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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 TEST YOURSELF

Solution 1(a)

Correct option: (ii) 5pr2

Total surface area of the solid

= Curved surface area of cylinder + Curved surface area of cone + Base of the cylinder

= 2prh + prl + pr2

= 2pr(r) + pr(2r) + pr2

= 2pr2 + 2pr2 + pr2

= 5pr2

Solution 1(b)

Correct option: (iii) 1 : 8

Let the radii of two solid spheres be r­1 and r2 respectively.

Then, r1 = 10 cm and r2 = 20 cm

The ratio between their volumes

Solution 1(c)

Correct option: (iv) 20 cm

Let the radius and height of cone be r­1 and h respectively.

And, the radius of sphere be r2.

Then, h = 10 cm and r2 = 10 cm

Volume of cone = Volume of sphere

*The information provided in question is about cone and sphere and not about cylinder. Hence, as per information, the radius of cone is to be calculated and not of cylinder. Error in question.

Solution 1(d)

Correct option: (i) 378

For a cuboid, l = 11 cm, b = 18 cm, h = 8 cm

For a sphere r = 1 cm

Solution 1(e)

Correct option: (i) only 1

If a wire is extended four times along its length with same width all around, its volume will remain constant in both the cases.

But, the surface area of solid will increase. Hence, it will be not same in both the cases.

Solution 2

Let the number of solid metallic spheres be 'n'

Volume of 1 sphere

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G1-1

Volume of metallic cone

= ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G1-2

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G1-3

 

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

Solution 3

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G2

 

 

 

 

 

 

 

 

Solution 4

Let the number of cones be 'n'.

Volume of the cylinder = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G3-1


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


                                     ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G3-2

 

 

 

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G3-3

 

Hence, number of cones required = 10

Solution 5

 Volume of the solid

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G4

 

 

 

 

 

 

Solution 6

Diameter of a sphere = 6 cm

Radius = 3 cm

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G5-1

Diameter of cylindrical wire = 0.2 cm

Therefore, radius of wire = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G5-2

Let length of wire = h

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G5-3

From (i) and (ii)

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G5-4

Hence, length of the wire = 36 m

Solution 7

Let edge of the cube = a

volume of the cube = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G6-1

The sphere, which exactly fits in the cube, has radius = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G6-2

Therefore, volume of sphere = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G6-3

Volume of cube : volume of sphere

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G6-4

Solution 8

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G7-1

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 = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G7-2

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G7-3

Weight of 1 cm3 = 8 gm

Therefore, total weight = 12780 ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G7-4 8 = 102240 gm = 102.24 kg

Solution 9

Side of square = 7 m

Radius of semicircle = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G9-1

Length of the tunnel = 80 m

Area of cross section of the front part = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G9-2

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G9-3

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G9-4

(ii) Circumference of the front of tunnel

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G9-5

Therefore, surface area of the inner part of the tunnel

= 25 ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G9-6 80

= 2000 m2

(iii) Area of floor = l ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G9-7 b = 7 ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G9-8 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 = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G10-1

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G10-2

Diameter of pipe = 7 cm

Therefore, radius (r) = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G10-3

Let length of water in the pipe = h1

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G10-4

From (i) and (ii)

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G10-5

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G10-6

Solution 11

Rate of flow of water = 9 km/hr

Water flow in 1 hour 15 minutes

i.e. in ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G11-1

Area of cross-section = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G11-2

Therefore, volume of waterICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G11-3

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G11-4

Solution 12

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G12-1

Diameter = 10 cm

Therefore, radius (r) = 5 cm

Height of the cone (h) = 12 cm

Height of the cylinder = 12 cm

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G12-2

(i) Total surface area of the solid

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G12-3

(ii) Total volume of the solid

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G12-4

(iii) Total weight of the solid = 1.7 kg

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G12-5

Solution 13

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G13-1

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 =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G13-2

Volume of water displaced = volume of cone + volume of

hemisphere

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G13-3

Therefore, volume of water which is left

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G13-4

Solution 14

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G14-1

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G14-2

(ii) Volume of the container = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G14-3

