h = 10$ m.
Radius of the base, $r = 24$ m.
Cost of canvas = $\textsf{₹}$ 70 per m$^2$.
Solution:
Step 1: Find the slant height ($l$).
Using the Pythagoras theorem for the cone:
$l^2 = r^2 + h^2$
$l^2 = (24)^2 + (10)^2 = 576 + 100 = 676$
$l = \sqrt{676} = 26$ m
... (1)
The slant height of the tent is 26 m.
Step 2: Find the area of the canvas required.
The canvas is used to make the curved surface of the tent (the ground is not covered). So, we need to find the Curved Surface Area (CSA) of the cone.
$\text{CSA} = \pi r l$
Using $\pi = \frac{22}{7}$:
$\text{CSA} = \frac{22}{7} \times 24 \times 26 = \frac{13728}{7} \text{ m}^2$
... (2)
Step 3: Calculate the total cost of the canvas.
Total Cost = Area of canvas $\times$ Cost per m$^2$.
$\text{Cost} = \frac{13728}{7} \times 70$
Cancel out the common factor 7:
$\text{Cost} = 13728 \times \frac{\cancel{70}^{10}}{\cancel{7}_1} = 13728 \times 10$
$\text{Cost} = \textsf{₹ } 1,37,280$
... (3)
Answer: The slant height of the tent is 26 m and the cost of the canvas required is $\textsf{₹}$ 1,37,280.
Example 3. Find the volume and total surface area of a hemisphere of radius 21 cm. (Use $\pi = \frac{22}{7}$)
Answer:
To Find:
1. Volume of the hemisphere.
2. Total Surface Area (TSA) of the hemisphere.
Given:
Radius of the hemisphere, $r = 21$ cm.
Solution:
Step 1: Find the Volume of the hemisphere.
The formula for the volume of a hemisphere is $V = \frac{2}{3}\pi r^3$.
$V = \frac{2}{3} \times \frac{22}{7} \times (21)^3$
$V = \frac{2}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 \text{ cm}^3$
Cancel out common factors (3 and 7 from the denominator with one 21 from the numerator):
$V = 2 \times 22 \times \frac{\cancel{21}}{\cancel{3} \times \cancel{7}} \times 21 \times 21 = 44 \times 441 \text{ cm}^3$
$V = 19404 \text{ cm}^3$
... (1)
Step 2: Find the Total Surface Area (TSA) of the hemisphere.
The formula for the TSA of a hemisphere is $\text{TSA} = 3\pi r^2$.
$\text{TSA} = 3 \times \frac{22}{7} \times (21)^2 = 3 \times \frac{22}{7} \times 441 \text{ cm}^2$
Cancel out the common factor 7:
$\text{TSA} = 3 \times 22 \times \frac{\cancel{441}^{63}}{\cancel{7}_1} = 66 \times 63 \text{ cm}^2$
$\text{TSA} = 4158 \text{ cm}^2$
... (2)
Answer: The volume of the hemisphere is 19,404 cm$^3$ and its total surface area is 4,158 cm$^2$.
Surface Area of a Combination of Solids
In many real-world objects, basic solids are combined to form more complex shapes, such as a tanker truck (a cylinder with two hemispheres), a toy top (a cone on a hemisphere), or an ice cream cone. To find the surface area of such a combination of solids, we must only consider the surfaces that are visible on the exterior of the new object.
A common mistake is to simply add the total surface areas of the individual solids. This is incorrect because when solids are joined, the surfaces where they meet are covered and are no longer part of the exterior surface area.
Calculating Surface Area of Combined Solids
The correct principle is to add the areas of only the exposed surfaces of the individual solids.
Steps to Calculate Surface Area of a Combination of Solids:
- Identify the basic solids that make up the combined solid (e.g., a cylinder and two hemispheres).
- Visualize the combined shape and determine which surfaces are exposed. For example, in a toy made of a cone on a hemisphere, the exposed parts are the curved surface of the cone and the curved surface of the hemisphere. The flat bases are hidden where they join.
- Calculate the areas of these exposed surfaces individually. This often means using the formulas for Curved or Lateral Surface Area (CSA/LSA).
- Sum the areas of all the exposed surfaces to find the total surface area of the combined solid.
Example 1. A toy is in the form of a cone mounted on a hemisphere of the same radius. The diameter of the base of the cone is 10 cm and its height is 12 cm. Find the total surface area of the toy. (Use $\pi = 3.14$)
Answer:
To Find:
The total surface area of the toy.
