# NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2

Question 1.
The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder.
Solution:
We know that

Curved surface area of cylinder = 2πrh
⇒ 88 = 2 × $$\frac{22}{7}$$ × 14 × r
⇒ 88 = 88 × r
⇒ r = 1 cm
Diameter of cylinder = 2r = 2 × 1 = 2 cm

Question 2.
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet is required for the same?
Solution:
Height of the cylinder = 1 m
and diameter of cylinder = 140 cm = 1.4 m

∴ radius of cylinder = $$\frac{14}{2}$$ = 0.7 m
∴ Total surface are of cylinder = 2r(h + r)
= 2 × $$\frac{22}{7}$$ × 0.7(1 + 0.7)
= 4.4 × 1.7
= 7.48 m2
Therefore, 7.48 m2 of sheets are required to make the closed circular cylinder.

Question 3.
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 fig. 13.11) find its.
(i) inner curved surface area.
(ii) Outer curved surface area.
(iii) Total surface area.

Solution:
(i) We have given,
Height of cylinder = 77 cm

and inner diameter = 4 cm
inner radius = $$\frac{4}{2}$$ = 2 cm
∴ Inner curved surface area = 2πrh
= 2 × $$\frac{22}{7}$$ × 2 × 77
= 968 cm2.
(ii) We have given
Height of cylinder = 77 cm
and Outer diameter = 4.4 cm
∴ Outer curved surface area of cylinder = 2πrh
= 2 × $$\frac{22}{7}$$ × 2.2 × 77
= 1064.8 cm2.
(iii) Total surface ara of pipe = Inner curved surface area + Outer curved surface area + Area of the two bases.
= 2πrh + 2πRh + 2π(R2 – r2)
= 2 × π × 2 × 77 + 2 × π × 2.2 × 77 + 2π((2.2)2 – (2)2)
= 2 × π × 77(2 + 2.2) + 2 × π × (4.84 – 4)
= 2 × $$\frac{22}{7}$$ × 88 × 4.2 + 2 × $$\frac{22}{7}$$ × 0.84
= 2032.8 + 5.28
= 2038.08 cm2

Question 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 m2.
Solution:
Clearly, the roller is a right circular cylinder of height h = 120 cm = 1.2 m
and diameter of its base = 84 cm = 0.84 m
∴ radius of its base = $$\frac{84}{2}$$ = 0.42 m.
∴ Area covered by roller in one revolution = Curved surface area of the roller = 2πrh

= 2 × $$\frac{22}{7}$$ × 0.42 × 1.2
= 3.168 m2
So, Area covered by roller in 500 revolutions = 3.168 × 500 = 1584 m2.
Hence, Area of playground = 1584 m2.

Question 5.
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface area of the pillar at the rate of Rs. 12.50 per m2.
Solution:
We have given that,
the height of pillar = 3.5m
and the diameter of pillar = 50 cm = 0.5 m
∴ Radious of the pillar = $$\frac{0.5}{2}$$ = 0.25 m
∴ Curved surface area of pillar = 2πrh
= 2 × $$\frac{22}{7}$$ × 0.25 × 3.5
= 5.5 m2
Now, Rate of painting the pillar = Rs. 12.50 per m2.
∴ Total cost of painting = 5.5 × 12.50 = Rs. 68.75.

Question 6.
The curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.
Solution:
We have given that
The curved surface area of cylinder = 4.4 m2
and radius of its base = 0.7 m
Height =?
We know that
Curved surface area of cylinder = 2πrh
⇒ 4.4 = 2 × $$\frac{22}{7}$$ × 0.7 × h (∵ π = $$\frac {22}{7}$$)
⇒ h = $$\frac{4.4 \times 7}{2 \times 22 \times 0.7}$$ = 1 m
∴ Height of cylinder = 1 m.

Question 7.
The inner diameter of the circular well is 3.5 m. It is a 10 m deep find.
(i) It’s inner curved surface area.
(ii) The cost of plastering this curved surface at the rate of Rs. 40 per m2.
Solution:
(i) We have given that
The inner diameter of circular well = 3.5 m

Inner radius circular well = $$\frac{3.5}{2}$$ m
and height of circular well = 10 m.
Now, we know that,
Curved surface area of well = 2πrh
= $$2 \times \frac{22}{7} \times \frac{3.5}{2} \times 10$$
= 110 m2
So, inner curved surface area of well = 110 m2
(ii) Rate of plastering = Rs. 40 per m2
Total cost of plastering its curved surface area = 110 × 40 = Rs. 4400

Question 8.
In a hot water heating system, there is a cylindrical pipe of length 28m and diameter 5 cm. Find the total radiating surface in the system.
Solution:
We have given that
Length of the cylindrical pipe (h) = 28m
and diameter of its base = 5 cm = 0.05 m
Radius of its base = $$\frac{.05}{2}$$ m

∴ Curved surface area of cylindrical pipe = 2πrh
= $$2 \times \frac{22}{7} \times \frac{.05}{2} \times 28$$
= 4.4 m2

Question 9.
Find
(i) The lateral or curved surface area of a cylindrical petrol Storage tank that is 4.2 m in diameter and 1.5 m high.
(ii) How much steel was actually used if $$\frac{1}{12}$$ of the steel actually used was wasted in making the closed tank.
Solution:
(i) We have given that,
Height of cylindrical tank = 4.5 m

Diameter of cylindrical tank = 4.2 m
Radius of cylindrical tank = 2.1 m
So, Curved surface area of tank = 2πrh
= 2 × $$\frac{22}{7}$$ × 4.5 × 21
= 59.4 m2
(ii) Let the actual area of steel used be x m2.
Since $$\frac{1}{12}$$ of the actual steel used was wasted = $$\frac{1}{12}$$ of x = $$\frac{x}{12}$$
∴ Area of steel which has gone into the tank = x – $$\frac{x}{12}$$ = $$\frac{11x}{12}$$
Total surface area of tank = 2πr(r + h)
= 2 × $$\frac{22}{7}$$ × 2.1(2.1 + 4.5)
= 87.12 m2
The actual area of steel used be
$$\frac{11x}{12}$$ = 87.12
⇒ x = $$\frac{87.12 \times 12}{11}$$
⇒ x = 95.04 m2
Hence, total steel was used in tank = 95.04 m2

Question 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 a height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and button of the frame. Find how much cloth is required for covering the lampshade.

Solution:
We have given that
Diameter of lampshade = 20 cm
and height of lampshade = 30 + 2.5 + 2.45 = 35 cm
Therefore, Curved surface area of cylindrical lampshade = 2πrh
= 2 × $$\frac{22}{7}$$ × 10 × 35
= 2200 cm2
Therefore, the cloth is required for covering the lampshade is 2200 cm2.

Question 11.
The students of a 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?
Solution:
Total surface area of one penholder is 2πrh + πr2
= 2 × $$\frac{22}{7}$$ × 3 × 10.5 + $$\frac{22}{7}$$ × 3 × 3
= 198 + 28.28
= 226.28 cm2

∴ Total cardboard is used to making one penholder is 226.28 cm2
Therefore, total cardboard is used to making 35 penholders is = 226.28 × 35 = 7919.8 cm2.

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