AFCAT Numerical Ability · Surface Area and Volume

Melting, Recasting and Displacement

A solid melted and recast, a liquid poured, a body sunk in water, earth dug and spread or water pumped through a pipe: the volume stays the same, so set the two volumes equal.

All Surface Area and Volume notes

Why this matters

Every question here is one equation, volume before equals volume after, and π cancels whenever both shapes are round. The marks go on units (cm³ against m³ and litres, km/h against a pipe measured in cm²) and on spreading earth over the field less the pit, not the whole field.

Concept 1 of 4: One solid melted into another

Melting changes the shape, never the amount of metal. Write the volume of the old solid, set it equal to the volume of the new one, and solve for the length you want.

Definition

$V_{\text{before}} = V_{\text{after}}$ . With round solids on both sides, $\pi$ cancels at once.
A liquid poured from one vessel to another keeps its volume in the same way.
Into a hollow shape: a hollow cylinder holds $\pi h(R^2 - r^2)$ and a hollow sphere $\tfrac43 \pi(R^3 - r^3)$ . Find the inner radius, then the thickness $R - r$ .
Carved and then recast: only the material left after carving is melted.

Volume is kept

V_{\text{before}} = V_{\text{after}}

Worked example

A metal cone of radius 10 cm and height 18 cm is melted and recast into a cylinder of radius 5 cm. Find the cylinder's height.

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Example 1 · Surface Area and Volume · Melting, Recasting and Displacement

If the height and diameter of a right circular cylinder are 32 cm and 6 cm respectively, then the radius of the sphere whose volume is equal to the volume of the cylinder is

Melt only what is left

If a sphere is first carved from a block and the rest is recast, the new solid holds the block minus the sphere, not the whole block.

Radius in, diameter out

Stems give diameters for one solid and radii for the other, and some ask for a diameter at the end. Halve going in and double coming out.

Drill 6 more on one solid melted into another

Concept 2 of 4: Many small solids from one, or many into one

Small balls from a big one: the number is how many small volumes fit in the big volume. For spheres every

\tfrac43 \pi

cancels, so the count is just the ratio of the radii cubed. Merging works the other way: the cubes of the radii add.

Definition

Number of small spheres of radius $r$ from one of radius $R$ : $\left(\dfrac Rr\right)^3$ .
Several spheres melted into one: $R^3 = r_1^3 + r_2^3 + \dots$
$n$ equal drops merging: $R = r\sqrt[3]{n}$ .
Merging keeps the volume but cuts the surface: $n$ drops have $\sqrt[3]{n}$ times the surface of the big drop they form.
'Can these shots be made?' Compare the sum of their cubed radii with the original's cube.

Spheres recast

n = \left(\dfrac{R}{r}\right)^3,\quad R^3 = r_1^3 + r_2^3 + \dots

Worked example

How many lead shots of radius 0.3 cm can be made from a sphere of radius 1.5 cm?

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Example 2 · Surface Area and Volume · Melting, Recasting and Displacement

What is the number of wax balls, each of radius 1 cm, that can be molded out of a sphere of radius 8 cm?

Radii do not add

Spheres of radii 3, 4 and 5 make one of radius 6, not 12. Add the cubes, then take the cube root.

Merging cuts the surface

64 drops merge into one drop of four times the radius, with 16 times one drop's surface. The 64 drops had 64 times, so the surface falls to a quarter.

Drill 4 more on many small solids from one, or many into one

Concept 3 of 4: A body sunk in water raises the level

A solid pushed fully under water takes up its own volume, and the water has nowhere to go but up. The rise is that volume spread over the floor of the vessel.

Definition

Rise = volume sunk ÷ base area of the vessel (for a vessel with straight sides).
People diving, stones dropped and balls sunk all work the same way; several of them: multiply first.
'How many balls before it spills?' The empty space above the water ÷ one ball's volume.
Keep one unit: a rise in cm needs the volume in cm³ and the area in cm².

Rise in level

\text{rise} = \dfrac{V_{\text{sunk}}}{\text{base area of the vessel}}

Worked example

A stone of volume 1,540 cm³ is sunk in a cylindrical jar of radius 7 cm partly full of water. How far does the water rise? (Use

\pi = \tfrac{22}{7}

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Example 3 · Surface Area and Volume · Melting, Recasting and Displacement

A cube of 9 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base are 15 cm and 12 cm, then the rise in water level in the vessel is

Divide by the vessel's base, not the object's

A cube sunk in a tank raises the water by its volume over the tank's floor. The cube's own face area plays no part.

Match the units before dividing

0.08 m³ over a pool of 200 m² is a rise of 0.0004 m, which is 0.04 cm. Convert at the end, not halfway.

Drill 3 more on a body sunk in water raises the level

Concept 4 of 4: Earth dug and spread, water flowing and stored

Earth dug from a pit is a cuboid of soil, and spread out it becomes a thin layer of the same volume. Water through a pipe is a long cylinder: each hour the pipe delivers its cross-section times the distance the water travels.

Definition

Earth spread over a field: rise = earth's volume ÷ the area it covers. If it is spread on the rest of the field, take the field's area less the pit's.
Flow: volume per hour = cross-section × speed. Put both in metres first.
Units: 1 m³ = 1,000 litres; 1 m² = 10,000 cm²; 1 km = 1,000 m.
Water lasting a town: stored litres ÷ litres used a day.

Flow in a pipe

\text{volume per hour} = \text{cross-section} \times \text{speed}

Worked example

A pit 10 m × 8 m × 2.4 m is dug in a field 50 m × 40 m and the earth is spread evenly over the rest of the field. By how much is the field raised?

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Example 4 · Surface Area and Volume · Melting, Recasting and Displacement

Water is distributed to a town of 50000 inhabitants from a rectangular reservoir consisting of 3 equal compartments. Each compartment has length and breadth 200 m, 100 m, respectively and 12 m depth of water in the beginning. The allowance is 20 L per head per day. For how many days will the supply of water hold out?

Spread over the field less the pit

With 192 m³ of earth on a 2,000 m² field holding an 80 m² pit, dividing by 2,000 gives 9.6 cm. The earth covers only 1,920 m², so the rise is 10 cm.

cm² to m² is ten thousand, not a hundred

A pipe of cross-section 25 cm² is 0.0025 m². Dividing by 100 gives 0.25 m², a pipe a hundred times too wide.

Drill 4 more on earth dug and spread, water flowing and stored

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

One solid melted into another
Volume is kept
$V_{\text{before}} = V_{\text{after}}$
Many small solids from one, or many into one
Spheres recast
$n = \left(\dfrac{R}{r}\right)^3,\quad R^3 = r_1^3 + r_2^3 + \dots$
A body sunk in water raises the level
Rise in level
$\text{rise} = \dfrac{V_{\text{sunk}}}{\text{base area of the vessel}}$
Earth dug and spread, water flowing and stored
Flow in a pipe
$\text{volume per hour} = \text{cross-section} \times \text{speed}$

Watch out for (8)

Melt only what is left
→ One solid melted into another
Radius in, diameter out
→ One solid melted into another
Radii do not add
→ Many small solids from one, or many into one
Merging cuts the surface
→ Many small solids from one, or many into one
Divide by the vessel's base, not the object's
→ A body sunk in water raises the level
Match the units before dividing
→ A body sunk in water raises the level
Spread over the field less the pit
→ Earth dug and spread, water flowing and stored
cm² to m² is ten thousand, not a hundred
→ Earth dug and spread, water flowing and stored

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