RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2 (Updated for 2024)

RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2: RD Sharma Solutions Class 9 Maths Chapter 21 Surface Area And Volume of Sphere Exercise 21.2. These solutions are created by our experts. They provided a very helpful answer to their Class 9 exam. The problem is solved in an organized way. These solutions are available for free PDF forms, which can be easily stored for a long time.

• RD Sharma Class 9 Maths Solutions

Download RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2

 

RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2

Access RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2

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 = 4/3πr³ Cubic Units

Where, r = radius of a sphere

(i) Radius = 2 cm

Volume = 4/3 × 22/7 × (2)³

= 33.52

Volume = 33.52 cm³

(ii) Radius = 3.5cm

Therefore volume = 4/3×22/7×(3.5)³

= 179.666

Volume = 179.666 cm³

(iii) Radius = 10.5 cm

Volume = 4/3×22/7×(10.5)³

= 4851

Volume = 4851 cm³

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 = 4/3πr³ 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 cm³

(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 dm³

(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 m³

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 πr³

=2/3×22/7×(2.8)³ m³

= 45.997 m³

[Using 1m³ = 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 cm³

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 cm³ = 10648 cm³

And,

The volume of each bullet which will be spherical in shape = 4/3πr³

= 4/3 × 22/7 × (1)³ cm³

= 4/3 × 22/7 cm³

= 88/21 cm³

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 cm³

Also, the volume of the laddoo having a radius of 2.5 cm (V2) = 4/3πr³

= 4/3×22/7×(2.5)³ cm³

= 1375/21 cm³

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 = 4/3πr³

= 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

The 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 = 4/3πr³

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 = πr²h

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

So, πr²h = 4/3πr³

π r² × 5/3 = 4/3 × π × (5)³

or r² = 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)

1/3πr²h = 2/3 πr³

(Using respective formulas)

r²h = 2r³

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 ( r₁) = 3.5cm

The inner radius of cylinder (r₂) = 7cm

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

2/3πr₁³ = πr₂²h

[Using respective formulas]

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

2/3π(3.5)³ = π(7)²h

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

πr²h = 4/3πR³

π × r² × (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 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 R³ = πr² 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

4/3πr³ = πR²h

4/3 r³ = (16)² × 9

or r³ = 1728

or r = 12

Therefore, the radius of the ball = 12 cm.

We have included all the information regarding CBSE RD Sharma Class 9 Solutions Chapter 21 Exercise 21.2. If you have any queries feel free to ask in the comment section. 

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