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

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 19 Volume and Surface Areas of Solids 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 Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Maths book pdf download. Now you will get step by step solution to each question.

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

RS Aggarwal Solutions Class 10 Chapter 19 Exercise 19A

Question 1:

Radius of the cylinder = 14 m
And its height = 3 m
Radius of cone = 14 m
And its height = 10.5 m
Let l be the slant height

Curved surface area of tent
= (curved area of cylinder + curved surface area of cone)

Hence, the curved surface area of the tent = 1034
Cost of canvas = Rs.(1034 × 80) = Rs. 82720

Read More:
Surface Area and Volume of a Cuboid
Surface Area and Volume of a Cube
Surface Area of a Sphere and a Hemisphere

Question 2:

For the cylindrical portion, we have radius = 52.5 m and height = 3 m
For the conical portion, we have radius = 52.5 m
And slant height = 53 m
Area of canvas = 2rh + rl = r(2h + l)

More Resources

Question 3:

Height of cylinder = 20 cm
And diameter = 7 cm and then radius = 3.5 cm
Total surface area of article
= (lateral surface of cylinder with r = 3.5 cm and h = 20 cm)

Question 4:

Radius of wooden cylinder = 4.2 cm
Height of wooden cylinder = 12 cm
Lateral surface area

Radius of hemisphere = 4.2 cm
Surface area of two hemispheres

Total surface area = (100.8 + 70.56) π cm²
= 538.56 cm²
= 171.36 π
= 171.36 × (frac { 22 }{ 7 }  )  cm²
= 538.56 cm²
Further, volume of cylinder = πr²h = 4.2 × 4.2 × 12 π cm²
= 211.68 π cm²
Volume of two hemispheres = 2 × (frac { 2 }{ 3 }  ) πr³ cu.units
= (frac { 4 }{ 3 }  ) π  × 4.2 × 4.2 × 4.2
= 98.784 cm³
Volume of wood left = (211.68 – 98.784) π
= 112.896 π cm³
= 112.896 × (frac { 22 }{ 7 }  )  cm³
= 354.816 cm³

Question 5:
Radius o f cylinder = 2.5 m
Height of cylinder = 21 m
Slant height of cone = 8 m
Radius of cone = 2.5 m
Total surface area of the rocket = (curved surface area of cone + curved surface area of cylinder + area of base)

Question 6:

Height of cone = h = 24 cm
Its radius = 7 cm

Total surface area of toy

Question 7:

Height of cylindrical container h₁ = 15 cm
Diameter of cylindrical container = 12 cm
Volume of container =
Height of cone r₂ = 12 cm
Diameter = 6 cm
Radius of r₂ = 3 cm

Radius of hemisphere = 3 cm
Volume of hemisphere =
Volume of cone + volume of hemisphere
= 36π + 18π = 54π
Number of cones

Number of cones that can be filled = 10

Question 8:

Diameter of cylindrical gulabjamun = 2.8 cm
Its radius = 1.4 cm
Total height of gulabjamun = AC + CD + DB = 5 cm
1.4 + CD + 1.4 = 5
2.8 + CD = 5
CD = 2.2 cm
Height of cylindrical part h = 2.2 cm
Volume of 1 gulabjamun = Volume of cylindrical part + Volume of two hemispherical parts

Volume of 45 gulabjamuns = 45 × 25.07 cm³
Quantity of syrup = 30% of volume of gulabjamuns
= 0.3 × 45 × 25.07 = 338.46 cm³

Question 9:
Diameter = 7cm, radius = = 3.5 cm
Height of cone = 14.5 cm – 3.5 cm = 11 cm

Total surface area of toy =

Question 10:
Diameter of cylinder = 24 m
Radius of cylinder = (frac { 24 }{ 2 }  ) = 12 cm
Height of the cylinder = 11 m
Height of cone = (16 – 11) cm = 5 cm
Slant height of the cone l =
Area of canvas required = (curved surface area of the cylindrical part) + (curved surface area of the conical part)

Question 11:
Radius of hemisphere = 10.5 cm
Height of cylinder = (14.5 – 10.5) cm = 4 cm
Radius of cylinder = 10.5 cm
Capacity = Volume of cylinder + Volume of hemisphere

Question 12:

Height of cylinder = 6.5 cm
Height of cone = h₂ = (12.8-6.5) cm = 6.3 cm
Radius of cylinder = radius of cone
= radius of hemisphere
= (frac { 7 }{ 2 }  ) cm
Volume of solid = Volume of cylinder + Volume of cone + Volume of hemisphere

Question 13:

Radius of each hemispherical end = (frac { 28 }{ 2 }  ) = 14 cm
Height of each hemispherical part = Its Radius
Height of cylindrical part = (98 – 2 × 14) = 70 cm
Area of surface to be polished = 2(curved surface area of hemisphere) + (curved surface area of cylinder)

Cost of polishing the surface of the solid
= Rs. (0.15 × 8624)
= Rs. 1293. 60

Question 14:

Radius of cylinder r₁ = 5 cm
And height of cylinder h₁ = 9.8 cm
Radius of cone r = 2.1 cm
And height of cone h₂ = 4 cm
Volume of water left in tub = (volume of cylindrical tub – volume of solid)

Question 15:
(i) Radius of cylinder = 6 cm
Height of cylinder = 8 cm

Volume of cylinder

Volume of cone removed

(ii) Surface area of cylinder = 2π = 2π × 6 × 8 cm² = 96 π cm²

Question 16:

Diameter of spherical part of vessel = 21 cm

Question 17:

Height of cylindrical tank = 2.5 m
Its diameter = 12 m, Radius = 6 m
Volume of tank =
Water is flowing at the rate of 3.6 km/ hr = 3600 m/hr
Diameter of pipe = 25 cm, radius = 0.125 m
Volume of water flowing per hour

