**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.**

Contents

**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

**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)

**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 cm^{2}

= 538.56 cm^{2}

= 171.36 π

= 171.36 × cm^{2}

= 538.56 cm^{2}

Further, volume of cylinder = πr^{2}h = 4.2 × 4.2 × 12 π cm^{2}

= 211.68 π cm^{2}

Volume of two hemispheres = 2 × πr^{3} cu.units

= π × 4.2 × 4.2 × 4.2

= 98.784 cm^{3}

Volume of wood left = (211.68 – 98.784) π

= 112.896 π cm^{3}

= 112.896 × cm^{3}

= 354.816 cm^{3}

**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 h_{1} = 15 cm

Diameter of cylindrical container = 12 cm

Volume of container =

Height of cone r_{2} = 12 cm

Diameter = 6 cm

Radius of r_{2} = 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 cm^{3}

Quantity of syrup = 30% of volume of gulabjamuns

= 0.3 × 45 × 25.07 ^{ }= 338.46 cm^{3}

**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 = = 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 = h_{2} = (12.8-6.5) cm = 6.3 cm

Radius of cylinder = radius of cone

= radius of hemisphere

= cm

Volume of solid = Volume of cylinder + Volume of cone + Volume of hemisphere

**Question 13:**

Radius of each hemispherical end = = 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 r_{1} = 5 cm

And height of cylinder h_{1} = 9.8 cm

Radius of cone r = 2.1 cm

And height of cone h_{2} = 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 cm^{2} = 96 π cm^{2}

**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 = πr^{2}h cu.units = 3.14 × 2.5 × 2.5 × 10 cm^{3 }= 196.25 cm^{3}

Apparent capacity of glass = 196.25

Radius of hemisphere = 2.5 cm

Volume of hemisphere

Actual capacity of glass = ( 196.25 – 32.608 ) cm^{3} = 163.54 cm^{3}

**Exercise 19B**

**Question 1:**

Radius of the cone = 12 cm and its height = 24 cm

Volume of cone = 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 = 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 = 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 = cm

Volume of sphere = πr^{3} =

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) = πr^{3} = (\frac { 4 }{ 3 } \times 1\times 1\times 1) π cm^{3}

Volume of sphere (when r = 8 cm) = 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 =

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

Radius of small sphere = 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 = cm = 21 cm

Volume of sphere = 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 = 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 = cm = 3 cm

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

Radius of wire = 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

∴ cm

Length of wire = m = 36 m

**Question 15:**

Diameter of sphere = 18 cm

Radius of copper sphere = 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.

**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 = cm = 28 cm and r = 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) + πr^{2}

Cost of metal sheet used = Rs. = 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 = = 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 l_{1} of frustum

Radius of the cone = EB = 7 m

Slant height l_{2} 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

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 **

- Real Numbers
- Polynomials
- Linear equations in two variables
- Triangles
- Trigonometric Ratios
- T-Ratios of Some Particular Angles
- Trigonometric Identities
- Trigonometric Ratios of Complementary Angles
- Mean, Median, Mode of Grouped Data, of Grouped Data Ex.9A Ex.9B Ex.9C and 9D Ex.9E
- Quadratic Equations Ex.10A Ex.10B Ex.10C Ex.10D
- Arithmetic Progressions
- Circles
- Constructions
- Height and Distance
- Probability
- Coordinate Geometry
- Perimeter and Areas of Plane Figures
- Areas of Circle, Sector and Segment
- Volume and Surface Areas of Solids