**For previous exercise – Surface Area and Volumes – Exercise 13.5 – Class IX**

Assume π = ^{22}/_{7} , unless stated otherwise.

**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:

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 = πr^{2}h = ^{22}/_{7} x 21^{2} x 25 = 34650 cm^{3}

Amount of water cylindrical vessel can hold = ^{34650}/_{1000} = 34.65 litres

**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 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(R^{2} – r^{2 })h

= ^{22}/_{7} x (14^{2} – 12^{2}) x 35

= 5720 cm^{3}

Mass of the pipe for 1 cm^{3} of wood = 0.6 g

Then, the mass of the pipe for 22880cm^{3} of wood = 5720 x 0.6 = 3432g

**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 cm^{3}

^{ }

(ii)d = 7cm⇒r = 3.5 cm ; h = 10 cm

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

= ^{22}/_{7} x 3.5^{2} x 10

= 385 cm^{3}

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

= 385 – 300

= 85 cm^{3}

Therefore, plastic cylinder has the greater capacity.

- I
**f 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:

Lateral surface area of cylinder = 94.2cm^{2} ; 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

(i) radius = 3 cm

(ii) volume of cylinder = πr^{2}h

= 3.14 x 3^{2} x 5

= 141.3 cm^{3}

**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:

h = 10 m

(i) inner curved surface area of the vessel = ^{total cost}/_{cost of painting per m}^{2} = ^{2200}/_{20} = 110 m^{2}

(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 = πr^{2}h

= ^{22}/_{7} x (^{7}/_{4})^{2} x (10)

= 96.25 kg [1m^{3} = 1000]

**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 = πr^{2}h

^{15.4}/_{1000 }= ^{22}/_{7} x r^{2} x 1

r^{2} = ^{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 cm^{2}

Therefore, 0.4708m^{2} of metal sheet would be needed.

**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 = πr^{2}h = ^{22}/_{7} x (0.05)^{2} x 14 = 0.11cm^{3}

volume of the wood = π(R^{2} – r^{2})h

= ^{22}/_{7} x(0.35^{2} – 0.05^{2}) x 14

= 5.28 cm^{3}

**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 = πr^{2}h

= ^{22}/_{7} x 3.5^{2} x 4

= 154 cm^{3}

Hence the soup consumed by 250 patients per day = 250 x 154 = 38500 cm^{3}

For next exercise – Surface Area and Volumes – Exercise 13.7 – Class IX

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