How Many Small Laddoos Can You Make From One Giant Laddoo?

When metal is melted and recast, or a solid is dropped into water, one thing is conserved: volume. The shape changes, but the amount of material or displaced water does not.

Surface area is different. It depends on shape, so it usually changes when a solid is reshaped, and parts hidden where two solids join do not count.

This part covers combined solids, recasting, hollow solids and changing water levels, using $\pi =\frac{22}{7}$.

Add the volumes of the parts, but for surface area add only the surfaces that are exposed, leaving out the faces where the solids are joined.

Worked example 1 — tent. A tent is a cylinder of radius $7$ m and height $4$ m topped by a cone of the same radius and height $24$ m, so $l=25$ m.

$$\text{canvas}=2\times \frac{22}{7}\times 7\times 4+\frac{22}{7}\times 7\times 25=176+550=726{\text{ m}}^2$$
$$\text{volume}=\frac{22}{7}\times 49\times 4+\frac{1}{3}\times \frac{22}{7}\times 49\times 24=616+1232=1848{\text{ m}}^3$$

Worked example 2 — hemisphere on a cylinder. A solid has radius $3.5$ cm, with a cylinder $10$ cm tall and a hemisphere on top.

$$\text{TSA}=2\pi (3.5)(10)+2\pi (3.5{)}^2+\pi (3.5{)}^2=220+77+38.5=335.5{\text{ cm}}^2$$
$$V=\pi (3.5{)}^2(10)+\frac{2}{3}\pi (3.5{)}^3=385+89.83=474.83{\text{ cm}}^3$$

An everyday example. A temple with a hemispherical dome on a cylindrical base needs paint only for the visible surfaces.

The substance. The joining circle is counted once, as the base, and not at all where the two solids meet.

Set the volume of the original solid equal to the total volume of the new solid or solids, and solve for the unknown dimension or number.

Worked example 1 — sphere into cylinder. A sphere of radius $6$ cm is melted into a cylinder of radius $4$ cm.

$$\frac{4}{3}\pi \times 216=\pi \times 16\times h288=16hh=18\text{ cm}$$

Worked example 2 — small spheres. How many spheres of radius $1$ cm can be made from one of radius $5$ cm?

$$n=\frac{\frac{4}{3}\pi \times 5^3}{\frac{4}{3}\pi \times 1^3}=125$$

Worked example 3 — wire. A copper sphere of radius $3$ cm is drawn into a wire of radius $0.1$ cm.

$$\frac{4}{3}\pi \times 27=\pi \times 0.01\times L36=0.01LL=3600\text{ cm}=36\text{ m}$$

An everyday example. **A giant laddoo of diameter $12$ cm is re-rolled into small laddoos of diameter $3$ cm.**

$$n={\left(\frac{6}{1.5}\right)}^3=4^3=64\text{ small laddoos}$$

The substance. The total surface area grows when one solid is recast into many smaller ones, even though the volume stays fixed.

**The volume of material is the outer volume minus the inner volume, using the external radius $R$ and internal radius $r$; the surface area adds the outer and inner surfaces, plus the rim rings for a pipe.

Worked example 1 — pipe.** A pipe has $R=4$ cm, $r=3$ cm and length $14$ cm.

$$\text{material}=\pi (R^2-r^2)h=\frac{22}{7}\times 7\times 14=308{\text{ cm}}^3$$
$$\text{outer CSA}=2\pi \times 4\times 14=352\text{inner CSA}=2\pi \times 3\times 14=264$$
$$\text{two rings}=2\pi (R^2-r^2)=44\text{total}=352+264+44=660{\text{ cm}}^2$$

Worked example 2 — hollow sphere. $R=7$ cm and $r=3.5$ cm.

$$\text{material}=\frac{4}{3}\pi (343-42.875)=\frac{4}{3}\times \frac{22}{7}\times 300.125=1257.67{\text{ cm}}^3$$
$$\text{outer surface}=616{\text{ cm}}^2\text{inner surface}=154{\text{ cm}}^2$$

An everyday example. An iron water pipe is a hollow cylinder, and its weight depends on the volume of iron, not on the water space inside.

The trap. **Use $R^2-r^2$, not $(R-r{)}^2$**: here $16-9=7$, while $(4-3{)}^2=1$.

**The volume of water displaced equals the volume of the immersed solid, and it spreads over the vessel's circular cross-section, so rise in level $=$ volume of solid $÷$ area of the vessel's base.

Worked example 1.** A sphere of radius $3$ cm is dropped into a cylinder of radius $6$ cm.

$$\frac{4}{3}\pi \times 27=\pi \times 36\times h36=36hh=1\text{ cm}$$

Worked example 2. Five spheres of radius $3$ cm are placed in a cylinder of radius $10$ cm.

$$5\times 36\pi =180\pi h=\frac{180\pi }{100\pi }=1.8\text{ cm}$$

Worked example 3 — counting marbles. Water in a cylinder of radius $7$ cm rises by $5.6$ cm when marbles of diameter $1.4$ cm are dropped in.

$$n=\frac{\pi \times 49\times 5.6}{\frac{4}{3}\pi \times 0.7^3}=\frac{274.4}{0.45733\dots }=600$$

An everyday example. In a school laboratory, the volume of an irregular stone is found by lowering it into a measuring cylinder and reading how far the water rises.

The substance. The rise depends on the vessel's width, not its depth — a narrow vessel shows a bigger rise for the same solid.