固体转化率–表面积和体积

转换或更改现在已成为我们日常生活的正常特征。例如，金匠将一条金条融化成宝石。同样，一个孩子用黏土玩耍，将其制成各种玩具，木匠用木原木塑造各种物品/家具用于家务。同样，出于不同的目的，必须将固体从一种形式转化为另一种形式。每个存在的固体都占据一定体积。无论新形状多么独特，当您将一个实体形状转换为另一个实体形状时，体积都保持不变。实际上，如果将一根大蜡烛融化成十个小蜡烛，则小蜡烛的体积总和将等于大蜡烛的体积。在本节中，我们将看到与之相关的一些问题。

固体转化的样品问题

示例1：将一个半径为10厘米的银球形球融化并重新铸造成每个半径为5厘米的小球形，以便儿童玩耍。可以获得多少个小球体?

解决方案：

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 * r³ = (4/3) * π * R³

Hence we get: x = $\frac{R^3}{r^3} \\$

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

示例2：我们有一个直径为14厘米，长度为2厘米的圆柱形蜡烛。熔化形成尺寸为7厘米×11厘米×1厘米的长方体蜡烛。可以获得多少个长方体蜡烛?

解决方案：

Dimnesions of the cylindrical Candle:

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

Height/Thichkness = 2 cm

Volume of one cylindrical candle = πr²h = π x 7 x 7 x (2) cm³ = 308 cm³

Volume of cuboid candle = 7 x 11 x 1 = 77 cm³

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.

例3：我们给了一个圆锥形的圆锥形，它由30厘米高的粘土制成。我们必须使其具有相同半径的圆柱形状。找到圆柱体的高度?

解决方案 ：

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.

πr²h₁ = (1/3) πr²h₂

h₂ = 30 cm

h₁= (30/3) = 10 cm

Therefore, the height of the cylinder is 10 cm

实施例4：将直径为2cm，长度为2cm的圆柱形铜棒拉入长度为72m的均匀厚度的线材中。确定电线的粗细?

解决方案：

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 cm³

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 × r² × 7200

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

⇒ π × r² × 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.