NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6

NCERT Solutions Class 9 Maths
Chapter – 13 (Surface Areas and Volumes)

The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 13 Surface Areas and Volumes Exercise 13.6 has been provided here to help the students in solving the questions from this exercise.

Chapter 13: Surface Areas and Volumes

Exercise – 13.6

Assume π = $\mathbf{\frac{22}{7}}$ , unless stated otherwise.

1. The circumference of the base of cylindrical vessel is 132cm and its height is 25cm. How many litres of water can it hold? (1000 cm³= 1 l)

Answer –
Let the radius of the base = r
Height of the cylinder, h = 25 cm
Circumference of the base = 132 cm
2πr = 132 cm
r = 132/2π
= $\frac{132}{2\times \frac{22}{7}}$

= $\frac{132\times 7}{2\times 22}$
= 21 cm
The capacity of the cylindrical vessel = πr²h
= $\frac{22}{7}$ × 21 cm × 21 cm × 25 cm
= 34650 cm³= 34650/1000 (Since 1000 cm³ = 1 l)
= 34.65 l
Therefore, the vessel can hold 34.65 litres of water.

2. The inner diameter of a cylindrical wooden pipe is 24cm and its outer diameter is 28 cm. The length of the pipe is 35cm. Find the mass of the pipe, if 1cm³ of wood has a mass of 0.6g.

Answer –
Outer diameter of the pipe, D = 28 cm
Outer radius of the pipe, R = 28/2 = 14 cm
Inner diameter of the pipe, d = 24 cm
Inner radius of the pipe, r = 24/2 = 12cm
Length of the pipe, h = 35 cm
Volume of the outer cylinder, V₁ = πR²h
V₁ = $\frac{22}{7}$ × 14cm × 14cm × 35cm
= 21560 cm³Volume of the inner cylinder, V₂ = πr²h
V₂ = $\frac{22}{7}$ × 12cm × 12cm × 35cm
= 15840 cm³The volume of the wood used = Volume of the outer cylinder – Volume of the inner cylinder
= 21560 cm³ – 15840 cm³= 5720 cm³Mass of 1 cm³ wood is 0.6 g
Mass of 5720 cm³ wood = 5720 × 0.6g
= 3432 g
= (3432/1000) kg
= 3.432 kg
Thus, the mass of the wooden pipe is 3.432 kg

3. A soft drink is available in two packs –
(i) a tin can with a rectangular base of length 5cm and width 4cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7cm and height 10cm.
Which container has greater capacity and by how much?

Answer – The diagram of the plastic cylinder and rectangular base .
NCERT Class 9 Solutions Maths
Dimensions of the plastic cylinder with a circular base are:
The diameter of the cylindrical plastic can, d = 7 cm
The radius of the cylindrical plastic can, r = 7/2 cm
Height of the cylindrical plastic can, h = 10cm
The volume of the cylindrical plastic can = πr²h
= $\frac{22}{7}$ × $\frac{7}{2}$ cm × $\frac{7}{2}$ cm × 10cm
= 385 cm³

Dimensions of tin can with a rectangular base are:
The Length of the cuboidal tin can, l = 5 cm
The breadth of the cuboidal tin can, b = 4 cm
Height of the cuboidal tin can, h = 15cm
The volume of the cuboidal tin can = l × b × h
= 5 cm × 4 cm × 15 cm
= 300 cm³

Clearly, the plastic cylinder with a circular base has greater capacity than the tin container.
Difference = 385 cm³ – 300 cm³ = 85 cm³The plastic cylindrical has more capacity than the tin can by 85 cm³.

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

Answer –
(i) Let the radius of the cylinder = r
Height of the cylinder, h = 5 cm
Lateral surface area = 92.4 cm²2πrh = 92.4
r = 92.4 / 2hπ
r = $\frac{92.4}{2\times 5\times 3.14}$
r = 3 cm (approx.)

(ii) Volume of cylinder = πr²h
= 3.14 × 3 cm × 3 cm × 5 cm
= 141.3 cm³Thus, the radius of the base is 3 cm and the volume is 141.3 cm³.

5. It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10m 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

Answer –
Total cost to paint inner CSA = ₹ 2200
Rate of painting = ₹ 20 per m²Height of the vessel, h = 10 m
(i) Inner CSA of the cylindrical vessel = 2200/20 = 110 m²

(ii) radius
Inner CSA of the vessel = 110 m²2πrh = 110 m²r = 110 / 2πh
= $\frac{110}{2\times 10\times \frac{22}{7}}$

= $\frac{110\times 7}{2\times 10\times 22}$
= 1.75 m

(iii) Capacity of the vessel
Volume of the vessel = πr²h
= $\frac{22}{7}$ × 1.75 m × 1.75 m × 10 m
= 96.25 m³

6. The capacity of a closed cylindrical vessel of height 1m is 15.4 liters. How many square meters of metal sheet would be needed to make it?

Answer –
Let the radius of the vessel = r
Height of the cylindrical vessel, h = 1 m
Capacity of the vessel = 15.4 litres
= 15.4 / 1000 m³ (Since, 1000 l = 1 m³)
= 0.0154 m³Volume of the vessel = 0.0154 m³πr²h = 0.0154 m³r² = 0.0154 / πh
r² = $\frac{0.0154}{\frac{22}{7}\times 1}$ = $\frac{0.0154 \times 7}{22\times 1}$
r² = 0.0049 m²r = 0.07 m
TSA of the cylinder = 2πr(r + h)
= 2 × $\frac{22}{7}$ × 0.07m × (0.07m + 1m)
= 0.44 m × 1.07 m
= 0.4708 m²Therefore, 0.4708 m² of the metal sheet would be required to make the cylindrical vessel.

7. A lead pencil consists of a cylinder of wood with 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.

Answer –
For graphite:
Diameter of the graphite, d = 1 mm
Radius (r) = d/2 = 1 mm /2 = 0.5 mm = 0.5/10 cm = 0.05 cm
h = 14 cm
Volume of the graphite = πr²h
= $\frac{22}{7}$ × 0.05 cm × 0.05 cm × 14 cm
= 0.11 cm³

For pencil:
Diameter of the pencil, D = 7 mm
Radius (R) = 7 / 2 mm = 3.5/10 cm = 0.35 cm
h = 14 cm
Volume of the pencil = π R² h
= $\frac{22}{7}$ × 0.35 cm × 0.35 cm × 14 cm
= 5.39 cm³Volume of wood = Volume of the pencil – Volume of the graphite
= 5.39 cm³ – 0.11cm³= 5.28 cm³The volume of the wood is 5.28 cm³ and the volume of graphite is 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?

Answer –
Diameter of the bowl, d = 7 cm
Radius of the bowl, r = 7/2 cm = 3.5cm
Height of the bowl, h = 4 cm
The volume of soup in each bowl = πr²h
= $\frac{22}{7}$ × 3.5 cm × 3.5 cm × 4 cm
= 154 cm³The volume of soup for 1 patient = 154 cm³Thus, the volume of soup for 250 patients
= 250 × 154 cm³= 38500 cm³= 38500 / 1000 (Since, 1000 cm³ = 1 l)
= 38.5 l
The hospital has to prepare 38.5 litres of soup daily to serve 250 patients.

NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6