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Class 10 SELINA Solutions Maths Chapter 20 - Cylinder, Cone and Sphere (Surface Area and Volume)

Cylinder, Cone and Sphere (Surface Area and Volume) Exercise Ex. 20(A)

Solution 1

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A1

Solution 2

Inner radius of pipe = 2.1 cm

Length of the pipe = 12 m = 1200 cm

Solution 3

Circumference of cylinder = 8 cm

Therefore, radius =

Length of the cylinder (h)= 21 cm

(i) If distance covered in one revolution is 8 cm, then distance covered in 9 revolutions = 9 8 = 72 cm or distance covered in 1 second = 72 cm.

Therefore, distance covered in seconds =

(ii) Curved surface area =

Area covered in one revolution =

Area covered in 9 revolutions =

Therefore, area covered in 1 second =

Hence, area covered in seconds =

Solution 4

Radius of the well =

Depth of the well = 28 m

Therefore, volume of earth dug out =

Area of curved surface =

Cost of plastering at the rate of Rs 4.50 per sq m

= Rs 246.40 4.50

= Rs 1108.80

Solution 5

External diameter of hollow cylinder = 20 cm

Therefore, radius = 10 cm

Thickness = 0.25 cm

Hence, internal radius = (10 - 0.25) = 9.75 cm

Length of cylinder (h) = 15 cm

Diameter = 2 cm

Therefore, radius (r) = 1 cm

Let h be the length

then,

Now, according to given condition:

Length of cylinder = 74.0625 cm

Solution 6

Diameter of the cylinder = 20 cm

Hence, Radius (r) = 10 cm

Height = h cm

(i) Curved surface area =

(ii) Volume of the cylinder =

or

Solution 7

Solution 8

Diameter of cylindrical container = 42 cm

Therefore, radius (r) = 21 cm

Dimensions of rectangular solid = 22cm 14cm 10.5cm

Volume of solid =

Let height of water = h

Therefore, volume of water in the container =

 

From (i) and (ii)

Solution 9

Internal radius of the cylindrical container = 10 cm

Height of water = 7 cm

Therefore, surface area of the wetted surface =

Solution 10

Length of an open pipe = 50 cm

External diameter = 20 cm External radius (R) = 10 cm

Internal diameter = 6 cm Internal radius (r) = 3 cm

Surface area of pipe open from both sides =

Area of upper and lower part =

Total surface area =

Solution 11

Ratio between height and radius of a cylinder = 3:1

Volume = …….(i)

Let radius of the base = r

then height = 3r

from (i) and (ii)

Therefore, radius = 7 cm and height = 3 x 7 = 21 cm

Now, total surface area =

Solution 12

Solution 13

left parenthesis ii right parenthesis space Curved space surface space area space left parenthesis Original right parenthesis space of space straight a space solid space right space circular space cylinder
space space space space space space equals 2 πrh
space space space space space space equals 2 straight pi cross times 100 cross times cross times 100
space space space space space space equals 20000 straight pi space cm squared
space space space space space space Curved space surface space area space left parenthesis New right parenthesis space of space straight a space solid space right space circular space cylinder
space space space space space space equals 2 πr apostrophe straight h apostrophe
space space space space space space equals 2 straight pi cross times 80 cross times cross times 110
space space space space space space equals 17600 straight pi space cm squared
space space space space space space space Decrease space in space curved space surface space area equals Original space CSA minus New space CSA equals left parenthesis 20000 straight pi minus 17600 straight pi right parenthesis space cm squared equals 2400 straight pi space cm squared
space space space space space space space Percentage space change space in space curved space surface space area equals fraction numerator Decrease space in space curved space surface space area over denominator Original space curved space surface space area end fraction cross times 100 percent sign
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 2400 straight pi space cm squared over denominator 20000 straight pi space cm squared end fraction cross times 100 percent sign
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 12 percent sign

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Total surface area of a hollow cylinder = 3575 cm2

Area of the base ring = 357.5 cm2

Height = 14 cm

Let external radius = R and internal radius = r

Let thickness of the cylinder = d = (R-r)

