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Geometry

Mensuration

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CAT 2025 Lesson : Mensuration - Torus

3.10 Torus

A ring-shaped figure formed by revolving a circle about an axis that is in the same plane as that of the circle. A close example to a torus is an vehicle tyre tube, where the tube becomes circular upon inflation.

In the adjacent figure,
r is the minor radius of the circle that is revolved.
R is the major radius i.e. the distance between the centre of the torus and the centre of the circle that is revolved.

Volume of Torus =

2 \pi R \times \pi r^2

2 \pi^2 R r^2

Curved Surface Area (CSA) = 2

\pi R \times 2 \pi r

4 \pi^2 Rr

Flat Surface Area (FSA) = 0
Total Surface Area (TSA) =

4 \pi^2 Rr

Example 18

John's bike tyre has a tube inside which is in the shape of a Torus. When inflated, the tube's outer diameter is 32 cm while its inner diameter is 24 cm. If the thickness of the rubber tube is negligible, then what is the volume of the air in the tube (in

cm^3

)? (1) 96

\pi

(2) 112

\pi

(3) 256

\pi

(4) 352

\pi

Solution

It is given that the diameter of the smaller circle is 24 cm and the outer diameter is 32 cm. In figure 1 we can see that the remaining 8 cm of the outer diameter can be split equally to fill in the remaining portion.

Volume of Torus =

2 \pi^2 R r^2

-----(1)
Figure 2 shows the values of r and R. Where r =

\dfrac{4}{2}

= 2 cm
and R = 12 +

\dfrac{4}{2}

= 14 cm.

So applying

r

and

R

values in (1)

Volume =

2 \pi^2 \times 14 \times 2^2

112\pi^2

But, since

112\pi^2

is not there in the options, we can expand one value of

\pi

and check. [

\pi

\dfrac{22}{7} approximately

]

Volume =

112 \times \dfrac{22}{7} \times \pi

352\pi

Answer: (4) 352

\pi

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