Class 10 Class 10 Maths RS Aggarwal Solutions Class 10

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

RS Aggarwal Solutions Class 10 Chapter 19
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RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids is a part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids.

RS Aggarwal Solutions Class 10 Chapter 19  Exercise 19A

Question 1:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q1-1


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
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q1-2
Curved surface area of tent
= (curved area of cylinder + curved surface area of cone)
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q1-3
Hence, the curved surface area of the tent = 1034
Cost of canvas = Rs.(1034 × 80) = Rs. 82720

Question 2:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q2-1
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)
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q2-2

Question 3:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q3-1
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)
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q3-2

Question 4:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q4-1
Radius of wooden cylinder = 4.2 cm
Height of wooden cylinder = 12 cm
Lateral surface area
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q4-2
Radius of hemisphere = 4.2 cm
Surface area of two hemispheres
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q4-3
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)
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q5

Question 6:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q6-1
Height of cone = h = 24 cm
Its radius = 7 cm
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q6-2
Total surface area of toy
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q6-3

Question 7:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q7-1
Height of cylindrical container h1 = 15 cm
Diameter of cylindrical container = 12 cm
Volume of container = rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q7-2
Height of cone r2 = 12 cm
Diameter = 6 cm
Radius of r2 = 3 cm
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q7-3
Radius of hemisphere = 3 cm
Volume of hemisphere = rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q7-4
Volume of cone + volume of hemisphere
= 36π + 18π = 54π
Number of cones
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q7-5
Number of cones that can be filled = 10

Question 8:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q8-1
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
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q8-2
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
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q9-1
Total surface area of toy = rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q9-2
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q9-3

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 = rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q10-1
Area of canvas required = (curved surface area of the cylindrical part) + (curved surface area of the conical part)
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q10-2

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
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q11

Question 12:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q12-1
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
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q12-2

Question 13:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q13-1
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)
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q13-2
Cost of polishing the surface of the solid
= Rs. (0.15 × 8624)
= Rs. 1293. 60

Question 14:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q14-1
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)
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q14-2

Question 15:
(i) Radius of cylinder = 6 cm
Height of cylinder = 8 cm
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q15-1
Volume of cylinder
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q15-2
Volume of cone removed
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q15-3
(ii) Surface area of cylinder = 2π = 2π × 6 × 8 cm2 = 96 π cm2
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q15-4

Question 16:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q16-1
Diameter of spherical part of vessel = 21 cm
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q16-2

Question 17:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q17-1
Height of cylindrical tank = 2.5 m
Its diameter = 12 m, Radius = 6 m
Volume of tank = rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q17-2
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
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q17-3

Question 18:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q18-1
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  cm= 196.25 cm3
Apparent capacity of glass = 196.25
Radius of hemisphere = 2.5 cm
Volume of hemisphere
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19a-q18-2
Actual capacity of glass = ( 196.25 – 32.608 ) cm3 = 163.54 cm3

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 12\times 12\times 24) π cm3  = (48 × 24 )π cm3
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q1

Question 2:
Internal radius = 3 cm and external radius = 5 cm
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q2
Hence, height of the cone = 4 cm

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

rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q3-1
Radius of each cylindrical bottle = 2.5 cm and its height = 6 cm
Volume of each cylindrical bottle
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q3-2
Required number of bottles = rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q3-3
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q3-4
Hence, bottles required = 60

Question 4:
Radius of the sphere = \frac { 21 }{ 2 }   cm
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q4-1
Let the number of cones formed be n, then
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q4-2
Hence, number of cones formed = 504

Question 5:
Radius of the cannon ball = 14 cm
Volume of cannon ball = rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q5-1
Radius of the cone = \frac { 35 }{ 2 }   cm
Let the height of cone be h cm
Volume of cone = rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q5-2
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q5-3
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-class-10-solutions-volume-and-surface-areas-of-solids-19b-q6

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-class-10-solutions-volume-and-surface-areas-of-solids-19b-q7-2
Hence diameter of the base of the cylinder = 12 cm

Question 8:
Volume of hemisphere of radius 9 cm
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q8-1
Volume of circular cone (height = 72 cm)
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q8-2
Volume of cone = Volume of hemisphere
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q8-3
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
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q9
Hence, number of cubes is 4851.

