SSC CGL Maths · Mensuration 3D

Spheres, Hemispheres and Shells

A sphere holds four thirds of π r³ and its surface is 4πr². A hemisphere is half of each, plus a flat circle when it is solid, and a hollow ball is the difference of two balls.

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Why this matters

Twenty-one questions in the bank, seventeen EASY and four MODERATE, fourteen of them from the 2025 shifts. Six of them fit a sphere inside a cube, a box or a cone. Learn the five formulas in the first box by heart: every row on this page starts from them.

Concept 1 of 4: Volume and surface area of a sphere and a hemisphere

A ball's skin is exactly four of the circles you get by cutting it through the centre. A hemisphere is half the ball: half the volume and half the curved skin, plus a flat circle if it is solid.

Definition

Sphere: volume $\frac{4}{3}\pi r^3$ , surface $4\pi r^2$ . In terms of the diameter the surface is $\pi d^2$ .
Hemisphere: half the sphere's volume, $\frac{2}{3}\pi r^3$ , and half its skin, $2\pi r^2$ . Being solid adds the flat circle $\pi r^2$ , which brings the total to $3\pi r^2$ .
A thin bowl painted inside and outside: $2 \times 2\pi r^2 = 4\pi r^2$ .
A dome over a circle of radius $r$ has curved surface $2\pi r^2$ , twice the circle it covers.
Working back: from the surface, $r^2 = \frac{S}{4\pi}$ ; from the volume, $r^3 = \frac{3V}{4\pi}$ .

Sphere and hemisphere

V_{\text{sphere}} = \frac{4}{3}\pi r^3 \qquad S_{\text{sphere}} = 4\pi r^2 \qquad V_{\text{hemi}} = \frac{2}{3}\pi r^3 \qquad \text{CSA}_{\text{hemi}} = 2\pi r^2 \qquad \text{TSA}_{\text{hemi}} = 3\pi r^2

Worked example

A solid hemisphere has radius 21 cm. Find its volume, its curved surface area and its total surface area. (Take

\pi = \frac{22}{7}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 19 Sep 2025, Shift 2 · Q62Easy

Example 1 · Mensuration 3D · Spheres, Hemispheres and Shells

If the curved surface area of a hemisphere is 154

\text{cm}^2

find its diameter.

Curved or total surface of a hemisphere

A solid hemisphere has a flat face too, so its total surface is

3\pi r^2

. The curved part alone is

2\pi r^2

. Read which one the question names.

Diameter put in as the radius

A ball 7 cm across has radius 3.5 cm. Putting 7 into

\frac{4}{3}\pi r^3

gives eight times the true volume.

Drill 7 more on volume and surface area of a sphere and a hemisphere

Concept 2 of 4: Hollow spheres and hemispherical shells

A hollow ball is a big ball with a smaller ball taken out of its centre. The metal is the difference of the two volumes, never the volume of a ball whose radius is the thickness.

Definition

Spherical shell, outer radius $R$ , inner radius $r$ : metal $= \frac{4}{3}\pi(R^3 - r^3)$ .
Hemispherical shell (a bowl with thickness): metal $= \frac{2}{3}\pi(R^3 - r^3)$ .
With thickness $t$ : $R = r + t$ . If the outer diameter is given, halve it first, then $r = R - t$ .
The outer and inner curved surfaces compare as $R^2 : r^2$ .

Metal in a shell

V_{\text{shell}} = \frac{4}{3}\pi(R^3 - r^3) \qquad V_{\text{hemi shell}} = \frac{2}{3}\pi(R^3 - r^3)

$R$ outer radius
$r$ inner radius

Worked example

A hollow ball has outer radius 6 cm and inner radius 5 cm. Find the volume of material in terms of

\pi

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 · Q54Moderate

Example 2 · Mensuration 3D · Spheres, Hemispheres and Shells

A spherical shell has an outer diameter of 20 cm and is 1 cm thick. What is the volume of the material in the shell? (Use

\pi \approx 3.14

Cubing the thickness

The metal is

\frac{4}{3}\pi(R^3 - r^3)

, not

\frac{4}{3}\pi(R - r)^3

. For radii 3 and 2 the bracket is 19 in the first and only 1 in the second.

Drill 4 more on hollow spheres and hemispherical shells

Concept 3 of 4: When the volume is a multiple of the surface area

Divide a sphere's volume by its surface and nearly everything cancels: four thirds of π r³ over 4πr² leaves r over 3. A hemisphere's volume over its curved surface leaves the same r over 3.