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G14-4

Solution 15

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G15-1

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) ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G15-2

Total surface area of the tent = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G15-3

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G15-4

Since 20% extra is needed for folds and stitching,

total area of canvas needed

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G15-5

Solution 16

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G16-1

Volume of water filled in the test tube = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G16-2

Volume of water filled up to 4 cm = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G16-3

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G16-4

and

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G16-5

Dividing (i) by (ii)

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G16-6

Subtracting (ii) from (i)

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G16-7

Substituting the value of r in (iii)

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G16-8

Hence, Height = 20 cm and radius = 3.5 cm

Solution 17

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G17-1

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 =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G17-2

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

Height (H) = 10 cm

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G17-3

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

Solution 18

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)

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G19

Solution 19

According to the condition in the question,

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G20-1

We know that,

l2 = r2 + h2

l2 = (7)2 + (24)2

l2 = 49 + 576

l2 = 625

l = 25 m

Curved Surface Area = πrl = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20G20-2× 7× 25 = 550m2

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

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

Solution 2

Inner radius of pipe = 2.1 cm

Length of the pipe = 12 m = 1200 cm

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A2

Solution 3

Radius of the well = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A4-1

Depth of the well = 28 m

Therefore, volume of earth dug out = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A4-2

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A4-3

Area of curved surface = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A4-4

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A4-5

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

= Rs 246.40 ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A4-6 4.50

= Rs 1108.80

Solution 4

External diameter of hollow cylinder = 20 cm

Therefore, external radius, R = 10 cm

Thickness = 0.25 cm

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

Length of cylinder (h) = 15 cm

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A5-1


For solid cylinder,

Diameter = 2 cm

Therefore, radius (r) = 1 cm

Let h be the length

then, ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A5-2

Now, according to given condition:

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A5-3

Length of cylinder = 74.06 cm

Solution 5

Diameter of the cylinder = 20 cm

Hence, Radius (r) = 10 cm

Height = h cm

(i) Curved surface area = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A6-1

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A6-2

(ii) Volume of the cylinder = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A6-3

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A6-4

or

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A6-5

Solution 6

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A7

Solution 7

Diameter of cylindrical container = 42 cm

Therefore, radius (r) = 21 cm

Dimensions of rectangular solid = 22cm × 14cm × 10.5cm

Volume of solid =22 × 14 ×10.5 cm3 ...... (i) 

Let height of water = h

Therefore, volume of water in the container =πr2 

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A8-1

From (i) and (ii)

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A8-2

Solution 8

Internal radius of the cylindrical container = 10 cm

Height of water = 7 cm

Therefore, surface area of the wetted surface =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A9

Solution 9

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 =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A10-1

Area of upper and lower part =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A10-2

Total surface area =4085.71+572=4657.71 cm2 

Solution 10

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

Volume = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A11-1 …….(i)

Let radius of the base = r

then height = 3r

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A11-2

from (i) and (ii)

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A11-3

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

Now, total surface area =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A11-4

Solution 11

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A13

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 12

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A14

Solution 13

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A15

Solution 14

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A16

Solution 15

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A17

Solution 16

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A18

Solution 17

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A21

Solution 18

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A22

Solution 19

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A23

Solution 20

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A24

Solution 21

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 = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A25-1

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A25-2

and Area of base =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A25-3

Dividing (i) by (ii)

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A25-4

Hence, thickness of the cylinder = 3.5 cm

Solution 22

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A27

Solution 23

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20A28

Solution 1(a)

Correct option: (iii) 4prh + 2pr2

Total surface area of open hollow cylinder with radius r and height h

= External curved surface area + Internal curved surface area + 2(Area of cross section)

= 2prh + 2prh + 2(pr2)

= 4prh + 2pr2

Solution 1(b)

Correct option: (iii) S ÷ C

Curved surface area of a solid cylinder = Circumference of its base × Height

S = C × Height

Height = S ÷ C

Solution 1(c)

Correct option: (iii)

Volume of cube submerged = Volume of water that rises

a3 = pr2h

Solution 1(d)

