These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths 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.

Solution:

We know that

Curved surface area of cylinder = 2πrh

⇒ 88 = 2 × \(\frac{22}{7}\) × 14 × r

⇒ 88 = 88 × r

⇒ r = 1 cm

Diameter of cylinder = 2r = 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 is required for the same?

Solution:

Height of the cylinder = 1 m

and diameter of cylinder = 140 cm = 1.4 m

∴ radius of cylinder = \(\frac{14}{2}\) = 0.7 m

∴ Total surface are of cylinder = 2r(h + r)

= 2 × \(\frac{22}{7}\) × 0.7(1 + 0.7)

= 4.4 × 1.7

= 7.48 m^{2}

Therefore, 7.48 m^{2} of sheets are required to make the closed circular cylinder.

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 fig. 13.11) find its.

(i) inner curved surface area.

(ii) Outer curved surface area.

(iii) Total surface area.

Solution:

(i) We have given,

Height of cylinder = 77 cm

and inner diameter = 4 cm

inner radius = \(\frac{4}{2}\) = 2 cm

∴ Inner curved surface area = 2πrh

= 2 × \(\frac{22}{7}\) × 2 × 77

= 968 cm^{2}.

(ii) We have given

Height of cylinder = 77 cm

and Outer diameter = 4.4 cm

∴ Outer curved surface area of cylinder = 2πrh

= 2 × \(\frac{22}{7}\) × 2.2 × 77

= 1064.8 cm^{2}.

(iii) Total surface ara of pipe = Inner curved surface area + Outer curved surface area + Area of the two bases.

= 2πrh + 2πRh + 2π(R^{2} – r^{2})

= 2 × π × 2 × 77 + 2 × π × 2.2 × 77 + 2π((2.2)^{2} – (2)^{2})

= 2 × π × 77(2 + 2.2) + 2 × π × (4.84 – 4)

= 2 × \(\frac{22}{7}\) × 88 × 4.2 + 2 × \(\frac{22}{7}\) × 0.84

= 2032.8 + 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}.

Solution:

Clearly, the roller is a right circular cylinder of height h = 120 cm = 1.2 m

and diameter of its base = 84 cm = 0.84 m

∴ radius of its base = \(\frac{84}{2}\) = 0.42 m.

∴ Area covered by roller in one revolution = Curved surface area of the roller = 2πrh

= 2 × \(\frac{22}{7}\) × 0.42 × 1.2

= 3.168 m^{2}

So, Area covered by roller in 500 revolutions = 3.168 × 500 = 1584 m^{2}.

Hence, Area of playground = 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 area of the pillar at the rate of Rs. 12.50 per m^{2}.

Solution:

We have given that,

the height of pillar = 3.5m

and the diameter of pillar = 50 cm = 0.5 m

∴ Radious of the pillar = \(\frac{0.5}{2}\) = 0.25 m

∴ Curved surface area of pillar = 2πrh

= 2 × \(\frac{22}{7}\) × 0.25 × 3.5

= 5.5 m^{2}

Now, Rate of painting the pillar = Rs. 12.50 per m^{2}.

∴ Total cost of painting = 5.5 × 12.50 = Rs. 68.75.

Question 6.

The 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:

We have given that

The curved surface area of cylinder = 4.4 m^{2}

and radius of its base = 0.7 m

Height =?

We know that

Curved surface area of cylinder = 2πrh

⇒ 4.4 = 2 × \(\frac{22}{7}\) × 0.7 × h (∵ π = \(\frac {22}{7}\))

⇒ h = \(\frac{4.4 \times 7}{2 \times 22 \times 0.7}\) = 1 m

∴ Height of cylinder = 1 m.

Question 7.

The inner diameter of the circular well is 3.5 m. It is a 10 m deep find.

(i) It’s inner curved surface area.

(ii) The cost of plastering this curved surface at the rate of Rs. 40 per m^{2}.

Solution:

(i) We have given that

The inner diameter of circular well = 3.5 m

Inner radius circular well = \(\frac{3.5}{2}\) m

and height of circular well = 10 m.

Now, we know that,

Curved surface area of well = 2πrh

= \(2 \times \frac{22}{7} \times \frac{3.5}{2} \times 10\)

= 110 m^{2}

So, inner curved surface area of well = 110 m^{2}

(ii) Rate of plastering = Rs. 40 per m^{2}

Total cost of plastering its curved surface area = 110 × 40 = Rs. 4400

Question 8.

In a hot water heating system, there is a cylindrical pipe of length 28m and diameter 5 cm. Find the total radiating surface in the system.

Solution:

We have given that

Length of the cylindrical pipe (h) = 28m

and diameter of its base = 5 cm = 0.05 m

Radius of its base = \(\frac{.05}{2}\) m

∴ Curved surface area of cylindrical pipe = 2πrh

= \(2 \times \frac{22}{7} \times \frac{.05}{2} \times 28\)

= 4.4 m^{2}

Question 9.

Find

(i) The lateral or curved surface area of a cylindrical petrol Storage tank that is 4.2 m in diameter and 1.5 m high.

(ii) How much steel was actually used if \(\frac{1}{12}\) of the steel actually used was wasted in making the closed tank.

Solution:

(i) We have given that,

Height of cylindrical tank = 4.5 m

Diameter of cylindrical tank = 4.2 m

Radius of cylindrical tank = 2.1 m

So, Curved surface area of tank = 2πrh

= 2 × \(\frac{22}{7}\) × 4.5 × 21

= 59.4 m^{2}

(ii) Let the actual area of steel used be x m^{2}.

Since \(\frac{1}{12}\) of the actual steel used was wasted = \(\frac{1}{12}\) of x = \(\frac{x}{12}\)

∴ Area of steel which has gone into the tank = x – \(\frac{x}{12}\) = \(\frac{11x}{12}\)

Total surface area of tank = 2πr(r + h)

= 2 × \(\frac{22}{7}\) × 2.1(2.1 + 4.5)

= 87.12 m^{2}

The actual area of steel used be

\(\frac{11x}{12}\) = 87.12

⇒ x = \(\frac{87.12 \times 12}{11}\)

⇒ x = 95.04 m^{2}

Hence, total steel was used in tank = 95.04 m^{2}

Question 10.

In Fig. 13.12, 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 a height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and button of the frame. Find how much cloth is required for covering the lampshade.

Solution:

We have given that

Diameter of lampshade = 20 cm

Radius of lampshade = 10 cm

and height of lampshade = 30 + 2.5 + 2.45 = 35 cm

Therefore, Curved surface area of cylindrical lampshade = 2πrh

= 2 × \(\frac{22}{7}\) × 10 × 35

= 2200 cm^{2}

Therefore, the cloth is required for covering the lampshade is 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:

Total surface area of one penholder is 2πrh + πr^{2}

= 2 × \(\frac{22}{7}\) × 3 × 10.5 + \(\frac{22}{7}\) × 3 × 3

= 198 + 28.28

= 226.28 cm^{2}

∴ Total cardboard is used to making one penholder is 226.28 cm^{2}

Therefore, total cardboard is used to making 35 penholders is = 226.28 × 35 = 7919.8 cm^{2}.