Question 18:

Diameter of cylinder = 5 cm
Radius = 2.5 cm
Height of cylinder = 10 cm
Volume of cylinder = πr²h  cu.units = 3.14 × 2.5 × 2.5 × 10  cm³ = 196.25 cm³
Apparent capacity of glass = 196.25
Radius of hemisphere = 2.5 cm
Volume of hemisphere

Actual capacity of glass = ( 196.25 – 32.608 ) cm³ = 163.54 cm³

RS Aggarwal Solutions Class 10 Chapter 19 Exercise 19B

Question 1:

Radius of the cone = 12 cm and its height = 24 cm
Volume of cone = (frac { 1 }{ 3 }  ) πr³ h = (frac { 1 }{ 3 } times 12times 12times 24) π cm³  = (48 × 24 )π cm³

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 = (frac { 21 }{ 2 }  ) 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 = (frac { 35 }{ 2 }  ) 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 = πr²h = ( πr² × 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 = (frac { 19 }{ 2 }  ) cm
Volume of sphere = (frac { 4 }{ 3 }  ) πr³ = ((frac { 4 }{ 3 } times frac { 22 }{ 7 } times frac { 21 }{ 2 } times frac { 21 }{ 2 } times frac { 21 }{ 2 } )   )
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) = (frac { 4 }{ 3 }  ) πr³ = (frac { 4 }{ 3 } times 1times 1times 1) π cm³
Volume of sphere (when r = 8 cm) = (frac { 4 }{ 3 }  ) πr³ = (frac { 4 }{ 3 } times 8times 8times 8) π cm³
Let the number of balls = n

Question 11:
Radius of marbles = (frac { Diameter }{ 2 } =frac { 1.4 }{ 2 } cm  )

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 = (frac { 4 }{ 3 }  ) πr³ = (frac { 4 }{ 3 } times 3times 3times 3) π cm³  = 36π cm³
Radius of small sphere = (frac { 0.6 }{ 2 }  ) cm = 0.3 cm
Volume of small sphere = (frac { 4 }{ 3 } times 0.3times 0.3times 0.3) π cm³

Let number of small balls be n

Hence, the number of small balls = 1000.

Question 13:
Diameter of sphere = 42 cm
Radius of sphere = (frac { 42 }{ 2 }  ) cm = 21 cm
Volume of sphere = (frac { 4 }{ 3 }  ) πr³ = (frac { 4 }{ 3 } times 21times 21times 21) π cm³
Diameter of cylindrical wire = 2.8 cm
Radius of cylindrical wire = (frac { 2.8 }{ 2 }  ) 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

Hence length of the wire 63 m.

Question 14:
Diameter of sphere = 6 cm
Radius of sphere = (frac { 6 }{ 2 }  ) cm = 3 cm
Volume of sphere = (frac { 4 }{ 3 }  ) πr³ = (frac { 4 }{ 3 } times 3times 3times 3) π cm³  = 36π cm³
Radius of wire = (frac { 2 }{ 2 }  ) 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=frac { 36 }{ 0.01 } =3600   ) cm
Length of wire = (frac { 3600 }{ 100 }  ) m = 36 m

Question 15:
Diameter of sphere = 18 cm
Radius of copper sphere = (frac { 3600 }{ 100 }  ) m = 36 m

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

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.

RS Aggarwal Solutions Class 10 Chapter 19 Exercise 19C

Question 1:
Here h = 42 cm, R = 16 cm, and r = 11 cm
Capacity =

Question 2:
Here R = 33 cm, r = 27 cm and l = 10 cm

Capacity of the frustum


Total surface area
=

Question 3:
Height = 15 cm, R = (frac { 56 }{ 2 }  ) cm = 28 cm and r = (frac { 42 }{ 2 }  ) cm = 21 cm
Capacity of the bucket =


Quantity of water in bucket = 28.49 litres

Question 4:
R = 20 cm, r = 8 cm and h = 16 cm

Total surface area of container = πl (R+r) + πr²

Cost of metal sheet used = Rs. ((1959.36times frac { 15 }{ 100 } )   ) = Rs. 293.90

Question 5:
R = 15 cm, r = 5 cm and h = 24 cm

(i) Volume of bucket =


Cost of milk = Rs. (8.164 × 20) = Rs. 163.28
(ii) Total surface area of the bucket

Cost of sheet = (( 1711.3times frac { 10 }{ 100 } )   ) = Rs. 171.13

Question 6:

R = 10cm, r = 3 m and h = 24 m
Let l be the slant height of the frustum, then

Quantity of canvas = (Lateral surface area of the frustum) + (lateral surface area of the cone)

Question 7:

ABCD is the frustum in which upper and lower radii are EB = 7 m and FD = 13 m
Height of frustum = 8 m
Slant height l₁ of frustum

Radius of the cone = EB = 7 m
Slant height l₂ of cone = 12 m
Surface area of canvas required

Question 8:

In the given figure, we have
∠COD = 30°, OC = 10 cm, OE = 20 cm
Let CD = r cm and EB = R cm

Also, CE = 10 cm
Thus, ABDF is the frustum of a cone in which

Volume of wire of radius r and length l

Volume of wire = Volume of frustum

Length of the wire is 7964.44 m

Question 9:

Radii of upper and lower end of frustum are r = 8 cm, R = 32 cm
Height of frustum h = 18 cm

Cost of milk at Rs 20 per litre = Rs. 25.344 × 20 = Rs. 506. 88

All Chapter RS Aggarwal Solutions For Class 10 Maths

—————————————————————————–

All Subject NCERT Exemplar Problems Solutions For Class10

All Subject NCERT Solutions For Class 10

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rsaggarwalsolutions.in to your friends.