Correct option: (ii)

Volume of big solid cylinder = pR2H

Volume of each small solid cylinder = pr2h

Then, number of smaller solid cylinders

Solution 1(e)

Correct option: (iv) 2(pR2pr2) + 2pRh + 2prh

For an open pipe,

Length = H cm, External radius = R cm, Internal radius = r cm

Then, total surface area of an open pipe

= 2(Area of cross section) + External curved surface area + Internal curved surface area

= 2(pR2pr2) + 2pRh + 2prh

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

Solution 2

Slant height () = 17 cm

Radius (r) = 8 cm

But,

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B1-1

Now, volume of cone = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B1-2

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B1-3

Solution 3

Curved surface area = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B2-1

Radius of base (r) = 56 cm

Let slant height = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B2-2

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B2-3

Height of the cone =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B2-4

Solution 4

Circumference of the conical tent = 66 m

and height (h) = 12 m

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B3-1

Therefore, volume of air contained in it = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B3-2

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B3-3

Solution 5

The ratio between radius and height = 5:12

Volume = 5212 cubic cm

Let radius (r) = 5x, height (h) = 12x and slant height = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B4-1

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B4-2

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B4-3

Now Volume = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B4-4

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B4-5

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 =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B6-1

curved surface area of the second cone = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B6-2

Hence, ratio between them = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B6-3

Solution 7

Let slant height of the first cone = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B7-1

then slant height of the second cone = 2ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B7-2

Radius of the first cone = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B7-3

Radius of the second cone = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B7-4

Then, curved surface area of first cone = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B7-5

curved surface area of second cone = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B7-6

According to given condition:

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B7-7

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 = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B8-1

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B8-2

(ii) Slant height (ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B8-3) = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B8-4

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B8-5

Therefore, cloth required or curved surface area = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B8-6

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B8-7

Solution 9

Diameter of the tent = 48 m

Therefore, radius (r) = 24 m

Height (h) = 7 m

Slant height (ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B9-1) = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B9-2

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B9-3

Curved surface area = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B9-4

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B9-5

Canvas required for stitching and folding

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B9-6

Total canvas required (area)

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B9-7

Length of canvas

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B9-8

Rate = Rs 24 per meter

Total cost ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B9-9

Solution 10

Height of solid cone (h) = 8 cm

Radius (r) = 6 cm

Volume of solid cone = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B10-1

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B10-2

Height of smaller cone = 2 cm

and radius = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B10-3

Volume of smaller cone

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B10-4

Number of cones so formed

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B10-5

Solution 11

Total surface area of cone = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B11-1

slant height (l) = 13 cm

(i) Let r be its radius, then

Total surface area = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B11-2

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B11-3

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

or r-5=0, then r = 5

Therefore, radius = 5 cm

(ii) Now

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B11-4

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B11-5

Solution 12

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B12

Solution 13

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B13-1

Volume of vessel = volume of water = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B13-2

diameter = 25.2 cm, therefore radius = 12.6 cm

height = 32 cm

Volume of water in the vessel = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B13-3

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B13-4

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B13-5

Now, volume of each cone submerged

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B13-6

Solution 14

(i) Let r be the radius of the base of the conical tent, then area of the base floor = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B14-1

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B14-2

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 =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B14-3

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B14-4

Hence, the height of the tent = 24 m

(iii) Let l be the slant height of the conical tent, then ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B14-5

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B14-6

The area of the canvas required to make the tent = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B14-7

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B14-8

Length of the canvas required to cover the conical tent of its width 2 m = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20B14-9

Solution 1(a)

Correct option: (iii)

For a cone, radius = height = r cm

Then, volume of the cone

Solution 1(b)

Correct option: (ii)

For a cone, radius = r cm and height = h cm

Let the slant height = l cm

Then, its curved surface area = prl

Solution 1(c)

Correct option: (iii) 140 cm

For a conical tent,

Radius = 28 cm

Height = 21 cm

Area of sheet required = Curved surface area of the tent-house

= prl

Let the smallest length of the paper = x cm

Then, x × 22 = 3080

- x = 140 cm

Solution 1(d)

Correct option: (ii) 5.5 units

Let the radius of the cone = r units, height = h units and slant height = l units

Now,

Solution 1(e)

Correct option: (ii) 7 : 4

Let the radius of each cone = r units

Let the slant height of two cones be l1 and l2 respectively.

