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

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

Chapter Name	RD Sharma Chapter 21 Surface Area and Volumes of a Sphere Exercise 21.2
Book Name	RD Sharma Mathematics for Class 10
Other Exercises	Exercise 21.1
Related Study	NCERT Solutions for Class 10 Maths

Exercise 21.2 Solutions

1. Find the volume of a sphere whose radius is:
(i) 2 cm

(ii) 3.5 cm

(iii) 10.5 cm

Solution

(i) Radius (r) = 2 cm 
∴ Volume = 4/3πr³
= 4/3 × 22/7 × (2)³ = 33.52 cm³ 

(ii) Radius (r) = 3.5 cm 
∴ Volume = (3.5)³ × π × 4/3
= 4/3 × 22/7 × (3.5)³
= 179.666 cm³ 

(iii) Radius (r) = 10.5 cm 
Volume = 4/3πr³
= 4/3 × 22/7 × (10.5)³
= 4851 cm³

2. Find the volume of a sphere whose diameter is : 
(i) 14 cm

(ii) 3.5 dm

(iii) 2.1 m 

Solution

(i) Diameter = 14 cm, radius = 14/2 = 7 cm 

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

Solution

Radius of tank = 2.8 m 
∴ Capacity = 2/3 × 22/7 × (2.8)³ = 45.994 m³  
∴ Capacity in liters = 45994 liters [1m³ = 1000]

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 = 5 cm 
Outer radius = 5 + 0.25 = 5.25
Volume of steel used = outer volume - inner volume 

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

Cube edge = 22 cm 
∴ Volume of cube = (22)³   = 10648 cm³ 
And, 
Volume of each bullet  = 4/3 πr³ 

∴ No. of bullets = Volume of cube/Volume of bullet 
= 10648/88/21 = 2541

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

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 3/2 cm and 2 cm, find the diameter of the third ball. 

Solution

Volume of lead ball = 4/3 πr³ 
= 4/3 × 22/7 × (3/2)³ 
∴ According to 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

Volume of sphere = 4/3 πr³ = 4/3π(5)³ 
∴ Volume of water rise in cylinder = Volume of sphere 

Let r be the radius of the cylinder 

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 V₁ and V₂ be the volumes of first sphere and second sphere respectively 
Radius of 1st sphere = r 
2nd sphere radius = 2r 

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

Solution

Given that 

Volume of thee cone = Volume of the hemisphere 

∴ Ratio of the their height is 2 : 1.

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

Given that 

Volume of water in the hemisphere bowl = Volume of water in the cylinder  
Let n be the height to which water rises in the cylinder. 

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

Given that, 

Height of cylinder = 2/3 (diameter)
We know that, 

Diameter = 2 (radius) 
h = 2/3 × 2r = 4/3 r 
Volume of the cylinder = volume of the sphere 

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

It is given that, 

Volume of water is hemisphere bowl = Volume of cylinder 

⇒ 2/3 π(6)³ = π(4)² h
⇒ h = 2/3 × (6×6×6)/(4×4)

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 

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

Therefore, radius of the ball = 12 cm. 

15. A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. (Use π = 22/7)

Solution

Given that, Radius of cylinder = 12 cm = r₁ 
Raised in height = 6.75 cm = h 
⇒ Volume of water raised = Volume of the sphere 

16. The diameter of a coper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter.

Solution

Given that diameter of a coper sphere = 18 cm. 
Radius of the sphere = 9 cm 

Length of the wire = 108 m = 10800 cm 

Volume of cylinder = volume of sphere 

17. A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1 .5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimetres?

Solution

Given that, 

Radius of cylinder jar = 6 cm = r₁ 
Level to be rised = 2cm = h 
Radius of each iron sphere = 1.5 cm  =r₂ 
Number of sphere = Volume of cylinder/Volume of sphere 

18. A measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm each are dropped in it and they sink down in water completely. What will be the change in the level of water in the jar?

