## RD Sharma Solutions Class 9 Maths Chapter 21 – Free PDF Download

**RD Sharma Solutions for Class 9 Maths Chapter 21** is given here which consists of questions and answers related to Spheres. These questions are really helpful for the preparation of different Maths exams including Maths Olympiad and IIT-JEE. Here all solutions to questions in RD Sharma textbook Solutions Class 9 are given in a detailed and step by step format to help the students understand the concepts and clear all their doubts easily.

Here you will find several exercises of **Class 9 Maths chapter 21 – Surface Area and Volume of Sphere**. Click on the link below to download all the solutions.

In this chapter, students will learn to find the surface area and volume of the sphere. Let us have a brief look on what is a sphere, In **three-dimensional space**, these are a set of points at an equal distance â€˜râ€™ from a given point. The shape resembles a ball.

## Download PDF of RD Sharma Solutions for Class 9 Maths Chapter 21 Surface Area and Volume of A Sphere

### Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 21 Surface Area and Volume of A Sphere

### Exercise 21.1 Page No: 21.8

**Question 1: Find the surface area of a sphere of radius:**

**(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm**

**Solution:**

Surface area of a sphere = 4Ï€r^{2}

Where, r = radius of a sphere

**(i)** Radius = 10.5 cm

Surface area = 4 x 22/7 x (10.5)^{2}

= 1386

Surface area is 1386 cm^{2}

**(ii)** Radius= 5.6 cm

Surface area = 4Ã—22/7Ã—(5.6)^{2}

= 394.24

Surface area is 394.24 cm^{2}

**(iii)** Radius = 14 cm

Surface area = 4Ã—22/7Ã—(14)^{2}

= 2464

Surface area is 2464 cm^{2}

**Question 2: Find the surface area of a sphere of diameter:**

**(i) 14 cm (ii) 21 cm (iii) 3.5 cm**

**Solution:**

Surface area of a sphere = 4Ï€r^{2}

Where, r = radius of a sphere

**(i)** Diameter= 14 cm

So, Radius = Diameter/2 = 14/2 cm = 7 cm

Surface area = 4Ã—22/7Ã—(7)^{2}

= 616

Surface area is 616 cm^{2}

**(ii) **Diameter = 21cm

So, Radius = Diameter/2 = 21/2 cm = 10.5 cm

Surface area= 4Ã—22/7Ã—(10.5)^{2}

= 1386

Surface area is 1386 cm^{2}

**(iii) **Diameter= 3.5cm

So, Radius = Diameter/2 = 3.5/2 cm = 1.75 cm

Surface area = 4Ã—22/7Ã—(1.75)^{2}

= 38.5

Surface area is 38.5 cm^{2}

**Question 3: Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm. (Ï€=3.14)**

**Solution:**

Radius of a hemisphere = Radius of a solid hemisphere = 10 cm (Given)

Surface area of the hemisphere = 2Ï€r^{2}

= 2Ã—3.14Ã—(10)^{2 } cm^{2}

= 628 cm^{2}

And, surface area of solid hemisphere = 3Ï€r^{2}

= 3Ã—3.14Ã—(10)^{2 } cm^{2}

= 942 cm^{2}

**Question 4: The surface area of a sphere is 5544 cm ^{2}, find its diameter.**

**Solution:**

Surface area of a sphere is 5544 cm^{2}

Surface area of a sphere = 4Ï€r^{2}

So, 4Ï€r^{2} = 5544

4Ã—22/7Ã—(r)^{2} = 5544

r^{2} = (5544 Ã— 7)/88

r^{2} = 441

or r = 21cm

Now, Diameter=2(radius) = 2(21) = 42cm

**Question 5: A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin plating it on the inside at the rate of Rs.4 per 100 cm ^{2}.**

**Solution:**

Inner diameter of hemispherical bowl = 10.5 cm

So, radius = Diameter/2 = 10.5/2 cm = 5.25 cm

Now, Surface area of hemispherical bowl = 2Ï€r^{2}

= 2 Ã— 3.14 Ã— (5.25)^{2}

= 173.25

So, Surface area of hemispherical bowl is 173.25 cm^{2}

Find the cost:

Cost of tin plating 100 cm^{2} area= Rs.4 (given)

Cost of tin plating 173.25cm^{2} area = Rs. 4Ã—173.25100 = Rs. 6.93

Therefore, cost of tin plating the inner side of hemispherical bowl is Rs.6.93.

