# AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes Ex 10.2

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 m2
= 6.90 m2

Question 2.
The volume of a cylinder is 308 cm3 . Its height is 8 cm. Find its lateral surface area and total surface area.
Solution:
Volume of the cylinder V = πr2h
= 308 cm3
Height of the cylinder h = 8 cm
∴ 308 = $$\frac { 22 }{ 7 }$$ . r2 × 8
r2 = 308 × $$\frac { 7 }{ 22 }$$ x $$\frac { 1 }{ 8 }$$
r2 = 12.25
∴ r = $$\sqrt{12.25}$$ = 3.5cm
L.S.A. = 2πrh
= 2 × $$\frac { 22 }{ 7 }$$ × 3.5 × 8 = 176cm2
T.S.A. = 2πr (r + h)
2 × $$\frac { 22 }{ 7 }$$ × 3.5 (3.5 + 8)
= 2 × 22 × 0.5 × 11.5 = 253 cm2

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
∴ Volume of cuboid = Volume of cylinder
lbh = 2πr2h
⇒ 22 × 15 × 7.5 = $$\frac { 22 }{ 7 }$$ × r2 × 14
⇒ r2 = $$\frac{22 \times 15 \times 7.5 \times 7}{14 \times 22}$$
⇒ r2 = 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.
cagp)
Solution:
Volume of the cylinder, V = πr2h = 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)
∴ πr2 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 = 968cm2

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 cm2

iii) Total surface area Sol. Total surface area .
= Inner C.S.A + Outer C.S.A
= 968 + 1064.8
= 2032.8 cm2

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 m2.
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 m2
∴ C.S.A. of 16 pillars = 16 × 61.6 = 985.6 m2
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 m2
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 m2

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 m2.
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 m2
ii) Cost of plastering at the rate of
₹ 40 / m2 = 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 = 87.12m2

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 m2
$$\frac{11}{12}$$ x = 87.12m2
∴ x = 87.12 x $$\frac{12}{11}$$ = 95.04m2

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 = πr2h
= $$\frac{22}{7}$$ × 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 cm2 and its volume is 12320 cm3. Find its height.
Solution:
C.S.A of the cylinder = 2πrh = 1760 cm2
Volume of the cylinder = πr2h
= 12320 cm3
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 = 1760cm2
2 × $$\frac{22}{7}$$ × 14h = 1760
h = $$\frac{1760 \times 7}{2 \times 22 \times 14}$$ = 20cm
∴Height of the cylinder = 20cm