Telangana SCERT Class 9 Math Solution Chapter 10 Surface Areas and Volumes Exercise 10.2

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 m2

∴ Metal sheet required to make the tank = 6.89 m2

(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 cm2

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

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

= 22 x 11.5

= 253 cm2

 (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 cm2

∴ 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 m3

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 = πr2h

(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 cm2

(ii) Outer curved surface area = 2πrh

= 2 x 22/7 x 2.2 x 77

= 484 x 2.2

= 1064.8 cm2

(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 (πr2 – πr2)

= 2π (2.2 x 2.2 – 2 x 2)

= 2 x 22/7 x 0.84

= 44 x 0.12

= 5.28 cm2

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

= 2038.08 cm2

(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 m2

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 m2

Curved surface of 16 pillars = 61.6 x 16

= 985.6 m2

Now, cost of pain pole m2 = ₹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 m2 .

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 cm2

= 31680/10000 m2

= 3.168 m2

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

∴ Total surface area covered by the roller

= 3.168 m2

= 1584 m2

∴ The area of the playground = 1584 m2

 (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 m2

(ii) ∴ Cost of painting the curved surface at ₹40 per m2

= 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 m2

(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 = πr2h

= 22/7 x 282 x 210

= 22 x 784 x 30

= 517440 cm2

Now,

1 litre = 1000 cm2

∴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 cm2 and its volume is 12320 cm3 . Find its height.

Solution:


Updated: October 2, 2021 — 4:27 pm

4 Comments

Add a Comment
  1. Nice explanation,can you give your contact number.

    1. Tell Please

  2. 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

    1. Sir can You please mention us the wrong point please

Leave a Reply

Your email address will not be published.