Conversion of Solid From One Shape to Another With Examples & Problems

Conversion of Solid from One Shape to Another: Solids occur in various sizes, and we see them every day. Each solid that has been moulded into a specific shape may not have been in that shape previously. Take, for example, the creation of jewellery. How can perfectly round and circular pearls appear to emerge from the earth? It is only achievable when a cast has been converted to the necessary shape or size. 

In this article, we will discuss the conversion of solid shapes.

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Conversion of Solid From One Shape to Another

Every solid in the universe has a volume. The volume of a solid shape remains constant when converted to another solid shape, regardless of how different the new shape is.

Example: If you melt one large cylindrical candle into four small cylindrical candles, the sum of the smaller candles’ volumes equals the larger candle’s volume.

Let us consider one more example,

In the above example, the cube is recast into the cylinder, but the volume of both the solids remains the same. As a result, we must remember when converting one solid shape to another that both the original and the new solid have the same volume.

Examples of Conversion of Solid From One Shape to Another

1. In the example given below, a cuboid is converted into many spherical balls. In the other example, a cuboid is converted into many small cuboids.

2. In the below example, a big candle is melted into a small candle.

Some of the other examples are listed below:

1. When we slice a watermelon, we are converting a solid shape into another solid shape. One thing that remains true throughout the process, regardless of the size and shape of the slices, is that the total volume of the slices is the same as the original watermelon.

2. If we melt one large cylindrical candle into six small cylindrical candles, the sum of the smaller candles’ volumes equals the larger candle’s volume.

3. A copper rod with a diameter of \(3~{\text{cm}}\) is drawn into a wire of \(10~{\text{mm}}\) of uniform thickness.

Know Important Geometry Formulas

Proof of Conversion of Solid from One Shape to Another (Volume)

The volume remains the same even if the conversion is to a different shape. If we cut down a large iron cube into smaller balls having different radii, the solid’s total volume is the same.

Let’s look at an example to help us understand it.

A spherical iron ball of radius \(21~{\text{cm}}\) is recast into \(27\) smaller balls of diameter \(14~{\text{cm}}.\)

Let’s use the concept of the volume of the sphere to prove this statement.
The volume of the sphere is given by
The volume of sphere \( = \frac{4}{3}\pi {r^3}\)
The volume of the spherical ball of radius \(21~{\text{cm}}\)
\(V = \frac{4}{3} \times \frac{{22}}{7} \times 21 \times 21 \times 21\)
\( \Rightarrow V = 4 \times 22 \times 21 \times 21\)
\( \Rightarrow V = 38808~{\text{c}}{{\text{m}}^3}\)

The volume of one spherical ball of diameter is \(14\,{\text{cm}}\) (Radius \({\text{=6}}\,{\text{cm}}{\text{.}}\))
\(V = \frac{4}{3} \times \frac{{22}}{7} \times {\left({\frac{{14}}{2}} \right)^3}\)
Hence, the volume of the \(27\) spherical balls of diameter is \({\text{=14}}\,{\text{cm}}{\text{.}}\)
\(V = \frac{4}{3} \times \frac{{22}}{7} \times {\left({\frac{{14}}{2}} \right)^3} \times 27\)
\( \Rightarrow V = 38808~{\text{c}}{{\text{m}}^3}\)

Therefore, the volume of the spherical ball of radius \(21~{\text{cm=}}\) the volume of \(27\) spherical balls of diameter \(14~{\text{cm}} = 38808~{\text{c}}{{\text{m}}^3}.\)
The volume remains the same even if the conversion is made into a different shape.

Hence, it is proved.

Solved Examples – Conversion of Solid From One Shape to Another

Q.1. Two solid metallic cuboids of dimensions \(15~{\text{cm}} \times 8~{\text{cm}} \times 5~{\text{cm}}\) and \(20~{\text{cm}} \times 5~{\text{cm}} \times 4~{\text{cm,}}\) are melted together and recast into solid cubes each of side \({\text{5}}\,{\text{cm}}{\text{.}}\) Find the number of solid cubes so formed.
Ans: Given: Dimension of the first cuboid \( = 15~{\text{cm}} \times 8~{\text{cm}} \times 5~{\text{cm}}\)
Dimension of second cuboid \( = 20~{\text{cm}} \times 5~{\text{cm}} \times 4~{\text{cm}}\)
Side of a cube \( = 5~{\text{cm}}\)
Number of solid cubes \( = \frac{{{\text{ the volume of a first cuboid }} + {\text{ the volume of a second cuboid }}}}{{{\text{ the volume of a cube }}}}\)

\( = \frac{{\left( {15 \times 8 \times 5} \right) + \left( {20 \times 5 \times 4} \right)}}{{\left( {5 \times 5 \times 5} \right)}}\)

\( = \frac{{600 + 400}}{{125}} = \frac{{1000}}{{125}}\)

\( = 8\)

Hence, \(8\) cubes are formed.

