**RD Sharma Solutions for Class 9 Maths Chapter 21 Surface Area and Volume of a Sphere Exercise 21.2** are provided here. These solutions are designed by the subject experts according to the questions in the RD Sharma textbook. This exercise is based on the volume of a sphere, hemisphere and spherical shell. RD Sharma Class 9 Solutions will help students in constructive preparation and learning. Students can practise all the textbook questions and match their answers with the solutions given below. They can also download Exercise 21.2 in PDF for future reference by clicking on the given links below.

## RD Sharma Solutions for Class 9 Maths Chapter 21 Surface Area and Volume of a Sphere Exercise 21.2

### Access Answers to RD Sharma Solutions for Class 9 Maths Chapter 21 Surface Area and Volume of a Sphere Exercise 21.2 Page Number 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: **

The volume of a sphere = 4/3πr^{3} Cubic Units

Where, r = radius of a sphere

**(i)** Radius = 2 cm

Volume = 4/3 × 22/7 × (2)^{3}

= 33.52

Volume = 33.52 cm^{3}

**(ii)** Radius = 3.5cm

Therefore volume = 4/3×22/7×(3.5)^{3}

= 179.666

Volume = 179.666 cm^{3}

**(iii)** Radius = 10.5 cm

Volume = 4/3×22/7×(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:**

The volume of a sphere = 4/3πr^{3} Cubic Units

Where, r = radius of a sphere

**(i)** diameter =14 cm

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

Volume = 4/3×22/7×(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 = 4/3×22/7×(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 = 4/3×22/7×(1.05)^{3}

= 4.851

Volume = 4.851 m^{3}

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

**Solution:**

The radius of hemispherical tank = 2.8 m

Capacity of hemispherical tank = 2/3 πr^{3}

=2/3×22/7×(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:**

The inner radius of a hemispherical bowl = 5 cm

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

The volume of steel used = Outer volume – Inner volume

= 2/3×π×((5.25)^{3}−(5)^{3})

= 2/3×22/7×((5.25)^{3}−(5)^{3})

= 41.282

The 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, the radius of the bullet (r) = 1 cm

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

And,

The volume of each bullet which will be spherical in shape = 4/3πr^{3}

= 4/3 × 22/7 × (1)^{3 } cm^{3}

= 4/3 × 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:**

The volume of laddoo having a radius of 5 cm (V1) = 4/3×22/7×(5)^{3}

= 11000/21 cm^{3}

Also, the volume of the laddoo having a radius of 2.5 cm (V2) = 4/3πr^{3}

= 4/3×22/7×(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:**

The volume of the lead ball with the radius 3/2 cm = 4/3πr^{3}

= 4/3×π×(3/2)^{3}

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

The radius of the first ball (r1) = 3/4 cm

Diameter of the second ball (d2) = 2 cm

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

Diameter of the third ball (d3) = d

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

Now,

So, the diameter of the third ball is 2.5 cm.

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

**Solution:**

Radius of sphere = 5 cm (Given)

Let ‘r’ be the radius of the cylinder.

We know, Volume of the sphere = 4/3πr^{3}

By putting values, we get

= 4/3×π×(5)^{3}

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

The volume of water rises in the cylinder = πr^{2}h

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

So, πr^{2}h = 4/3πr^{3}

π r^{2} × 5/3 = 4/3 × π × (5)^{3}

or r^{2 }= 100

or r = 10

Therefore, the 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,

The ratio of the 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: **

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

1/3πr^{2}h = 2/3 πr^{3}

(Using respective formulas)

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

or h = 2r

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

h/r = 2

or h : r = 2 : 1

Therefore, the 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:**

The volume of water in the hemispherical bowl = Volume of water in the cylinder … (Given)

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

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

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

2/3πr_{1}^{3} = πr_{2}^{2}h

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

2/3π(3.5)^{3} = π(7)^{2}h

or h = 7/12

Therefore, 7/12 cm is 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 = 2/3× (2r) = 4r/3

The volume of the cylinder = Volume of the sphere

πr^{2}h = 4/3πR^{3}

π × r^{2} × (4r/3) = 4/3 π (4)^{3}

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

or r = 4

Therefore, the 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,

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

2/3 π R^{3} = πr^{2} h

2/3 π (6)^{3} = π(4)^{2} h

or h = 9

Therefore, the height of the water in the cylinder is 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 the 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,

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

4/3πr^{3} = πR^{2}h

4/3 r^{3} = (16)^{2 }× 9

or r^{3} = 1728

or r = 12

Therefore, the radius of the ball = 12 cm.

## RD Sharma Solutions for Class 9 Maths Chapter 21 Surface Area and Volume of a Sphere Exercise 21.2

**RD Sharma Solutions Class 9 Maths Chapter 21 Surface Area and Volume of a Sphere Exercise 21.2** are based on the following topics:

- Volume of a sphere
- Volume of a hemisphere and
- Volume of a spherical shell

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