CUET UG 2025 — General Test Geometry

2025mcqmedium

A drainage tile is a cylindrical shell 21cm long. The inside and outside diameters are 4.5 cm and 5.1 cm, respectively.

What is the volume of clay required for making a tile?

Official previous-year question

Held on 3 Jun 2025 · Verified 13 Jul 2026.

Options

A
30.24 π cm³
B
5.76 π cm³
C
6.76 π cm³
D
6.96 π cm³

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Solution

The drainage tile is a cylindrical shell with:

Length = 21 cm
Outside diameter = 5.1 cm
Inside diameter = 4.5 cm

The outside radius $R = \dfrac{5.1}{2} = 2.55$ cm

The inside radius $r = \dfrac{4.5}{2} = 2.25$ cm

The volume of clay is the volume of the outer cylinder minus the volume of the inner cylinder.

Volume of clay $= \pi R^2 h - \pi r^2 h$

$= \pi h (R^2 - r^2)$

Calculate $R^2$ :

$R^2 = (2.55)^2 = 6.5025$

Calculate $r^2$ :

$r^2 = (2.25)^2 = 5.0625$

Find the difference:

$R^2 - r^2 = 6.5025 - 5.0625 = 1.44$

Volume $= \pi \times 21 \times 1.44$

$= 30.24\pi$ cm³

Therefore, the volume of clay required is $30.24\pi$ cm³.

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Solution

The drainage tile is a cylindrical shell with:

Length = 21 cm
Outside diameter = 5.1 cm
Inside diameter = 4.5 cm

The outside radius $R = \dfrac{5.1}{2} = 2.55$ cm

The inside radius $r = \dfrac{4.5}{2} = 2.25$ cm

The volume of clay is the volume of the outer cylinder minus the volume of the inner cylinder.

Volume of clay $= \pi R^2 h - \pi r^2 h$

$= \pi h (R^2 - r^2)$

Calculate $R^2$ :

$R^2 = (2.55)^2 = 6.5025$

Calculate $r^2$ :

$r^2 = (2.25)^2 = 5.0625$

Find the difference:

$R^2 - r^2 = 6.5025 - 5.0625 = 1.44$

Volume $= \pi \times 21 \times 1.44$

$= 30.24\pi$ cm³

Therefore, the volume of clay required is $30.24\pi$ cm³.

CUET UG 2025 — General Test Geometry

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CUET UG 2025 — General Test Geometry

Options

Did you get this right?

Solution

Your note

CUET UG Geometry in other years

More CUET UG Geometry questions

Other General Test topics for CUET UG

General Knowledge

Logical Reasoning

Arithmetic

Quantitative Reasoning

Current Affairs

Keep practicing Geometry