Mensuration (3D)

medium importance~2 Q in Tier 121 formulas⚡ 15 shortcuts5 subtopics

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Cylinder

⏱ 3 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

A cylinder of radius r and height h has volume pi r squared h and curved surface two pi r h. The total surface adds the two circular ends. Most questions either substitute directly or reverse one dimension from the volume.

The three formulas

V=\pi r^2h,\qquad \text{CSA}=2\pi rh,\qquad \text{TSA}=2\pi r(h+r)

TSA is the curved surface plus both circles: $2\pi rh+2\pi r^2$ .

Radius $7$ , height $10$ : volume $\dfrac{22}{7}\times49\times10=1540$ ; CSA $2\times\dfrac{22}{7}\times7\times10=440$ ; TSA $440+308=748$ . Check TSA the second way: $2\pi r(h+r)=2\times\dfrac{22}{7}\times7\times17=748$ .

Rule: Fix the radius first and keep it factored. The $r$ cancels out of most reverse questions.

Reverse: find the missing dimension

Volume $1540$ with radius $7$ : $h=\dfrac{1540}{154}=10$ , because $\pi r^2=\dfrac{22}{7}\times49=154$ . One division, and the same trick answers every reverse question.

The same move finds the radius when the height is given: $r^2=\dfrac{V}{\pi h}$ , then recall the square. And the CSA reverses the same way: $440=2\times\dfrac{22}{7}\times7\times h$ gives $h=10$ .

Tip: $\pi r^2$ for $r=7$ is $154$ ; for $r=14$ it is $616$ ; for $r=21$ it is $1386$ . Halves appear too: $r=3.5$ gives $38.5$ . These four end up everywhere.

Pipes, wells and capacity

A hollow pipe or a well uses the same cylinder; a pipe's length plays the height's role. Well of radius $1.4$ m and depth $5$ m:

V=\frac{22}{7}\times1.96\times5=30.8\text{ m}^3=30{,}800\text{ litres}

Cost of digging multiplies the volume by the rate per cubic metre, exactly like tank questions. At Rs $25$ per m $^3$ , that well costs $25\times30.8=$ Rs $770$ .

A one-litre bottle holds $1000$ cu cm; the tank and bottle rules are one rule.

Watch: Radius $1.4$ m is $140$ cm if the question mixes units. Pick one unit before the first line.

Changing the dimensions

Radius doubled, height same: volume $4$ times, curved surface $2$ times.
Radius doubled, height halved: volume $2$ times.
Height alone doubled: volume $2$ times, curved surface $2$ times.
Radius and height both doubled: volume $8$ times.

The exponent tracks the factor: $r$ appears squared in the volume, once in the curved surface.

Remember: Ask which symbols the change touches. Volume $V=\pi r^2h$ : scale $r^2$ by $4$ and $h$ by $\dfrac12$ , net $2$ .

Melting into a cylinder

Volume survives a melt. The shape changes; the amount of metal does not. A cone of radius $7$ , height $12$ has volume $\dfrac13\times154\times12=616$ ; poured into a cylinder of radius $7$ it reaches height $\dfrac{616}{154}=4$ cm.

Tip: Melted questions never need the second solid's shape knowledge beyond its volume formula. Equate volumes, solve for one length.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common3 practice Q

Volume, CSA and TSA of a cylinder

How to spot it:

Radius and height given, a quantity asked.

Method

Volume: $\pi r^2h$ .
CSA: $2\pi rh$ .
TSA: add $2\pi r^2$ for the ends.

Why it works:

Three direct substitutions cover the whole pattern; only the arithmetic changes.

Try this

Find the volume of a cylinder of radius 7 cm and height 10 cm.

Show solution

$V=\dfrac{22}{7}\times49\times10$ .
$V=1540$ cu cm.

Answer

1540 cu cm

Type 2common2 practice Q

Reverse: solve for the missing dimension

How to spot it:

Volume plus one dimension, the other asked.

Method

Write $V=\pi r^2h$ with the knowns in.
Compute $\pi r^2$ (or $\pi h$ ) as one number.
Divide to isolate the unknown.

Why it works:

One equation with one hidden quantity; the pi terms cancel to a clean division.

Try this

The volume of a cylinder is 1540 cu cm and its radius is 7 cm. Its height is:

Show solution

$\pi r^2=154$ .
$h=\dfrac{1540}{154}=10$ cm.

Answer

10 cm

Type 3common2 practice Q

Hollow pipe and well capacity

How to spot it:

A pipe or well with radius and length; litres or cost asked.

Method

Treat it as a cylinder: $V=\pi r^2h$ .
Convert cubic metres to litres by $1000$ .
For cost, multiply by the rate.

