## RD Sharma Solutions Class 9 Maths Chapter 19 – Free PDF Download

**RD Sharma Solutions for Class 9 Maths Chapter 19 Surface Area and Volume of A Right Circular Cylinder** provide in-depth knowledge of each concept, keeping in mind the different grasping abilities of students. In this chapter, students will learn how to find the surface area and volume of a cylinder. RD Sharma Solutions for Class 9 Chapter 19 are given here to help students understand the important concepts with ease and prepare for the annual exams more effectively.

Class 9 is a stage where several important topics are introduced. These crucial topics are discussed here and are formulated by BYJU’S experts in Maths. The **RD Sharma Solutions** are provided in accordance with the latest syllabus of CBSE updated for 2023-24 which, in turn, helps students to build a strong foundation of the basics and secure excellent marks in their board exams.

## RD Sharma Solutions for Class 9 Maths Chapter 19 Surface Area and Volume of a Right Circular Cylinder

### Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 19 Surface Area and Volume of A Right Circular Cylinder

### Exercise 19.1 Page No: 19.7

**Question 1: Curved surface area of a right circular cylinder is 4.4 m ^{2}. If the radius of the base of the cylinder is 0.7 m. Find its height.**

**Solution:**

Radius of the base of the cylinder = r = 0.7 m (Given)

Curved surface area of cylinder = C.S.A = 4.4m** ^{2 }** (Given)

Let ‘h’ be the height of the cylinder.

We know, curved surface area of a cylinder = 2πrh

Therefore,

2πrh = 4.4

2 x 3.14 x 0.7 x h = 4.4

[using π=3.14 ]or h = 1

*Therefore the height of the cylinder is 1 m.*

**Question 2: In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.**

**Solution:**

Height of cylinder (h) = Length of cylindrical pipe = 28 m or 2800 cm (Given)

[1 m = 100 cm]Diameter of circular end of pipe = 5 cm (given)

Let ‘r’ be the radius of circular end, then r = diameter/2 = 5/2 cm

We know, Curved surface area of cylindrical pipe = 2πrh

= 2 x 3.14 x 5/2 x 2800

[using π = 3.14]= 44000

*Therefore, the area of radiating surface is 44000 cm ^{2}.*

**Question 3: 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}.**

**Solution:**

Height of cylindrical pillar (h) = 3.5 m

Radius of circular end of pillar ( r) = 50/2 cm = 25 cm = 0.25 m

[As radius = half of the diameter] and [1 m = 100 cm]Curved surface area of cylindrical pillar = 2πrh

= 2 x 3.14 x 0.25 x 3.5

= 5.5

Curved surface area of cylindrical pillar is 5.5 m.

Find the cost:

Cost of whitewashing 1m^{2} is Rs 12.50 (Given)

Cost of whitewashing 5.5 m^{2} area = Rs. 12.50 x 5.5 = Rs. 68.75

*Thus the cost of whitewashing the pillar is Rs 68.75.*

**Question 4: It is required to make a closed cylindrical tank of height 1 m and the base diameter of 140 cm from a metal sheet. How many square meters of the sheet are required for the same? **

**Solution:**

Height of cylindrical tank (h) = 1 m

Base radius of cylindrical tank (r) = diameter/2 = 140/2 cm = 70 cm = 0.7 m

[1 m = 100 cm]Now,

Area of sheet required = Total surface area of tank (TSA) = 2πr(h + r)

=2 x 3.14 x 0.7(1 + 0.7)

= 7.48

*Therefore, 7.48 m ^{2} metal sheet is required to make required closed cylindrical tank.*

**Question 5: A solid cylinder has a total surface area of 462 cm ^{2}. Its curved surface area is one-third of its total surface area. Find the radius and height of the cylinder. **

**Solution:**

Total surface area of a cylinder = 462 cm^{2 } (Given)

As per given statement:

Curved or lateral surface area = 1/3 (Total surface area)

⇒ 2πrh = 1/3(462)

⇒ 2πrh = 154

⇒ h = 49/2r ….(1)

[Using π = 22/7]Again,

Total surface area = 462 cm^{2}

2πr(h + r) = 462

2πr(49/2r + r) = 462

or 49 + 2r^{2} = 147

or 2r^{2} = 98

or r = 7

Substitute the value of r in equation (1), and find the value of h.

h = 49/2(7) = 49/14 = 7/2

Height (h) = 7/2 cm

*Answer: Radius = 7 cm and height = 7/2 cm of the cylinder*

**Question 6: The total surface area of a hollow cylinder which is open on both the sides is 4620 sq.cm and the area of the base ring is 115.5 sq.cm and height is 7 cm. Find the thickness of the cylinder.**

**Solution:**

Given:

Total surface area of hollow cylinder = 4620 cm^{2}

Height of cylinder (h) = 7 cm

Area of base ring = 115.5 cm^{2}

To find: Thickness of the cylinder

Let ‘r_{1}’ and ‘r_{2}’ are the inner and outer radii of the hollow cylinder respectively.