Definition

Sphere: $\frac{V}{S} = \frac{\frac{4}{3}\pi r^3}{4\pi r^2} = \frac{r}{3}$ .
Hemisphere, volume over curved surface: $\frac{\frac{2}{3}\pi r^3}{2\pi r^2} = \frac{r}{3}$ .
So "the volume is numerically $k$ times the surface" means $\frac{r}{3} = k$ , that is $r = 3k$ .
The shortcut holds only for the surface named. Against the total surface $3\pi r^2$ of a solid hemisphere the ratio is $\frac{2r}{9}$ .

Volume over surface for a sphere

\frac{V}{S} = \frac{r}{3}

Worked example

The volume of a sphere is numerically equal to twice its surface area. Find its radius.

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 21 Sep 2025, Shift 2 · Q55Easy

Example 3 · Mensuration 3D · Spheres, Hemispheres and Shells

If the numerical value of the volume of a sphere is equal to the numerical value of its surface area, find the radius of the sphere.

The wrong surface for a hemisphere

The

\frac{r}{3}

shortcut for a hemisphere uses its curved surface

2\pi r^2

. With the total surface

3\pi r^2

the ratio is

\frac{2r}{9}

, so check which surface the question means.

Drill 1 more on when the volume is a multiple of the surface area

Concept 4 of 4: The largest sphere inside a cube, a box or a cone

A ball inside a box is stopped by the closest pair of walls. So the largest ball that fits has a diameter equal to the box's smallest edge.

Definition

In a cube of edge $a$ : the largest ball has diameter $a$ , radius $\frac{a}{2}$ . It fills $\frac{\pi}{6}$ of the cube, about 52%, and leaves about 48% empty.
In a cuboid $l \times b \times h$ : the diameter is the smallest of $l$ , $b$ and $h$ .
Empty space $=$ volume of the box $-$ volume of the ball.
In a cone of base radius $r$ , height $h$ and slant height $l$ : the ball resting on the base and touching the curved side has radius $\frac{rh}{r + l}$ . Cut the cone through its tip and this is the circle inside the triangle you see: its radius is the triangle's area $rh$ over half its perimeter $r + l$ .

Radius of the largest ball

\rho_{\text{cube}} = \frac{a}{2} \qquad \rho_{\text{cuboid}} = \frac{\text{smallest edge}}{2} \qquad \rho_{\text{cone}} = \frac{rh}{r + l}

$\rho$ radius of the ball
$r, h, l$ base radius, height and slant height of the cone

Worked example

A box is 20 cm long, 15 cm wide and 10 cm high. The largest ball that fits is put inside it. Find the space left empty.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 24 Sep 2025, Shift 1 · Q55Easy

Example 4 · Mensuration 3D · Spheres, Hemispheres and Shells

A rectangular tank measures 12 cm long, 10 cm wide, and 6 cm high. If a spherical ball of the largest possible size is placed inside this tank, what is the volume of the space not occupied by the ball?

The longest edge limits the ball

A ball is stopped by the smallest edge of the box. In a box 9 cm by 8 cm by 4 cm the largest ball has diameter 4 cm, not 9 cm.

Edge of the cube used as the radius

A ball that just fits a cube of edge 14 cm has radius 7 cm. Using 14 as the radius gives eight times the true volume.

Drill 5 more on the largest sphere inside a cube, a box or a cone

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)

Volume and surface area of a sphere and a hemisphere
Sphere and hemisphere
$V_{\text{sphere}} = \frac{4}{3}\pi r^3 \qquad S_{\text{sphere}} = 4\pi r^2 \qquad V_{\text{hemi}} = \frac{2}{3}\pi r^3 \qquad \text{CSA}_{\text{hemi}} = 2\pi r^2 \qquad \text{TSA}_{\text{hemi}} = 3\pi r^2$
Hollow spheres and hemispherical shells
Metal in a shell
$V_{\text{shell}} = \frac{4}{3}\pi(R^3 - r^3) \qquad V_{\text{hemi shell}} = \frac{2}{3}\pi(R^3 - r^3)$
When the volume is a multiple of the surface area
Volume over surface for a sphere
$\frac{V}{S} = \frac{r}{3}$
The largest sphere inside a cube, a box or a cone
Radius of the largest ball
$\rho_{\text{cube}} = \frac{a}{2} \qquad \rho_{\text{cuboid}} = \frac{\text{smallest edge}}{2} \qquad \rho_{\text{cone}} = \frac{rh}{r + l}$

Watch out for (6)

Curved or total surface of a hemisphere
→ Volume and surface area of a sphere and a hemisphere
Diameter put in as the radius
→ Volume and surface area of a sphere and a hemisphere
Cubing the thickness
→ Hollow spheres and hemispherical shells
The wrong surface for a hemisphere
→ When the volume is a multiple of the surface area
The longest edge limits the ball
→ The largest sphere inside a cube, a box or a cone
Edge of the cube used as the radius
→ The largest sphere inside a cube, a box or a cone

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.

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