Class 9 RD Sharma Solutions – Chapter 21 Surface Area and Volume of a Sphere – Exercise 21.2 | Set 2

Question 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 = r1 = 6 cm,

Level to be raised = 2 cm,

Radius of each iron sphere = r2 = 1.5 cm,

Number of sphere = (Volume of Cylinder) / (Volume of Sphere)

= (π x (r1)^2 x h) / (4/3 x π x (r2)^3)

= ((r1)^2 x h) / (4/3 x (r2)^3) = (6 x 6×2) / (43 x 1.5 x 1.5 x 1.5)

Hence Number of Sphere is 16.

Question 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 = 10 cm,

Radius of jar = 5 cm,

Diameter of the spherical bowl = 2 cm,

Radius of the ball = 1 cm.

Let the level of water be raised by h

Volume of jar = 4 x (Volume of spherical ball)

π x (r1)^2 x h = 4 x (4/3 x π x (r2)^3)

(r1)^2 x h = 4 x (4/3 x (r2)^3)

5 x 5 x h = 4 x 4/3 x 1 x 1 x 1

h = 16/75 cm

Hence Height of water in jar = 16/75 cm.

Question 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 = r1 = 6/2 = 3 cm,

Diameter of the wire = 0.2 cm,

Radius of the wire = r2 = 0.1 cm,

Volume of sphere = Volume of wire

4/3 x π x (r1)^3 = π(r2)^2 x h

4/3 x 3 x 3 x 3 = 0.1 x 0.1 x h

h = 3600 cm = 36m

Hence length of wire = 36 m.

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

Solution: 

Given that,

Internal radius of the sphere = r1 = 3 cm,

External radius of the sphere = r2 = 5 cm,

Height of the cylinder = h = 8/3cm,

Volume of the spherical shell = Volume of cylinder

4/3π(r2³ – r1³) = π x (r3)^2 x h

4/3(5³ – 3³) = 8/3 x (r3)^2

(r3)^2 = √49

r3 = 7 cm

Hence Diameter of the cylinder is 2 x radius = 14 cm.

Question 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 that,

Radius of the hemisphere = Volume of cone

2/3 π(r1)^3 = 1/3 π(r2)^2 x h

2/3 x 7^3 = 1/3 x (r2)^2 x 49

(r2)^2 = (2x7x7x7x3) / (3×49)

(r2)^2 = 2058147

r2 = 3.47 cm

Hence radius of the base is 3.74 cm.

Question 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 radius = r2 = 4 cm,

Internal radii = r1 = 2 cm,

Cone base radius (R) = 4 cm.

Height = h

Volume of cone = Volume of sphere (Given)

1/3 π(r)^2 x h = 4/3 x π x ((r2)^2 – (r1)^2)

4^2 x h = 4(4³ – 2³)

h = 14 cm

As we know that Slant Height = √(r)² + (h)²

l = √(4)² + (14)² = √16 + 196 = √212 = 14.56 cm

Hence Height of cone is 14 cm and Slant Height is 14.56 cm.

Question 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 = 10.5 cm,

Cone radius = 3.5 cm,

Height of radius = 3 cm,

Let the number of cones obtained be x

V_s = X x Volume of cone

4/3 x π x (r)^3 = X x 1/3 π (r)^2 h

X = (4 x 10.5 x 10.5 x 10.5) / (3.5 x 3.5 x 3)

X = 126

Hence number of cones is 126.

Question 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.

Volume of cone = Volume of hemisphere

1/3πr²h = 2/3πr³

r²h = 2r³

h = 2r

h x r = 2/1

h : r = 2:1

Therefore the ratio is 2 : 1

Question 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

hence,

Volume of cone : Volume of hemisphere : Volume of cylinder

1/3 πr²h : 2/3 πr³ : πr²h

Multiplying by 3

πr²h : 2πr³ : 3πr²h

πr³ : 2πr³ : 3πr³ (Given that r = h and r²h = r³)

Hence, the ratio is 1:2:3.

Question 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: 

Given that,

Radius of cylindrical tub = 12 cm,

Depth = 20 cm,

Let r be the radius of the ball then,

Volume of the ball = Volume of water raised

4/3 x π x (r)^3 = π x (r)^2 x h

r^3 = (3.14x(12)2 x 6.75 x 3) / 4

r^3 = 729

r = 9 cm

Hence radius of the ball is 9 cm.

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

Solution:

Given that,

Side of cube = 10.5 cm,

Diameter of the largest sphere = 10.5 cm so r = 5.25 cm

Volume of sphere = 4/3πr³ = 4/3 × 22/7 × 5.25 × 5.25 × 5.25 = 606.375 cm³

Hence Volume of Sphere is 606.375 cm³.

Question 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 us assume that r be the common radius and,

Height of the cone = Height of the cylinder = 2r (Given)

Let’s assume that,

v1 = Volume of sphere = 4/3 πr³

v2 = Volume of cylinder = πr²h = πr² × 2r

v3 = Volume of cone = 1/3πr²h = 1/3πr³

Now

v1 : v2 : v3 = 4/3πr³ : 2πr³ : 2/3πr³

= 4 : 6 : 2 = 2 : 3 : 1

Hence the ratio of their volumes is 2 : 3 : 1.

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

Solution:

Given that,

Cube side = 4cm

Volume of cube = (4 cm)³ = 64 cm³

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

Therefore radius of the sphere = 2cm

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

Volume of gap = Volume of cube – Volume of sphere

= 64 – 33.52 = 30.48 cm³

Hence the volume of the gap in between is 30.48 cm³.

Question 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 of the hemispherical tank = 1 m = r1,

Thickness of the hemispherical tank = 1 cm = 0.01 m,

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

Volume of iron used to make the tank = 2/3 π x ((r2)³ – (r1)³)

= 2/3 x 22/7 x [(1.01)³ – (1)³]

= 0.06348 m^3

Hence volume of the iron used to make the tank is 0.06348 m³.

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

Solution:

Given that,

Diameter of capsule = 3.5 mm

Radius = 3.5/2 = 1.75 mm

Volume of spherical sphere = 4/3πr³

= 4/3 × 22/7 × (1.75)³

= 22.458 mm³

Therefore 22.46 mm³ of medicine is required.

Question 32. The diameter of the moon is approximately one-fourth of the diameter of the earth. What is the earth the volume of the moon?

Solution:

Given that,

Diameter of moon = 1/4th Diameter of earth.

Let us assume, diameter of earth is d, so radius = d/2

Then diameter of moon = d/4

Radius = d/8

Volume of moon = 4/3πr³ = 4/3π(d/8)3 = 4/3 x (1/512) x πd³

Volume of earth = 4/3πr³ = 4/3π(d/2)3 = 4/3 x (1/8) x πd³

(Volume of moon) / (Volume of earth) = 1/64

Hence the volume of the moon is 1/64 of volume of earth.