Therefore, Total surface area =

and Area of base =

Dividing (i) by (ii)

Hence, thickness of the cylinder = 3.5 cm

Solution 26

Solution 27

Solution 28

Cylinder, Cone and Sphere (Surface Area and Volume) Exercise Ex. 20(B)

Solution 1

Slant height () = 17 cm

Radius (r) = 8 cm

But,

Now, volume of cone =

Solution 2

Curved surface area =

Radius of base (r) = 56 cm

Let slant height =

Height of the cone =

Solution 3

Circumference of the conical tent = 66 m

and height (h) = 12 m

Therefore, volume of air contained in it =

Solution 4

The ratio between radius and height = 5:12

Volume = 5212 cubic cm

Let radius (r) = 5x, height (h) = 12x and slant height =

Now Volume =

Solution 5

Let radius of cone y = r

Therefore, radius of cone x = 3r

Let volume of cone y = V

then volume of cone x = 2V

Let h1 be the height of x and h2 be the height of y.

Therefore, Volume of cone =

Volume of cone x =

Volume of cone y =

Solution 6

Let radius of each cone = r

Ratio between their slant heights = 5:4

Let slant height of the first cone = 5x

and slant height of second cone = 4x

Therefore, curved surface area of the first cone =

curved surface area of the second cone =

Hence, ratio between them =

Solution 7

Let slant height of the first cone =

then slant height of the second cone = 2

Radius of the first cone =

Radius of the second cone =

Then, curved surface area of first cone =

curved surface area of second cone =

According to given condition:

Solution 8

Diameter of the cone = 16.8 m

Therefore, radius (r) = 8.4 m

Height (h) = 3.5 m

(i) Volume of heap of wheat =

(ii) Slant height () =

Therefore, cloth required or curved surface area =

Solution 9

Diameter of the tent = 48 m

Therefore, radius (r) = 24 m

Height (h) = 7 m

Slant height () =

Curved surface area =

Canvas required for stitching and folding

Total canvas required (area)

Length of canvas

Rate = Rs 24 per meter

Total cost

Solution 10

Height of solid cone (h) = 8 cm

Radius (r) = 6 cm

Volume of solid cone =

Height of smaller cone = 2 cm

and radius =

Volume of smaller cone

Number of cones so formed

Solution 11

Total surface area of cone =

slant height (l) = 13 cm

(i) Let r be its radius, then

Total surface area =

Either r+18 = 0, then r = -18 which is not possible

or r-5=0, then r = 5

Therefore, radius = 5 cm

(ii) Now

Solution 12

Solution 13

Volume of vessel = volume of water =

diameter = 25.2 cm, therefore radius = 12.6 cm

height = 32 cm

Volume of water in the vessel =

On submerging six equal solid cones into it, one-fourth of the water overflows.

Therefore, volume of the equal solid cones submerged

= Volume of water that overflows

Now, volume of each cone submerged

Solution 14

(i) Let r be the radius of the base of the conical tent, then area of the base floor =

Hence, radius of the base of the conical tent i.e. the floor = 7 m

(ii) Let h be the height of the conical tent, then the volume =

Hence, the height of the tent = 24 m

(iii) Let l be the slant height of the conical tent, then

The area of the canvas required to make the tent =

Length of the canvas required to cover the conical tent of its width 2 m =

Cylinder, Cone and Sphere (Surface Area and Volume) Exercise Ex. 20(C)

Solution 1

Surface area of the sphere = 2464 cm2

Let radius = r, then

Volume =

Solution 2

Volume of the sphere = 38808 cm3

Let radius of sphere = r

Solution 3

Let the radius of spherical ball = r

Radius of smaller ball =

Therefore, number of smaller balls made out of the given ball =

Solution 4

Diameter of bigger ball = 8 cm

Therefore, Radius of bigger ball = 4 cm

Radius of small ball = 1 cm

Number of balls =

Solution 5

Radius of metallic sphere = 2 mm =

Volume of 8 spheres =

Let radius of new sphere = R

From (i) and (ii)

Solution 6

Volume of first sphere = 27 volume of second sphere

Let radius of first sphere =

and radius of second sphere =

Therefore, volume of first sphere =

and volume of second sphere =

(i) Now, according to the question

(ii) Surface area of first sphere =

and surface area of second sphere =

Ratio in surface area =

Solution 7

Let r be the radius of the sphere.