**Question 6: The dome of a building is in the form of a hemisphere. Its radius is 63 dm. Find the cost of painting it at the rate of Rs. 2 per sq m.**

**Solution: **

Radius of hemispherical dome = 63 dm or 6.3 m

Inner surface area of dome = 2Ï€r^{2}

=2Ã—3.14Ã—(6.3)^{2}

= 249.48

So, Inner surface area of dome is 249.48 m^{2}

Now find the cost:

Cost of painting 1m^{2 }= Rs.2 (given)

Therefore, cost of painting 249.48 m^{2}= Rs. (249.48Ã—2) = Rs.498.96.

### Exercise 21.2 Page No: 21.19

**Question 1: Find the volume of a sphere whose radius is:**

**(i) 2 cm (ii) 3.5 cm (iii) 10.5 cm.**

**Solution: **

Volume of a sphere = 4/3Ï€r^{3} Cubic Units

Where, r = radius of a sphere

**(i)** Radius = 2 cm

Volume = 4/3 Ã— 22/7 Ã— (2)^{3}

= 33.52

Volume = 33.52 cm^{3}

**(ii)** Radius = 3.5cm

Therefore volume = 4/3Ã—22/7Ã—(3.5)^{3}

= 179.666

Volume = 179.666 cm^{3}

**(iii)** Radius = 10.5 cm

Volume = 4/3Ã—22/7Ã—(10.5)^{3}

= 4851

Volume = 4851 cm^{3}

**Question 2: Find the volume of a sphere whose diameter is: **

**(i) 14 cm (ii) 3.5 dm (iii) 2.1 m**

**Solution:**

Volume of a sphere = 4/3Ï€r^{3} Cubic Units

Where, r = radius of a sphere

**(i)** diameter =14 cm

So, radius = diameter/2 = 14/2 = 7cm

Volume = 4/3Ã—22/7Ã—(7)^{3 }

= 1437.33

Volume = 1437.33 cm^{3}

**(ii)** diameter = 3.5 dm

So, radius = diameter/2 = 3.5/2 = 1.75 dm

Volume = 4/3Ã—22/7Ã—(1.75)^{3}

= 22.46

Volume = 22.46 dm^{3}

**(iii)** diameter = 2.1 m

So, radius = diameter/2 = 2.1/2 = 1.05 m

Volume = 4/3Ã—22/7Ã—(1.05)^{3}

= 4.851

Volume = 4.851 m^{3}

**Question 3: A hemispherical tank has the inner radius of 2.8 m. Find its capacity in liters.**

**Solution:**

Radius of hemispherical tank = 2.8 m

Capacity of hemispherical tank = 2/3 Ï€r^{3}

=2/3Ã—22/7Ã—(2.8)^{3 } m^{3}

= 45.997 m^{3}

^{3 }= 1000 liters]

Therefore, capacity in litres = 45997 litres

**Question 4: A hemispherical bowl is made of steel 0.25 cm thick. The inside radius of the bowl is 5 cm. Find the volume of steel used in making the bowl.**

**Solution:**

Inner radius of a hemispherical bowl = 5 cm

Outer radius of a hemispherical bowl = 5 cm + 0.25 cm = 5.25 cm

Volume of steel used = Outer volume – Inner volume

= 2/3Ã—Ï€Ã—((5.25)^{3}âˆ’(5)^{3})

= 2/3Ã—22/7Ã—((5.25)^{3}âˆ’(5)^{3})

= 41.282

Volume of steel used is 41.282 cm^{3}

**Question 5: How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?**

**Solution:**

Edge of a cube = 22 cm

Diameter of bullet = 2 cm

So, radius of bullet (r) = 1 cm

Volume of the cube = (side)^{3} = (22)^{3} cm^{3 }= 10648 cm^{3}

And,

Volume of each bullet which will be spherical in shape = 4/3Ï€r^{3}

= 4/3 Ã— 22/7 Ã— (1)^{3 } cm^{3}

= 4/3 Ã— 22/7 cm^{3}

= 88/21 cm^{3}

Number of bullets = (Volume of cube) / (Volume of bullet)

= 10648/88/21

= 2541

Therefore, 2541 bullets can be made.

