**What is the Radius of a Right Circular Cylinder**

If r and h denote respectively the radius of the base and height of a **right circular cylinder**, then –

- Area of each end = πr
^{2} - Curved surface area = 2πrh = (circumference) height
- Total surface area = 2πr (h + r) sq. units.
- Volume = πr
^{2}h = Area of the base × height

where, π = or 3.14

If R and r (R > r) denote respectively the external and internal radii of a **hollow right circular cylinder**, then –

Let external radius = R, Internal radius = r, height = h. Then,

- Outer curved surface area = 2πRh
- Inner curved surface area = 2πrh
- Area of each end = π (R
^{2}– r^{2}) - Curved surface area of hollow cylinder = 2π (R + r) h
- Total surface area = 2π (R + r) (R + h – r)
- Volume of material = πh (R
^{2}– r^{2})

Read more about Area of a Right Circular Cone

**Right Circular Cylinder Example Problems with Solutions**

**Example 1: **The inner diameter of a circular well in 2 m and its depth is 10.5 m. Find :

(i) the inner curved surface area of the well.

(ii) the cost of plastering this curved surface area at the rate of Rs. 35 per m^{2}.

**Solution: **Given : Radius = = 1m i.e., r = 1 m and depth = 10.5 m i.e., h = 10.5 m

(i) The inner curved surface area of the well = 2πrh = 2 × m^{2} = 66 m^{2}

(ii) The cost of plastering

= Area to be plastered × Rate

= 66 × Rs. 35 = Rs. 2,310

**Example 2: **In a hot water heating system there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.** **

**Solution: **Given : Length of the cylindrical pipe = 28 m i.e., h = 2800 cm and, its radius = × diameter = × 5 cm = 2.5 cm.

∴ The total radiating surface in the system

= curved surface area of the pipe

= 2πrh = 2 × × 2.5 × 2800 cm^{2}

= 44000 cm^{2}

It is not clear from the question that how the cylindrical pipe is used. We can take the radiating surface of the system

= Total surface area of the pipe

= 2πr (r + h) = 2 × (2.5 + 2800) cm^{2}

= 44039.29 cm^{2}

**Example 3: **Find :(i) the lateral or curved surface area of a cylindrical patrol storage tank that is 4.2 m in diameter and 4.5 m high.

(ii) how much steel was actually used if of the steel actually used was wasted in making the closed tank.

**Solution: **(i) Given : r = = 2.1 m and h = 4.5 m

∴ Curved surface area of the tank

= 2πrh = 2 × × 4.5 m^{2} = 59.4 m^{2}

(ii) Let the steel actually used be x m^{2}

∴ x – = 59.4 ⇒ = 59.4

⇒ x = 59.4 × = 64.8 m^{2}

∴ Actual amount of steel used = 64.8 m^{2}

**Example 4: **The figure, given alongside, shows the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame . Find how much cloth is required for covering the lampshade.

**Solution: **Given : The height of the lampshade = 30 cm.

∵ A margin of 2.5 cm is required for folding its over and bottom of the frame ; the resulting height of the cloth required in the shape of the cylinder = (30 + 2.5 + 2.5) cm = 35 cm.

∴ For the cloth, which must be in the

shape of a cylinder with height 35 cm and

radius = 10 cm.

i.e., h = 35 cm and r = 10 cm.

∴ Area of the cloth required for covering the lampshade.

= 2πrh = 2 × × 10 × 35 cm^{2}

= 2200 cm^{2 }.

**Example 5: **The total surface area and the curved surface area of a cylinder are in the ratio 5 : 3. Find the ratio between its height and its diameter.-

**Solution: **Required =

According to the given statement :

⇒

⇒ 5h = 3h + 3r

i.e., 2h = 3r

⇒

and = 3 : 4

**Example 6: **A soft drink is available in two packs of different shapes : (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

**Solution: **(i) Volume of soft drink in the tin can

= Volume of the tin can

= Its length × breadth × height

= 5 cm × 4 cm × 15 cm = 300 cm^{3}

(ii) Volume of soft drink in the plastic cylinder

= πr^{2}h =

[ ∵ r = ]

= 385 cm^{3}

Clearly, plastic container has greater capacity by

385 cm^{3} – 300 cm^{3} = 85 cm^{3}

**Example 7: **The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many liters of water can it hold? (1000 cm^{3} = 1 ℓ)

**Solution: **Let the radius of the base of the cylindrical vessel be r cm.

∴ 2πr = 132 [∵ circumference = 2πr]

⇒ 2×

⇒ r =

Now, radius (r) = 21 cm and height (h) = 25 cm

⇒ Volume of the cylindrical vessel = πr^{2}h

= × 21 × 21× 25 cm^{2} = 34560 cm^{3}

∴ Vol. of water which this vessel can hold

= Volume of the vessel = 34560 cm^{3}

= [∵ 1000 cm^{3} = 1 ℓ]

= 34.650 ℓ

**Example 8: **A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

**Solution: **Radius of cylindrical bowl = i.e., r = and the height of the soup in a bowl h = 4 cm

∴ Volume of soup required for 1 patient

= πr^{2}h =

= 154 cm^{3}.