Q.2. A solid metallic sphere of diameter \(28\,{\text{cm}}\) is melted and recast into several cones, each of diameter \(7\,{\text{cm}}\) and height \(4\,{\text{cm}}.\) Find the number of cones so formed.
Ans: Given that a solid metallic sphere is melted and recast into several cones.
Diameter of sphere \( = 28\,{\text{cm}}\)
So, its radius \({r_1} = \frac{1}{2} \times 28 = 14~{\text{cm}}\)
Diameter of cone \( = 7\,{\text{cm}}\)
So, its radius \({r_2} = \frac{1}{2} \times 7 = 3.5~{\text{cm}}\)
And, the height of cone \( = 4~{\text{cm}}\)
So, number of cones formed \( = \) the volume of the sphere \( = \frac{{{\text{ the volume of the sphere }}}}{{{\text{ the volume of the cone }}}}\)

\( = \frac{{\frac{4}{3}\pi {r_1}^3}}{{\frac{1}{3}\pi {r_2}^2h}}\)

\( = \frac{{4r_1^3}}{{r_2^2h}}\)

\( = \frac{{4 \times {{14}^3}}}{{{{3.5}^2} \times 4}}\)

\( = \frac{{10976}}{{49}}\)

\( = 224\)

Therefore, the number of cones so formed is \(224.\)

Q.3. Find the volume of the largest cone that can be carved out of a cube of side \(16.8\,{\text{cm}}.\)
Ans: From the given information, we have, the side of a cube \( = 16.8\,{\text{cm,}}\)
the height of the cone \(\left( h \right) = \) side of a cube \( = 16.8\,{\text{cm}}\)
We have the diameter of the cone \( = 16.8\,{\text{cm}}\)
Then, the radius of the cone \(\left( r \right) = \frac{{16.8}}{2} = 8.4~{\text{cm}}\)
We know that the volume of a cone \(V = \frac{1}{3} \times \pi \times {r^2} \times h\)
\(\Rightarrow {\text{V}} = \frac{1}{3} \times \frac{{22}}{7} \times {\left({8.4} \right)^2} \times 16.8\)
\( \Rightarrow V = 1241.86~{\text{c}}{{\text{m}}^3}\)
Therefore, the volume of the largest cone that can be carved out of a cube of side \(16.8~{\text{cm}}\) is \(1241.86~{\text{c}}{{\text{m}}^3}.\)

Q.4. A solid metallic cylinder of radius \(14~{\text{cm}}\) and height is \(21~{\text{cm}}\) melted and recast into several spheres, each of radius \(3.5~{\text{cm}}{\text{.}}\) Find the number of spheres formed.
Ans: Let the number of spheres be \(N.\)
From the given information we get,
The volume of the cylinder \( = N \times \) the volume of the spere.
The radius of the cylinder \(r = 14~{\text{cm}}\)
The height of cylinder \(h = 21~{\text{cm}}\)
The volume of cylinder \( = \pi {r^2}h\)
\( \Rightarrow V = \frac{{22}}{7} \times 21 \times 14 \times 14\)
\( = 12936~{\text{c}}{{\text{m}}^3}\)
The radius of sphere \( = 3.5~{\text{cm}}\)
The volume of sphere \( = \frac{4}{3}\pi {r^2}\)
\(V = \frac{4}{3} \times \frac{{22}}{7} \times 3.5 \times 3.5 \times 3.5\)
\( = 179.67~{\text{c}}{{\text{m}}^3}\)
So, \(N = \frac{{{\text{ Volume of cylinder }}}}{{{\text{ Volume of Sphere }}}} = \frac{{12936}}{{179.67}} = 72\)
Therefore, \(72\) spheres can be made.

