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

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

**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 cm^{2}

= 538.56 cm^{2}

= 171.36 π

= 171.36 × \(\frac { 22 }{ 7 } \) 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 × \(\frac { 2 }{ 3 } \) πr^{3} cu.units

= \(\frac { 4 }{ 3 } \) 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 × \(\frac { 22 }{ 7 } \) 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 = \(\frac { 24 }{ 2 } \) = 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 = h_{2} = (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 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 = \(\frac { 1 }{ 3 } \) 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 = \(\frac { 21 }{ 2 } \) 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 = 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 = \(\frac { 19 }{ 2 } \) cm

Volume of sphere = \(\frac { 4 }{ 3 } \) πr^{3} = \((\frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times \frac { 21 }{ 2 } \times \frac { 21 }{ 2 } \times \frac { 21 }{ 2 } ) \)

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

Volume of sphere (when r = 8 cm) = \(\frac { 4 }{ 3 } \) 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 = \(\frac { Diameter }{ 2 } =\frac { 1.4 }{ 2 } cm \)

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

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 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 = \(\frac { 42 }{ 2 } \) cm = 21 cm

Volume of sphere = \(\frac { 4 }{ 3 } \) 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 = \(\frac { 2.8 }{ 2 } \) 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 = \(\frac { 6 }{ 2 } \) cm = 3 cm

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

Radius of wire = \(\frac { 2 }{ 2 } \) 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

∴ \(l=\frac { 36 }{ 0.01 } =3600 \) 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

= 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 = \(\frac { 56 }{ 2 } \) 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) + πr^{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

(i) Volume of bucket =

Cost of milk = Rs. (8.164 × 20) = Rs. 163.28

(ii) Total surface area of the bucket

Cost of sheet = \(( 1711.3\times \frac { 10 }{ 100 } ) \) = 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 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

Hope given RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids are helpful to complete your math homework.

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