Given:
The toy is a combination of a cone and a hemisphere with a common radius.
Diameter of the common base = 10 cm.
Height of the cone, $h = 12$ cm.
Solution:
The radius of the cone's base is half the diameter: $r = 10 / 2 = 5$ cm. The hemisphere shares this radius, so $r=5$ cm for both solids.
The total surface area of the toy is the sum of the exposed surfaces, which are the curved surface area of the cone and the curved surface area of the hemisphere.
TSA of Toy = (CSA of Cone) + (CSA of Hemisphere)
Step 1: Calculate the Curved Surface Area (CSA) of the cone.
The formula for the CSA of a cone is $\pi r l$. We first need to find the slant height, $l$.
$l = \sqrt{r^2 + h^2} = \sqrt{(5)^2 + (12)^2}$
$l = \sqrt{25 + 144} = \sqrt{169}$
$l = 13$ cm
... (1)
Now, calculate the CSA of the cone:
CSA of Cone $= \pi r l = 3.14 \times 5 \times 13$
CSA of Cone $= 3.14 \times 65 = 204.1 \text{ cm}^2$
... (2)
Step 2: Calculate the Curved Surface Area (CSA) of the hemisphere.
The formula for the CSA of a hemisphere is $2\pi r^2$.
CSA of Hemisphere $= 2 \times 3.14 \times (5)^2 = 2 \times 3.14 \times 25$
CSA of Hemisphere $= 50 \times 3.14 = 157 \text{ cm}^2$
... (3)
Step 3: Calculate the total surface area of the toy.
Add the two curved surface areas.
Total Surface Area $= 204.1 \text{ cm}^2 + 157 \text{ cm}^2$
Total Surface Area $= 361.1 \text{ cm}^2$
... (4)
Answer: The total surface area of the toy is 361.1 cm$^2$.
Example 2. 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.
Answer:
To Find:
The total surface area of the capsule.
Given:
The capsule is a combination of a cylinder and two hemispheres.
Total length of the capsule = 14 mm.
Diameter of the capsule = 5 mm.
Solution:
The surface area of the capsule is the sum of the curved surface area of the central cylinder and the curved surface areas of the two hemispherical ends.
TSA of Capsule = (CSA of Cylinder) + 2 $\times$ (CSA of a Hemisphere)
Note: The two hemispherical ends together form a complete sphere. So, the formula can also be written as:
TSA of Capsule = (CSA of Cylinder) + (Surface Area of a Sphere)
Step 1: Determine the dimensions of the cylinder and hemispheres.
Diameter = 5 mm, so the radius is $r = 5 / 2 = 2.5$ mm.
The length of the cylindrical part ($h$) is the total length of the capsule minus the lengths of the two hemispherical ends. Each hemispherical end has a length equal to its radius.
$h = \text{Total Length} - (\text{radius of end 1}) - (\text{radius of end 2})$
$h = 14 - 2.5 - 2.5 = 14 - 5 = 9$ mm
... (1)
Step 2: Calculate the Curved Surface Area (CSA) of the cylinder.
Using the formula CSA = $2\pi rh$ with $r=2.5$ mm and $h=9$ mm.
CSA of Cylinder $= 2 \times \frac{22}{7} \times 2.5 \times 9$
$= 2 \times 2.5 \times 9 \times \frac{22}{7} = 5 \times 9 \times \frac{22}{7} = 45 \times \frac{22}{7} = \frac{990}{7} \text{ mm}^2$
... (2)
Step 3: Calculate the surface area of the two hemispherical ends.
The area of the two hemispherical ends is equal to the surface area of one full sphere with the same radius, $r = 2.5$ mm. Formula is SA = $4\pi r^2$.
SA of two ends $= 4 \times \frac{22}{7} \times (2.5)^2 = 4 \times \frac{22}{7} \times 6.25$
$= (4 \times 6.25) \times \frac{22}{7} = 25 \times \frac{22}{7} = \frac{550}{7} \text{ mm}^2$
... (3)
Step 4: Calculate the total surface area of the capsule.
Add the areas from Step 2 and Step 3.
Total SA $= \frac{990}{7} + \frac{550}{7} = \frac{990 + 550}{7} = \frac{1540}{7} \text{ mm}^2$
Total SA $= 220 \text{ mm}^2$
... (4)
Answer: The surface area of the medicine capsule is 220 mm$^2$.
