Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.6 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.6

Assume π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.

The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm^{3} = 11)

Solution:

We have,

Height of cylindrical vessel (h) = 25 cm

Circumference of the base of the cylindrical vessel = 132 cm …..(1)

Let the radius of the base of the cylindrical vessel be r сm, then

Circumference of the base of the cylindrical vessel = 2πr

⇒ 132 = 2 × \(\frac{22}{7}\) × r [Using (1)]

⇒ \(\frac{132 \times 7}{2 \times 22}=r\)

⇒ r = 21 cm

∴ volume of the water in the vessel = πr2h

\(\frac{22}{7}\) × (21)^{2} × 25

= 34650 cm^{3}

= \(\frac{34650}{1000}\)

= 34.65 litres

Hence, volume of the water in the vessel = 34.65 litres.

Question 2.

The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm^{3} of wood has a mass of 0.6 g.

Solution:

We have,

Length of the cylindrical pipe (h) = 35 cm

Inner diameter of the cylindrical pipe = 24 cm

∴ Inner radius of the cylindrical pipe (r) = \(\frac{24}{2}\) = 12 cm

and outer diameter of the cylindrical pipe = 28 cm

∴ Outer radius of the cylindrical pipe (R) = \(\frac{28}{2}\) = 14 cm

∴ Volume of wooden used in making the pipe

= π(R^{2} – r^{2}) × h

= \(\frac{22}{7}\)(14^{2} – 12^{2}) × 35

= \(\frac{22}{7}\)(14 + 12)(14 – 12) × 35

= 110 × 26 × 2 = 5720 cm^{3}

∵ Weight of 1 cm^{3} wood = 0·6 gram

∴ Weight of 5720 cm^{3} wood = 0·6 × 5720 gram

= 3432 gram

= \(\frac{3432}{1000}\) kg = 3.432

Hence,mass of the wooden pipe = 3.432 kg.

Question 3.

A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Solution:

(i) We have,

Length of rectangular tin (l) = 5 cm

Breadth of rectangular tin (b) = 4 cm

Height of rectangular tin (h) = 15 cm

∴ Volume of the rectangular tin = l × b × h

= 5 × 4 × 15 = 300 cm^{3}

(ii) Diameter of the base of the plastic cylinder (d) = 7 cm

∴ Radius of the base of the plastic cylinder (r) = \(\frac{7}{2}\) = 3.5 cm

and height of the plastic cylinder (h) = 10 cm

∴ Volume of the plastic cylinder = πr^{2}h

= \(\frac{22}{7}\) × (3.5)^{2} × 10

= 385 cm^{3}

Hence, plastic cylinder has greater capacity and it is equal (385 – 300) cm^{3} = 85 cm^{3}.

Question 4.

If the lateral surface of a cylinder is 94.2 cm^{2} and its height is 5 cm, then find :

(i) radius of its base

(ii) its volume. (Use π = 3.14)

Solution:

We have,

Height of the cylinder (h) = 5 cm

Let radius of the cylinder be r сm.

and lateral surface of the cylinder = 94.2 cm^{2}

⇒ 2πrh = 94.2

⇒ 2 × 3.14 × r × 5 = 94.2

⇒ 31.4 × r = 94.2

⇒ r = \(\frac{94.2}{31.4}\)

⇒ r = 3 cm

(ii) Volume of the cylinder = πr^{2}h

= 3.14 × (3)^{2} × 5

= 141.3 cm^{3}

Hence, (i)Radius of the cylinder = 3 cm

(ii) Volume of the cylinder = 141.3 cm^{3}.

Question 5.

It costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs. 20 per m^{2}, find :

(i) inner curved surface area of the vessel,

(ii) radius of the base,

(iii) capacity of the vessel.

Solution:

We have,

Depth of the cylindrical vessel (h) = 10 m

cost of painting = Rs. 20 per m^{2}

Total cost of painting = Rs. 2200

(i) Inner curved surface area of the vessel = \(\frac{\text { Total cost }}{\text { Cost of painting }}=\frac{2200}{20}\) = 110 m^{2}

(ii) Let the radius of the base of vessel ber m.

