# Conversion of solids – Surface Areas and Volumes

Conversions or changes are now a normal feature of our everyday lives. A goldsmith, for instance, melts a strip of gold to turn it into a gem. Likewise, a kid plays with clay forms it into various toys, a carpenter uses the wooden logs to shape various items/furniture for housekeeping. Likewise, for different purposes, the conversion of solids from one form to another is necessary. **Each and every solid that exists occupies some volume. No matter how distinctive the new shape is, as you transform one solid shape to another, the volume stays the same. **In fact, if you melt one large candle into ten small candles, the sum of the volume of the smaller candles would be equal to the volume of the larger candles. **In this section, we will see some problems related to it.**

**Sample Problems on **Conversion of Solids

**Example 1: A silver sphere ball of radius 10 cm is melted and recast into small spheres each of radius 5 cm so that children can play. How many small spheres can be obtained?**

**Solution:**

Say the number of small spheres obtained be x.

We know that the total volume will remain the same.

R = 10 cm (Radius of the big ball).

r = 5 cm (radius of the small ball).

Now, n × (Volume of a small ball) = Volume of a big metallic ball

x * (4/3) * π * r

^{3 }= (4/3) * π * R^{3}

Hence we get: x =

Therefore x is 8. Hence we can make 8 small balls out of 1 big ball.

**Example 2: We have a cylindrical candle, 14 cm in diameter and of length 2cm.** **It is melted to form a cuboid candle of dimensions 7 cm × 11 cm×1 cm. How many Cuboidal candles can be obtained?**

**Solution:**

Dimensions of the cylindrical Candle:

Radius of cylindrical candle = 14/2 cm = 7 cm

Height/Thickness = 2 cm

Volume of one cylindrical candle = πr

^{2}h = π x 7 x 7 x (2) cm^{3 }= 308 cm^{3}Volume of cuboid candle = 7 x 11 x 1 = 77 cm

^{3}

Hence, the number of Cuboidal candles = Volume of cuboid candle/Volume of one cylindrical candle = 308/77 = 4. Hence we can get 4 Cuboidal shaped candles.

**Example 3: We are given a cone made up of clay of height 30 cm. We have to make it in the shape of a cylinder of the same radius. Find the height of the cylinder?**

**Solution :**

Let h1 and h2 be the heights of a cylinder and cone respectively. Let r be the radius of the cone and also the radius of the cylinder as given in the question.

As we know that:

The volume of cylinder = Volume of cone.

πr

^{2}h_{1 }= (1/3) πr^{2}h_{2}h

_{2}= 30 cm

h_{1}= (30/3) = 10 cm

Therefore, the height of the cylinder is 10 cm

**Example 4: A cylindrical copper rod with a diameter of 2 cm and length of 2 cm is drawn into a wire of length 72 m of uniform thickness. Determine the thickness of the wire?**

**Solution:**

Given that, the diameter of the Copper rod = 2 cm.

Radius = 1 cm.

Length of the copper rod = 2 cm

The volume of the copper Cylindrical material = π × (1)

^{2}× 2 = 2π cm^{3}Length of new wire = 72 m = 18 × 100 = 7200 cm.

We know that the wire should be in a cylindrical shape.

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

The volume of the wire = π × r

^{2}× 7200Since the volume of the copper rod and the volume of the new wire should be equal, then we can write

⇒ π × r

^{2}× 7200 = 2π⇒ r = 1/60 cm.

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

Thickness = (1/60) x 2 = 1/30 cm. Thus, the thickness of the wire is approximately equal to 0.0334 cm.

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