Question 10:
Volume of sphere (when r = 1 cm) = \frac { 4 }{ 3 }   πr3 = (\frac { 4 }{ 3 } \times 1\times 1\times 1) π cm3
Volume of sphere (when r = 8 cm) = \frac { 4 }{ 3 }   πr3 = (\frac { 4 }{ 3 } \times 8\times 8\times 8) π cm3
Let the number of balls = n
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q10

Question 11:
Radius of marbles = \frac { Diameter }{ 2 } =\frac { 1.4 }{ 2 } cm  
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q11-1
Let the number of marbles be n
∴ n × volume of marble = volume of rising water in beaker
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q11-2

Question 12:
Radius of sphere = 3 cm
Volume of sphere = \frac { 4 }{ 3 }   πr3 = (\frac { 4 }{ 3 } \times 3\times 3\times 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 0.3\times 0.3\times 0.3) π cm3
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q12-1
Let number of small balls be n
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q12-2
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 21\times 21\times 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
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q13
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 3\times 3\times 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
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q15-1
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-class-10-solutions-volume-and-surface-areas-of-solids-19b-q15-2
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-class-10-solutions-volume-and-surface-areas-of-solids-19b-q16-1
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q16-2
Let r be the radius of sphere whose volume is equal to the total volume of three spheres.
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q16-3

Exercise 19C

Question 1:
Here h = 42 cm, R = 16 cm, and r = 11 cm
Capacity = rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19c-q1-1
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19c-q1-2

Question 2:
Here R = 33 cm, r = 27 cm and l = 10 cm
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19c-q2-1
Capacity of the frustum
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19c-q2-2
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19c-q2-3
Total surface area
= rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19c-q2-4
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19c-q2-5

Question 3:
Height = 15 cm, R = \frac { 56 }{ 2 }   cm = 28 cm and r = \frac { 42 }{ 2 }   cm = 21 cm
Capacity of the bucket =
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q3-1
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q3-2
Quantity of water in bucket = 28.49 litres

Question 4:
R = 20 cm, r = 8 cm and h = 16 cm
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q4-1
Total surface area of container = πl (R+r) + πr2
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q4-2
Cost of metal sheet used = Rs. (1959.36\times \frac { 15 }{ 100 } )   = Rs. 293.90

Question 5:
R = 15 cm, r = 5 cm and h = 24 cm
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q5-1
(i) Volume of bucket =
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q5-2
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q5-3
Cost of milk = Rs. (8.164 × 20) = Rs. 163.28
(ii) Total surface area of the bucket
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q5-4
Cost of sheet = ( 1711.3\times \frac { 10 }{ 100 } )   = Rs. 171.13

Question 6:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q6-1
R = 10cm, r = 3 m and h = 24 m
Let l be the slant height of the frustum, then
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q6-2
Quantity of canvas = (Lateral surface area of the frustum) + (lateral surface area of the cone)
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q6-3

Question 7:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q7-1
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
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q7-2
Radius of the cone = EB = 7 m
Slant height l2 of cone = 12 m
Surface area of canvas required
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q7-3

Question 8:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q8-1
In the given figure, we have
∠COD = 30°, OC = 10 cm, OE = 20 cm
Let CD = r cm and EB = R cm
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q8-2
Also, CE = 10 cm
Thus, ABDF is the frustum of a cone in which
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q8-3
Volume of wire of radius r and length l
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q8-4
Volume of wire = Volume of frustum
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19c-q8-5
Length of the wire is 7964.44 m

Question 9:
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q9-1
Radii of upper and lower end of frustum are r = 8 cm, R = 32 cm
Height of frustum h = 18 cm
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q9-2
Cost of milk at Rs 20 per litre = Rs. 25.344 × 20 = Rs. 506. 88

We hope the given Solutions of RS Aggarwal Solutions Class 10 Chapter 19 are helpful to complete your math homework. If you have any doubts, please comment below. We will try to provide online math tutoring for you.


RS Aggarwal Solutions Class 10

  1. Real Numbers
  2. Polynomials
  3. Linear equations in two variables
  4. Triangles
  5. Trigonometric Ratios
  6. T-Ratios of Some Particular Angles
  7. Trigonometric Identities
  8. Trigonometric Ratios of Complementary Angles
  9. Mean, Median, Mode of Grouped Data, of Grouped Data Ex.9A Ex.9B Ex.9C and 9D Ex.9E
  10. Quadratic Equations Ex.10A Ex.10B Ex.10C Ex.10D
  11. Arithmetic Progressions
  12. Circles
  13. Constructions
  14. Height and Distance
  15. Probability
  16. Coordinate Geometry
  17. Perimeter and Areas of Plane Figures
  18. Areas of Circle, Sector and Segment
  19. Volume and Surface Areas of Solids

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