Students can Download Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus 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 }= 1 l)

Solution:

The circumference of the base of cylindrical vessel, C = 132 cm.

height, h = 25 cm

C = 2πr = 132

r = 21 cm

∴ Volume of Cylinder, V = πr^{2}h

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 cm3 of wood has a mass of 0.6 g.

Solution:

Let the inner diameter of a cylindrical vessel be d_{1} and outer diameter be d_{2}.

height of pipe, h = 35 cm.

∴ volume of pipe, V = \(\pi r_{2}^{2} h-\pi r_{1}^{2} h\)

= 110 × 52

∴ V = 5720 cm^{3}

Mass of pipe of 1 ccm is 0.6 gm.

∴ Mass of 5720 cm^{3}. … ? …

= 5720 × 0.6

= 3432 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:

Length of rectangular tin, l = 5 cm,

breadth, b = 4 cm,

height, h = 15 cm.

∴ Volume of rectangular tin, V = l × b × h

V = 5 × 4 × 15

= 300 cm^{3}

Diameter of plastic cylinder, d = 7 cm,

∴ r = \(\frac{7}{2}\)

Volume of cylinder, v = πr^{2}h

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

= 11 × 7 × 5

= 385 cm^{3}.

∴ Plastic cylinder’s volume is great.

= Volume of cylindrical tin – Volume of rectangular tin.

= 385 – 300

= 85 cm^{3}.

∴ Volume of plastic tin is greater than rectangular tin by 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:

Curved surface area of cylinder = 94.2 cm^{2}

height, h = 5 cm.

r = ?

V = ?

(i) Curved Surface area of cylinder, L.S.A.

A= 2πrh

94.2 = 2 × 3.14 × 5 × r

= 3.01 cm.

= 3 cm.

(ii) Volume of cylinder, V = πr^{2}h

= 3.14 × (3)2 × 5

= 3.14 × 9 × 5

= 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:

(i) Cost of painting for cylindrical vessel = Rs. 2200

Cost of painting is at the rate of Rs. 20 per m^{2}.

∴ Curved Surface area of vessel = \(\frac{2200}{20}\)

= 110 m^{2}

(ii) Depth of vessel, h = 10 m

r = ?

Curved surface area of vessel, V = 2πrh

(iii) Capacity of the Vessel = Volume of Vessel

V = πr^{2}h

V = 96.25 m^{3}.

Question 6.

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

Solution:

Volume of a cylindrical vessel = 15.4 litres

1000 c.cm. = 1 litre.

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

height, h = 1 m,

radius, r = ?

Curved surface area =?

Volume of cylinder, V = πr^{2}h

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:

Diameter of pencil, d = 7 mm = 0.7 m.

∴ radius r_{1} = \(\frac{0.7}{2}\) = 0.35 cm

length of pencil, h = 14 cm

diameter of pencil, d = 1 mm = 0.1 cm

radius, r_{2} = \(\frac{0.1}{2}\) = 0.05 cm.

Volume of wood in pencil, V

Let radius of pencil = R Diameter(2R) 7mm

Volume of the wood Volume of a pencil – Volume of graphite (5.39 – 0.11) = 5.28 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:

Diameter of pencil, d = 7 cm

∴ radius, r = \(\frac{7}{2}\) cm

height of soup, h = 4 cm

Volume of soup given for 1 patient is 154 cm^{3}

Volume of soup given for 250 patients .. ? ….

= 154 × 250 = 38500 cm^{3}

For 1000 cm^{3}, 1 litre

For 38500 cm^{3} ….. ? …

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

= 38.5 litres soup.