Surface area = and volume =

According to the condition:

Diameter of sphere = 2 3 cm = 6 cm

Solution 8

Diameter of sphere =

Therefore, radius of sphere =

Total curved surface area of each hemispheres =

Solution 9

External radius (R) = 14 cm

Internal radius (r) =

(i) Internal curved surface area =

(ii) External curved surface area =

(iii) Total surface area =

(iv) Volume of material used =

Solution 10

Let the radius of the sphere be 'r1'.

Let the radius of the hemisphere be 'r2'

TSA of sphere = 4∏r12

TSA of hemisphere = 3∏r22

TSA of sphere = TSA of hemi-sphere

  

Volume of sphere, V1 =

Volume of hemisphere, V2 =

  

 

Dividing V1 by V2,

 

 

Solution 11

Let radius of the larger sphere be 'R'

Volume of single sphere

= Vol. of sphere 1 + Vol. of sphere 2 + Vol. of sphere 3

 

Surface area of the sphere

 

 

 

 

Solution 12

Let the radius of the sphere be 'r'.

Total surface area the sphere, S   

New surface area of the sphere, S'

 

(i) Let the new radius be r1

 

Percentage change in radius



Percentage change in radius = 10%


(ii) Let the volume of the sphere be V

Let the new volume of the sphere be V'.



Percentage change in volume =33.1%

Cylinder, Cone and Sphere (Surface Area and Volume) Exercise Ex. 20(D)

Solution 1

Solution 2

External diameter = 8 cm

Therefore, radius (R) = 4 cm

Internal diameter = 4 cm

Therefore, radius (r) = 2 cm

Volume of metal used in hollow sphere =

Diameter of cone = 8 cm

Therefore, radius = 4 cm

Let height of cone = h

From (i) and (ii)

Height of the cone = 14 cm

Solution 3

Internal radius = 3cm

External radius = 5 cm

Volume of spherical shell

 

 

Volume of solid circular cone

 

Vol. of Cone = Vol. of sphere


Hence, diameter = 2r = 7 cm

 

 

Solution 4

Let the radius of the smaller cone be 'r' cm.

Volume of larger cone

 

 

Volume of smaller cone

 

 

Volume of larger cone=3× Volume of smaller cone

  

rightwards double arrow straight r equals 3 1 third space cm

 

 

 

 

Solution 5

Volume of rectangular block =

Let r be the radius of sphere

From (i) and (ii)

Radius of sphere = 21 cm

Solution 6

Radius of hemispherical bowl = 9 cm

Diameter each of cylindrical bottle = 3 cm

Radius = 3 over 2cm, and height = 4 cm

Solution 7

Diameter of the hemispherical bowl = 7.2 cm

Therefore, radius = 3.6 cm

Volume of sauce in hemispherical bowl =

Radius of the cone = 4.8 cm

Volume of cone =

Now, volume of sauce in hemispherical bowl = volume of cone

Height of the cone = 4.05 cm

Solution 8

Radius of a solid cone (r) = 5 cm

Height of the cone = 8 cm

Volume of a cone

Solution 9

Total area of solid metallic sphere = 1256 cm2

(i)Let radius of the sphere is r then

(ii) Volume of sphere = Volume of right circular cone =

Number of cones

Solution 10

Volume of the whole cone of metal A

 

 

Volume of the cone with metal B

 

 

 

 

Final Volume of cone with metal A=120  -12  =108

 

 

 

 

 

 

 

 

Solution 11

Let the number of small cones be 'n'

Volume of sphere

 

 

 

 

Volume of small spheres

 

 

Volume of sphere = n× Volume of small sphere

 

 

The number of cones = 37

Solution 12

Cylinder, Cone and Sphere (Surface Area and Volume) Exercise Ex. 20(E)

Solution 1

Height of cone = 15 cm

and radius of the base = cm

Therefore, volume of the solid = volume of the conical part + volume of hemispherical part.