Then, πr_{2}^{2} – πr_{1}^{2} = 115.5 …….(1)

And,

2πr_{1}h +2πr_{2}h+ 2(πr_{2}^{2} – πr_{1}^{2}) = 4620

Or 2πh (r_{1} + r_{2} ) + 2 x 115.5 = 4620

(Using equation (1) and h = 7 cm)

or 2π7 (r_{1} + r_{2} ) = 4389

or π (r_{1} + r_{2} ) = 313.5 ….(2)

Again, from equation (1),

πr_{2}^{2} – πr_{1}^{2} = 115.5

or π(r_{2} + r_{1}) (r_{2} – r_{1}) = 115.5

Using result of equation (2),

313.5 (r_{2} – r_{1}) = 115.5

or r_{2} – r_{1 }= 7/19 = 0.3684

*Therefore, thickness of the cylinder is 7/19 cm or 0.3684 cm.*

**Question 7: Find the ratio between the total surface area of a cylinder to its curved surface area, given that height and radius of the tank are 7.5 m and 3.5 m.**

**Solution:**

Height of cylinder (h) = 7.5 m

Radius of cylinder (r) = 3.5 m

We know, Total Surface Area of cylinder (T.S.A) = 2πr(r+h)

And, Curved surface area of a cylinder(C.S.A) = 2πrh

Now, Ratio between the total surface area of a cylinder to its curved surface area is

T.S.A/C.S.A = 2πr(r+h)/2πrh

= (r + h)/h

= (3.5 + 7.5)/7.5

= 11/7.5

= 22/15 or 22:15

*Therefore the required ratio is 22:15.*

### Exercise 19.2 Page No: 19.20

**Question 1: A soft drink is available in two packs- (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm, Which container has greater capacity and by how much?**

**Solution:**

(i) Dimensions of a cubical tin can:

Length (L) = 5 cm

Breadth (B) = 4 cm

Height (H) = 15 cm

Capacity of the tin can = Volume of Tin Can = l x b x h cubic units = (5 x 4 x 15) cm^{3 }= 300 cm^{3}

(ii) Radius of the circular end of the plastic cylinder (R) = diametr/2 = 7/2 cm = 3.5 cm

Height of plastic cylinder (H) = 10 cm

Capacity of plastic cylinder = Volume of cylindrical container = πR^{2}H = 22/7 × (3.5)^{2} × 10 cm^{3} = 385 cm^{3}

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

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

**Question 2: The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius 20 cm and height 10 m. How much concrete mixture would be required to build 14 such pillars?**

**Solution:**

In this case, we have to find the volume of the cylinders.

Given:

Radius of the base of a cylinder = 20 cm

Height of cylinder = 10 m = 1000 cm

[1m = 100 cm]Volume of the cylindrical pillar = πR^{2}H

= (22/7×20^{2}×1000) cm^{3}

= 8800000/7 cm^{3} or 8.87 m^{3}

*Therefore, volume of 14 pillars = 14 x 8.87 m ^{3} = 17.6 m^{3}*

**Question 3: The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm ^{3} of wood has a mass of 0.6 gm.**

**Solution:**

Let r and R be the inner and outer radii of cylindrical pipe.

Inner radius of a cylindrical pipe (r) = 24/2 = 12 cm

Outer radius of a cylindrical pipe (R) = 24/2 = 14 cm

Height of pipe (h) = length of pipe = 35 cm

Mass of pipe = volume x density = π(R^{2} – r^{2})h

= 22/7(14^{2} – 12^{2})35

= 5720

Mass of pipe is 5720 cm^{3}

Mass of 1 cm^{3} wood = 0.6 gm (Given)

*Therefore, mass of 5720 cm ^{3} wood = 5720 x 0.6 = 3432 gm = 3.432 kg*

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

**i) radius of its base (ii) volume of the cylinder**

** [Use π = 3.141]**

**Solution:**

Lateral surface of the cylinder = 94.2 cm^{2}

Height of the cylinder = 5 cm

Let ‘r’ be the radius.

(i) Lateral surface of the cylinder = 94.2 cm^{2}

2 πrh = 94.2

or 2 x 3.14 x r x 5 = 94.2

or r = 3 cm

(ii) Volume of the cylinder = πr^{2}h

= (3.14 x 3^{2} x 5) cm^{3}

*= 141.3 cm ^{3}*

**Question 5: The capacity of a closed cylindrical vessel of height 1 m is 15.4 liters. How many square meters of the metal sheet would be needed to make it?**

**Solution:**

Given, The capacity of a closed cylindrical vessel of height 1 m is 15.4 liters.

