Conversion Of Solids From One Shape Into Another

In our daily life, we come across various shapes of objects. For example, we see the objects of wax (candle) in various shapes like cylindrical, conical, circular etc. Have you ever thought how these objects are made?

First of all, the wax is melted and after that the liquid wax is poured into containers which have the special shape in which we want to mould the wax. Then after cooling the wax we get the desired shape.

We also convert solids from one shape into another. For example, a metallic wire of cylindrical shape is melted and recast into a spherical shape; earth dug out from a well is uniformly distributed to form an embankment around it.

The main concept in such conditions is that the amount of material before the conversion remains the same as the amount of material after the conversion or we can say that the volume before the conversion and after the conversion remains constant.

Let us consider a situation where we have a ball which is made up of wax. We measured the radius of the ball and it came out to be 3 cm. Now, we want to make a candle of length 16 cm from the wax ball. What should be the radius of the mould so that no wax is wasted?

We use this concept for solving the problems related to conversion of solids. Let us discuss some examples based on the above idea.

Example 1: A solid right circular metallic cone of radius 10 cm and height 5 cm is melted and recast into a sphere. Find the radius of the sphere.

Solution:

Let us consider the above information geometrically.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_20a8dd2c.png

It is given that radius of cone, r = 10 cm

Height of cone, h = 5 cm

Let R be the radius of sphere.

Since the cone is melted to form a sphere,

Volume of metal before conversion = volume of metal after conversion

∴ Volume of cone = Volume of sphere

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_1c61040d.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_5140832b.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_m7ced8436.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_m5c91abdc.gif cm

Thus, the radius of the sphere is 5 cm.

Example 2: If the volumes of a solid hemisphere, a solid right circular cone and a solid right circular cylinder of same base are equal, then find the ratio of their heights.

Solution:

Let us consider the above information by drawing figures.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_m3fb6f655.png

All the three solids have equal base, i.e. their radii are equal.

Let r be the radius of the above three figures. The height of the hemisphere will also be r. Let the height of the cone and cylinder be h and H respectively.

But it is given that,

Volume of hemisphere = Volume of cone = Volume of cylinder

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_m33d89628.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_m46f92d79.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_6f8c9cb1.gif

The LCM of 2, 1, and 3 is 6.

Now, let us write

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_6f8c9cb1.gif= 6x

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_m6894b6d9.gif

Hence,https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_4103f590.gif

Now, r: h: H = 3x: 6x: 2x

r: h: H = 3: 6: 2

∴ Ratio of their heights = 3: 6: 2

Example 3: A solid right circular metallic cone of height 5 cm and radius 21 cm is melted to form some coins of diameter 2 cm and width 6 mm and some spherical balls of diameter 3 cm. If the ratio of the number of coins to the number of balls is 2:3, then find the number of coins and balls.

Solution:

For the cone,

Height, h = 5 cm and radius, r = 21 cm

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_7ded96c2.png

For cylindrical coin,

Diameter = 2 cm

So, radius, r1 = 1 cm

Height, h1 = 6 mm = 0.6 cm

For spherical balls,

Diameter = 3 cm

So, radius, r2 =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_79f5f82c.gif cm

Let number of spherical coins = 2x and number of spherical balls = 3x

Metallic cone is melted and recast to form coins and balls.

∴ Volume of coins + Volume of balls = Volume of Cone

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_620bf729.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_1405cc4.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_e033566.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_m2db18ed2.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_m4f8893c7.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_664e7eac.gif

∴ Number of coins = 2x = 2 × 50 = 100

And number of spherical balls = 3x = 3 × 50 = 150

Example 4: A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 10 cm and the diameter of the base is 14 cm. The toy is placed in a cylinder which is half filled with water. Find the height of increase in water level in the cylinder, if the radius of the cylinder is 14 cm.

Solution:

Let us draw the figure for the given information.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_1e0e2aed.png

For the toy,

Diameter of the hemisphere = 14 cm

Radius of hemisphere, r = 7 cm (Also radius for cone)

Height of the cone, h = 10 cm

For the cylinder, Radius, R = 14 cm

Let height of increase in the level of water = H

When the toy is kept in the cylindrical vessel, the water will rise uniformly taking the shape of a cylinder, i.e. the increase in the level of water will be in the shape of a cylinder of radius 14 cm.

Volume of water raised in the cylinder = Volume of the toy

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_m35fbbd8.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_4653533e.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_26cc8c98.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_62838bcc.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/140/190/471/535/10.1.13.2.2_ok_html_4d87a4f0.gifcm

Thus, the water level will become 2 cm high after keeping the toy in the cylindrical vessel.