SSC CGL Maths · Mensuration 3D

Melting, Recasting and Water

Melting a solid into new shapes, pouring a liquid from one vessel into others, or sinking a stone in water never changes the volume, so you set the volume before equal to the volume after.

All Mensuration 3D notes

Why this matters

Twenty-one questions in the bank, twelve EASY and nine MODERATE, seventeen of them from the 2025 shifts. Every one rests on the single fact that the volume does not change. Write volume before equal to volume after, cancel the common factors such as four thirds of π, and the arithmetic is short.

Concept 1 of 4: Melting and recasting keeps the volume the same

Metal melted into new shapes is the same metal. However the pieces look, their volumes add up to the volume you started with, unless the question says some is wasted.

Definition

Volume before $=$ volume after. Write both sides with their formulas and cancel what is common.
Several spheres melted into one sphere: $\frac{4}{3}\pi$ cancels, so $R^3 = r_1^3 + r_2^3 + \dots$ . The same works with diameters, and with cubes and their edges.
Know the cubes up to $20^3 = 8000$ , so the final cube root is quick.
Different shapes: write each formula in full. A sphere into a cone: $\frac{4}{3}\pi R^3 = \frac{1}{3}\pi r^2 h$ , and $\frac{1}{3}\pi$ cancels to $4R^3 = r^2 h$ .
A hollow sphere has metal $\frac{4}{3}\pi(R^3 - r^3)$ , and that is what gets recast.
Two equal hemispheres make exactly one sphere of the same radius.

Volume is conserved

V_{\text{before}} = V_{\text{after}} \qquad R^3 = r_1^3 + r_2^3 + r_3^3

Worked example

A solid metal sphere of radius 6 cm is melted and recast into a cone whose base radius is also 6 cm. Find the height of the cone.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 20 Sep 2025, Shift 1 · Q57Easy

Example 1 · Mensuration 3D · Melting, Recasting and Water

Three metallic spheres with radii 3 cm,4 cm, and 5 cm respectively are melted together and recast into a single solid sphere. What is the radius of the new sphere?

Adding the radii

Spheres of radii 1 cm, 6 cm and 8 cm melt into a sphere of radius 9 cm, not 15 cm. Volumes add, so it is the cubes of the radii that add:

1 + 216 + 512 = 729 = 9^3

Forgetting the one third in a cone

A cylinder melted into a cone on the same base gives a cone 3 times as tall as the cylinder, because the cone's formula carries

\frac{1}{3}

Drill 10 more on melting and recasting keeps the volume the same

Concept 2 of 4: How many smaller solids, or how many containers

When one big solid becomes many identical small ones, divide: the number is the big volume over one small volume. The same division counts the containers a tank can fill.

Definition

$n = \frac{V_{\text{big}}}{V_{\text{small}}}$ .
Same shape with every length scaled (spheres, hemispheres, cubes): $n = \left(\frac{R}{r}\right)^3$ .
Different shapes: cancel $\pi$ and simplify the fraction before you multiply.
Litres: $1\ \text{m}^3 = 1000$ litres, so buckets of $b$ litres fill a $1\ \text{m}^3$ tank in $\frac{1000}{b}$ trips.

Number of pieces

n = \frac{V_{\text{big}}}{V_{\text{small}}} \qquad n_{\text{same shape}} = \left(\frac{R}{r}\right)^3

Worked example

How many lead balls of radius 1 cm can be made from a solid lead sphere of radius 5 cm?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 23 Sep 2025, Shift 3 · Q54Easy

Example 2 · Mensuration 3D · Melting, Recasting and Water

A spherical tank is filled with water. The radius of the sphere is 6 meters. The tank is used to fill cylindrical containers, each with a radius of 1 meter and a height of 12 meters. How many containers can be filled?

Dividing the radii instead of the volumes

A ball of radius 4 cm makes 64 balls of radius 1 cm, not 4. The count is the ratio of the volumes,

\left(\frac{4}{1}\right)^3

Drill 3 more on how many smaller solids, or how many containers

Concept 3 of 4: Surface area after recasting

Volume survives melting, but surface does not. Turn one ball into many small ones and there is far more skin, which is why crushed ice melts faster than a block.

