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

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 11.

NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.6 contains only 8 questions which are based on the formula of the geometrical objects. NCERT Solutions help students in their studies. Here, we have provided the solutions to all the questions in PDF format with detailed explanations. Students can download the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.6 PDF, through the links provided further below on this page.

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Access Other Exercise Solutions of Class 9 Maths Chapter 13 – Surface Areas and Volumes

Exercise 13.1 solution (9 questions)

Exercise 13.2 solution (8 questions)

Exercise 13.3 solution (9 questions)

Exercise 13.4 solution (5 questions)

Exercise 13.5 solution (5 questions)

Exercise 13.7 solution (9 questions)

Exercise 13.8 solution (10 questions)

Exercise 13.9 solution (3 questions)

Access Answers to NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.6

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= 1L) (Assume π = 22/7)

Solution:

Circumference of the base of cylindrical vessel = 132 cm

Height of vessel, h = 25 cm

Let r be the radius of the cylindrical vessel.

Step 1: Find the radius of vessel

We know that circumference of the base = 2πr, so

2πr = 132 (given)

r = (132/(2 π))

r = 66×7/22 = 21

Radius is 21 cm

Step 2: Find the volume of vessel

Formula: Volume of cylindrical vessel = πr²h

= (22/7)×21²×25

= 34650

Therefore, the volume is 34650 cm³

Since 1000 cm³ = 1L

So, Volume = 34650/1000 L= 34.65L

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. (Assume π = 22/7)

Solution:

Inner radius of cylindrical pipe, say r₁ = diameter₁/ 2 = 24/2 cm = 12cm

Outer radius of cylindrical pipe, say r₂ = diameter₂/ 2 = 28/2 cm = 14 cm

Height of pipe, h = Length of pipe = 35cm

Now, the Volume of pipe = π(r₂²-r₁²)h cm³

Substitute the values.

Volume of pipe = 110×52 cm³ = 5720 cm³

Since Mass of 1 cm³ wood = 0.6 g

Mass of 5720 cm³ wood = (5720×0.6) g = 3432 g or 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? (Assume π=22/7)

Solution:

(i) Tin can will be cuboidal in shape

The dimensions of the tin can are

Length, l = 5 cm

Breadth, b = 4 cm

Height, h = 15 cm

Capacity of tin can = l×b×h= (5×4×15) cm³ = 300 cm³

(ii) Plastic cylinder will be cylindrical in shape.

Dimensions of the plastic can are:

Radius of the circular end of the plastic cylinder, r = 3.5cm

Height , H = 10 cm

Capacity of the  plastic cylinder = πr²H

Capacity of the plastic cylinder = (22/7)×(3.5)²×10 = 385

Capacity of the plastic cylinder is 385 cm³

From results (i) and (ii), the plastic cylinder has more capacity.

Difference in capacity = (385-300) cm³ = 85cm³

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]

Solution:

CSA of cylinder = 94.2 cm²

Height of cylinder, h = 5cm

(i) Let the radius of the cylinder be r.

Using the CSA of the cylinder, we get

2πrh = 94.2

2×3.14×r×5 = 94.2

r = 3

radius is 3 cm

(ii) Volume of cylinder

Formula for volume of cylinder = πr²h

Now, πr²h = (3.14×(3)²×5) (using the value of r from (i))

= 141.3

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

(Assume π = 22/7)

Solution:

(i) Rs 20 is the cost of painting 1 m² area.

Rs 1 is the cost to paint 1/20 m² area

So, Rs 2200 is the cost of painting = (1/20×2200) m²

= 110 m² area

The inner surface area of the vessel is 110m².

(ii) Radius of the base of the vessel, let us say r.

Height (h) = 10 m and

Surface area formula = 2πrh

Using the result of (i)

2πrh = 110 m²

2×22/7×r×10 = 110

r = 1.75

Radius is 1.75 m.

(iii) Volume of vessel formula = πr²h

Here r = 1.75 and h = 10

Volume = (22/7)×(1.75)²×10 = 96.25

Volume of the vessel is 96.25 m³

Therefore, the capacity of the vessel is 96.25 m³ or 96250 litres.

6. The capacity of a closed cylindrical vessel of height 1m is15.4 liters. How many square meters of metal sheet would be needed to make it? (Assume π = 22/7)

Solution:

Height of the cylindrical vessel, h = 1 m

Capacity of  the cylindrical vessel = 15.4 litres = 0.0154 m³

Let r be the radius of the circular end.

Now,

Capacity of the vessel = (22/7)×r²×1 = 0.0154

After simplifying, we get r = 0.07 m

Again, the total surface area of vessel = 2πr(r+h)

= 2×22/7×0.07(0.07+1)

= 0.44×1.07

= 0.4708

Total surface area of the vessel is 0.4708 m²

Therefore, 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. (Assume π = 22/7)

Solution:

Radius of pencil, r₁ = 7/2 mm = 0.7/2 cm = 0.35 cm

Radius of graphite, r₂ = 1/2 mm = 0.1/2 cm = 0.05 cm

Height of pencil, h = 14 cm

Formula to find the volume of wood in pencil = (r₁²-r₂²)h cubic units

Substituting the values we have

= [(22/7)×(0.35²-0.05²)×14]

= 44×0.12

= 5.28

This implies the volume of wood in pencil = 5.28 cm³

Again,

The volume of graphite = r₂²h cubic units

Substituting the values, we have

= (22/7)×0.05²×14

= 44×0.0025

= 0.11

So, the volume of graphite is 0.11 cm³.

8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7cm. If the bowl is filled with soup to a height of 4cm, how much soup the hospital has to prepare daily to serve 250 patients? (Assume π = 22/7)

Solution:

Diameter of the cylindrical bowl = 7 cm

Radius of cylindrical bowl, r = 7/2 cm = 3.5 cm

Bowl is filled with soup to a height of4cm, so h = 4 cm

Volume of soup in one bowl= πr²h

(22/7)×3.5²×4 = 154

Volume of soup in one bowl is 154 cm³

Therefore,

Volume of soup given to 250 patients = (250×154) cm³= 38500 cm³

= 38.5litres.

We hope this information on “NCERT Solution for Class 9 Maths Chapter 13 – Surface Areas and Volumes is useful for students preparing for exams. Stay tuned for further updates on CBSE and other competitive exams. To access interactive Maths and Science Videos, download the BYJU’S App and subscribe to YouTube Channel.

Key Features of NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volume Exercise 13.6

Solving these solutions help students to:

• Students can self-assess their knowledge of the concept.
• It helps to solve the cylinder concept problems.
• It helps the student to find the inner curved surface area of the vessel, the radius of the base of the vessel and determine the capacity of the vessel.
• Improve efficiency in solving problems.
• Remember the formulas in an easy way.