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

^{3}= 1000 litres]

Let ‘r’ be the radius of the circular ends of the cylinders, then

πr^{2}h = 0.0154 m^{3}

3.14 x r^{2} x 1 = 0.0154 m^{3}

or r = 0.07 m

Again,

Total surface area of a vessel = 2πr(r+h)

= 2(3.14(0.07)(0.07+1)) m^{2}

*= 0.470 m ^{2}*

**Question 6: 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?**

**Solution:**

Radius of cylindrical bowl (R) = diameter/2 = 7/2 cm = 3.5 cm

Height = 4 cm

Now,

Volume of soup in 1 bowl = πr^{2}h

= 22/7×3.5^{2}×4 cm^{3}

= 154 cm^{3}

Volume of soup in 250 bowls = (250 x 154) cm^{3}

= 38500 cm^{3}

= 38.5 liters

*Thus, hospital has to prepare 38.5 liters of soup daily in order to serve 250 patients.*

**Question 7: A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4 cm thick iron. Find the volume of the iron.**

**Solution:**

The outer circumference of the roller = 440 cm

Thickness of the roller = 4 cm and

Its height (h) = 63 cm

Let ‘R’ be the external radius and ‘r’ be the inner radius of the roller.

Circumference of roller = 2πR = 440

Or 2πR = 440

2×22/7 x R = 440

or R = 70

And, inner radius ‘r’ is given as

⇒ r = R – 4

⇒ r = 70 – 4

⇒ r = 66

Inner radius is 66 cm

Now, volume of the iron is given as

V = π(R^{2}−r^{2})h

V = 22/7 (70^{2}−66^{2})63

V = 107712

*Therefore, required volume is 107712 cm ^{3}.*

**Question 8: A solid cylinder has a total surface area of 231 cm ^{2}. Its curved surface area is 2/3 of the total surface area. Find the volume of the cylinder.**

**Solution:**

Total surface area = 231 cm^{2}

As per given statement: Curved surface area = 2/3(Total surface area)

Curved surface area = 2/3 x 231 = 154

So, Curved surface area = 154 cm^{2} …(1)

We know, Curved surface area of cylinder = 2πrh + 2πr^{2}

Or 2πrh + 2πr^{2} = 231 …..(2)

Here 2πrh is the curved surface area, so using (1), we have

⇒ 154 + 2πr^{2} = 231

⇒ 2πr^{2} = 231- 154

⇒ 2 x 22/7 x r^{2} = 77

⇒ r^{2} = 49/4

or r = 7/2

Find the value of h:

CSA = 154 cm^{2}

⇒ 2πrh = 154

⇒ 2 x 22/7 x 7/2 x h = 154

⇒ h = 154/22

⇒ h = 7

Now,

Find Volume of the cylinder:

V = πr^{2}h

= 22/7 x 7/2 x 7/2 x 7

= 269.5

*The volume of the cylinder is 269.5 cm ^{3}*

**Question 9: The cost of painting the total outside surface of a closed cylindrical oil tank at 50 paise per square decimetre is Rs 198. The height of the tank is 6 times the radius of the base of the tank. Find the volume corrected to 2 decimal places.**

**Solution:**

Let ‘r’ be the radius of the tank.

As per given statement: Height (h) = 6(Radius) = 6r dm

Cost of painting for 50 paisa or Rs 1/2 per dm^{2} = Rs 198 (Given)

⇒ 2πr(r+h) × 1/2 = 198

⇒ 2×22/7×r(r+6r) × 1/2 = 198

⇒ r = 3 dm

And, h = (6 x 3) dm = 18 dm

Now,

*Volume of the tank = πr ^{2}h = 22/7×9×18 = 509.14 dm^{3}*

**Question 10: The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their volumes and the ratio of their curved surfaces.**

**Solution:**

Let the radius of the cylinders be 2x and 3x and the height of the cylinders be 5y and 3y.

**Question 11: The ratio between the curved surface area and the total surface area of a right circular cylinder is 1:2. Find the volume of the cylinder, if its total surface area is 616 cm ^{2}.**

**Solution:**

Total surface area (T.S.A) = 616 cm^{2} (given)

Let r be the radius of cylinder and h be the radius of cylinder.