Definition

Use the volume first to find the new size, then work out the new surface.
One solid split into $n$ equal copies of its own shape: each length becomes $\frac{1}{\sqrt[3]{n}}$ of the old, and the total surface becomes $\sqrt[3]{n}$ times the old.
So 27 equal balls have three times the surface of the ball they came from, and 64 have four times.

Total surface after splitting into n equal copies

S_{\text{after}} = \sqrt[3]{n} \times S_{\text{before}}

Worked example

A metal sphere of radius 9 cm is melted into 27 equal small spheres. Find the ratio of the total surface area of the small spheres to the surface area of the big one.

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 15 Sep 2025, Shift 1 · Q64Moderate

Example 3 · Mensuration 3D · Melting, Recasting and Water

A metal sphere having a radius of 10 centimeters is melted down and molded into 8 identical smaller solid spheres. What is the ratio of the surface area of the original sphere to the total surface area of all 8 smaller spheres?

Assuming the surface stays the same

Only the volume is kept. Splitting a ball into 27 equal balls triples the total surface, so treating the two surfaces as equal is wrong.

Drill 1 more on surface area after recasting

Concept 4 of 4: Water displaced and the rise in level

A stone sunk in a tank pushes aside exactly its own volume of water. That water has nowhere to go but up, so it spreads over the tank's base as a thin slab.

Definition

A fully sunk solid displaces its own volume. The size of the tank does not change that, as long as the solid fits and is fully under water.
Rise in level $= \frac{\text{volume sunk}}{\text{base area of the tank}}$ .
The base is $\pi R^2$ for a cylindrical tank and length $\times$ breadth for a rectangular one. The water already in the tank does not enter the sum.
Several identical solids: multiply one volume by the number first.

Rise in water level

\text{rise} = \frac{V_{\text{sunk}}}{\text{base area}}

Worked example

A cylindrical jar of radius 6 cm holds some water. A solid ball of radius 3 cm is dropped in and sinks completely. By how much does the water rise?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 24 Sep 2025, Shift 2 · Q55Moderate

Example 4 · Mensuration 3D · Melting, Recasting and Water

A rectangular aquarium (100 cm L

\times

50 cm W

\times

60 cm H) contains water with an initial level of 20 cm. 10 identical metal spheres, each with a diameter of 10 cm, are gently lowered into the aquarium. Assuming the spheres are fully submerged, what is the rise in the water level?

Using the water already in the tank

The rise depends only on the volume sunk and the base area. The starting depth of the water never enters the sum.

Dividing by the tank's volume

Divide the volume sunk by the base area of the tank, not by its volume: the answer is a length.

Drill 3 more on water displaced and the rise in level

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)

Melting and recasting keeps the volume the same
Volume is conserved
$V_{\text{before}} = V_{\text{after}} \qquad R^3 = r_1^3 + r_2^3 + r_3^3$
How many smaller solids, or how many containers
Number of pieces
$n = \frac{V_{\text{big}}}{V_{\text{small}}} \qquad n_{\text{same shape}} = \left(\frac{R}{r}\right)^3$
Surface area after recasting
Total surface after splitting into n equal copies
$S_{\text{after}} = \sqrt[3]{n} \times S_{\text{before}}$
Water displaced and the rise in level
Rise in water level
$\text{rise} = \frac{V_{\text{sunk}}}{\text{base area}}$

Watch out for (6)

Adding the radii
→ Melting and recasting keeps the volume the same
Forgetting the one third in a cone
→ Melting and recasting keeps the volume the same
Dividing the radii instead of the volumes
→ How many smaller solids, or how many containers
Assuming the surface stays the same
→ Surface area after recasting
Using the water already in the tank
→ Water displaced and the rise in level
Dividing by the tank's volume
→ Water displaced and the rise in level

Test yourself on Mensuration 3D

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Mensuration 3D chapter test Drill all 21 questions