**Question 6: A shopkeeper has one laddoo of radius 5 cm. With the same material, how many laddoos of radius 2.5 cm can be made?**

**Solution:**

Volume of laddoo having radius 5 cm (V1) = 4/3Ã—22/7Ã—(5)^{3}

= 11000/21 cm^{3}

Also, Volume of laddoo having radius 2.5 cm (V2) = 4/3Ï€r^{3}

= 4/3Ã—22/7Ã—(2.5)^{3 } cm^{3}

= 1375/21 cm^{3}

Therefore,

Number of laddoos of radius 2.5 cm that can be made = V1/V2 = 11000/1375 = 8

**Question 7: A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be 3/2cm and 2 cm, find the diameter of the third ball.**

**Solution:**

Volume of lead ball with radius 3/2 cm = 4/3Ï€r^{3}

= 4/3Ã—Ï€Ã—(3/2)^{3}

Let, Diameter of first ball (d1) = 3/2cm

Radius of first ball (r1) = 3/4 cm

Diameter of second ball (d2) = 2 cm

Radius of second ball (r2) = 2/2 cm = 1 cm

Diameter of third ball (d3) = d

Radius of third ball (r3) = d/2 cm

Now,

So, diameter of third ball is 2.5 cm.

**Question 8: A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises 5/3 cm. Find the radius of the cylinder.**

**Solution:**

Radius of sphere = 5 cm (Given)

Let â€˜râ€™ be the radius of cylinder.

We know, Volume of sphere = 4/3Ï€r^{3}

By putting values, we get

= 4/3Ã—Ï€Ã—(5)^{3}

Height (h) of water rises is 5/3 cm (Given)

Volume of water rises in cylinder = Ï€r^{2}h

Therefore, Volume of water rises in cylinder = Volume of sphere

So, Ï€r^{2}h = 4/3Ï€r^{3}

Ï€ r^{2} Ã— 5/3 = 4/3 Ã— Ï€ Ã— (5)^{3}

or r^{2 }= 100

or r = 10

Therefore, radius of the cylinder is 10 cm.

**Question 9: If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere?**

**Solution: **

Let r be the radius of the first sphere then 2r be the radius of the second sphere.

Now,

Ratio of volume of the first sphere to the second sphere is 1:8.

**Question 10: A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.**

**Solution: **

Volume of the cone = Volume of the hemisphere (Given)

1/3Ï€r^{2}h = 2/3 Ï€r^{3}

(Using respective formulas)

r^{2}h = 2r^{3}

or h = 2r

Since, cone and a hemisphere have equal bases which implies they have the same radius.

h/r = 2

or h : r = 2 : 1

Therefore, Ratio of their heights is 2:1

**Question 11: A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm respectively. Find the height to which the water will rise in the cylinder.**

**Solution:**

Volume of water in the hemispherical bowl = Volume of water in the cylinder â€¦ (Given)

Inner radius of the bowl ( r_{1}) = 3.5cm

Inner radius of cylinder (r_{2}) = 7cm

Volume of water in the hemispherical bowl = Volume of water in the cylinder

2/3Ï€r_{1}^{3} = Ï€r_{2}^{2}h

Where h be the height to which water rises in the cylinder.

2/3Ï€(3.5)^{3} = Ï€(7)^{2}h

or h = 7/12

Therefore, 7/12 cm be the height to which water rises in the cylinder.

