Surface Areas and Volumes – Exercise 13.6 – (MATHEMATICS) – 9th Class

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Exercise 13.6

Assume  π = {{22} \over 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 = 1 l).

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Sol. Circumference of base = 132 cm.

 \Rightarrow 2\pi r = 132 \Rightarrow r = {{132 \times 7} \over {2 \times 22}} = 21cm.

∴  Volume of cylindrical vessel = πr²h

 = {{22} \over 7} \times 21 \times 21 \times 25c{m^3} = 64650c{m^3}

∴ Capacity of the vessel = {{34650} \over {1000}} l = 34.65 l

Hence, vessel can hold 34.65 l of water.


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.

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Sol. Volume of the wood = π(r12 – r22)h
 = {{22} \over 7} \times \left\{ {{{(14)}^2} - {{(12)}^2}} \right\} \times 35c{m^3}

= 110 (196 – 144) cm³ = 5720 cm³

Mass of the pipe (or wood) = 0.6 × 5720 g = 3432 g

= 3.432 kg.


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?

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Sol.
(i) Capacity of tin can = area of base × height
= (5 × 4) × 15 cm³ = 300 cm³

(ii) Capacity of the plastic cylinder

 = {{22} \over 7} \times {7 \over 2} \times {7 \over 2} \times 10c{m^3} = 385c{m^3}

Hence, the plastic cylinder has greater capacity by 85 cm³.


4. If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find
(i) radius of its base
(ii) its volume.          (Use π = 3.14)

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Sol. Given lateral surface = 94.2 cm².
(i)  2πrh = 94.2     ⇒   2 × 3.14 ×  r × 5  = 94.2

⇒                  r = {{94.2} \over {10 \times 3.14}} = 3cm

(ii) Volume = πr²h = 3.14 × (3)² × 5 cm³ = 3.14 × 45 cm³ = 141.3 cm³.


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 m², find
(i) inner curved surface area of the vessel,
(ii) radius of the base,
(iii) capacity of the vessel.

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Sol. Cost of the painting is Rs.2200, if rate is Rs.20 per m²

(i) Inner curved surface area = {{2200} \over {20}} = 100{m^2}

(ii) 2\pi rh = 110 \Rightarrow 2 \times {{22} \over 7} \times r \times 10 = 110

 \Rightarrow r = {7 \over 4}m = 1.75m

(iii) Capacity of the vessel = \pi {r^2}h = {{22} \over 7} \times {7 \over 4} \times {7 \over 4} \times 10{m^3}

= 96.25 m³


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?

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Sol. Capacity of closed cylindrical vessel
= 15.4 l
= 15.4 × 1000 cm³ = 15400 cm³.

Height = 1 m = 100 cm.

∴     \pi {r^2}h = 15400 \Rightarrow {{22} \over 7} \times {r^2} \times 100 = 15400

Area of metal sheet needed = 2πr(h + r)

 = 2 \times {{22} \over 7} \times 7 \times (100 + 7)c{m^2}

= 4708 cm² = 0.4708 m².


7. A lead pencil consists of a cylinder of mm wood with a solid cylinder of graphite filled in the interim: 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.

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Sol. Volume of graphite
 = \pi {\left( {{1 \over 2}} \right)^2} \times 14c{m^3}

 = {{22} \over 7} \times {1 \over {400}} \times 14c{m^3}

= 0.11 cm3

Volume of wood = volume of cylinder – volume of graphite

 = \left[ {{{22} \over 7} \times {{\left( {{7 \over {20}}} \right)}^2} \times 14 - 0.11} \right]c{m^3}

= (5.39 – 0.11) cm3

= 5.28 cm3


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?

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Sol. Volume of soup for one patient =  {{{22} \over 7} \times {{\left( {{7 \over 2}} \right)}^2} \times 4c{m^3} = 154c{m^3}}

Volume of soup for 250 patients = 250 × 154 cm³
= 38500 cm³ = 38.5 l.


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