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

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

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

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 13 Surface Areas and Volumes Ex 13.2.


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


Ex 13.2 Class 9 Maths 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.
Let the radius of the base be r cm.
Given, height of the cylinder (h) = 14 cm


Ex 13.2 Class 9 Maths 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 metres of the sheet are required for the same?

Solution:
Here, height of the cylindrical tank (h) = 1 m
Diameter of the base = 140 cm = 1.40 m
Radius of the base (r) = 1.40/2m = 0.70 m
Total surface area of the cylinder = 2πr(h + r)
= 2 × 22/7 × 0.70(1 + 0.70) m2
= 2 × 22 × 0.10 × 1.70 m2
= 7.48 m2
Hence, the area of required sheet is 7.48 m2.

 

Ex 13.2 Class 9 Maths 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 figure). Find its:
(i) inner curved surface area.
(ii) outer curved surface area.
(iii) total surface area.


Solution:
Length of the metal pipe = 77 cm
It is in the form of a cylinder.
Height of the cylinder (h) = 77 cm
Inner diameter = 4 cm
Inner radius (r) = 4/2 cm = 2 cm
Outer diameter = 4.4 cm
Outer radius (R) = 4.4/2 cm = 2.2 cm

(i) Inner curved surface area = 2πrh
= 2 × 22/7 × 2 × 77 cm2
= 2 × 22 × 2 × 11 cm2 = 968 cm2

(ii) Outer curved surface area = 2πRh

(iii) Total surface area = [Inner curved surface area] + [Outer curved surface area] + [Area of two circular ends]
= [2πrh] + [2πRh] + 2[π(R2 – r2)]


Ex 13.2 Class 9 Maths 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:
The roller is in the form of a cylinder of diameter = 84 cm
Radius of the roller (r) = 84/2 cm = 42 cm
Length of the roller (h) = 120 cm
Curved surface area of the roller = 2πrh
= 2 × 22/7 × 42 × 120 cm2
= 2 × 22 × 6 × 120 cm2 = 31680 cm2
Now, area of the playground levelled in one revolution of the roller = 31680 cm2
31680/10000 m2
Area of the playground levelled in 500 revolutions = 500 × 31680/10000 m2 

                                                                                             = 1584 m2

 

Ex 13.2 Class 9 Maths Question 5.
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2.

Solution:
Diameter of the pillar = 50 cm
Radius (r) = 50/2 cm = 25 cm = 25/100 m = ¼ m
and height (h) = 3.5 m
Curved surface area of a pillar = 2πrh

Curved surface area to be painted = 11/2 m2
Cost of painting of 1 m2 pillar = ₹12.50
Cost of painting of 11/2 m2 pillar
= ₹ (11/2 × 12.50 )
= ₹ 68.75

 

Ex 13.2 Class 9 Maths Question 6.
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:
Radius of base (r) = 0.7 m
Let the height of the cylinder be h m.
Curved surface area of a cylinder = 2πrh
= 2 × 22/7 × 7/10 × h m2
But the curved surface area is 4.4 m2. [Given]


Ex 13.2 Class 9 Maths Question 7.

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 ₹40 per m2.

Solution:
Inner diameter of the well = 3.5 m
Inner radius of the well (r) = 3.5/2
Depth of the well (h) = 10 m
(i) Inner curved surface area = 2πrh

(ii) Cost of plastering per m2 = ₹40
Total cost of plastering the area 110 m2
= ₹ (110 × 40) = ₹ 4400

 

Ex 13.2 Class 9 Maths Question 8.
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Solution:
Length of the cylindrical pipe (h) = 28 m
Diameter of the pipe = 5 cm
Radius of the pipe (r) = 5/2 cm = 5/200 m
Curved surface area of a cylinder = 2πrh

Thus, the total radiating surface is 4.4 m2.

 

Ex 13.2 Class 9 Maths Question 9.
Find:
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank.

Solution:
The storage tank is in the form of a cylinder.
Diameter of the tank = 4.2 m
Radius (r) = 4.2/2 = 2.1 m
Height (h) = 4.5 m

Now,
(i) Lateral (or curved) surface area of the tank = 2πrh
= 2 × 227 × 2.1 × 4.5 m2
= 2 × 22 × 0.3 × 4.5 m2 = 59.4 m2

(ii) Total surface area of the tank = 2πr(r + h)
= 2 × 22/7 × 2.1(2.1 + 4.5) m2
= 44 × 0.3 × 6.6 m2 = 87.13 m2
Let actual area of the steel used be x m2
Area of steel that was wasted = 1/12 × x m = x/12 m2
Area of steel used = x – x/12 m2

Thus, the required area of the steel that was actually used is 95.04 m2.

 

Ex 13.2 Class 9 Maths Question 10.
In the given figure, 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 height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.


Solution:
The lampshade is in the shape of a cylinder, where radius (r) = 20/2 cm = 10 cm
and height = 30 cm.
A margin of 2.5 cm is to be added to the top and bottom of the frame.
Total height of the cylinder, (h) = 30 cm + 2.5 cm + 2.5 cm = 35 cm
Now, curved surface area = 2πrh
= 2 × 22/7 × 10 × 35 cm2
= 2200 cm2
Thus, the required area of the cloth = 2200 cm2

 

Ex 13.2 Class 9 Maths 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:
Here, the penholders are in the shape of a cylinder.
Radius of a penholder (r) = 3 cm
Height of a penholder (h) = 10.5 cm
Since, a penholder must be open from the top.
Surface area of a penholder = [Lateral surface area] + [Base area]
= [2πrh] + [πr2]

Surface area of 35 penholders = 35 × 1584/7 cm2 = 7920 cm2
Thus, 7920 cm2 of cardboard was required to be bought.


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