The ratio of curved surface areas of two cones

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

Solution 2

Volume of the sphere = 38808 cm3

Let radius of sphere = r

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C2

Solution 3

Let the radius of spherical ball = r

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C3-1

Radius of smaller ball = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C3-2

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C3-3

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C3-4

Solution 4

Diameter of bigger ball = 8 cm

Therefore, Radius of bigger ball = 4 cm

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C4-1

Radius of small ball = 1 cm

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C4-2

Number of balls = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C4-3

Solution 5

Volume of first sphere = 27 ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C6-1 volume of second sphere

Let radius of first sphere =ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C6-2

and radius of second sphere = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C6-3

Therefore, volume of first sphere = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C6-4

and volume of second sphere = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C6-5

(i) Now, according to the question

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C6-6

(ii) Surface area of first sphere = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C6-7

and surface area of second sphere = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C6-8

Ratio in surface area = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C6-9

Solution 6

Let r be the radius of the sphere.

Surface area = 4πrand volume = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C7-1

According to the condition:

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C7-2

Diameter of sphere = 2 × 3 cm = 6 cm

Solution 7

Diameter of sphere = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C8-1

Therefore, radius of sphere = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C8-2

Total curved surface area of each hemispheres =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C8-3

Solution 8

External radius (R) = 14 cm

Internal radius (r) = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C9-1

(i) Internal curved surface area =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C9-2

(ii) External curved surface area =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C9-3

(iii) Total surface area =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C9-4

(iv) Volume of material used =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C9-5

Solution 9

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C10-1 

Volume of sphere, V1 = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C10-2

Volume of hemisphere, V2 = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C10-3

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C10-4 

 

Dividing V1 by V2,

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C10-5

 

 

Solution 10

Let radius of the larger sphere be 'R'

Volume of single sphere

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C11-1

 

Surface area of the sphere

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20C11-2

 

 

 

 

Solution 1(a)

Correct option: (iii) 512

Radius of big sphere, R = 16 cm

Radius of each small sphere, r = 2 cm

Then, number of spheres formed

Solution 1(b)

Correct option: (ii) 3pR2 pr2

Outer surface area of the bowl

= Total surface area of hemispherical bowl – pr2

= 3pR2 pr2

Solution 1(c)

Correct option: (iv) 9 : 4

Let the radii of two spherical solids be r1 and r2 respectively.

Then,

Now, ratio of their curved surface areas

Solution 1(d)

Correct option: (i) 4 cm

Let the diameter of each small sphere = 2r cm

Then, radius of each small sphere = r cm

Now, Volume of big sphere = Volume of 64 small spheres

Therefore, diameter = 2r = 4 cm

Solution 1(e)

Correct option: (iv)

Let the radius of big sphere formed = R

Now, Volume of big sphere = Volume of three small spheres

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

Solution 2

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D1

Solution 3

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 = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D2-1

Diameter of cone = 8 cm

Therefore, radius = 4 cm

Let height of cone = h

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D2-2

From (i) and (ii)

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D2-3

Height of the cone = 14 cm

Solution 4

Internal radius = 3cm

External radius = 5 cm

Volume of spherical shell

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D3-1

 

 

Volume of solid circular cone

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D3-2

 

Vol. of Cone = Vol. of sphere

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D3-3


Hence, diameter = 2r = 7 cm

 

 

Solution 5

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

Volume of larger cone

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D4-1

 

 

Volume of smaller cone

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D4-2

 

 

Volume of larger cone=3× Volume of smaller cone

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D4-3 

rightwards double arrow straight r equals 3 1 third space cm

 

 

 

 

Solution 6

Volume of rectangular block = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D5-1

Let r be the radius of sphere

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D5-2

From (i) and (ii)

