## EXERCISE 13.2

#### Page No 216:

#### Question 1:

The curved surface area of a right circular cylinder of height 14 cm is 88 cm^{2}. Find the diameter of the base of the cylinder. Assume π =

#### Answer:

Height (*h*) of cylinder = 14 cm

Let the diameter of the cylinder be *d*.

Curved surface area of cylinder = 88 cm^{2}

⇒ 2π*rh* = 88 cm^{2} (*r *is the radius of the base of the cylinder)

⇒ π*dh* = 88 cm^{2} (*d* = 2*r*)

⇒

⇒ *d *= 2 cm

Therefore, the diameter of the base of the cylinder is 2 cm.

#### Question 2:

It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?

#### Answer:

Height (*h*) of cylindrical tank = 1 m

Base radius (*r*) of cylindrical tank

Therefore, it will require 7.48 m^{2 }area of sheet.

#### Question 3:

A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm.

(i) Inner curved surface area,

(ii) Outer curved surface area,

(iii) Total surface area.

#### Answer:

Inner radius of cylindrical pipe

Outer radius of cylindrical pipe

Height (*h*) of cylindrical pipe = Length of cylindrical pipe = 77 cm

(i) CSA of inner surface of pipe

(ii) CSA of outer surface of pipe

(iii) Total surface area of pipe = CSA of inner surface + CSA of outer surface + Area of both circular ends of pipe

Therefore, the total surface area of the cylindrical pipe is 2038.08 cm^{2}.

#### Page No 217:

#### Question 4:

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m^{2}?

#### Answer:

It can be observed that a roller is cylindrical.

Height (*h*) of cylindrical roller = Length of roller = 120 cm

Radius (*r*) of the circular end of roller =

CSA of roller = 2π*rh*

Area of field = 500 × CSA of roller

= (500 × 31680) cm^{2}

= 15840000 cm^{2}

= 1584 m^{2}

#### Question 5:

A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs.12.50 per m^{2}.

#### Answer:

Height (*h*) cylindrical pillar = 3.5 m

Radius (*r*) of the circular end of pillar =

= 0.25 m

CSA of pillar = 2π*rh*

Cost of painting 1 m^{2 }area = Rs 12.50

Cost of painting 5.5 m^{2} area = Rs (5.5 × 12.50)

= Rs 68.75

Therefore, the cost of painting the CSA of the pillar is Rs 68.75.

#### Question 6:

Curved surface area of a right circular cylinder is 4.4 m^{2}. If the radius of the base of the cylinder is 0.7 m, find its height.

#### Answer:

Let the height of the circular cylinder be *h*.

Radius (*r*) of the base of cylinder = 0.7 m

CSA of cylinder = 4.4 m^{2}

2π*rh* = 4.4 m^{2}

*h* = 1 m

Therefore, the height of the cylinder is 1 m

#### Question 7:

The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find

(i) Its inner curved surface area,

(ii) The cost of plastering this curved surface at the rate of Rs 40 per m^{2}.

#### Answer:

Inner radius (*r*) of circular well

Depth (*h*) of circular well = 10 m

Inner curved surface area = 2π*rh*

= (44 × 0.25 × 10) m^{2}

= 110 m^{2}

Therefore, the inner curved surface area of the circular well is 110 m^{2}.

Cost of plastering 1 m^{2} area = Rs 40

Cost of plastering 110 m^{2} area = Rs (110 × 40)

= Rs 4400

Therefore, the cost of plastering the CSA of this well is Rs 4400.

#### Question 8:

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.

#### Answer:

Height (*h*) of cylindrical pipe = Length of cylindrical pipe = 28 m

Radius (*r*) of circular end of pipe = = 2.5 cm = 0.025 m

CSA of cylindrical pipe = 2π*rh*

= 4.4 m^{2}

The area of the radiating surface of the system is 4.4 m^{2}

#### Question 9:

Find

(i) The lateral or curved surface area of a closed cylindrical petrol 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 tank.

#### Answer:

Height (*h*) of cylindrical tank = 4.5 m

Radius (*r*) of the circular end of cylindrical tank =

(i) Lateral or curved surface area of tank = 2π*rh*

= (44 × 0.3 × 4.5) m^{2}

= 59.4 m^{2}

Therefore, CSA of tank is 59.4 m^{2}.

(ii) Total surface area of tank = 2π*r *(*r + h*)

= (44 × 0.3 × 6.6) m^{2}

= 87.12 m^{2}

Let A m^{2} steel sheet be actually used in making the tank.

Therefore, 95.04 m^{2} steel was used in actual while making such a tank.

#### Question 10:

In the given figure, you see 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.

#### Answer:

Height (*h*) of the frame of lampshade = (2.5 + 30 + 2.5) cm = 35 cm

Radius (*r*) of the circular end of the frame of lampshade =

Cloth required for covering the lampshade** = **2π*rh*

*= *2200 cm^{2}

Hence, for covering the lampshade, 2200 cm^{2} cloth will be required.

#### Question 11:

The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

#### Answer:

Radius (*r*) of the circular end of cylindrical penholder = 3 cm

Height (*h*) of penholder = 10.5 cm

Surface area of 1 penholder = CSA of penholder + Area of base of penholder

= 2π*rh* + π*r*^{2}

Area of cardboard sheet used by 1 competitor

Area of cardboard sheet used by 35 competitors

*= *= 7920 cm^{2}

Therefore, 7920 cm^{2} cardboard sheet will be bought.