# Surface Areas and Volumes : Exercise 13.2 (Mathematics NCERT Class 9th)

**Assume , unless stated otherwise.
Q.1Â Â Â Â The curved surface area of a right circular cylinder ofÂ heightÂ 14 cm is 88 . Find the diameter of theÂ base of the cylinder.
**

**Sol.**Let r be the radius of the base and h = 14 cm be the height of the cylinder. Then,

Curved surface area of cylinder

Therefore Diameter of the base = 2r

**Q.2Â Â Â It is required to make a closed cylindrical tank of height 1m and base diameter 140 cm from a metal sheet. How many squareÂ metersÂ of the sheet are required for the same?**

**Sol. **

Let r be the radius of the base and h be the height of the cylinder.

Base diameter = 140 cm, radius of base = = 70 cm = .70 m

Height = 1m

Metal sheet required to make a closed cylindrical tank

= Its total surface area

[Because h = 1 m, r = cm = 70 cm = 0.70 cm]

Hence, the sheet required

**Q.3Â Â Â Â Â 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.Â Â Â Â Â Â Â Â Â Â Â **

**(i) Inner curved surface area,**

Â Â Â Â Â Â Â Â Â Â Â Â Â

**(ii) Outer curved surface area,**

Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â

**(iii) Total surface area.**

**Sol.**We have, R = external radius

r = internal radius

h = length of the pipe = 77 cm **
(i)** Inner curved surface

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

**(ii)** Outer curved surface

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

**(iii)** Total surface area of a pipe

= Inner curved surface area + outer curved surface area + areas of two bases

=

= cm^{2}

=

=

**Q.4Â Â Â Â 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 .**

**Sol. **

The length of the roller is 120 cm i.e., h = 1.2 m and ,

radius of the cylinder (i.e., roller) .

Distance covered by roller in one revolution

= Its curved surface area =

=

=

Area of the playground = Distance covered by roller in 500 revolution.

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =

Hence , the area of playground is

**Q.5Â Â Â Â A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curve surface of the pillar at the rate of Rs 12.50 per .**

**Sol.**

* *Let r be the radius of the base and h be the height of the pillar.

Therefore cm = 25 cm = .25 m and h = 3.5 m.

Curved surface =

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Cost of painting the curved surface @ Rs. 12. 50 per

=Â Rs. 68.75

**Q.6Â Â Â Â Â Curved surface area of a right circular cylinder is 4.4 . If the radius of the base of the cylinder 0.7 m, Find its height.**

**Sol. **

Let r be the radius of the base and h be the height of the cylinder. Then,

Curved surface area = 4.4

[since r = 0.7]

Thus, the height of the cylinder = 1 metre.

**Q.7Â Â Â Â The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find. Â Â Â Â Â Â Â Â Â Â Â **

**(i) Its inner curved surface area.**

Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â

**(ii) The cost of plastering this curved surface at the rate of Rs 40 per**

**Sol.****(i)** Let r be the radius of the face and h be depth of the well. Then,

Curved surface =

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =

**(ii)** Cost of plastering is Rs 40 per

Therefore , cost of plastering the curved surface = Rs (110 Ã— 40) = Rs 4400.

**Q.8Â Â Â Â 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.**

**Sol. **

Total radiating surface in the system

= Curved surface area of the pipe =

[where ]

**Q.9Â Â Â Â Find Â Â Â Â Â Â Â Â Â Â Â (i) the lateral or curved surface area of cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high. **

Â Â Â Â Â Â Â Â Â Â Â (ii) How much steel was actually used, if of the steel actually used was wasted in making the closed tank.

Â Â Â Â Â Â Â Â Â Â Â (ii) How much steel was actually used, if of the steel actually used was wasted in making the closed tank.

**Sol.****(i)** Here,

Lateral surface area

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

**(ii)** Since of the actual steel used was wasted,

the area of the steel which has gone into the tank = of

Steel used = Letarl surface area + 2 x Area of base

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = =

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =

Therefore,Â Â actual steel used =

Hence, the actual area of the steel used =

**Q.10Â Â Â In 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. ** **Sol. **

Here , cm = 10 cm and h = 30 cm + 2 Ã— 2.5 cm (i.e., margin) = 35 cm

Cloth required for covering the lampshade

= Its curved surface area =

=

=

**Q.11Â Â The students of a Vidyalaya were asked to participate in a competition for making and decoratingÂ pen holdersÂ in theÂ shape of a cylinder with a base, ****using cardboard. EachÂ pen holderÂ was to be of radius 3 cm and heightÂ 10.5 cm. The Vidyalaya was to supply theÂ competitorsÂ with cardboard. If there were 35 competitors , howÂ much cardboard was required to be bought for the competition?**

**Sol. **

Cardboard required by each competitor

= Curved surface area of one penholder + base area =

[ ]

=

= =

Cardboard required for 35 competitors =

=