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D5-3

Radius of sphere = 21 cm

Solution 7

Radius of hemispherical bowl = 9 cm

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D6-1

Diameter each of cylindrical bottle = 3 cm

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D6-2

Solution 8

Total area of solid metallic sphere = 1256 cm2

(i)Let radius of the sphere is r then

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D9-1

(ii) Volume of sphere = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D9-2Volume of right circular cone =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D9-3

Number of cones

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D9-4

Solution 9

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20D12

Solution 1(a)

Correct option: (ii) 40 cm

Let the height of cone = h cm

Volume of cone = Volume of sphere

Solution 1(b)

Correct option: (iv) 1 : 4

Volume of cone = Volume of sphere

Solution 1(c)

Correct option: (iii) 8 : 1

For a cone, radius : height = r : h = 2 : 1

Solution 1(d)

Correct option: (i) 3

Solution 1(e)

Correct option: (ii) 4 × p m

For a cone, radius, r = 10 cm and height, h = 12 cm

Volume of cone

Volume of cylinder = Volume of cone

Area of cross-section × length = Volume of cone

1 × L = 400p

L = 400 × p cm = 4 × p m

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

Solution 2

Height of cone = 15 cm

and radius of the base = ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E1-1cm

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

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E1-2

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E1-3

Solution 3

Radius of hemispherical part (r) = 3.5 m = ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E2-1

Therefore, Volume of hemisphere = ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E2-2

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E2-3

Volume of conical part = ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E2-4 (2/3 of hemisphere)

Let height of the cone = h

Then,

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E2-5

Height of the cone = 4.67 m

Surface area of buoy = ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E2-6

But ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E2-7

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E2-8

Therefore, Surface area =

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E2-9

Surface Area = 141.17 m2

Solution 4

(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 = ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E3-1

Area of circular baseICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E3-2

Area of curved surface area of coneICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E3-3

Surface area of remaining part = 5400 + 550 - 154 =5796 cm2

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

volume = (42 × 30 × 20) = 25200 cm3

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E3-4

Radius of conical cavity (r) =7 cm

height (h) = 24 cm

Volume of cone = ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E3-5

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E3-6

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

(iii) Weight of material drilled out

=1232 × 7 g = 8624g = 8.624 kg

Solution 5

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 = ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E4-1

Its curved surface area =

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E4-2

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

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E4-3

Surface area of the cube = 5 × (side)2 = 5 × 49 = 245 cm2 .........(iii)

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

Solution 6

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E5-1

Height of cone = 8 cm

Radius = 5 cm

Volume = ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E5-2

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E5-3

Therefore, volume of water that flowed out =

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E5-4

Radius of each ball = 0.5 cm = ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E5-5

Volume of a ball = ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E5-6

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E5-7

Therefore, No. of balls = ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E5-8

Hence, number of lead balls = 100

Solution 7

Let r be the radius of the bowl.

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E6-1

Capacity of the bowl =

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E5-2

Solution 8

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E7-1

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

Volume of the cone

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E7-2

 


Solution 9

Let the height of the solid cones be 'h'

Volume of solid circular cones

  ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E8-1

Volume of sphere

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E8-2

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


ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E8-3


Solution 10

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

Volume of cone, Vc

  ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E10-1

Volume of hemisphere, Vh

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E10-2

ICSE class 10 selina solutions Cylinder Cone And Sphere-Ex20E10-3

Solution 1(a)

Correct option: The given options are not applicable.

