- Plane Figure
- |
- Solid Figure
- |
- Perimeter
- |
- Surface Area
- |
- Volume
- |
- Surface Area and Volume of Cube and Cuboid
- |
- Surface Area and Volume of Right circular cylinder
- |
- Surface Area and Volume of Right circular cone
- |
- Surface Area and Volume of sphere and hemisphere
- |
- How the Surface area and Volume are determined

In this page we have ** NCERT book Solutions for Class 9th Maths:Surface area and volume** for
EXERCISE 2 . Hope you like them and do not forget to like , social share
and comment at the end of the page.

**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.

**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?

**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

**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}

**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 Rs 12.50 per m^{2}

**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.

**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 Rs 40 per m^{2}

**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.

**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?

**Question 10)**

In below 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.

**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?

__We are assuming the value of π=22/7 in all the solutions__

**Solution 1:**

Height (h) of cylinder = 14 cm

Let the diameter of the cylinder be d.

Curved surface area of cylinder = 88 cm^{2}

2πrh = 88 cm^{2}

r=1 cm

So d=2cm

**Solution 2**

Height of closed cylindrical tank =h=1 m

Base diameter of cylindrical tank =d=140 cm =1.4 m

Radius of cylindrical tank =

r= d/2= 0.7 m

Total surface area of cylindrical tank =2π.r(r+h)=2×(22/7)×0.7(0.7+1)

=7.48 m^{2}

**Solution 3:**

Inner radius of the pipe(a) =2 cm

Outer radius of the pipe(b) =2.2cm

Height of the cylindrical pipe(h)=77 cm

Inner curved area of cylindrical pipe=2 πaH=968 cm^{2}

Outer curved area of cylindrical pipe=2 πbH=1064. cm^{2}

Surface area of whole pipe=Inner curved area + outer curved area +circular area at both ends of the pipe

=2 πaH+2 πbH +2π(b^{2} –a^{2})

=2038.08 cm^{2}

**Solution 4 **

It can be observed that a roller is cylindrical.

Height (h) of cylindrical roller = Length of roller = 120 cm

Radius (r) of the circular end of roller = 84/2=42 cm

Curved surface area of roller = 2πrh

=31680 cm^{2}

Area of field = 500 × CSA of roller

= (500 × 31680) cm^{2}

= 15840000 cm^{2}

= 1584 m^{2}

**Solution 5:**

Height of the cylindrical pillar =h=3.5 m

Diameter of the cylindrical pillar =d=50 cm

So Radius of the cylindrical pillar =r=50/2=25 cm =0.25 m

So Curved surface of the cylindrical pillar =2π.r.h==5.5 m^{2}

Cost of painting the curved surface of the pillar at the rate of Rs 12.50 per m^{2}

=12.5×5.5= Rs 68.75

**Solution 6:**

Let the height of the cylinder be h.

Radius of the base of the cylinder = 0.7 m

Curved surface of cylinder = 4.4m^{2}

2π.r.h=4.4

We know the value of r,so substituting and calculating

h = 1 m

**Solution 7:**

Inner radius (r) of circular well = 1.75 m

Depth (h) of circular well = 10 m

Inner curved surface area = 2πrh

= (44 × 0.25 × 10) m^{2}

= 110 m^{2}

Therefore, the inner curved surface area of the circular well is 110 m^{2}.

Cost of plastering 1 m^{2} area = Rs 40

So Cost of plastering 100 m^{2} area = Rs (110 × 40)

= Rs 4400

**Solution 8:**

Length of cylindrical pipe =h=28 m

Diameter of cylindrical pipe =d=5 cm

Radius of cylindrical pipe =r=2.5 cm =0.025 m

Total radiating surface in the system=Curved Surface Area of cylindrical pipe =2π.r.h

=2×(22/7)×0.025×28=4.4 m^{2}

**Solution 9**:

Height of tank=4.5 m

Radius of the circular end of the tank=4.2/2=2.1 m

Lateral curved area of the tank=2π.r.h=59.4 m^{2}

Total surface of tank= Lateral curved area + area of the circular end

=2πr(r+h)=87.12 m^{2}

Let x m^{2} of the sheet is used in making the tank

Then actual sheet in making the tank = x(1-1/12)=11x/12

Now

11x/12=87.12

Or x=95.04 m^{2}

**Solution 10:**

Height (h) of the frame of lampshade after taking the margin at the top and bottom of the lamppost = (2.5 + 30 + 2.5) cm = 35 cm

Radius (r) of the circular end of the frame of lampshade = 10 cm

Cloth required for covering the lampshade = 2πrh

= 2200 cm^{2}

**Solution 11:**

Radius of each pen holder =r=3 cm

Height of each pen holder =h=10.5 cm

Cardboard needed for 1 pen holder = Curved Surface area of 1 pen holder + Area of base of 1 pen holder

=2π.r.h + π.r^{2}

= π.r(2h+r)

=72 π cm^{2}

Cardboard required for 35 pen holders =35 × (22/7)× 3 × 24=7920 cm^{2}

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