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## Chapter 13: Surface Areas and Volumes

#### TEXTBOOK’S EXERCISE – 13.6

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)

Sol. Here, h = 5 cm, 2πrh = 94.2 cm2.

(i) 2πrh = 94.2

⇒ 2 × 3.14 × r × 5 = 94.2

⇒  r = 94.2 /2x 3.14x 5 = 3

Hence, the base radius of the cylinder = 3 cm

(ii) The volume of the cylinder

= πr2h = 3.14 × 3 × 3 × 5 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) the inner curved surface area of the vessel,

(ii) the radius of the base,

(iii) the capacity of the vessel.

Sol. Here, h = 10 m

(i) Inner curved surface area

(ii) We have, 2πrh = 110

⇒ 2 × 22/7 × r × 10 = 110

⇒   r = 110x 7/2x 22 x10 = 1.75 m

(iii) Capacity of the vessel = πr2h

= 22/ 7 × 1.75 × 1.75 × 10 m3 = 96.25 m3

= 96.25 kl [1 m3 = 1 kl]

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?

Sol. Here, h = 1 m, volume = 15.4 litres

= 15.4/1000 m3 = 0.0154 m3

Also, volume of the cylindrical vessel = πr2h

⇒ 0.0154 = 22 7 × r2 × 1

= 0.0049 ⇒ r = 0.07 m

∴ Total surface area of the cylinder

= 2πr (h + r)

= 2 × 22/7 × 0.07 (1 + 0.07) m2

= 44 × 0.01 × 1.07 m2 = 0.4708 m2

Hence, 0.4708 m2 of metal sheet would be needed.

#### TEXTBOOK’S EXERCISE – 13.6

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.

Sol. Here, h = 14 cm.

Radius of the pencil (R) = 72 mm = 0.35 cm.

Radius of the graphite (r) = 12 mm = 0.05 cm.

The volume of the graphite

= πr2h = 22/7 × 0.05 × 0.05 × 14 cm3

= 0.11 cm3

The volume of the wood = π (R2 – r2)h

= 22/7 × [(0.35)2 – (0.05)2] × 14 cm3

= 22/7 × 0.4 × 0.3 × 14 cm3 = 5.28 cm3

Hence, the volume of the wood = 5.28 cm3 and volume of the graphite

= 0.11 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?

Sol. Here, r = 72 cm = 3.5 cm, h = 4 cm

Capacity of 1 cylindrical bowl = πr2h

= 22/7 × 3.5 × 3.5 × 4 cm3 = 154 cm3

Hence, soup consumed by 250 patients per day = 250 × 154 cm3

= 38500 cm3

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