**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 cm

^{2}. 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 = 2Ï€r(h + r)

= 2 × 22/7 × 0.70(1 + 0.70) m

^{2}

= 2 × 22 × 0.10 × 1.70 m

^{2}

= 7.48 m

^{2}

Hence, the area of required sheet is 7.48 m

^{2}.

^{}

**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 = 2Ï€rh

= 2 × 22/7 × 2 × 77 cm^{2}

= 2 × 22 × 2 × 11 cm^{2} = 968 cm^{2}

**(ii)** Outer curved surface area = 2Ï€Rh

**(iii) **Total surface area = [Inner curved surface
area] + [Outer curved surface area] + [Area of two circular ends]

= [2Ï€rh] + [2Ï€Rh] + 2[Ï€(R^{2} – r^{2})]

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}.

**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 = 2Ï€rh

= 2 × 22/7 × 42 × 120 cm

^{2}

= 2 × 22 × 6 × 120 cm

^{2}= 31680 cm

^{2 }

Now, area of the playground levelled in one revolution of the roller = 31680 cm

^{2}

= 31680/10000 m

^{2}

∴ Area of the playground levelled in 500 revolutions = 500 × 31680/10000 m

^{2}

= 1584 m^{2}

**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 m

^{2}.

**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 = 2Ï€rh

^{2}

∴ Cost of painting of 1 m

^{2}pillar = ₹12.50

∴ Cost of painting of 11/2 m

^{2}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 m

^{2}. 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 = 2Ï€rh

= 2 × 22/7 × 7/10 × h m

^{2}

But the curved surface area is 4.4 m

^{2}. [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 m^{2}.

**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 = 2Ï€rh

**(ii)** Cost of plastering per m^{2} = ₹40

∴ Total cost of plastering the area 110 m^{2}

= ₹ (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 = 2Ï€rh

Thus, the total radiating surface is 4.4 m

^{2}.

**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 = 2Ï€rh

= 2 × 227 × 2.1 × 4.5 m^{2 }

= 2 × 22 × 0.3 × 4.5 m^{2} = 59.4 m^{2}

**(ii)**
Total surface area of the tank = 2Ï€r(r + h)

= 2 × 22/7 × 2.1(2.1 + 4.5) m^{2}

= 44 × 0.3 × 6.6 m^{2} = 87.13 m^{2 }

Let actual area of the steel used be x m^{2}

∴ Area of steel that was wasted = 1/12 × x m = x/12 m^{2}

Area of steel used = x – x/12 m^{2}

Thus, the required area of the steel that was actually used is 95.04 m^{2}.

**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 = 2Ï€rh

= 2 × 22/7 × 10 × 35 cm

^{2}

= 2200 cm

^{2 }

Thus, the required area of the cloth = 2200 cm

^{2}

**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]

= [2Ï€rh] + [Ï€r

^{2}]

∴ Surface area of 35 penholders = 35 × 1584/7 cm^{2} = 7920 cm^{2}

Thus, 7920 cm^{2} of cardboard was required to be bought.

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