Q.5. A spherical shell of lead, whose external diameter is \(16\,{\text{cm}},\) is melted and recast into a cylinder of height \(9\frac{1}{3}{\text{cm}}\) and diameter \(16\,{\text{cm}}.\) Find the internal diameter of the shell.
Ans: The external radius of the spherical shell \({r_2} = \frac{d}{2} = \frac{{16}}{2} = 8~{\text{cm}}\)
Internal radius of the spherical shell \({r_1} = ?\)
The radius of cylinder \(r = \frac{{16}}{2} = 8~{\text{cm}}\) and height \(\left( h \right) = 9\frac{1}{3} = \frac{{28}}{3}~{\text{cm}}\)
The volume of the spherical shell \( = \) volume of the cylinder
\( \Rightarrow \frac{4}{3}\pi \left({r_2^3 – r_1^3} \right) = \pi {r^2}~{\text{h}}\)
\( \Rightarrow 4\left({512 – r_1^3} \right) = 64 \times 28\)
\( \Rightarrow 512 – r_1^3 = 448\)
\( \Rightarrow r_1^3 = 512 – 448\)
\( \Rightarrow r_1^3 = 64\)
\( \Rightarrow {r_1} = 4\)
Then, the internal diameter of the shell \(d = 2 \times {r_1} = 2 \times 4 = 8~{\text{cm}}\)
Therefore, \(8~{\text{cm}}\) is the internal diameter of the shell.

Q.6. A hemisphere of lead of radius \(7~{\text{cm}}\) is cast into a cone of base radius \(3.5~{\text{cm}}{\text{.}}\) Find the height of the cone.
Ans: Height of the cone \(\left( h \right) = ?\)
The radius of the cone \( = 3.5\,{\text{cm}}\)
The radius of the hemisphere \(r = 7\,{\text{cm}}\)
We know that the volume of a cone \( = \) the volume of a hemisphere
\( \Rightarrow \frac{1}{3} \times \pi \times{r^2} \times h = \frac{2}{3} \times \pi \times {r^3}\)
\( \Rightarrow {\left({3.5} \right)^2} \times h = 2 \times {7^3}\)
\( \Rightarrow 12.25 \times h = 2 \times 343\)
\( \Rightarrow 12.25~h = 686\)
\( \Rightarrow h = \frac{{686}}{{12.25}} = 56~{\text{cm}}\)
Hence, the height of a cone is \(56~{\text{cm}}{\text{.}}\)

Summary

Solids occur in a variety of sizes, and we see them daily. Each solid that has been formed into a specific shape may not have been in that shape previously. But the volume of the solid shape remains the same. This article explains the meaning of solid conversion from one shape to another, examples, proof, and problems based on it.

This article helps in better understanding the topic of the conversion of solid from one shape to another. This article’s outcome helps learn how solids are converted from one shape to another and solve various problems based on it.

Conversion of Solids – Frequently Asked Questions (FAQs)

Q.1. What is the conversion of solid from one shape to another?
Ans: The volume of a solid shape remains constant when converted to another solid shape, regardless of how different the new shape is.

Q.2. Does surface area change with shape?
Ans: If we convert one solid object to another, the new solid’s surface area changes.

Q.3. Give examples of the conversion of solids from one shape to another.
Ans: Few examples of conversion of a solid from one shape to another are:
Example.1: If you melt one large cylindrical candle and make five small cylindrical candles, the sum of the smaller candles’ volumes equals the larger candle’s volume.
Example.2: A copper rod with a diameter of \(2~{\text{cm}}\) and a length of \(16~{\text{cm}}\) is drawn into a wire of \(1~{\text{mm}}\) of uniform thickness.

Q.4. When a solid is converted to another solid shape, which options below are true?
(i) Surface area remains same (ii) Volume remains same
Ans: We know, when we convert one solid shape to another, the new solid’s surface area changes, but the volume of a solid shape remains constant.
So, option (ii) is true.

Q.5. What will happen to the volume of the new shape when a solid is converted from one shape to another?
Ans: The volume of a solid shape remains constant when it is converted to another solid shape.

Q.6. What is it called when one solid is converted to another solid shape?
Ans: Every solid in the universe has some volume. The volume of a solid shape remains constant when converted to another shape, regardless of how different the new shape is.
Example: If you melt one large cylindrical candle into two small cylindrical candles, the sum of the smaller candles’ volumes equals the larger candle’s volume.