You might have come across several instances in your day to day life when an object is required to be converted into a different shape or size. Assume that, your mom has a gold bar. She wants to get a ring cast out of it. What does she do? She takes it to a gold merchant to melt the gold bar and mould it into a ring. Similarly, you see wax candles of different shapes. If you want to convert a cylindrical candle into a cube, you melt the wax and pour it into a cube-shaped mould. Likewise, the conversion of solids from one shape to another is required for various purposes in daily life.

## Conversion From One Shape to Another

Each and every solid that exists occupies some volume. When you convert one solid shape to another, its volume remains the same, no matter how different the new shape is. In fact, if you melt one big cylindrical candle to 5 small cylindrical candles, the sum of the volumes of the smaller candles is equal to the volume of the bigger candle.

Hence, when you convert one solid shape to another, all you need to remember is that the volume of the original, as well as the new solid, remains the same. Let us discuss some examples to understand this better.

### Conversion of Shapes Examples

**Example 1: **

An iron ball of radius 21 cm is melted and recast into 27 spherical balls of the same radius. Find the radius of each spherical ball.

**Solution:**

Volume of the iron ball = (4/3) Ï€r^{3} = (4/3) x (22/7 )x 21 x 21 x 21 = 38,808 cm^{3}

Let us assume that the radius of the smaller balls is *r.*

So volume of 27 smaller balls = 27 x (4/3) Ï€*r*^{3} = 36Ï€*r*^{3}

Thus we have, the volume of the big iron ball = volume of 27 smaller balls

â‡’38,808 cm^{3 }= 36Ï€*r*^{3}

*â‡’ r*^{3 }= 343 cm^{3}

So *r* = 7 cm

Therefore, the radius of each spherical ball is 7 cm.

**Example 2: **

How many cylindrical candles, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid candle of dimensions 5.5 cm Ã— 10 cm Ã— 3.5 cm?

**Solution:**

Radius of cylindrical candle = 1.75/2 = 0.875 cm

Volume of one cylindrical candle = Ï€r^{2}h = Ï€ x (0.875)^{2} x (0.02) cm^{3}

= Ï€ x 0.0153125 = 0.048125 cm^{3}

Volume of cuboid candle = 5.5 x 10 x 3.5 = 192.5 cm^{3}

Thus, number of cylindrical candles = Volume of cuboid candle/Volume of one cylindrical candle

= 192.5/0.048125 = 4000

**Example 3:Â **

A copper rod with a diameter 1 cm and length 8 cm is drawn into a wire of length 18mÂ of uniform thickness. Determine the thickness of the wire.

**Solution:**

Given that, Copper rod diameter = 1 cm.

Length of the copper rod = 8 cm

Length of new wire = 18 m = 18 Ã— 100 = 1800 cm.

We know that the rod should be in the cylindrical shape.Â

Hence, the volume of the rod = Ï€ Ã— (Â½)^{2} Ã— 8Â

= 2Ï€ cm^{3}

Therefore, the volume of the copper rod = 2Ï€ cm^{3}

If “r” is the radius of cross-section of the wire, then the volume of the wire is given as:

The volume of the wire =Â Ï€ Ã— r^{2} Ã— 1800

Since the volume of the copper rod and the volume of the new wire should be equal, then we can write

â‡’ Ï€ Ã— r^{2} Ã— 1800 = 2Ï€

â‡’ r^{2} = 2Ï€/1800Ï€

â‡’ r^{2} = 1/900

â‡’ r = 1/30

Hence, the thickness of the wire should be the diameter of the cross-section of the new wire.

Thickness = (1/30) x 2 = 1/15 cm.

Thus, the thickness of wire is approximately equal to 0.067 cm.

### Practice Problems

- The metallic spheres of radii 6cm, 8cm and 10 cm, respectively are melted to form a single solid sphere. Find the radius of the resulting solid sphere.
- Water in the canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 kilometers per hour. How much area will it irrigate in 30 minutes, if 8cm of standing water is required?
- The hemispherical tank full of water is emptied by a pipe at the rate of 3(4/7) litres per second. How much time will it take to empty half the tank, if it is 3 m in diameter? (use Ï€ = 22/7)Â

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