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
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Q1
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

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

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
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Q2
∴ 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.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Q3
Solution:
(i) We have given,
Height of cylinder = 77 cm
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Q3.1
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

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

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
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Q4
= 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.

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

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
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Q7
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

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

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
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Q8
∴ 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
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Q9
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

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

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.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Q10
Solution:
We have given that
Diameter of lampshade = 20 cm
Radius of lampshade = 10 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.

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

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
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Q11
∴ 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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