RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19 B

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TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 19
Chapter NameVolume and Surface Areas of Solids
Exercise19 B
CategoryRS Aggarwal Solutions

RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19 B

Question 1:

Radius of the cone = 12 cm and its height = 24 cm
Volume of cone = 13 πr3 h = (\frac { 1 }{ 3 } \times 12\times 12\times 24) π cm3  = (48 × 24 )π cm3
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 1.1

Question 2:
Internal radius = 3 cm and external radius = 5 cm
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 2.1
Hence, height of the cone = 4 cm

Question 3:
Inner radius of the bowl = 15 cm
Volume of liquid in it =

RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 3.1
Radius of each cylindrical bottle = 2.5 cm and its height = 6 cm
Volume of each cylindrical bottle
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 3.2
Required number of bottles = RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 3.3
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 3.4
Hence, bottles required = 60

Question 4:
Radius of the sphere = 212 cm
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 4.1
Let the number of cones formed be n, then
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 4.2
Hence, number of cones formed = 504

Question 5:
Radius of the cannon ball = 14 cm
Volume of cannon ball = RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 5.1
Radius of the cone = 352 cm
Let the height of cone be h cm
Volume of cone = RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 5.2
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 5.3
Hence, height of the cone = 35.84 cm

Question 6:
Let the radius of the third ball be r cm, then,
Volume of third ball = Volume of spherical ball – volume of 2 small balls
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 6.1

Question 7:
External radius of shell = 12 cm and internal radius = 9 cm
Volume of lead in the shell = rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q7-1
Let the radius of the cylinder be r cm
Its height = 37 cm
Volume of cylinder = πr2h = ( πr2 × 37 )
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 7.1
Hence diameter of the base of the cylinder = 12 cm

Question 8:
Volume of hemisphere of radius 9 cm
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 8.1
Volume of circular cone (height = 72 cm)
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 8.2
Volume of cone = Volume of hemisphere
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 8.3
Hence radius of the base of the cone = 4.5 cm

Question 9:
Diameter of sphere = 21 cm
Hence, radius of sphere = 192 cm
Volume of sphere = 43 πr3 = (43×227×212×212×212)
Volume of cube = a3 = (1 × 1 × 1)
Let number of cubes formed be n
∴ Volume of sphere = n × Volume of cube
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 9.1
Hence, number of cubes is 4851.

Question 10:
Volume of sphere (when r = 1 cm) = 43 πr3 = (\frac { 4 }{ 3 } \times 1\times 1\times 1) π cm3
Volume of sphere (when r = 8 cm) = 43 πr3 = (\frac { 4 }{ 3 } \times 8\times 8\times 8) π cm3
Let the number of balls = n
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 10.1

Question 11:
Radius of marbles = Diameter2=1.42cm
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 11.1
Let the number of marbles be n
∴ n × volume of marble = volume of rising water in beaker
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 11.2

Question 12:
Radius of sphere = 3 cm
Volume of sphere = 43 πr3 = (\frac { 4 }{ 3 } \times 3\times 3\times 3) π cm3  = 36π cm3
Radius of small sphere = 0.62 cm = 0.3 cm
Volume of small sphere = (\frac { 4 }{ 3 } \times 0.3\times 0.3\times 0.3) π cm3
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 12.1
Let number of small balls be n
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 12.2
Hence, the number of small balls = 1000.

Question 13:
Diameter of sphere = 42 cm
Radius of sphere = 422 cm = 21 cm
Volume of sphere = 43 πr3 = (\frac { 4 }{ 3 } \times 21\times 21\times 21) π cm3
Diameter of cylindrical wire = 2.8 cm
Radius of cylindrical wire = 2.82 cm = 1.4 cm
Volume of cylindrical wire =  πr2h = ( π × 1.4 × 1.4 × h ) cm3 = ( 1.96πh ) cm3
Volume of cylindrical wire = volume of sphere
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 13.1
Hence length of the wire 63 m.

Question 14:
Diameter of sphere = 6 cm
Radius of sphere = 62 cm = 3 cm
Volume of sphere = 43 πr3 = (\frac { 4 }{ 3 } \times 3\times 3\times 3) π cm3  = 36π cm3
Radius of wire = 22 mm = 1 mm = 0.1 cm
Volume of wire = πr2l = ( π × 0.1 × 0.1 × l ) cm2 = ( 0.01 πl ) cm2
36π = 0.01 π l
∴ l=360.01=3600 cm
Length of wire = 3600100 m = 36 m

Question 15:
Diameter of sphere = 18 cm
Radius of copper sphere = 3600100 m = 36 m
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 15.1
Length of wire = 108 m = 10800 cm
Let the radius of wire be r cm
= πr2l cm3 = ( πr2 × 10800 ) cm3
But the volume of wire = Volume of sphere
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 15.2
Hence the diameter = 2r = (0.3 × 2) cm = 0.6 cm

Question 16:
The radii of three metallic spheres are 3 cm, 4 cm and 5 cm respectively.
Sum of their volumes RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 16.1
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 16.2
Let r be the radius of sphere whose volume is equal to the total volume of three spheres.
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9b 16.3

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