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# Class 9 NCERT Solutions- Chapter 13 Surface Areas And Volumes – Exercise 13.2

• Last Updated : 28 Dec, 2020

### 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:

Given:

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i) Height of cylinder(h)=14 cm

ii)Curved Surface Area of cylinder=88 cm2

Curved Surface Area = 2πrh

88 = 2*(22/7)*r*14

Hence, r= = 1 cm

The diameter of the base of cylinder =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 are required for the same?

Solution:

Given: i) Height of cylindrical tank (h)=1 m

ii) Base diameter of cylinder (d)=140 cm

iii) Radius of cylinder (r)=d/2=140/2=70cm=0.7 m

As metal sheet requirement for closed cylindrical tank is asked,

Area of metal sheet required =Total surface area of the cylindrical tank

= 2πr(r+h)

= 2*(22/7)*0.7*(0.7+1)

= 7.48 m2

### 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. Find its (i) inner curved surface area, (ii) outer curved surface area,  (iii) total surface area.

Solution:

We know that pipe has a hollow cylindrical structure

1) Height of Cylindrical metal pipe (h) =77 cm

2) The inner diameter of metal pipe (d)=4cm,

3)The outer diameter of metal pipe (D)=4.4cm,

i) Inner curved surface area=2πrh =2*(22/7)*2*77

=968 cm2

ii) Outer curved surface area=2πrh =2*(22/7)*2.2*77

=1064.8 cm2

iii) Total Surface Area=Inner curved surface area+Outer curved surface area+area of two bases

As pipe is hollow, area of two bases is a ring area given by=2(πR2-πr2) = 2π(R2-r2)

= 2* (22/7)*((2.2)2-(2)2)

= 5.28 cm2

Hence, Total surface area=968+1064.8+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

Roller is cylindrical in shape.

Diameter of a roller(d)=84 cm,

Hence, base radius of roller (r)=d/2=42cm =0.42 m

Height of cylinder(h)=Length of roller=120cm=1.2 m

Curved surface area of cylinder=2πrh =2*(22/7)*0.42*1.2 =3.168 m2

Revolutions done by roller=500

Hence, Area of playground= Curved surface area of roller*revolutions done by roller

= 3.168*500

=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 of the pillar at the rate of ₹12.50 per m2.

Solution

Diameter of cylindrical pillar=50 cm

Height of cylindrical pillar (h)=3.5 m

Curved surface area of cylindrical pillar=2πrh=2*(22/7)*0.25*3.5

=5.50 m2

Rate of painting=₹12.50 per m2

Hence, cost of painting the curved surface of the pillar=5.50*12.50=₹68.75

### 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

Given: i) Radius of the base of cylinder(h)=0.7 m

ii)Curved Surface Area of cylinder=4.4 m2

Curved Surface Area =2πrh

4.4 =2*(22/7)*0.7*h

Hence, h= = 1 m

### (ii) the cost of plastering this curved surface at the rate of ₹40 per m2.

Solution

Well is cylindrical in nature.

Its inner diameter=3.5m, hence inner radius (r)=3.5/2=1.75 m

Its height (h)=10 m

i) Its inner curved surface area (A)=2πrh = 2*(22/7)*1.75*10

= 110 m2

ii) The cost of Plastering this inner curved surface = rate of plastering*A

= 40*110

= ₹4400

### 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 cylindrical pipe (h)= 28m = 2800 cm

Diameter of cylindrical pipe=5 cm

Hence, radius of cylindrical pipe=5/2=2.5 cm

Cylindrical pipe radiates from curved surface only. Hence, we have to calculating radiating surface we have to measure curved surface area of that pipe.

Total radiating surface in the system = Curved surface area of pipe

= 2πrh

= 2*(22/7)*2.5*2800

= 44000 cm2

= 4.4 m2

### (ii) how much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank.

Solution

Diameter of cylindrical tank=4.2 m

Hence, radius of cylindrical tank (r)=4.2/2=2.1 m

Height of cylindrical tank (h)=4.5 m

i) Curved surface area of tank=2πrh = 2*(22/7)*2.1*4.5

= 59.4 m2

ii) As it is closed cylindrical tank,

Area of steel required to make = Total surface area of cylindrical tank

= 2πr(r+h)

= 2*(22/7)*2.1*(2.1+4.5)

= 87.12 m2                                       (1)

Let actually used steel =x m2.  (1/2) of actually used steel was wasted.

Hence, tank is made up of 1-(1/12) = 11/12 part of steel.                  (2)

Hence, from (1) and (2), we get,

(11/12)x=87.12

i.e. x= = 95.04 m2

Hence, 95.04 m2 steel was actually used to make tank.

### Question 10. A cylindrical frame of a lampshade 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

Base diameter of frame=20 cm

Hence, base radius of frame=20/2=10 cm

Height of frame (including margins provided for folding frame over top and bottom of frame) = 30+2.5+2.5=35 cm

= 2πrh

= 2*(22/7)*10*35

= 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

Penholder is in shape of a cylinder with a base. That is with curved surface with one base is to be created using cardboard.

Height of penholder=10.5 cm

Area of cardboard required to make one penholder (A)

= Curved surface area of penholder + area of one base

= 2πrh+πr2

= 2*(22/7)*3*10.5+(22/7)*(3)2

= 198+28.28

= 226.28 cm2

Total area of Cardboard required to be bought for the competition

= Number of competitors* cardboard required for one penholder

= 35*226.28

= 7919.99 ≈ 7920 cm2

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