⇒ Volume of soup required for 250 patients.

= 250 × 154 cm^{3} = 38500 cm^{3}

**Example 9: **If the lateral surface of a cylinder is 94.2 cm^{2} and its height is 5 cm, then find:

(i) radius of its base

(ii) its volume. (Use π = 3.14)

**Solution: **Given. 2πrh = 94.2 cm^{2} and h = 5 cm

(i) 2πrh = 94.2 cm^{2}

⇒ 2 × 3.14 × r × 5 = 94.2

⇒ r = = 3 cm .

(ii) Its volume = πr^{2}h

= 3.14 × 3 × 3 × 5 cm^{2}

= 141.3 cm^{3}

**Example 10: **A hollow cylinder is 35 cm in length (height). Its internal and external diameters are 8 cm and 8.8 cm respectively. Find its :

(i) outer curved surface area

(ii) inner curved surface area

(iii) area of cross-section

(iv) total surface area.

**Solution: **The height of the cylinder h = 35 cm

The internal radius r = = 4 cm

The external radius R = = 4.4 cm

(i) Outer curved surface area = 2πRh

= 2 × = 968 cm^{2}

(ii) Inner curved surface area = 2πrh

= 2 × = 880 cm^{2}

(iii) The cross-section of a hollow cylinder is like a ring with external radius R = 4.4 cm and internal radius r = 4 cm.

∴ Area of cross-section = πR^{2} – πr^{2}

= π (R^{2} – r^{2}) =

=

=

= 10.56 cm^{2}

**Example 11: **Find the total surface area of a hollow cylindrical pipe of length 50 cm, external diameter 12 cm and internal diameter 9 cm.

**Solution: **Given : Height (h) = 50 cm, external radius (R) = 6 cm and internal radius (r) = 4.5 cm

∵ Total surface area of the hollow cylinder

= External C.S.A + Internal C.S.A + 2 × Area of cross-section

= 2πRh + 2πrh + 2π (R^{2} – r^{2})

= 2π (R + r)h + 2π (R + r) (R – r)

= 2 × × (6 + 4.5) × 50 + 2 × × (6 + 4.5) (6 – 4.5) cm^{2}

= × 50 +

= 3300 cm^{2} + 99 cm^{2} = 3399 cm^{2}

**Example 12: **The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm^{3} of wood has a mass of 0.6 g.

**Solution: **Since, inner diameter = 24 cm

⇒ inner radius (r) = = 12 cm

Since, outer diameter = 28 cm

⇒ outer radius (R) = = 14 cm

Also, given that height (h) = 35 cm

∴ Volume of wood in the pipe = π (R^{2} – r^{2})h

= (14^{2} –12^{2}) × 35 cm^{3} = 5720 cm^{3}

Since, mass of 1 cm^{3} of wood = 0.6 gm

⇒ Mass of 5720 cm^{3} of wood = 0.6 × 5720 gm = 3432 gm

∴ Mass of the pipe = mass of wood

= 3432 gm

= 3.432 kg

**Example 13: **A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

**Solution: **Clearly, for wooden part, which is the form of a hollow cylinder :

External radius (R) = = 3.5 mm

= 0.35 cm

Internal radius (r) = = 0.5 mm

= 0.05 cm

And, height (length) = 14 cm.

∴ Volume of the wood = π (R^{2} – r^{2})h

= × [(0.35)^{2} – (0.05)^{2}] × 14 cm^{3}

= × 0.12 × 14 cm^{3} = 5.28 cm^{3}

And, volume of graphite = π r^{2}h

= × (0.05)^{2} × 14 cm^{3 }= 0.11 cm^{3}

**Example 14: **The internal radius of a hollow cylinder is 8 cm and thickness of its wall is 2 cm. Find the volume of material in the cylinder, if its length is 42 cm.

**Solution: **Since, internal radius = 8 cm π r = 8 cm,

thickness of the wall of the cylinder = 2 cm.

∴ Its external radius = 8 cm + 2 cm = 10 cm i.e., R = 10 cm.

Also, length (height) of the cylinder = 42 cm i.e., h = 42 cm.

∴ Volume of material in the hollow cylinder.

= π (R^{2} –r^{2})h

= [(10)^{2} – (8)^{2}] × 42 cm^{3}

= × 36 × 42 cm3 = 4752 cm^{3}

**Example 15: **The radii of two right circular cylinders are in the ratio 3 : 4 and their heights are in the ratio 6 : 5. Find the ratio between their curved (lateral) surface areas.

**Solution: **If the radii of two cylinders be r_{1} and r_{2},

let r_{1} = 3x and r_{2} = 4x.

Similarly, if the heights of two cylinders be h_{1} and h_{2}, let h_{1} = 6y and h_{2} = 5y.

Ratio between their C.S.A.

= =

= 9 : 10

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