**Exercise 13.2 Page No: 216**

**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 =22/7 )**

**Solution:**

Height of cylinder, h = 14cm

Let the diameter of the cylinder be d

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

We know that, formula to find Curved surface area of cylinder is 2πrh.

So 2πrh =88 cm^{2} (r is the radius of the base of the cylinder)

2×(22/7)×r×14 = 88 cm^{2}

2r = 2 cm

d =2 cm

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

**2. It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? Assume π = 22/7**

**Solution:**

Let h be the height and r be the radius of a cylindrical tank.

Height of cylindrical tank, h = 1m

Radius = half of diameter = (140/2) cm = 70cm = 0.7m

Area of sheet required = Total surface are of tank = 2πr(r+h) unit square

= [2×(22/7)×0.7(0.7+1)]

= 7.48

Therefore, 7.48 square meters of the sheet are required.

**3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4cm. (see fig. 13.11). Find its**

**(i) inner curved surface area,**

**(ii) outer curved surface area**

**(iii) total surface area**

**(Assume π=22/7)**

**Solution:**

Let r_{1} and r_{2} Inner and outer radii of cylindrical pipe

r_{1 }= 4/2 cm = 2 cm

r_{2 }= 4.4/2 cm = 2.2 cm

Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm

(i) curved surface area of outer surface of pipe = 2πr_{1}h

= 2×(22/7)×2×77 cm^{2}

= 968 cm^{2}

(ii) curved surface area of outer surface of pipe = 2πr_{2}h

= 2×(22/7)×2.2×77 cm^{2}

= (22×22×2.2) cm^{2}

= 1064.8 cm^{2}

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

= 2r_{1}h+2r_{2}h+(r_{1}^{2}-r_{2}^{2})

= 9668+1064.8+2(2.2^{2}-2^{2})

= 2031.8+5.28

= 2038.08 cm^{2}

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

**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}? (Assume π = 22/7)**

**Solution:**

A roller is shaped like a cylinder.

Let h be the height of the roller and r be the radius.

h = Length of roller = 120 cm

Radius of the circular end of roller = r = (84/2) cm = 42 cm

Now, CSA of roller = 2πrh

= 2×(22/7)×42×120

= 31680 cm^{2}

Area of field = 500×CSA of roller

= (500×31680) cm^{2}

= 15840000 cm^{2}

= 1584 m^{2}.

Therefore, area of playground is 1584 m^{2}. Answer!

**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}.**

**(Assume π = 22/7)**

**Solution:**

Let h be the height of a cylindrical pillar and r be the radius.

Given:

Height cylindrical pillar = h = 3.5 m

Radius of the circular end of pillar = r = diameter/2 = 50/2 = 25cm = 0.25m

CSA of pillar = 2πrh

= 2×(22/7)×0.25×3.5

= 5.5 m^{2}

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 curved surface of the pillar at the rate of Rs. 12.50 per m^{2} is Rs 68.75.

**6. Curved surface area of a right circular cylinder is 4.4 m ^{2}. If the radius of the base of the base of the cylinder is 0.7 m, find its height. (Assume π = 22/7)**

**Solution:**

Let h be the height of the circular cylinder and r be the radius.

Radius of the base of cylinder, r = 0.7m

CSA of cylinder = 2πrh

CSA of cylinder = 4.4m^{2}

Equating both the equations, we have

2×(22/7)×0.7×h = 4.4

Or h = 1

Therefore, the height of the cylinder is 1 m.

**7. The inner diameter of a circular well is 3.5m. It is 10m 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}.**

**(Assume π = 22/7)**

**Solution:**

Inner radius of circular well, r = 3.5/2m = 1.75m

Depth of circular well, say h = 10m

(i) Inner curved surface area = 2πrh

= (2×(22/7 )×1.75×10)

= 110

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

(ii)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 curved surface of the well is Rs. 4400.

**8. In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find**

**the total radiating surface in the system. (Assume π = 22/7)**

**Solution:**

Height of cylindrical pipe = Length of cylindrical pipe = 28m

Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m

Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of the cylinder

= 2×(22/7)×0.025×28 m^{2}

= 4.4m^{2}

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

**9. Find**

**(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in**

**diameter and 4.5m high.**

**(ii) How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank. (Assume π = 22/7)**

**Solution:**

Height of cylindrical tank, h = 4.5m

Radius of the circular end , r = (4.2/2)m = 2.1m

(i) the lateral or curved surface area of cylindrical tank is 2πrh

= 2×(22/7)×2.1×4.5 m^{2}

= (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 = 2πr(r+h)

= 2×(22/7)×(2.1+4.5)

= 44×0.3×6.6

= 87.12 m^{2}

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

S(1 -1/12) = 87.12 m^{2}

This implies, S = 95.04 m^{2}

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

**10. In fig. 13.12, 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. (Assume π = 22/7)**

**Solution:**

Say h = height of the frame of lampshade, looks like cylindrical shape

r = radius

Total height is h = (2.5+30+2.5) cm = 35cm and

r = (20/2) cm = 10cm

Use curved surface area formula to find the cloth required for covering the lampshade which is 2πrh

= (2×(22/7)×10×35) cm^{2}

= 2200 cm^{2}

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

**11. The students of 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? (Assume π =22/7)**

**Solution:**

Radius of the circular end of cylindrical penholder, r = 3cm

Height of penholder, h = 10.5cm

Surface area of a penholder = CSA of pen holder + Area of base of penholder

= 2πrh+πr^{2}

= 2×(22/7)×3×10.5+(22/7)×3^{2}= 1584/7

Therefore, Area of cardboard sheet used by one competitor is 1584/7 cm^{2}

So, Area of cardboard sheet used by 35 competitors = 35×1584/7 = 7920 cm^{2}

Therefore, 7920 cm^{2} cardboard sheet will be needed for the competition.