**Telangana**** SCERT Solution Class IX (9) Math Chapter 10 Surface Areas and Volumes Exercise 10.2**

**(1) A closed cylindrical tank of height 1.4 m. and radius of the base is 56 cm. is made up of a thick metal sheet. How much metal sheet is required (Express in square meters)**

Given,

Height of cylindrical tank = 1.4 m

Radius of base = 56 cm

= 0.56 m

∴ Total surface area of cylindrical tank = 2πr (r + h)

= 2 x 22/7 x 0.56 (0.56 + 1.4)

= 2 x 22 x 0.08 x 1.96

= 44 x 0.1568

= 6.89 m^{2}

∴ Metal sheet required to make the tank = 6.89 m^{2}

**(2) The volume of a cylinder is 308 cm.3 . Its height is 8 cm. Find its lateral surface area and total surface area.**

Solution:

∴ Lateral surface area of the cylinder = 2πrh

= 2 x 22/7 x 7/2 x 8

= 176 cm^{2}

Total surface area of the cylinder = 2πr (r + h)

= 2 x 22/7 x 7/2 (3.5 + 8)

= 22 x 11.5

= 253 cm^{2}

** (3) A metal cuboid of dimension 22 cm. × 15 cm. × 7.5 cm. was melted and cast into a cylinder of height 14 cm. What is its radius?**

Solution: Given,

Dimensions of cuboid

Length = 22 cm

Breadth = 15 cm

Height = 7.5 cm

Dimensions of cylinder,

Height = 14 cm

Let, r be the radius of cylinder’s base

Volume of **cuboid** = l x b x

= 22 x 15 x 7.5

= 2475 cm^{2}

∴ Volume of **cuboid** = Volume of cylinder when casted

**(4) An overhead water tanker is in the shape of a cylinder has capacity of 61.6 cu.mts. The diameter of the tank is 5.6 m. Find the height of the tank**

Solution: Given,

Volume of cylindrical tank = 61.6 m^{3}

Diameter of its base = 5.6 m

∴ Radius of its base = 5.6/2 = 2.8 m

Let, height of the tank be h

Volume = πr^{2}h

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

**(i) inner curved surface area**

** (ii) outer curved surface area **

**(iii) Total surface area**

**Solution:**

Given,

Height of pipe = 77 cm

h = 77 cm

Inner diameter = 4 cm

Let, inner radius r = 4/2 = 2 cm

Outer diameter = 4.4 cm

Let, outer radius R be = 4.4/2 = 2.2 cm

(i) Inner curved surface area = 2πrh

= 2 x 22/7 x 2 x 77

= 4 x 242

= 968 cm^{2}

(ii) Outer curved surface area = 2πrh

= 2 x 22/7 x 2.2 x 77

= 484 x 2.2

= 1064.8 cm^{2}

(iii) Now, Total surface area of the pipe is the sum of curved surface area of the inner and outer cylinder and the two rings on top mud base of the pipe since the pipe is hollow in the middle.

∴The area of the two rings = 2 (πr^{2} – πr^{2})

= 2π (2.2 x 2.2 – 2 x 2)

= 2 x 22/7 x 0.84

= 44 x 0.12

= 5.28 cm^{2}

∴Total surface area of the pipe = 968 + 1064.8 + 5.28

= 2038.08 cm^{2}

**(6) A cylindrical piller has a diameter of 56 cm and is of 35 m high. There are 16 pillars around the building. Find the cost of painting the curved surface area of all the pillars at the rate of `5.50 per 1 m**^{2}

**Solution:**

Given,

No of pillars = 16

Diameter of base = 56 cm

Radius of base = 56/2 = 28 cm = 28/100 m = 0.28 m

Height of the pillars = 35 m

Curved surface area of each pillar = 2πrh

= 2 x 22/7 x 0.28 x 35

= 44 x 1.4

= 61.6 m^{2}

Curved surface of 16 pillars = 61.6 x 16

= 985.6 m^{2}

Now, cost of pain pole m^{2} = ₹5.5

Total cost of painting the curved surface area of the pillars = 5.5 x 985.6

= ₹ 5420.8

**(7) The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to roll once over the play ground to level. Find the area of the play ground in m ^{2} .**

**Solution:**

Given,

Diameter of roller = 84 cm

∴ Radius of roller = 42 cm

Length of roller = 120 cm

Now, curved surface area of roller

= 2πrh

= 2 x 22/7x 42 x 120

= 31680 cm^{2}

= 31680/10000 m^{2}

= 3.168 m^{2}

Total no of revolutions it takes to cover the play grow = 500

∴ Total surface area covered by the roller

= 3.168 m^{2}

= 1584 m^{2}

∴ The area of the playground = 1584 m^{2}

** (8) 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**

**Solution:** Given,

Inner diameters of the well = 3.5 cm

Inner radius of the well = 3.5/2 m

Depth of the well = 10 m

(i) ∴ Inner curved surface area of the well = 2πrh

= 2 x 22/7 x 3.5/2 x 10

= 22 x 5

= 110 m^{2}

(ii) ∴ Cost of painting the curved surface at ₹40 per m^{2}

= 110 x 40

= Rs. 4400

**(9) Find (i) The total surface area of a closed cylindrical petrol storage tank whose diameter 4.2 m. and height 4.5 m.**

(i) Given,

Diameter of cylindrical tank = 4.2 m

∴ Radius of cylindrical tank = 4.2/2 = 2.1 m

Height of the cylindrical tank = 4.5 m

∴ Total surface area of the cylindrical tank

= 2πr (r + h)

= 2 x 22/7 x 2.1 x (2.1 + 4.5)

= 44 x 0.3 x 6.6

= 87.12 m^{2}

**(10) A one side open cylinderical drum has inner radius 28 cm. and height 2.1 m. How much water you can store in the drum. Express in litres. (1 litre = 1000 cc.)**

**Solution:** Given,

Radius of drum = 28 cm

Height of drum = 2.1 m

= 210 cm

∴Volume of drum = πr^{2}h

= 22/7 x 28^{2} x 210

= 22 x 784 x 30

= 517440 cm^{2}

Now,

1 litre = 1000 cm^{2}

∴Total amount of water that can be stored in the drum = 517440/1000 litres

= 517.44 litres

**(11) The curved surface area of the cylinder is 1760 cm ^{2} and its volume is 12320 cm^{3} . Find its height.**

**Solution:**

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In maths telangana syllabus in text book in 10th chapter exercise 10.2 in 9(ii) in text answers there were given wrong answer plz correct it otherwise it will be difficult for upcoming children to find it and it is so complicated

Sir can You please mention us the wrong point please