Solution 2

Radius of hemispherical part (r) = 3.5 m =

Therefore, Volume of hemisphere =

Volume of conical part = (2/3 of hemisphere)

Let height of the cone = h

Then,

Height of the cone = 4.67 m

Surface area of buoy =

But

Therefore, Surface area =

Surface Area = 141.17 m2

Solution 3

(i) Total surface area of cuboid = 2(b + bh + h)

=2(42 30 + 30 20 + 20 42)

=2(1260 + 600 + 840)

=2 2700

=5400 cm2

Diameter of the cone = 14 cm

Radius of the cone =

Area of circular base

Area of curved surface area of cone

Surface area of remaining part

(ii) Dimensions of rectangular solids = (42 30 20) cm

volume = (42 30 20) = 25200 cm3

Radius of conical cavity (r) =7 cm

height (h) = 24 cm

Volume of cone =

Volume of remaining solid = (25200 - 1232) = 23968 cm3

(iii) Weight of material drilled out

=1232 7 g = 8624g = 8.624 kg

Solution 4

The diameter of the largest hemisphere that can be placed on a face of a cube of side 7 cm will be 7 cm.

Therefore, radius =

Its curved surface area =

Surface area of the top of the resulting solid = Surface area of the top face of the cube - Area of the base of the hemisphere

Surface area of the cube =

Total area of resulting solid = 245 + 10.5 + 77 = 332.5 cm2

Solution 5

Height of cone = 8 cm

Radius = 5 cm

Volume =

Therefore, volume of water that flowed out =

 

Radius of each ball = 0.5 cm =

Volume of a ball =

Therefore, No. of balls =

Hence, number of lead balls = 100

Solution 6

Let r be the radius of the bowl.

Capacity of the bowl =

Solution 7

For the volume of cone to be largest, h = r cm

Volume of the cone

 


Solution 8

Let the height of the solid cones be 'h'

Volume of solid circular cones

 

Volume of sphere

Volume of sphere = Volume of cone 1 + volume of cone 2



Solution 9

Volume of the solid hemisphere

 

Volume of 1 cone


No. of cones formed


 


Solution 10

Let the radius of base be 'r' and the height be 'h'

Volume of cone, Vc

 

Volume of hemisphere, Vh

Cylinder, Cone and Sphere (Surface Area and Volume) Exercise Ex. 20(F)

Solution 1

Height of the cylinder (h) = 10 cm

and radius of the base (r) = 6 cm

Volume of the cylinder =

Height of the cone = 10 cm

Radius of the base of cone = 6 cm

Volume of the cone =

Volume of the remaining part

Solution 2

Radius of solid cylinder (R) = 12 cm

and Height (H) = 16 cm

Radius of cone (r) = 6 cm, and height (h) = 8 cm.

(i) Volume of remaining solid

(ii) Slant height of cone

Therefore, total surface area of remaining solid = curved surface area of cylinder + curved surface area of cone + base area of cylinder + area of circular ring on upper side of cylinder

Solution 3

Radius of the cylindrical part of the tent (r) =

Slant height () = 80 m

Therefore, total curved surface area of the tent =

Width of canvas used = 1.5 m

Length of canvas =

Total cost of canvas at the rate of Rs 15 per meter

Solution 4

Height of the cylindrical part = H = 8 m

Height of the conical part = h = (13-8)m = 5 m

Diameter = 24 m radius = r = 12 m

 

Slant height of the cone = l

 

Slant height of cone = 13 m

 

(i) Total surface area of the tent =

(ii)Area of canvas used in stitching = total area

Solution 5

Diameter of cylindrical boiler = 3.5 m

Height (h) = 2 m

Radius of hemisphere (R) =

Total volume of the boiler =

Solution 6

Diameter of the base = 3.5 m

Therefore, radius =

Height of cylindrical part =

(i) Capacity (volume) of the vessel =

(ii)Internal curved surface area =

Solution 7

Height of the cone = 24 cm

Height of the cylinder = 36 cm

Radius of the cone = twice the radius of the cylinder = 10 cm

Radius of the cylinder = 5 cm

Slant height of the cone =

Now, the surface area of the toy = curved area of the conical point + curved area of the cylinder