Volume of a Combination of Solids
When basic solid shapes are joined together to form a new, composite solid, calculating the total volume is generally simpler than calculating the surface area. This is because volume is a measure of the space occupied by the matter of the solid. Unlike surface area, where joining surfaces are hidden, the internal space of each solid contributes directly to the total space of the combined object. Therefore, their individual volumes simply add up.
Calculating Volume of Combined Solids
To find the volume of a solid made by combining two or more basic solids, we calculate the volume of each component solid separately and then sum them up. The volume of the combined solid is the sum of the volumes of the individual solids.
Steps to Calculate Volume of a Combination of Solids:
- Identify the basic solids that make up the combined object (e.g., a cone, a cylinder, a hemisphere).
- Determine the dimensions (radius, height, length, etc.) of each individual solid from the information given about the combined object.
- Calculate the volume of each individual solid using the appropriate formulas.
- Sum the volumes of all the constituent solids to find the total volume of the combined solid.
Example 1. A solid is in the shape of a cone standing on a hemisphere, both with radii 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$.
Answer:
To Find:
The volume of the solid in terms of $\pi$.
Given:
The solid is a combination of a cone standing on a hemisphere.
Radius of cone = Radius of hemisphere = 1 cm.
Height of cone = Radius of cone = 1 cm.
Solution:
The total volume of the solid is the sum of the volume of the hemisphere and the volume of the cone.
Total Volume = Volume of Hemisphere + Volume of Cone
Step 1: Calculate the volume of the hemisphere.
The formula for the volume of a hemisphere is $V = \frac{2}{3} \pi r^3$. Given $r = 1$ cm:
Volume of Hemisphere $= \frac{2}{3} \pi (1)^3 = \frac{2}{3}\pi \text{ cm}^3$
... (1)
Step 2: Calculate the volume of the cone.
The formula for the volume of a cone is $V = \frac{1}{3} \pi r^2 h$. Given $r = 1$ cm and $h = 1$ cm:
Volume of Cone $= \frac{1}{3} \pi (1)^2 (1) = \frac{1}{3}\pi \text{ cm}^3$
... (2)
Step 3: Calculate the total volume of the solid.
Add the volumes from Step 1 and Step 2.
Total Volume $= \frac{2}{3}\pi + \frac{1}{3}\pi = \left(\frac{2}{3} + \frac{1}{3}\right)\pi$
Total Volume $= \frac{3}{3}\pi = \pi \text{ cm}^3$
... (3)
Answer: The volume of the solid is $\pi$ cm$^3$.
Example 2. 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 a total length of 5 cm and a diameter of 2.8 cm.
Answer:
To Find:
The approximate volume of sugar syrup in 45 Gulab Jamuns.
Given:
Number of Gulab Jamuns = 45.
Syrup volume is 30% of the total volume.
Shape: Cylinder with two hemispherical ends.
Total length = 5 cm, Diameter = 2.8 cm.
Solution:
First, we will find the volume of a single Gulab Jamun. Then, we'll find the total volume for 45 Gulab Jamuns. Finally, we'll calculate 30% of that total volume to find the amount of syrup.
Volume of one Gulab Jamun = (Volume of central Cylinder) + (Volume of 2 Hemispherical ends)
Step 1: Determine the dimensions of the cylinder and hemispheres.
Diameter = 2.8 cm, so the radius is $r = 2.8 / 2 = 1.4$ cm.
The length of the cylindrical part ($h$) is the total length minus the radius of each of the two hemispherical ends.
$h = \text{Total Length} - 2 \times r = 5 - 2(1.4) = 5 - 2.8 = 2.2$ cm
... (1)
Step 2: Calculate the volume of one Gulab Jamun.
The volume of the two hemispherical ends is equal to the volume of one full sphere with radius $r = 1.4$ cm.
Volume of 2 hemispheres $= \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (1.4)^3$
$= \frac{4}{3} \times \frac{22}{7} \times 2.744 \approx 11.50 \text{ cm}^3$
Volume of the cylindrical part:
Volume of cylinder $= \pi r^2 h = \frac{22}{7} \times (1.4)^2 \times 2.2$
$= \frac{22}{7} \times 1.96 \times 2.2 = 22 \times 0.28 \times 2.2 \approx 13.55 \text{ cm}^3$
Total volume of one Gulab Jamun:
Volume$_{1} = 11.50 + 13.55 = 25.05 \text{ cm}^3$
... (2)
Step 3: Calculate the total volume of 45 Gulab Jamuns.