**Question 12: A cylinder whose height is two thirds of its diameter, has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.**

**Solution:**

Radius of a sphere (R)= 4 cm (Given)

Height of the cylinder = 2/3 diameter (given)

We know, Diameter = 2(Radius)

Let h be the height and r be the base radius of a cylinder, then

h = 2/3Ã— (2r) = 4r/3

Volume of the cylinder = Volume of the sphere

Ï€r^{2}h = 4/3Ï€R^{3}

Ï€ Ã— r^{2} Ã— (4r/3) = 4/3 Ï€ (4)^{3}

(r)^{3} = (4)^{3}

or r = 4

Therefore, radius of the base of the cylinder is 4 cm.

**Question 13: A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of water in the cylinder.**

**Solution:**

Radius of a bowl (R)= 6 cm (Given)

Radius of a cylinder (r) = 4 cm (given)

Let h be the height of a cylinder.

Now,

Volume of water in hemispherical bowl = Volume of cylinder

2/3 Ï€ R^{3} = Ï€r^{2} h

2/3 Ï€ (6)^{3} = Ï€(4)^{2} h

or h = 9

Therefore, height of water in the cylinder 9 cm.

**Question 14: A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball?**

**Solution:**

Let r be the radius of the iron ball.

Radius of the cylinder (R) = 16 cm (Given)

A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 9 cm. So, height (h) = 9 cm

From statement,

Volume of iron ball = Volume of water raised in the hub

4/3Ï€r^{3} = Ï€R^{2}h

4/3 r^{3} = (16)^{2 }Ã— 9

or r^{3} = 1728

or r = 12

Therefore, radius of the ball = 12cm.

### Exercise VSAQs Page No: 21.25

**Question 1: Find the surface area of a sphere of radius 14 cm.**

**Solution: **

Radius of a sphere (r) = 14 cm

Surface area of a sphere = 4Ï€r^{2}

= 4 Ã— (22/7) Ã— 14^{2} cm^{2}

= 2464 cm^{2 }

**Question 2: Find the total surface area of a hemisphere of radius 10 cm.**

**Solution: **

Radius of a hemisphere (r) = 10 cm

Total surface area of a hemisphere = 3Ï€r^{2}

= 3 Ã— (22/7) Ã— 10^{2} cm^{2}

= 942 cm^{2}

**Question 3: Find the radius of a sphere whose surface area is 154 cm ^{2}.**

**Solution: **

Surface area of a sphere = 154 cm^{2}

We know, Surface area of a sphere = 4Ï€r^{2}

So, 4Ï€r^{2} = 154

4 x 22/7 x r^{2} = 154

r^{2} = 49/4

or r = 7/2 = 3.5

Radius of a sphere is 3.5 cm.

**Question 4: The hollow sphere, in which the circus motor cyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding.**

**Solution: **

Diameter of hollow sphere = 7 m

So, radius of hollow sphere = 7/2 m = 3.5 cm

Now,

Area available to the motorcyclist for riding = Surface area of a sphere = 4Ï€r^{2}

= 4 Ã— (22/7) Ã— 3.5^{2 } m^{2}

= 154 m^{2 }

**Question 5: Find the volume of a sphere whose surface area is 154 cm ^{2}.**

**Solution:**

Surface area of a sphere = 154 cm^{2}

We know, Surface area of a sphere = 4Ï€r^{2}

So, 4Ï€r^{2 } = 154

4 x 22/7 x r^{2} = 154

or r^{2} = 49/4

or r = 7/2 = 3.5

Radius (r) = 3.5 cm

Now,

Volume of sphere = 4/3 Ï€ r^{3}

= (4/3) Ï€ Ã— 3.5^{3}

= 179.66

Therefore, Volume of sphere is 179.66 cm^{3}.

### RD Sharma Solutions for Class 9 Maths Chapter 21 Surface Area and Volume of A Sphere

In the 21st chapter of Class 9Â RD Sharma Solutions students will study important concepts listed below:

- Sphere Introduction
- Section of a sphere by a plane
- Surface Area of a sphere, hemisphere and spherical shell
- Volume of a sphere, hemisphere and spherical shell

Students will be seeing questions just like the ones below and many more

1. How to find the surface area of a sphere when the radius is given.

2. How to find the surface area of a sphere when diameter is given.