# R.D. Sharma Solutions Class 9th: Ch 21 Surface Area and Volume of a Sphere Exercise 21.2

#### Chapter 21 Surface Area and Volume of a Sphere R.D. Sharma Solutions for Class 9th Exercise 21.2

**Exercise 21.2**

**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)= 2cm

Therefore volume=4/3πr

^{3}

= 4/3×22/7×(2)

^{3}

= 33.52cm

^{3}

(ii) Radius(r)= 3.5cm

Therefore volume=4/3πr

^{3}

= 4/3×22/7×(3.5)

^{3}= 179.666cm

^{3}

(iii) Radius(r)= 10.5cm

Therefore volume=4/3πr

^{3}

= 4/3×22/7×(10.5)

^{3}= 4851cm

^{3}

2. Find the volume of a sphere whose diameter is:

(i) 14 cm

(ii) 3.5 dm

(iii) 2.1 m

**Solution**

(i) Diameter=14cm, Radius(r)= 14/2= 7cm

Therefore volume=4/3πr

^{3}

= 4/3×22/7×(7)

^{3}= 1437.33cm

^{3}

(ii) Diameter=3.5dm, Radius(r)= 3.5/2= 1.75dm

Therefore volume=4/3πr

^{3}

=4/3×22/7×(1.75)

^{3}

=22.46dm

^{3}

(iii) Diameter = 2.1m, Radius(r)= 2.1/2 = 1.05m

Therefore volume=4/3πr

^{3}

= 4/3×22/7×(1.05)

^{3}= 4.851m

^{3}

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

**Solution**

Radius of the tank= 2.8m

Therefore Capacity= 2/3πr

^{3}

=2/3×22/7×(2.8)

^{3}= 45.994m

^{3}

1m

^{3}=1000

*l*

Therefore capacity in litres = 45994 litres .

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 = 5cm

Outer radius = 5 + 0.25 = 5.25

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})

= 41.282cm

^{3}

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**

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**

**Solution**

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**

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

Solution

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

**Solution**

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**

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**

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**

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**

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**

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 copper sphere = 18cm

Radius of the sphere = 9cm

Length of the wire =108m =10800cm

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 centimeters?

**Solution**

Given that,

Radius of the cylinder jar = 6cm = r

_{1}

Level to be rised = 2cm Radius of each iron sphere =1.5cm = r

_{2}

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 be raised by h

Diameter of the spherical bowl = 2cm

Radius of the ball = 1cm

Volume of jar = 4(Volume of spherical ball)

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**

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 22/3cm. Find the diameter of the cylinder.

**Solution**

Given that,

Internal radius of the sphere =3cm = r_{1}

External radius of the sphere =5cm = r

_{2}

Height of the cylinder = 8/3cm = h

Volume of the spherical shell = Volume of cylinder .

21. A hemisphere of the lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.

**Solution**

Given,

Radius of the hemisphere = Volume of cone

**Solution**

Given that

Hollow sphere external radii = r2 = 4cm

Internal radii = r1= 2cm

Cone base radius(R) = 4cm

Height = h

Volume of cone = Volume of sphere

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

**Solution**

Given that,

Metallic sphere of radius = 10.5cm

Cone radius =3.5cm

Height of radius = 3cm

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 volumes

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, a hemisphere and a cylinder stand on one equal bases and have the same weight

We know that

**Solution**

Radius of cylindrical tub =12cm

Depth =20cm

Let r be the radius of the ball

Then

Volume of the ball= Volume of water raised

**Solution**

Side of cube =10.5cm

Volume of sphere = vDiameter of the largest sphere =10.5cm

2r = 10.5

r = 5.25cm

**Solution**

Let r be the common radius

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 = 4cmVolume of cube = (4cm)

^{3}= 64cm

^{3}

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

Therefore radius of the sphere = 2cm .

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**

Inner radius of the hemispherical tank = 1m = r1

Thickness of the hemispherical tank = 1cm = 0.01m

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

Volume of iron used to make the tank =

^{3}) is needed to fill this capsule?

**Solution**

Given that,

Diameter of capsule = 3.5mm

Radius =3.5/2 = 1.75mm

Volume of spherical sphere =

**Solution**