In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19 B Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19 B Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19 B Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 19 |

Chapter Name | Volume and Surface Areas of Solids |

Exercise | 19 B |

Category | RS 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 πr^{3} h = (\frac { 1 }{ 3 } \times 12\times 12\times 24) π cm^{3} = (48 × 24 )π cm^{3}

**Question 2:**

Internal radius = 3 cm and external radius = 5 cm

Hence, height of the cone = 4 cm

**Question 3:**

Inner radius of the bowl = 15 cm

Volume of liquid in it =

Radius of each cylindrical bottle = 2.5 cm and its height = 6 cm

Volume of each cylindrical bottle

Required number of bottles =

Hence, bottles required = 60

**Question 4:**

Radius of the sphere = 212 cm

Let the number of cones formed be n, then

Hence, number of cones formed = 504

**Question 5:**

Radius of the cannon ball = 14 cm

Volume of cannon ball =

Radius of the cone = 352 cm

Let the height of cone be h cm

Volume of cone =

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

**Question 7:**

External radius of shell = 12 cm and internal radius = 9 cm

Volume of lead in the shell =

Let the radius of the cylinder be r cm

Its height = 37 cm

Volume of cylinder = πr^{2}h = ( πr^{2} × 37 )

Hence diameter of the base of the cylinder = 12 cm

**Question 8:**

Volume of hemisphere of radius 9 cm

Volume of circular cone (height = 72 cm)

Volume of cone = Volume of hemisphere

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 πr^{3} = (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

Hence, number of cubes is 4851.

**Question 10:**

Volume of sphere (when r = 1 cm) = 43 πr^{3} = (\frac { 4 }{ 3 } \times 1\times 1\times 1) π cm^{3}

Volume of sphere (when r = 8 cm) = 43 πr^{3} = (\frac { 4 }{ 3 } \times 8\times 8\times 8) π cm^{3}

Let the number of balls = n

**Question 11:**

Radius of marbles = Diameter2=1.42cm

Let the number of marbles be n

∴ n × volume of marble = volume of rising water in beaker

**Question 12:**

Radius of sphere = 3 cm

Volume of sphere = 43 πr^{3} = (\frac { 4 }{ 3 } \times 3\times 3\times 3) π cm^{3} = 36π cm^{3}

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 cm^{3}

Let number of small balls be n

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 πr^{3} = (\frac { 4 }{ 3 } \times 21\times 21\times 21) π cm^{3}

Diameter of cylindrical wire = 2.8 cm

Radius of cylindrical wire = 2.82 cm = 1.4 cm

Volume of cylindrical wire = πr^{2}h = ( π × 1.4 × 1.4 × h ) cm^{3} = ( 1.96πh ) cm^{3}

Volume of cylindrical wire = volume of sphere

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 πr^{3} = (\frac { 4 }{ 3 } \times 3\times 3\times 3) π cm^{3} = 36π cm^{3}

Radius of wire = 22 mm = 1 mm = 0.1 cm

Volume of wire = πr^{2}l = ( π × 0.1 × 0.1 × l ) cm^{2} = ( 0.01 πl ) cm^{2}

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

Length of wire = 108 m = 10800 cm

Let the radius of wire be r cm

= πr^{2}l cm^{3} = ( πr^{2} × 10800 ) cm^{3}

But the volume of wire = Volume of sphere

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

Let r be the radius of sphere whose volume is equal to the total volume of three spheres.

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