Volume of whole body = Volume of cone + Volume of hemisphere

Solution 1(b)

Correct option: (iii) 2p cm3

Volume of the body = Volume of cone + Volume of cylinder

Solution 1(c)

Correct option: (ii) pr2 + prl + 2prh

Total surface area of the remaining solid

= Base of the cylinder + Curved surface area of cone + Curved surface area of cylinder

= pr2 + prl + 2prh

Solution 1(d)

Correct option: (i) 2prh + pr2 + prl

Wetted surface area of the whole body

= Curved surface area of cylinder + Base of the cylinder + Curved surface area of cone

= 2prh + pr2 + prl

Solution 1(e)

Correct option: (iii)

Volume of air left in the cylinder

= Volume of cylinder – Volume of sphere

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

Solution 2

Height of the cylinder (h) = 10 cm

and radius of the base (r) = 6 cm

Volume of the cylinder = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F1-1

Height of the cone = 10 cm

Radius of the base of cone = 6 cm

Volume of the cone = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F1-2

Volume of the remaining part

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F1-3

Solution 3

 

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F2-1

Radius of solid cylinder (R) = 12 cm

and Height (H) = 16 cm

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F2-2

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F2-3

(i) Volume of remaining solid

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F2-4

(ii) Slant height of cone ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F2-5

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F2-6

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F2-7

Solution 4

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F3-1

Radius of the cylindrical part of the tent (r) = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F3-2

Slant height (ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F3-3) = 80 m

Therefore, total curved surface area of the tent = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F3-4

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F3-5

Width of canvas used = 1.5 m

Length of canvas = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F3-6

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F3-7

Solution 5

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F4-1

Height of the cylindrical part = H = 8 m

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

Diameter = 24 m ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F4-2 radius = r = 12 m

 

Slant height of the cone = l

 

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F4-3

Slant height of cone = 13 m

 

(i) Total surface area of the tent = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F4-4

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F4-5

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F4-6

Solution 6

Diameter of cylindrical container = 42 cm

Therefore, radius (r) = 21 cm

Dimensions of rectangular solid = 22cm ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F8-1 14cm ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F8-2 10.5cm

Volume of solid = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F8-3

Let height of water = h

Therefore, volume of water in the container = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F8-4

 

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F8-5

From (i) and (ii)

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F8-6

Solution 7

Diameter of spherical marble = 1.4 cm

Therefore, radius = 0.7 cm

Volume of one ball ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F9-1

Diameter of beaker = 7 cm

Therefore, radius = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F9-2

Height of water = 5.6 cm

Volume of water = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F9-3

No. of balls dropped

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F9-4

Solution 8

Length = 21 cm, Breadth = 7 cm

Radius of semicircle = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F12-1

Area of cross section of the water channel = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F12-2

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F12-3

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F12-4Length of the water column = 20 ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F12-5 60 = 1200 cm

Therefore, volume of water =

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F12-6

Solution 9

Diameter of the base of the cylinder = 7 cm

Therefore, radius of the cylinder = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F13-1

Volume of the cylinder ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F13-2

Diameter of the base of the cone = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F13-3

Therefore, radius of the cone = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F13-4

Volume of the cone ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F13-5

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F13-6

Height of new cone = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F13-7

Radius = 2 cm

Therefore, volume of new cone

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F13-8

Volume of water which comes down = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F13-9

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

Radius = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F13-10

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F13-11

From (i) and (ii)

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F13-12

Drop in water level = ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F13-13

Solution 10

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F14-1

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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F14-2

(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

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F14-3

Solution 11

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F15-1

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

Volume of water in the cylinder

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F15-2

 

 

Volume of the hemisphere

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F15-3

 

 

Volume of water in the cylinder

= Volume of water level - Volume of the hemisphere

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F15-4

 

 

ICSE Class 10 Selina solutions Cylinder Cone And Sphere-Ex20F15-5

 

 

 

 

Solution 1(a)

Correct option: (iii) 1 : 3

Radius of cylinder = Radius of cone = r

Let the heights of cylinder and cone be H and h.

Volume of cylinder = Volume of cone

Solution 1(b)

Correct option: (iii) 2pr3

Volume of the given solid

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

Solution 1(c)

Correct option: (ii) 2 cm

Let the radius of the solid sphere formed = R cm

Volume of 8 identical spheres = Volume of solid sphere formed

Solution 1(d)

Correct option: (i) 27 : 32

Let the radii of two cones be r­1 and r2 respectively.

And, let their heights be h­1 and h2 respectively.

Now, the ratio between their volumes

Solution 1(e)

Correct option: (iv) 300%

Percentage increase in volume

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