# Class 9 NCERT Solutions- 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 cm**^{2}. Find the diameter of the base of the cylinder.

^{2}. Find the diameter of the base of the cylinder.

**Solution**:

Given:

i) Height of cylinder(h)=14 cm

ii)Curved Surface Area of cylinder=88 cm

^{2}Curved Surface Area = 2πrh

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

Hence, r=

= 1 cm

As diameter=2*radius,

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

closedcylindrical 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 m

^{2}

**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

hollowcylindrical structure1) Height of Cylindrical metal pipe (h) =77 cm

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

Hence, inner radius(r)=d/2=2 cm

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

Hence, outer radius(R)=D/2=2.2 cm

i)Inner curved surface area=2πrh =2*(22/7)*2*77=968 cm

^{2}

ii)Outer curved surface area=2πrh =2*(22/7)*2.2*77=1064.8 cm

^{2}

iii)Total Surface Area=Inner curved surface area+Outer curved surface area+area of two basesAs pipe is hollow, area of

twobases is a ring area given by=2(πR^{2}-πr^{2}) = 2π(R^{2}-r^{2})= 2* (22/7)*((2.2)

^{2}-(2)^{2})= 5.28 cm

^{2}Hence, Total surface area=968+1064.8+5.28=2038.08 cm

^{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 m**^{2}.

^{2}.

**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 m

^{2}Revolutions done by roller=500

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

= 3.168*500

=1584 m

^{2}

**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 m**^{2}.

^{2}.

**Solution**:

Diameter of cylindrical pillar=50 cm

Hence radius (r)=25cm=0.25 m

Height of cylindrical pillar (h)=3.5 m

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

=5.50 m

^{2}Rate of painting=₹12.50 per m

^{2}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 m

^{2}Curved Surface Area =2πrh

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

Hence, h=

= 1 m

**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 m**^{2}.

^{2}.

**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 m

^{2}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 cm

^{2}= 4.4 m

^{2}

**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**:

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 m

^{2}

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 m

^{2 }(1)Let actually used steel =x m

^{2}. (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 m

^{2}Hence, 95.04 m

^{2}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

Cloth required for covering the lampshade= Curved surface area of lampshade

= 2πrh

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

= 2200 cm

^{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**:

Penholder is in shape of a cylinder with

a base. That is with curved surface with one base is to be created using cardboard.Radius of penholder=3 cm

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+πr

^{2}= 2*(22/7)*3*10.5+(22/7)*(3)

^{2}= 198+28.28

= 226.28 cm

^{2}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 cm

^{2}

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