AP State Syllabus AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.2 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 10th Lesson Surface Areas and Volumes Exercise 10.2

Question 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 metres).

Solution:

Radius of the tank r’ = 56 cm

= \(\frac { 56 }{ 100 }\) m = 0.56m

Height of the tank h = 1.4 m

T.S.A. of a cylinder = 2πr (r + h)

∴ Area of the metal sheet required = 2πr (r + h)

A = 2 × \(\frac { 22 }{ 7 }\) × 0.56 × (0.56 + 1.4)

= 2 × 22 × 0.08 × 1.96

= 6.8992 m^{2}

= 6.90 m^{2}

Question 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:

Volume of the cylinder V = πr^{2}h

= 308 cm^{3}

Height of the cylinder h = 8 cm

∴ 308 = \(\frac { 22 }{ 7 }\) . r^{2} × 8

r^{2} = 308 × \(\frac { 7 }{ 22 }\) x \(\frac { 1 }{ 8 }\)

r^{2} = 12.25

∴ r = \(\sqrt{12.25}\) = 3.5cm

L.S.A. = 2πrh

= 2 × \(\frac { 22 }{ 7 }\) × 3.5 × 8 = 176cm^{2}

T.S.A. = 2πr (r + h)

2 × \(\frac { 22 }{ 7 }\) × 3.5 (3.5 + 8)

= 2 × 22 × 0.5 × 11.5 = 253 cm^{2}

Question 3.

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

Solution:

Dimensions of the metal cuboid

= 22 cm × 15 cm × 7.5 cm

Height of the cylinder, h = 14 cm

Cuboid made as cylinder

∴ Volume of cuboid = Volume of cylinder

lbh = 2πr^{2}h

⇒ 22 × 15 × 7.5 = \(\frac { 22 }{ 7 }\) × r^{2} × 14

⇒ r^{2} = \(\frac{22 \times 15 \times 7.5 \times 7}{14 \times 22}\)

⇒ r^{2} = 7.5 × 7.5

r = 7.5 cm

Question 4.

An overhead water tanker is in the shape of a cylinder has capacity of 616 litres. The diameter of the tank is 5.6 m. Find the height of the tank.

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Solution:

Volume of the cylinder, V = πr^{2}h = 616

Diameter of the tank = 5.6 m

Thus its radius, r = \(\frac{d}{2}=\frac{5.6}{2}\) = 2.8 m

Height = h (say)

∴ πr^{2} h = 616

\(\frac{22}{7}\) × 2.8 × 2.8 × h = 616

h = \(\frac{616 \times 7}{22 \times 2.8 \times 2.8}\) = 25

∴ Height = 25 m

Question 5.

A metal pipe is 77 cm long. The inner diametre 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

i) Inner curved surface area

Solution:

Height of the pipe = 77 cm

Inner diameter = 4 cm

Inner radius = \(\frac{d}{2}=\frac{4}{2}\) = 2 cm

∴ Inner C.S.A. = 2πrh

= 2 × \(\frac{22}{7}\) × 2 × 77

= 88 × 11 = 968cm^{2}

ii) Outer curved surface area

Solution:

Outer diameter = 4.4 cm

∴ Outer radius, r = \(\frac{d}{2}=\frac{4.4}{2}\) = 2.2 cm

Height of the pipe, h = 77 cm

∴ Outer C.S.A. = 2πrh

= 2 × \(\frac{22}{7}\) × 2.2 × 77

= 96.8 × 11

= 1064.8 cm^{2}

iii) Total surface area Sol. Total surface area .

= Inner C.S.A + Outer C.S.A

= 968 + 1064.8

= 2032.8 cm^{2}

Question 6.

A cylindrical pillar 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:

Diametre of the cylindrical pillar = 56 cm

Thus its radius, r = \(\frac{d}{2}\)

= \(\frac{56}{2}\) = 28cm = \(\frac{28}{100}\)m = 0.28m

Height of the pillar, h = 35 m

Total number of pillars =16

Cost of painting = ₹ 5.50 per sq. m.

C.S.A. of each pillar = 2πrh

= 2 × \(\frac{22}{7}\) × 0.28 × 35

= 2 × 22 × 0.04 × 35 = 61.6 m^{2}

∴ C.S.A. of 16 pillars = 16 × 61.6 = 985.6 m^{2}

Cost of painting 16 pillars at the rate of ₹ 5.5 per sq.m. = 985.6 × 5.5

= ₹ 5420.8

Question 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}

Soi. Diameter of the roller = 84 cm

Thus radius = \(\frac{84}{2}\) = 42 cm

= \(\frac{42}{100}\)m = 0.42m

Length of the roller =120 cm

= \(\frac{120}{100}\) = 1.2m

It takes 500 complete revolutions to roll over the play ground.

Thus 500 × L.S.A. of the roller

= Area of the play ground

∴ Area of the play ground = 500 × 2πrh

= 500 × 2 × \(\frac{22}{7}\) × 0.42 × 1.2 = 1584 m^{2}

Question 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^{2}.

Solution:

Inner diameter of the circular well, d = 3.5 m

Thus its radius, r = \(\frac{d}{2}=\frac{3.5}{2}\) = 1. 75 m

Depth of the well (height) = 10 m

i) Inner C.S.A. = 2πrh

= 2 × \(\frac{22}{7}\) × 1.75 × 10

= 110 m^{2}

ii) Cost of plastering at the rate of

₹ 40 / m^{2} = 110 × 40 = ₹ 4400

Question 9.

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

Solution:

Diameter of the cylindrical tank ‘d’ = 4.2m

Thus its radius, r = \(\frac{\mathrm{d}}{2}=\frac{4.2}{2}\) = 2.1 m

Height of the tank, h = 4.5 m

T.S.A. of the tank = 2πr (r + h)

= 2 × \(\frac{22}{7}\) × 2.1 (2.1 + 4.5)

= 2 × 22 × 0.3 × 6.6 = 87.12m^{2}

ii) How much steel sheet was actually used, if \(\frac{1}{12}\) of the steel was wasted in making the tank ?

Solution:

\(\frac{1}{12}\) of the sheet was wasted.

=> 1 – \(\frac{1}{12}\) = \(\frac{11}{12}\) of the sheet was used

in making the tank.

Let the metal sheet originally brought was = x m^{2}

\(\frac{11}{12}\) x = 87.12m^{2}

∴ x = 87.12 x \(\frac{12}{11}\) = 95.04m^{2}

Question 10.

A one side open cylindrical 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 c.c)

Solution:

Inner radius of the cylindrical drum ‘r’ = 28 cm

. Its height, h = 2.1 m = 2.1 × 100 = 210 cm

Volume of the drum = πr^{2}h

= \(\frac{22}{7}\) × 28 × 28 × 210

= 22 × 4 × 28 × 210

= 517440 cc

= \(\frac{517440}{1000}\)

= 517.44 lit.

Question 11.

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

Solution:

C.S.A of the cylinder = 2πrh = 1760 cm^{2}

Volume of the cylinder = πr^{2}h

= 12320 cm^{3}

Height = h (say)

\(\frac{\text { Volume }}{\text { C.S.A. }}=\frac{\pi r^{2} h}{2 \pi r h}=\frac{12320}{1760}\)

⇒ \(\frac{r}{2}\) = 7

∴ r = 7 × 2 = 14cm

Now 2πrh = 1760cm^{2}

2 × \(\frac{22}{7}\) × 14h = 1760

h = \(\frac{1760 \times 7}{2 \times 22 \times 14}\) = 20cm

∴Height of the cylinder = 20cm