Gujarat Board GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6

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} = 1L)

Solution:

Let the base radius of the cylindrical vessel be r cm. Then, the circumference of the base of the cylindrical vessel = 2πr cm.

According to the question,

2πr = 132

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

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

⇒ r = 21 cm; h = 25 cm

∴ Capacity of the cylindrical vessel = πr^{2}h

= \(\frac {22}{7}\) (21)^{2} (25) cm^{3} = 34650 cm^{3}

= \(\frac {34650}{1000}\) L = 34.65 L

Hence, the cylindrical vessel can hold 34.65 L of water.

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:

∴ Inner diameter = 24 cm

∴ Inner radius (r) = \(\frac {24}{2}\) cm = 12 cm

∴ Outer diameter = 28 cm

∴ Outer radius (R) = \(\frac {28}{2}\) cm = 14 cm

Length of the pipe (h) = 35 cm

∴ Inner volume = πr^{2}h

= \(\frac {22}{7}\) x (12)^{2} x 35

= 15840 cm^{3}

and outer volume = πR^{2}h

= \(\frac {22}{7}\) x (14)^{2} x 35

= 21560 cm^{3}

∴ Volume of the wood used

= Outer volume – Inner volume = 21560 cm^{3} – 15840 cm^{3} = 5720 cm^{3}

∴ Mass of the pipe

= 5720 x 0.6 g = 3432 g = 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 a circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Solution:

(i) For tin can

1 = 5 cm

b = 4 cm

h = 15 cm

Capacity = l x b x h

= 5 x 4 x 15 cm^{3} = 300 cm^{3}

(ii) For plastic cylinder

Diameter = 7 cm

∴ Radius (r) = \(\frac {7}{2}\) cm

Height (h) = 10 cm

∴ Capacity = πr^{2}h

= \(\frac {22}{7}\) x (\(\frac {7}{2}\))^{2}x 10 = 385 cm^{2}

Clearly the second container i.e., a plastic cylinder has greater capacity than the first container i.e., a tin can by 385 – 300 = 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:

(i) Let the radius of the base of the cylinder be r cm.

h = 5 cm

Lateral surface = 94.2 cm^{2}

⇒ 2πrh = 94.2

⇒ 2 x 3.14 x r x 5 = 94.2

r = \(\frac{94.2}{2 \times 3.14 \times 5}\)

r = \(\frac{94.2}{31.4}\) = r = 3cm

Hence, the radius of the base is 3 cm.

(ii) r = 3 cm

h = 5 cm

∴ Volume of the cylinder = πr^{2}h

= 3.14 x (3)^{2} x 5 = 141.3 cm^{3}

Question 5.

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

(i) inner curved surface area of the vessel,

(ii) radius of the base,

(iii) capacity of the vessel.

Solution:

(i) Inner curved surface area of the vessel

= \(\frac{2200}{20}\) = 110 m^{2}

(ii) Let the radius of the base be r m.

h = 10 m

Inner curved surface area = 110 m^{2}

⇒ 2πrh = 110

⇒ 2 x \(\frac{22}{7}\) x r x 10 = 110

r = \(\frac{110 \times 7}{2 \times 22 \times 10}\)

r = \(\frac{7}{4}\)

r = 1.75 m

Hence, the radius of the base is 1.75 m.

(iii) r = 1.75 m

h = 10 m

∴ Capacity of the vessel = πr^{2}h

= \(\frac{22}{7}\) x (1.75)^{2} x 10

= 96.25 m^{3}

Hence, the capacity of the vessel is 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 the metal sheet would be needed to make it?

Solution:

h = 1 m

Capacity = 15.4 litres

\(\frac{15.4}{1000}\) m^{3} = 0.0154 m^{3}

Let the radius of the base be r m.

Capacity = 0.0154 m^{3}

⇒ πr^{2}h = 0.0154

\(\frac{22}{7}\) x r^{2} x 1 = 0.0154

⇒ r^{2} = \(\frac{0.0154 \times 7}{22}\)

⇒ r^{2} = 0.0049

⇒ r = \(\sqrt{0.0049}\)

⇒ r = 0.07 m

∴ Total surface area = 2πrh + 2πr^{2}

= 2 x \(\frac{22}{7}\) x 0.07 x 1 + 2 x \(\frac{22}{7}\) x (0.07)^{2}

= 0.44 + 0.0308 = 0.4708 m^{2}

Hence, 0.4708 m2 of the metal sheet should be needed.

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:

For solid cylinder of graphite

Diameter = 1 mm

∴ Radius (r) = \(\frac{1}{2}\) mm

Length of the pencil (h) = 14 cm = 140 mm

∴ Volume of the graphite = πr^{2} h

= \(\frac{22}{7}\) x (\(\frac{1}{2}\))^{2} x 140 = 110 mm^{3}

= \(\frac{110}{10 \times 10 \times 10}\)cm^{3} = 0.11 cm^{3}

For cylinder of wood

Diameter = 7 mm

∴ Radius (R) = \(\frac{7}{2}\) mm

Length of the pencil (h) = 14 cm = 140 mm

∴ Volume of the wood

= π(R^{2} – r^{2})h

= \(\frac{22}{7}\)[(\(\frac{7}{2}\))^{2} – (\(\frac{7}{2}\))^{2}] 140

= 5280 mm^{3}

Question 8.

A patient in a hospital is given soup daily in a cylindrical bowl of a 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:

Diameter = 7 cm

∴ Radius (r) = \(\frac{7}{2}\) cm

Height (h) = 4 cm

∴ Volume of soup in the cylindrical bowl = πr^{2}h

= \(\frac{22}{7}\) x (\(\frac{7}{2}\))^{2} x 4 cm^{3} = 154 cm^{3}

∴ Volume of soup to be prepared daily to serve 250 patients

= 154 x 250 cm^{3} = 38500 cm^{3} (or 38.5 L)

Hence, the hospital has to prepare 38500 cm^{3} (or 38.5 L) of soup daily to serve 250 patients.