Why it works:

Wells and pipes add only a units conversion on top of the standard volume.

Try this

A cylindrical well has radius 1.4 m and depth 5 m. Its capacity in litres is:

Show solution

$V=\dfrac{22}{7}\times1.96\times5=30.8$ m $^3$ .
$30.8\times1000=30{,}800$ litres.

Answer

30,800 litres

Type 4common2 practice Q

Dimension-change multipliers

How to spot it:

A radius or height scaled; the new volume or surface asked.

Method

Note the factor on each symbol.
Raise it to the symbol's power ( $r^2$ in volume).
Multiply the factors.

Why it works:

No recomputation needed: the formula's exponents convert scale factors directly.

Try this

The radius of a cylinder is doubled and its height is halved. Its volume becomes:

Show solution

$r^2$ gives $4$ , $h$ gives $\dfrac12$ .
$4\times\dfrac12=2$ times.

Answer

2 times

Type 5common

Melting another solid into a cylinder

How to spot it:

A cone, sphere or cubes melted and recast as a cylinder.

Method

Compute the source volume.
Set it equal to $\pi r^2h$ of the cylinder.
Solve for the asked length.

Why it works:

Melting conserves volume, so the cylinder formula becomes a one-unknown equation.

Try this

A cone of radius 7 cm and height 12 cm is melted into a cylinder of radius 7 cm. The cylinder's height is:

Show solution

$V_{\text{cone}}=\dfrac13\times154\times12=616$ cu cm.
$154h=616$ .
$h=4$ cm.

Answer

4 cm

Formula sheet

Cylinder

V=\pi r^2h,\quad \text{CSA}=2\pi rh,\quad \text{TSA}=2\pi r(h+r)

r = radius, h = height.

Missing height

h=\frac{V}{\pi r^2}

Reverse of the volume formula.

Capacity

\text{litres}=\text{m}^3\times1000

Same conversion as tanks.

Dimension change

V\propto r^2h,\quad \text{CSA}\propto rh

Scale each symbol by its own factor.

Shortcuts that save time

⚡ 154 and friends

pi r squared for r = 7, 14, 21 is 154, 616, 1386. Reverse questions then divide by a friendly number.

Example

The volume of a cylinder is 1540 cu cm and its radius is 7 cm. Its height is:

Show solution

$\pi r^2=\dfrac{22}{7}\times49=154$ .
$h=\dfrac{1540}{154}$ .
$h=10$ cm.

Answer

10 cm

⚡ Scale the symbols, not the shape

Radius doubled, height halved: volume factor is 4 times 1/2 = 2. Track r squared and h separately.

Example

The radius of a cylinder is doubled and its height is halved. Its volume becomes:

Show solution

$r^2$ factor: $2^2=4$ .
$h$ factor: $\dfrac12$ .
Net: $4\times\dfrac12=2$ times.

Answer

2 times

⚡ Melt: equate volumes

Melting conserves volume. Write both volume formulas equal, cancel, solve for the new length.

Example

A cone of radius 7 cm and height 12 cm is melted into a cylinder of radius 7 cm. The cylinder's height is:

Show solution

Cone: $\dfrac13\times\dfrac{22}{7}\times49\times12=616$ cu cm.
$616=\dfrac{22}{7}\times49\times h=154h$ .
$h=4$ cm.

Answer

4 cm

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using the diameter in $\pi r^2h$ .

Radius only. Diameter $14$ means $r=7$ .

Mistake 02

Calling the CSA the total surface.

TSA $=$ CSA $+$ two ends: $440+308=748$ , not 440.

Mistake 03

Multiplying cubic metres by 100 for litres.

The factor is $1000$ : $30.8$ m $^3$ is 30,800 L.

Mistake 04

Scaling volume by $2$ when only the radius doubles.

$r$ is squared: volume $\times4$ . Height halving then brings it to $\times2$ .

Mistake 05

Re-deriving $\pi r^2$ every line.

Factor it once ( $154$ for $r=7$ ) and reuse; errors drop sharply.

Quick revision

Read this the night before the exam.

$V=\pi r^2h$ , CSA $=2\pi rh$ , TSA $=2\pi r(h+r)$ .
Reverse: $h=\dfrac{V}{\pi r^2}$ ; know $154$ , $616$ , $1386$ for $r=7,14,21$ .
Wells and pipes are cylinders; litres $=$ m $^3\times1000$ .
Radius doubles: volume $\times4$ , CSA $\times2$ ; both double: volume $\times8$ .
Melting keeps volume: equate the two formulas, cancel, solve.

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 9 min · wrong answers go to your mistake notebook automatically.

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