# Surface Area and Volumes – Exercise 13.6 – Class IX

Assume π = 22/7 , unless stated otherwise.

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 cm3 = 1l)

Solution:

The circumference of the base of a cylindrical vessel = 132 cm = 2πr ; h = 25 cm

2πr = 132 cm

2 x 22/7 x r = 132

r = 132×7/2×22 = 21 cm

Volume of the cylinder = πr2h = 22/7 x 212 x 25 = 34650 cm3

Amount of water cylindrical vessel can hold = 34650/1000 = 34.65 litres

1. 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:

Inner diameter of cylindrical wooden pipe , d = 24 cm ⇒ r = 12cm

Outer diameter of cylindrical wooden pipe, D = 28 cm ⇒ r = 14cm

l = 35 cm

Volume of the wood used in the pipe = π(R2 – r2 )h

= 22/7 x (142 – 122) x 35

= 5720 cm3

Mass of the pipe for 1 cm3 of wood = 0.6 g

Then, the mass of the pipe for 22880cm3 of wood  = 5720 x 0.6 = 3432g

1. 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)l = 5 cm , b = 4 cm and h = 15 cm

Volume of the rectangular tin = lbh = 5 x 4 x 15 = 300 cm3

(ii)d = 7cm⇒r = 3.5 cm ; h = 10 cm

Volume of the plastic cylinder  = πr2h

= 22/7 x 3.52 x 10

= 385 cm3

Difference in the capacities of two packs = Volume of the plastic cylinder  – Volume of the rectangular tin

= 385 – 300

= 85 cm3

Therefore, plastic cylinder  has the greater capacity.

1. If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then find

(ii) its volume. (Use π = 3.14)

Solution:

Lateral surface area of cylinder  = 94.2cm2 ; h = 5 cm

Lateral surface area of cylinder = 2πrh

94.2 = 2 x 3.14 x r x 5

r = 94.2/2×3.14×5 = 3 cm

(ii) volume of cylinder = πr2h

= 3.14 x 32 x 5

= 141.3 cm3

1. 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 m2 , find

(i) inner curved surface area of the vessel,

(iii) capacity of the vessel.

Solution:

h = 10 m

(i) inner curved surface area of the vessel = total cost/cost of painting per m2 = 2200/20 = 110 m2

(ii) radius of the base = 2πrh = 110

2 x 22/7 x r x 10 = 110

r = 110×7/2x22x10 = 7/4

(iii) capacity of the vessel = πr2h

= 22/7 x (7/4)2 x (10)

= 96.25 kg           [1m3 = 1000]

1. 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:

h = 1m , volume 15.4 litres

Volume of the cylindrical vessel = πr2h

15.4/1000 = 22/7  x r2 x 1

r2 = 15.4/1000 x 7/22 = 0.0049

r =  0.07 m

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

= 2 x 22/7 x 0.07 x (0.07 + 1)

= 0.4708 cm2

Therefore, 0.4708m2 of metal sheet would be needed.

1. 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:

h = 14 cm

Diameter of the pencil = 7mm

Radius of the pencil = 7/2 = 3.5 mm = 0.35 cm

Diameter of the graphite = 1 mm

Radius of the graphite = 0.5 mm = 0.05 cm

Volume of the graphite = πr2h = 22/7­ x (0.05)2 x 14 = 0.11cm3

volume of the wood =  π(R2 – r2)h

= 22/7 x(0.352 – 0.052) x 14

= 5.28 cm3

1. 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:

d = 7 cm

r = d/2 = 3.5 cm

h = 4 cm

Capacity  of 1 cylindrical bowl = πr2h

= 22/7 x 3.52 x  4

= 154 cm3

Hence the soup consumed by  250 patients per day = 250 x 154 = 38500 cm3