#### Class nine Maths NCERT Sol Surface Area Volume

## Chapter 13: Surface Areas and Volumes

#### TEXTBOOK’S EXERCISE – 13.2

**1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm**^{2}**. Find the diameter of the base of the cylinder. **

**Sol.
**Here, h = 14 cm,
curved surface area

= 88 cm2, r = ?

The curved surface area of the cylinder = 2πrh

⇒ 88 = 2 × 22/7 × r × 14

⇒ 88 = 44 × 2 × r ⇒ r = 88/44× 2 = 1

Hence, the base diameter of the cylinder

= 1 × 2 cm = 2 cm

**2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet is required for the same?**

**Sol. **Here, *h *= 1 m, *r *=
140/ 2 cm

= 70 cm = 0.7 m

Total surface area of the cylinder

= 2π*r *(*h
*+ *r*)

= 2 × 22 7 ×
0.7 (1 + 0.7) m^{2}

= 44 × 0.1 × 1.7 m2
= 7.48 m^{2 }

Hence, 7.48 m2 of sheet is required

#### Class nine Maths NCERT Sol Surface Area Volume

**3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see figure). Find its: **

(i) inner curved surface area,

(ii) outer curved surface area,

(iii) total surface area.

**Sol.
**Here, h = 77 cm,

Outer radius (R) = 4 4 2. cm = 2.2 cm

Inner radius (r) = 42 cm = 2 cm

(i) Inner curved
surface area of the pipe = 2πrh = 2
× 22/7 × 2 × 77 cm^{2}

= 2 × 22 × 22 cm^{2}
= 968 cm^{2}

(ii) Outer curved surface area of the pipe

= 2πRh = 2 × 22 7
× 2.2 × 77 cm^{2}

=
44 × 24.2 cm^{2}

=
1064.80 cm^{2}

(iii) Total surface area of the pipe = inner curved surface area + outer curved surface area + areas of the two base rings.

= 2πrh + 2πRh + 2π (R^{2}
– r^{2})

= 968 cm^{2}
+1064.80 cm^{2} + 2 × 22/7 [(2.2)^{2} – 2^{2}]
cm^{2 }

= 2032.80 cm^{2} + 5.28 cm^{2} =
2038.08 cm^{2}

**4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m**^{2}**.**

**Sol.
**Radius of the roller (r)
= 84/2 cm = 42 cm

Length of the roller (h) = 120 cm

Curved surface area of the roller

= 2πrh = 2 × 22/ 7 ×
42 × 120 cm^{2}

= 44 × 720 cm^{2}
= 31680 cm^{2}

∴ Area covered by the roller in 1 revolution

= 31680 cm^{2}

∴ Area covered by the roller in 500 revolutions = 31680 × 500 cm^{2}
= 15840000 cm^{2}

Hence, area of the playground

= 15840000/100x 100 m^{2} **= **1584 m^{2}

**5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs 12.50 per m**^{2}**. **

**Sol. **Here, r = 50/2 cm
= 25 cm = 0.25 m, h = 3.5 m

Curved surface area of the pillar

= 2πrh = 2 × 22/7 ×
0.25 × 3.5 m2 = 5.5 m^{2}

Cost of painting 1 m^{2}
= Rs 12.50

∴ Total cost of painting the curved surface of the pillar = Rs 12.50 × 5.5 = Rs 68.75