Solution 8

Diameter of cylindrical container = 42 cm

Therefore, radius (r) = 21 cm

Dimensions of rectangular solid = 22cm 14cm 10.5cm

Volume of solid =

Let height of water = h

Therefore, volume of water in the container =

 

From (i) and (ii)

Solution 9

Diameter of spherical marble = 1.4 cm

Therefore, radius = 0.7 cm

Volume of one ball

Diameter of beaker = 7 cm

Therefore, radius =

Height of water = 5.6 cm

Volume of water =

No. of balls dropped

Solution 10

Breadth of the tunnel = 6 m

Height of the tunnel = 8 m

Length of the tunnel = 35 m

Radius of the semi-circle = 3 m

Circumference of the semi-circle =

Internal surface area of the tunnel

Rate of plastering the tunnel = Rs 2.25 per m2

Therefore, total expenditure

Solution 11

Length = 21 m

Depth of water = 2.4 m

Breadth = 7 m

Therefore, radius of semicircle =

Area of cross-section = area of rectangle + Area of semicircle

Therefore, Volume of water filled in gallons

Solution 12

Length = 21 cm, Breadth = 7 cm

Radius of semicircle =

Area of cross section of the water channel =

Flow of water in one minute at the rate of 20 cm per second

Length of the water column = 20 60 = 1200 cm

Therefore, volume of water =

Solution 13

Diameter of the base of the cylinder = 7 cm

Therefore, radius of the cylinder =

Volume of the cylinder

Diameter of the base of the cone =

Therefore, radius of the cone =

Volume of the cone

On placing the cone into the cylindrical vessel, the volume of the remaining portion where the water is to be filled

Height of new cone =

Radius = 2 cm

Therefore, volume of new cone

Volume of water which comes down =

Let h be the height of water which is dropped down.

Radius =

From (i) and (ii)

Drop in water level =

Solution 14

Radius of the base of the cylindrical can = 3.5 cm

(i) When the sphere is in can, then total surface area of the can =

Base area + curved surface area

(ii) Let depth of water = x cm

When sphere is not in the can, then volume of the can =

volume of water + volume of sphere

Solution 15

Let the height of the water level be 'h', after the solid is turned upside down.

Volume of water in the cylinder

 

 

Volume of the hemisphere

 

 

Volume of water in the cylinder

= Volume of water level - Volume of the hemisphere

 

 

 

 

 

 

Cylinder, Cone and Sphere (Surface Area and Volume) Exercise Ex. 20(G)

Solution 1

Let the number of solid metallic spheres be 'n'

Volume of 1 sphere

Volume of metallic cone

=

 

The least number of spheres needed to form the cone is 15

Solution 2

Radius of largest sphere that can be formed inside the cylinder should be equal to the radius of the cylinder.

 

 

Radius of the largest sphere = 7 cm

Volume of sphere

 

 

 

 

 

 

 

 

Solution 3

Let the number of cones be 'n'.

Volume of the cylinder =


Volume of ice-cream cone = Volume of cone + Volume of hemisphere


                                     equals 1 third straight pi left parenthesis 3 right parenthesis squared cross times 12 plus 2 over 3 straight pi left parenthesis 3 right parenthesis cubed
equals 36 straight pi plus 18 straight pi
equals 54 straight pi space cm cubed

 

 

 

Number space of space cones equals fraction numerator Volume space of space cylinder over denominator Volume space of space ice minus cream space cone end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator straight pi cross times 36 cross times 15 over denominator 54 straight pi end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 10

 

Hence, number of cones required = 10

Solution 4

 Volume of the solid

 

 

 

 

 

 

Solution 5

Diameter of a sphere = 6 cm

Radius = 3 cm

Diameter of cylindrical wire = 0.2 cm

Therefore, radius of wire =

Let length of wire = h

From (i) and (ii)