Total Volume $= 45 \times 25.05 = 1127.25 \text{ cm}^3$
... (3)
Step 4: Calculate the volume of the syrup.
The syrup is 30% of the total volume.
Volume of syrup $= 30\% \text{ of } 1127.25 = \frac{30}{100} \times 1127.25$
$= 0.3 \times 1127.25 = 338.175 \text{ cm}^3$
... (4)
Answer: Approximately 338 cm$^3$ of syrup would be found in 45 Gulab Jamuns.
Conversion of Solid from One Shape to Another
In various industrial and artisanal processes, materials are often reshaped. For instance, a block of metal might be melted down and cast into engine parts, or a lump of clay might be molded into a pot. The fundamental principle governing these transformations is the conservation of volume.
When a solid of a certain shape is converted into another solid shape (or multiple smaller solids), the total amount of material does not change. This means the volume of the material remains constant, assuming no material is wasted or lost in the process. This key concept allows us to create an equation relating the dimensions of the original shape to the dimensions of the new shape.
The core mathematical relationship for all such problems is:
Volume of the Original Solid(s) = Volume of the New Solid(s)
Example 1. 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.
Answer:
To Find:
The height of the cylinder.
Given:
A sphere is recast into a cylinder.
Radius of sphere, $r_s = 4.2$ cm.
Radius of cylinder, $r_c = 6$ cm.
Solution:
Since the metallic sphere is melted and recast into a cylinder, the volume of the material remains constant. Let the height of the cylinder be $h$.
Volume of Cylinder
= Volume of Sphere
Using the formulas for volume, we can write:
$\pi (r_c)^2 h = \frac{4}{3} \pi (r_s)^3$
Substitute the given values for the radii:
$\pi (6)^2 h = \frac{4}{3} \pi (4.2)^3$
We can cancel the common factor $\pi$ from both sides.
$36 \times h = \frac{4}{3} \times (4.2)^3$
Now, rearrange the equation to solve for $h$:
$h = \frac{4 \times (4.2)^3}{3 \times 36}$
Simplify the numerical constants by cancelling 4 into 36:
$h = \frac{\cancel{4} \times (4.2)^3}{3 \times \cancel{36}_9} = \frac{(4.2)^3}{27}$
We can rewrite the denominator as $27 = 3^3$.
$h = \frac{(4.2)^3}{3^3}$
Using the law of exponents, $\frac{a^m}{b^m} = \left(\frac{a}{b}\right)^m$:
$h = \left(\frac{4.2}{3}\right)^3 = (1.4)^3$
Calculate the final value of $h$:
$h = 1.4 \times 1.4 \times 1.4$
$h = 1.96 \times 1.4 = 2.744$ cm
... (1)
Answer: The height of the cylinder is 2.744 cm.
Example 2. 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 $\times$ 10 cm $\times$ 3.5 cm? (Use $\pi = \frac{22}{7}$)
Answer:
To Find:
The number of silver coins needed.
Given:
Coin (cylinder): diameter = 1.75 cm, thickness (height) = 2 mm.
Cuboid: dimensions = 5.5 cm $\times$ 10 cm $\times$ 3.5 cm.
Solution:
Let $N$ be the number of coins. The principle of volume conservation gives:
$N \times (\text{Volume of one coin}) = \text{Volume of Cuboid}$
Step 1: Calculate the volume of one silver coin.
The coin is a small cylinder. First, ensure all units are in cm.
Radius of coin, $r = \frac{\text{diameter}}{2} = \frac{1.75}{2}$ cm.
Height of coin, $h = 2 \text{ mm} = 0.2$ cm.
Volume of one coin $= \pi r^2 h$.
$V_{\text{coin}} = \frac{22}{7} \times \left(\frac{1.75}{2}\right)^2 \times 0.2$
$= \frac{22}{7} \times \frac{1.75 \times 1.75}{4} \times 0.2$
Simplify by canceling: $1.75 / 7 = 0.25$.
$= 22 \times 0.25 \times \frac{1.75}{4} \times 0.2$
$= 5.5 \times \frac{0.35}{4} = \frac{1.925}{4} = 0.48125 \text{ cm}^3$
... (1)
Step 2: Calculate the volume of the cuboid.
$V_{\text{cuboid}} = l \times b \times h = 5.5 \times 10 \times 3.5$
$= 55 \times 3.5 = 192.5 \text{ cm}^3$
... (2)
Step 3: Find the number of coins (N).