Curved surface area of the vessel = 110 m^{2},

[As solved in (i)]

⇒ 2πrh = 110

⇒ 2 × \(\frac{22}{7}\) × r × 10 = 110

⇒ \(\frac{440}{7}\) × r = 110

⇒ r = \(\frac{110 \times 7}{440}\)

⇒ r = 1.75 m.

(iii) Capacity of the vessel = πr^{2}h

= \(\frac{22}{7}\) × (175)^{2} × 10

= 96.25 m^{3} = 96.25 kl

Hence, (i) Inner curved surface area of the vessel = 110 m^{2}

(ii) Radius of the base = 1.75 m

(iii) Capacity of the vessel = 96.25 m^{3} or 96.25 kl.

Question 6.

The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

Solution:

We have,

Height of the cylindrical vessel (h) = 1 m

and volume of the cylindrical vessel = 15.4 litres …..(1)

Let the radius of the cylindrical vessel be r m, then

Volume of the cylindrical vessel = πr^{2}h

⇒ 15.4 litres = \(\frac{22}{7}\) × r^{2} × 1[Using (1)]

⇒ \(\frac{15.4}{1000} \mathrm{~m}^3=\frac{22}{7} r^2\)

⇒ \(\frac{15.4 \times 7}{1000 \times 22}=r^2\)

⇒ 0.0049 = r^{2}

⇒ r = \(\sqrt{0.0049}\)

⇒ r = \(\sqrt{\frac{49}{10000}}\)

r = \(\frac{7}{100}\)

Required area of the metal sheet = Total surface area of the cylindrical vessel

= 2πrh + 2πr^{2}

= 2 \(\frac{22}{7}\) × 1 + 2 × \(\frac{22}{7} \times\left(\frac{7}{100}\right)^2\)

= \(\frac{44}{100}\) + 0.0308

= 0.44 + 0.0308

= 0.4708 m^{2}

Hence, area of the metal sheet required to make the vessel = 0.4708 m^{2}.

Question 7.

A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Solution:

We have,

Length of the cylindrical pencil (h)= 14 cm

Diameter of the cylindrical pencil = 7 mm

= \(\frac{7}{10}\) cm

∴ Radius of the cylindrical pencil (R) = \(\frac{7}{20}\) cm

and diameter of the cylindrical graphite

= 1 mm = \(\frac{1}{10}\) cm

∴ Radius of the cylindrical graphite (r) = \(\frac{1}{10}\)

∴ Volume of the cylindrical pencil = πR^{2}h

= \(\frac{22}{7} \times\left(\frac{7}{20}\right)^2 \times 14\)

= 5.39 cm^{3}

and volume of the cylindrical graphite = πr^{2}h

= \(\frac{22}{7} \times\left(\frac{1}{20}\right)^2 \times 14\)

= 0·11 cm^{3}

Volume of the wood = Volume of the pencil – Volume of the graphite

= 5.39 – 0·11 = 5.28 cm^{3}

Hence, volume of the wood = 5.28 cm^{3} and volume of the graphite = 0·11 cm^{3}.

Question 8.

A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

Solution:

We have,

Height of the cylidrical bowl filled with soup (h) = 4 cm

Diameter of the cylindrical bowl (d) = 7 cm

∴ Radius of the cylindrical bowl (r) = \(\frac{7}{2}\) cm

Volume of the soup for 1 patient = πr^{2}h

= \(\frac{22}{7} \times\left(\frac{7}{2}\right)^2 \times 4\)

= 154 cm^{3}

Volume of the soup for 250 patients

= 154 × 250 = 38500 cm^{3}

= \(\frac{38500}{1000}\) litres

= 38.5 litres

Hence, volume of the soup for 250 patients = 38500 cm^{3} or 38.5 litres.