Hence, length of the wire = 36 m

Solution 6

Let edge of the cube = a

volume of the cube =

The sphere, which exactly fits in the cube, has radius =

Therefore, volume of sphere =

Volume of cube : volume of sphere

Solution 7

Radius of the base of poles (r) = 6 cm

Height of the cylindrical part (h1) = 110 cm

Height of the conical part (h2) = 9 cm

Total volume of the iron pole =

Weight of 1 cm3 = 8 gm

Therefore, total weight = 12780 8 = 102240 gm = 102.24 kg

Solution 8

Length of the platform = 22 m

Circumference of semicircle = 11 m

Therefore, breadth of the rectangular part =

And length =

Now area of platform =

Height of the platform = 1.5 m

Rate of construction = Rs 4 per m3

Total expenditure = Rs 4 223.125 = Rs 892.50

Solution 9

Side of square = 7 m

Radius of semicircle =

Length of the tunnel = 80 m

Area of cross section of the front part =

(i) Therefore, volume of the tunnel = area x length

(ii) Circumference of the front of tunnel

Therefore, surface area of the inner part of the tunnel

= 25 80

= 2000 m2

(iii) Area of floor = l b = 7 80 = 560 m2

Solution 10

Diameter of cylindrical tank = 2.8 m

Therefore, radius = 1.4 m

Height = 4.2 m

Volume of water filled in it =

Diameter of pipe = 7 cm

Therefore, radius (r) =

Let length of water in the pipe = h1

From (i) and (ii)

Therefore, time taken at the speed of 4 m per second

Solution 11

Rate of flow of water = 9 km/hr

Water flow in 1 hour 15 minutes

i.e. in

Area of cross-section =

Therefore, volume of water

Dimensions of water tank = 7.5m 5m 4m

Area of tank = l b = 7.5 5 = 37.5 m2

Let h be the height of water then,

37.5 h = 28.125

Solution 12

Diameter = 10 cm

Therefore, radius (r) = 5 cm

Height of the cone (h) = 12 cm

Height of the cylinder = 12 cm

(i) Total surface area of the solid

(ii) Total volume of the solid

(iii) Total weight of the solid = 1.7 kg

Solution 13

Radius of cylinder = 3 cm

Height of cylinder = 6 cm

Radius of hemisphere = 2 cm

Height of cone = 4 cm

Volume of water in the cylinder when it is full =

Volume of water displaced = volume of cone + volume of

hemisphere

Therefore, volume of water which is left

Solution 14

Radius of the cylinder = 3.5 m

Height = 7 m

(i) Total surface area of container excluding the base = Curved

surface area of the cylinder + area of hemisphere

(ii) Volume of the container =

Solution 15

Total height of the tent = 85 m

Diameter of the base = 168 m

Therefore, radius (r) = 84 m

Height of the cylindrical part = 50 m

Then height of the conical part = (85 - 50) = 35 m

Slant height (l)

Total surface area of the tent =

Since 20% extra is needed for folds and stitching,

total area of canvas needed

Solution 16

Volume of water filled in the test tube =

Volume of water filled up to 4 cm =

Let r be the radius and h be the height of test tube.

and

Dividing (i) by (ii)

Subtracting (ii) from (i)

Substituting the value of r in (iii)

Hence, Height = 20 cm and radius = 3.5 cm

Solution 17

Diameter of hemisphere = 7 cm

Diameter of the base of the cone = 7 cm

Therefore, radius (r) = 3.5 cm

Height (h) = 8 cm

Volume of the solid =

Now, radius of cylindrical vessel (R) = 7 cm

Height (H) = 10 cm

Volume of water required to fill = 1540 - 192.5 = 1347.5 cm3

Solution 18

Solution 19

Let the number of cones melted be n.

Let the radius of sphere be rs = 6 cm

Radius of cone be rc = 2 cm

And, height of the cone be h = 3 cm

Volume of sphere = n (Volume of a metallic cone)

Solution 20

According to the condition in the question,

We know that,

l2 = r2 + h2

l2 = (7)2 + (24)2

l2 = 49 + 576

l2 = 625

l = 25 m

Curved Surface Area = πrl =  × 7× 25 = 550m2

Therefore the height of the tent is 24m and it curved surface area is 550m2. 

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