#### Class nine Maths Surface Areas and Volumes NCERT

## Chapter 13: Surface Areas and Volumes

#### TEXTBOOK’S EXERCISE – 13.6

**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) **

**Sol.
**Here, h = 5 cm, 2πrh = 94.2 cm^{2}.

(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

= π*r*2*h
= *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 m**^{2}**, 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 = πr^{2}h

= 22/ 7 × 1.75 × 1.75 × 10 m^{3} = 96.25 m^{3}

= 96.25 kl [1 m^{3} = 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 m^{3}
= 0.0154 m^{3}

Also, volume of the
cylindrical vessel = πr^{2}h

⇒ 0.0154 = 22 7 × r^{2 }× 1

= 0.0049 ⇒ r = 0.07 m

∴ Total surface area of the cylinder

= 2πr (h + r)

= 2 × 22/7 × 0.07 (1 + 0.07) m^{2}

= 44 × 0.01 × 1.07 m^{2
}= 0.4708 m^{2}

Hence, 0.4708 m^{2} 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 cm^{3}

= 0.11 cm^{3 }

The volume of the wood = π (R2 – r2)h

= 22/7
× [(0.35)^{2} – (0.05)^{2}] × 14 cm^{3}

= 22/7 ×
0.4 × 0.3 × 14 cm^{3} = 5.28 cm^{3}

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

= 0.11 cm^{3}

**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 = πr^{2}h

= 22/7 ×
3.5 × 3.5 × 4 cm^{3} = 154 cm^{3}

Hence, soup consumed by 250 patients per day =
250 × 154 cm^{3}

= 38500 cm^{3}