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 in this page.

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### Download PDF of NCERT Solutions for Class 9 Maths Chapter 13 â€“ Surface Areas and Volumes Exercise 13.6

### Access Other Exercise Solutions of Class 9 Maths Chapter 13 â€“ 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 â€“ 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 cm ^{3}= 1L) (Assume Ï€ = 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 base = 2Ï€r, so

2Ï€r = 132 (given)

r = (132/(2 Ï€))

r = 66Ã—7/22 = 21

Radius is 21 cm

**Step 2: Find the volume of vessel**

Formula: Volume of cylindrical vessel = Ï€r^{2}h

= (22/7)Ã—21^{2}Ã—25

= 34650

Therefore, volume is 34650 cm^{3}

Since, **1000 cm ^{3 }= 1L**

So, Volume = 34650/1000 L= 34.65L

Therefore, 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 1cm ^{3} of wood has a mass of 0.6g. (Assume Ï€ = 22/7)**

**Solution:**

Inner radius of cylindrical pipe, say r_{1} = diameter_{1}/ 2 = 24/2 cm = 12cm

Outer radius of cylindrical pipe, say r_{2} = diameter_{2}/ 2 = 28/2 cm = 14 cm

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

Now, the Volume of pipe = Ï€(r_{2}^{2}-r_{1}^{2})h cm^{3}

Substitute the values.

Volume of pipe = 110Ã—52 cm^{3 }= 5720 cm^{3}

Since**, Mass of 1 cm ^{3} wood = 0.6 g**

Mass of 5720 cm^{3} wood = (5720Ã—0.6) g = 3432 g or 3.432 kg. Answer!

**3. A soft drink is available in two packs â€“ (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 Ï€=22/7)**

**Solution:**

- tin can will be cuboidal in shape

Dimensions of tin can are

Length, l = 5 cm

Breadth, b = 4 cm

Height, h = 15 cm

Capacity of tin can = lÃ—bÃ—h= (5Ã—4Ã—15) cm^{3 }= 300 cm^{3}

- Plastic cylinder will be cylindrical in shape.

Dimensions of plastic can are:

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

Height , H = 10 cm

Capacity of plastic cylinder = Ï€r^{2}H

Capacity of plastic cylinder = (22/7)Ã—(3.5)^{2}Ã—10 = 385

Capacity of plastic cylinder is 385 cm^{3}

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

Difference in capacity = (385-300) cm^{3} = 85cm^{3}

**4. If the lateral surface of a cylinder is 94.2cm ^{2} and its height is 5cm, then find **

**(i) radius of its base (ii) its volume.[Use Ï€= 3.14]**

**Solution:**

CSA of cylinder = 94.2 cm^{2}

Height of cylinder, h = 5cm

(i) Let radius of cylinder be r.

Using CSA of cylinder, we get

2Ï€rh = 94.2

2Ã—3.14Ã—rÃ—5 = 94.2

r = 3

radius is 3 cm

(ii) Volume of cylinder

Formula for volume of cylinder = Ï€r^{2}h

Now, Ï€r^{2}h = (3.14Ã—(3)^{2}Ã—5) (using value of r from (i))

= 141.3

Volume is 141.3 cm^{3}

**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 m ^{2}, find **

**(i) inner curved surface area of the vessel **

**(ii) radius of the base **

**(iii) capacity of the vessel **

**(Assume Ï€ = 22/7)**

**Solution:**

(i) Rs 20 is the cost of painting 1 m^{2 }area.

Rs 1 is the cost to paint 1/20 m^{2 }area

So, Rs 2200 is the cost of painting = (1/20Ã—2200) m^{2 }

= 110 m^{2} area

The inner surface area of the vessel is 110m^{2}.

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

Height (h) = 10 m and

Surface area formula = 2Ï€rh

Using result of (i)

2Ï€rh = 110 m^{2}

2Ã—22/7Ã—rÃ—10 = 110

r = 1.75

Radius is 1.75 m.

(iii) Volume of vessel formula = Ï€r^{2}h

Here r = 1.75 and h = 10

Volume = (22/7)Ã—(1.75)^{2}Ã— 10 = 96.25

Volume of vessel is 96.25 m^{3}

Therefore, the capacity of the vessel is 96.25 m^{3} 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 Ï€ = 22/7)**

**Solution:**

Height of cylindrical vessel, h = 1 m

Capacity of cylindrical vessel = 15.4 litres = 0.0154 m^{3}

Let r be the radius of the circular end.

Now,

Capacity of cylindrical vessel = (22/7)Ã—r^{2}Ã—1 = 0.0154

After simplifying, we get, r = 0.07 m

Again, total surface area of vessel = 2Ï€r(r+h)

= 2Ã—22/7Ã—0.07Ã—(0.07+1)

= 0.44Ã—1.07

= 0.4708

Total surface area of vessel is 0.4708 m^{2}

Therefore, 0.4708 m^{2} 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 Ï€ = 22/7)**

**Solution:**

Radius of pencil, r_{1 }= 7/2 mm = 0.7/2 cm = 0.35 cm

Radius of graphite, r_{2 }= 1/2 mm = 0.1/2 cm = 0.05 cm

Height of pencil, h = 14 cm

Formula to find, volume of wood in pencil = (r_{1}^{2}-r_{2}^{2})h cubic units

Substitute the values, we have

= [(22/7)Ã—(0.35^{2}-0.05^{2})Ã—14]

= 44Ã—0.12

= 5.28

This implies, volume of wood in pencil = 5.28 cm^{3}

Again,

Volume of graphite = r_{2}^{2}h cubic units

Substitute the values, we have

= (22/7)Ã—0.05^{2}Ã—14

= 44Ã—0.0025

= 0.11

So, the volume of graphite is 0.11 cm^{3}.

**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 does the hospital have to prepare daily to serve 250 patients? (Assume Ï€ = 22/7)**

**Solution:**

Diameter of 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= Ï€r^{2}h

(22/7)Ã—3.5^{2}Ã—4 = 154

Volume of soup in one bowl is 154 cm^{3}

Therefore,

Volume of soup given to 250 patients = (250Ã—154) cm^{3}= 38500 cm^{3}

= 38.5litres. Answer!

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### Key Features of NCERT Solutions for Class 9 Maths Chapter 13 â€“ Surface Areas and Volume Exercise 13.6

Solving these solutions help students to:

- Students can self-assess their knowledge about the concept
- 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 the problems
- Remember the formulas in an easy way

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