$N = \frac{\text{Volume of Cuboid}}{\text{Volume of one coin}} = \frac{192.5}{0.48125}$
$N = 400$
... (3)
Answer: 400 silver coins must be melted.
Example 3. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Answer:
To Find:
The time required to fill the tank.
Given:
Pipe (cylinder): internal diameter = 20 cm.
Tank (cylinder): diameter = 10 m, depth (height) = 2 m.
Speed of water flow = 3 km/h.
Solution:
This is a volume conservation problem where the volume of water flowing out of the pipe in a certain time must equal the volume of the tank.
Volume of water from pipe in time 't' = Volume of the tank
Step 1: Calculate the volume of the tank.
Ensure all units are consistent. Let's use meters (m).
Radius of tank, $R = \frac{10}{2} = 5$ m.
Height of tank, $H = 2$ m.
$V_{\text{tank}} = \pi R^2 H = \pi (5)^2 (2) = 50\pi \text{ m}^3$
... (1)
Step 2: Calculate the volume of water flowing from the pipe per hour.
Radius of pipe, $r = \frac{20 \text{ cm}}{2} = 10 \text{ cm} = 0.1$ m.
Speed of water = 3 km/h = 3000 m/h. This speed can be considered as the length (or height) of the cylindrical column of water that flows out of the pipe in one hour.
Let the length of the water column be $h_{\text{water}} = 3000$ m.
Volume of water flowing per hour:
$V_{\text{pipe/hr}} = \pi r^2 h_{\text{water}} = \pi (0.1)^2 (3000)$
$= \pi \times 0.01 \times 3000 = 30\pi \text{ m}^3/\text{hour}$
... (2)
Step 3: Calculate the time required to fill the tank.
Time = (Total volume required) / (Volume flow rate).
Time (in hours) $= \frac{\text{Volume of Tank}}{\text{Volume of water per hour}} = \frac{50\pi}{30\pi}$
Cancel the common factor $\pi$.
Time $= \frac{50}{30} = \frac{5}{3}$ hours
... (3)
To convert this into minutes, multiply by 60.
Time (in minutes) $= \frac{5}{3} \times 60 = 5 \times 20 = 100$ minutes.
Answer: The tank will be filled in 100 minutes (or 1 hour and 40 minutes).
Frustum of a Cone
A frustum of a cone is what you get when you slice off the top of a cone with a cut that is parallel to its base. The remaining bottom part is the frustum. Many everyday objects are in the shape of a frustum, such as a bucket, a drinking glass, or a lampshade.
A frustum has two parallel circular bases (a large one and a small one), a height ($h$), and a slant height ($l$).
- Height ($h$): The perpendicular distance between the two bases.
- Radius of the larger base: $R$
- Radius of the smaller base: $r$
- Slant height ($l$): The length of the slanted side.
The slant height $l$ can be found using the Pythagorean theorem on the right-angled triangle formed by the height $h$, the difference in radii $(R-r)$, and the slant height $l$ itself.
$l = \sqrt{h^2 + (R-r)^2}$
... (1)
Derivation of Frustum Formulas
The formulas for a frustum are derived by considering it as a large cone with a smaller cone removed from its top. Let the original large cone have total height $H$ and total slant height $L$. Let the small cone that is removed have height $h'$ and slant height $l'$.
By similar triangles, we have the crucial relationship:
$\frac{r}{R} = \frac{h'}{H} = \frac{l'}{L}$
1. Derivation of Curved Surface Area (CSA)
CSA of Frustum = CSA of Large Cone - CSA of Small Cone
$\text{CSA} = \pi R L - \pi r l'$
From the diagram, $L = l + l'$. From similar triangles, $\frac{r}{R} = \frac{l'}{L} = \frac{l'}{l+l'}$.
Solving for $l'$: $R l' = r(l+l') \implies R l' = rl + rl' \ $$ \implies l'(R-r) = rl \ $$ \implies l' = \frac{rl}{R-r}$.
And $L = l + l' = l + \frac{rl}{R-r} = \frac{l(R-r)+rl}{R-r} = \frac{Rl}{R-r}$.