As per given statement:

(curved surface area / (total surface area) = 1/2

or CSA = 12 TSA

CSA = 12 x 616 = 308

⇒ CSA = 308 cm^{2}

Now,

TSA = 2πrh + 2πr^{2}

⇒ 616 = CSA + 2πr^{2}

⇒ 616 = 308 + 2πr^{2}

⇒ 2πr^{2} = 616 – 308

⇒ 2πr^{2} = 308/2π

⇒ r^{2} = 49

or r = 7 cm …(1)

As, CSA = 308 cm^{2}

2πrh = 308

⇒ 2 x 22/7 x 7 x h = 308

(using (1))

⇒ h = 7 cm

Now,

Volume of cylinder = πr^{2}h

= 22/7 x 7 x 7 x 7

= 1078

*Therefore, Volume of cylinder is 1078 cm ^{3}.*

**Question 12: The curved surface area of a cylinder is 1320 cm ^{2} and its base had diameter 21 cm. Find the height and volume of the cylinder.**

**Solution:**

Curved surface area of a cylinder = 1320 cm^{2}

Let, r be the radius of the cylinder and h be the height of the cylinder.

⇒ r = diameter/2 = 21/2 cm = 10.5 cm

We know, Curved surface area(CSA) = 2πrh

So, 2πrh = 1320

⇒ 2x 22/7 x 10.5 x h = 1320

or h = 20 cm

Now,

Volume of cylinder = πr^{2}h

= 22/7 x 10.5 x 10.5 x 20

= 6930

*Thus, Volume of cylinder is 6930 cm ^{3}.*

**Question 13: The ratio between the radius of the base and the height of a cylinder is 2:3. Find the total surface area of the cylinder, if its volume is 1617cm ^{3}.**

**Solution:**

Let, r be the radius of the cylinder and h be the height of the cylinder.

As per statement: r:h = 2:3

Then, radius = 2x cm and height = 3x cm

Volume of cylinder = πr^{2}h

And Volume of cylinder= 1617 cm^3 (given)

So, 1617= 22/7 (2x)^{2 }3x

1617 = 22/7 (12 x^{3} )

x^{3 }= 343/8

or x = 7/2

or x = 3.5 cm

Now, radius, r = 2 x 3.5 = 7 cm and

Height = 3x = 3 x 3.5 = 10.5 cm

Now,

Total surface area of cylinder = 2πr(h+r)

= 2 x 22/7 x 7(10.5+7)

= 770

*Thus, Total surface area of cylinder is 770 cm ^{2}.*

**Question 14: A rectangular sheet of paper, 44 cm x 20 cm, is rolled along its length of form cylinder. Find the volume of the cylinder so formed.**

**Solution:**

Length of a rectangular sheet = 44 cm

Height of a rectangular sheet = 20 cm

Now, 2πr = 44

r = 44/2π

r = 44 x 1/2 x 7/22

or r = 7 cm

Now,

Volume of cylinder = π r^{2}h

= 22/7 x 7 x 7 x 20

= 3080

*So, Volume of cylinder is 3080 cm ^{3}.*

**Question 15: The curved surface area of cylindrical pillar is 264 m ^{2} and its volume is 924 m^{3}. Find the diameter and the height of the pillar.**

**Solution:**

Let, r be the radius of the cylindrical pillar and h be the height of the cylindrical pillar

Curved surface area of cylindrical pillar = CSA = 264 m^{2 } (Given)

So, 2πrh = 264

or πrh = 132 …(1)

Again,

Volume of the cylinder = 924 m^{3} (given)

πr^{2} h= 924

or πrh(r) = 924

Using equation (1)

132 r = 924

or r = 924/132

or r = 7m

Substitute value of r value in equation (1)

22/7 x 7 x h = 132

Or h = 6m

*Therefore, diameter = 2r = 2(7) = 14 m and height = 6 m*

### Exercise VSAQs Page No: 19.27

**Question 1: Write the number of surfaces of a right circular cylinder.**

**Solution: **

There are 3 surfaces in a cylinder.

**Question 2: Write the ratio of total surface area to the curved surface area of a cylinder of radius r and height h.**

**Solution: **

Ratio of total surface area to the curved surface area of a cylinder of radius r and height h can be written as:

### RD Sharma Solutions for Class 9 Maths Chapter 19 Surface Area and Volume of A Right Circular Cylinder

In the 19th Chapter of Class 9 RD Sharma Solutions students will study important concepts listed below:

- Introduction to Right Circular Cylinder
- Some important definitions of Terms – Base, Axis, Radius, Height and Lateral Surface.
- Surface Area of a Cylinder
- Volume of a Cylinder

Here is the formula which is applied to calculate the surface area and volume of a circular cylinder.

- Surface area of a cylinder = 2πr (r + h) and
- Volume of a cylinder = πr
^{2}h Where r = radius of the cylinder and h = height of the cylinder.

## Frequently Asked Questions on RD Sharma Solutions for Class 9 Maths Chapter 19

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