Solution 

Given that,

Diameter of jar = 10cm

Radius of jar = 5cm

Let the level of water raised by ‘h’

Diameter of spherical ball = 2cm

Radius of the ball =1cm

Volume of jar = 4(Volume of spherical)

19. The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire.

Solution

Given, that, 

Diameter of sphere = 6 cm 
Radius of sphere = d/2 = 6/2 cm = 3 cm = r₁ 
Diameter of the wire = 0.2 cm 

Radius of the wire = 0.1 cm  =r₂ 
Volume of sphere = Volume of wire 

20. The radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm  respectively. If it is melted and recast into a solid cylinder of height 8/3 cm. Find the diameter of the cylinder. 

Solution

Given that, 

Internal radius of the sphere = 3 cm  = r₁ 

External radius of the sphere = 5 cm  = r₂ 
Height of cylinder = 8/3 m = h 
Volume of spherical shell = Volume of the cylinder 

∴ Diameter of the cylinder = 2 (radius) = 14 cm 

21. A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 46 cm. Find the radius of the base.

Solution

Given radius of hemisphere = 7 cm = r₁ 
Height of cone h = 49 cm 

Volume of hemisphere = Volume of cone 

22. A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone. 

Solution

Given that 

Hollow sphere external radii = 4cm = r₂ 
Internal radii (r₁) = 2 cm 

Cone base radius (R) = 4 cm 

Height = ? 

Volume of cone = Volume of sphere 

23. A metallic sphere of radius 10.5 cm is melted and thus recast into small cones, each or radius 3.5 cm and height 3 cm. Find how many cones are obtained.

Solution

Given that,

Metallic sphere of radius  = 10.5 cm 

Cone radius = 3.5 cm

Height of radius = 3 cm

Let the number of cones obtained be x.

24. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

Solution

Given that 
A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights. 

25. A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1 : 2: 3. 

Solution

Given that, 

A cone, hemisphere and a cylinder stand one equal bases and have the same weight
We know that 

26. A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form ball

is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of

the ball?

Solution

A cylindrical tub of radius = 12 cm 

Depth = 20 cm. 

Let r cm be the radius of the ball 

Then, volume of ball = Volume of water raised 

27. The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of the sphere. 

Solution

Given that, 

The largest sphere is carved out of a cube of side = 10.5 cm 

Volume of the sphere = ? 

We have, 

Diameter of the largest sphere = 10.5 cm

2r = 10.5 

⇒ r = 5.25 cm 

28. A sphere, a cylinder and a cone have the same diameter. The height of the cylinder and also

the cone are equal to the diameter of the sphere. Find the ratio of their volumes.

Solution

Let r be the common radius thus, 

h = height of the cone = Height of the cylinder = 2r 
Let 

29. A cube of side 4 cm contains a sphere touching its side . Find the volume of the gap in between.

Solution

It is given that 

Cube side = 4 cm 

Volume of cube = (4 cm)³ = 64 cm³ 

Diameter of the sphere = Length of the side of the cube = 4 cm 

∴ Radius of sphere = 2 cm 

Volume of the sphere = 4/3 πr³ 
= 4/3 × 22/7 × (2)³ = 33.52 cm³ 
∴ Volume of gap = Volume of cub - Volume of sphere 

= 64 cm³ - 33.52 cm³ = 30.48 cm³ 

30. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then

find the volume of the iron used to make the tank.

Solution

Given that, 

Inner radius (r₁ ) of hemispherical tank = 1m = r₁ 

Thickness of hemispherical tank = 1 cm  = 0.01m

Outer radius (r₂ ) of the hemispherical = (1 + 0.01 m) = 1.01 m = r₂ 

31. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule ?

Solution

Given that, 

Diameter of capsule = 3.5mm

Radius = 3.5/2 = 1.75 mm

Volume of spherical capsule = 4/3 πr³ 
= 4/3 × 22/7 × (1.75)³ mm³ 

= 22.458 mm³ 
∴ 2.46 mm³ of medicine is required.

32. The diameter of the moon is approximately one fourth of the diameter of the earth. What

fraction of the volume of the earth is the volume of the moon?

Solution

Given that,

The diameter of the moon is approximately one fourth of the diameter of the earth.