Now substitute these into the CSA formula:
$\text{CSA} = \pi R \left(\frac{Rl}{R-r}\right) - \pi r \left(\frac{rl}{R-r}\right)$
$= \frac{\pi l}{R-r} (R^2 - r^2) = \frac{\pi l}{R-r} (R-r)(R+r)$
$\text{CSA of Frustum} = \pi (R+r)l$
... (2)
2. Derivation of Volume
Volume of Frustum = Volume of Large Cone - Volume of Small Cone
$V = \frac{1}{3}\pi R^2 H - \frac{1}{3}\pi r^2 h'$
From similar triangles, $\frac{r}{R} = \frac{h'}{H} = \frac{h'}{h+h'}$.
Solving for $h'$: $R h' = r(h+h') \implies R h' = rh + rh' \implies h'(R-r) = rh \ $$ \implies h' = \frac{rh}{R-r}$.
And $H = h + h' = h + \frac{rh}{R-r} = \frac{h(R-r)+rh}{R-r} = \frac{Rh}{R-r}$.
Substitute these into the volume formula:
$V = \frac{1}{3}\pi \left[ R^2\left(\frac{Rh}{R-r}\right) - r^2\left(\frac{rh}{R-r}\right) \right]$
$V = \frac{1}{3}\frac{\pi h}{R-r} [R^3 - r^3]$
Using the algebraic identity $a^3 - b^3 = (a-b)(a^2+ab+b^2)$:
$V = \frac{1}{3}\frac{\pi h}{R-r} [(R-r)(R^2 + Rr + r^2)]$
$\text{Volume of Frustum} = \frac{1}{3} \pi h (R^2 + Rr + r^2)$
... (3)
3. Total Surface Area (TSA)
The total surface area is simply the curved surface area plus the area of the two circular bases.
$\text{TSA} = \text{CSA} + (\text{Area of top circle}) + (\text{Area of bottom circle})$
$\text{TSA of Frustum} = \pi (R+r)l + \pi r^2 + \pi R^2$
... (4)
Example 1. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Answer:
To Find:
The curved surface area (CSA) of the frustum.
Given:
Slant height $l = 4$ cm.
Circumference of large end = 18 cm.
Circumference of small end = 6 cm.
Solution:
Let the radii of the large and small ends be $R$ and $r$ respectively.
We are given $2\pi R = 18$ cm and $2\pi r = 6$ cm.
The formula for the CSA of a frustum is $\text{CSA} = \pi (R+r)l$.
We can rewrite this as $\text{CSA} = (\pi R + \pi r)l$.
From the given circumferences, we can find $\pi R$ and $\pi r$:
$\pi R = \frac{18}{2} = 9$
... (1)
$\pi r = \frac{6}{2} = 3$
... (2)
Now, substitute these values and the slant height $l=4$ cm into the formula:
$\text{CSA} = (9 + 3) \times 4 \text{ cm}^2$
$\text{CSA} = 12 \times 4 = 48 \text{ cm}^2$
... (3)
Answer: The curved surface area of the frustum is 48 cm$^2$.
Example 2. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass. (Use $\pi = \frac{22}{7}$)
Answer:
To Find:
The capacity (volume) of the glass.
Given:
Shape of glass is a frustum of a cone.
Height $h = 14$ cm.
Diameter of the larger end = 4 cm.
Diameter of the smaller end = 2 cm.
Solution:
First, find the radii of the two circular ends from their diameters.
Radius of the larger end, $R = \frac{4}{2} = 2$ cm.
Radius of the smaller end, $r = \frac{2}{2} = 1$ cm.
The capacity of the glass is its volume. We use the formula for the volume of a frustum:
$V = \frac{1}{3} \pi h (R^2 + Rr + r^2)$
Substitute the given values $h=14, R=2, r=1$ and $\pi = \frac{22}{7}$:
$V = \frac{1}{3} \times \frac{22}{7} \times 14 \times ((2)^2 + (2)(1) + (1)^2)$
Simplify the expression. We can cancel 7 into 14.
$V = \frac{1}{3} \times 22 \times \frac{\cancel{14}^2}{\cancel{7}_1} \times (4 + 2 + 1)$
$V = \frac{1}{3} \times 22 \times 2 \times (7)$
$V = \frac{44 \times 7}{3} = \frac{308}{3} \text{ cm}^3$
... (1)
Convert the fraction to a mixed number or decimal:
$V = 102\frac{2}{3} \text{ cm}^3 \approx 102.67 \text{ cm}^3$
Answer: The capacity of the glass is $\frac{308}{3}$ cm$^3$ or approximately 102.67 cm$^3$.