Class 10 RD SHARMA Solutions Maths Chapter 14 - Surface Areas and Volumes

Ex. 14.1

Ex. 14.2

Ex. 14.3

Rev. 14

14.88

14.89

14.90

14.91

Surface Areas and Volumes Exercise Ex. 14.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

$table attributes columnalign left end attributes row cell L e t text end text t h e text end text n u m b e r text end text o f text end text c o n e s text end text b e apostrophe n apostrophe. end cell row cell R a d i u s text end text o f text end text t h e text end text s p h e r e equals 5.6 text end text c m end cell row cell R a d i u s text end text o f text end text t h e text end text c o n e equals 2.8 text end text c m end cell row cell H e i g h t text end text o f text end text t h e text end text c o n e equals 3.2 text end text c m end cell row cell V o l u m e text end text o f text end text t h e text end text s p h e r e text end text equals 4 over 3 pi r cubed end cell row cell equals 4 over 3 cross times pi cross times 5.6 cubed end cell row cell V o l u m e text end text o f text end text 1 text end text c o n e equals 1 third pi r squared h end cell row cell equals 1 third pi r squared h end cell row cell equals 1 third cross times pi cross times 2.8 squared cross times 3.2 end cell row blank row cell n equals fraction numerator V o l u m e text end text o f text end text t h e text end text s p h e r e over denominator V o l u m e text end text o f text end text 1 text end text c o n e end fraction end cell row cell equals fraction numerator 4 over 3 cross times pi cross times 5.6 cubed over denominator 1 third cross times pi cross times 2.8 squared cross times 3.2 end fraction end cell row cell equals fraction numerator 4 cross times 5.6 cross times 5.6 cross times 5.6 over denominator 2.8 cross times 2.8 cross times 3.2 end fraction end cell row cell equals fraction numerator 4 cross times 2 cross times 2 cross times 5.6 over denominator 3.2 end fraction end cell row cell equals fraction numerator 4 cross times 2 cross times 2 cross times 56 over denominator 32 end fraction end cell row cell equals 28 end cell row cell 28 text end text s u c h text end text c o n e s text end text c a n text end text b e text end text f o r m e d. end cell end table$

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

*Answer given in the book is incorrect.

Solution 47

$table attributes columnalign left end attributes row cell L e t text end text t h e text end text n u m b e r text end text o f text end text c o n e s text end text b e text end text apostrophe n apostrophe end cell row cell H e i g h t text end text o f text end text t h e text end text c o n e comma h equals 11 text end text c m end cell row cell R a d i u s text end text o f text end text t o p comma R equals 2.5 text end text c m end cell row cell V o l u m e text end text o f text end text t h e text end text c o n e equals 1 third pi R squared h end cell row cell equals 1 third cross times 22 over 7 cross times 2.5 squared cross times 11 end cell row cell R a d i u s text end text o f text end text t h e text end text s p h e r e comma r equals fraction numerator 0.5 over denominator 2 end fraction equals 0.25 c m end cell row cell V o l u m e text end text o f text end text 1 text end text s p h e r e equals 4 over 3 pi r cubed end cell row cell equals 4 over 3 cross times 22 over 7 cross times 0.25 cubed end cell row cell V o l u m e text end text o f text end text w a t e r text end text d i s p l a c e d equals T o t a l text end text v o l u m e text end text o f text end text t h e text end text s p h e r e s end cell row cell V o l u m e text end text o f text end text w a t e r text end text d i s p l a c e d equals n cross times text end text v o l u m e text end text o f text end text 1 text end text s p h e r e end cell row cell n equals fraction numerator V o l u m e text end text o f text end text t h e text end text c o n e over denominator V o l u m e text end text o f text end text 1 text end text s p h e r e end fraction end cell row cell equals fraction numerator 2 over 5 cross times 1 third cross times 22 over 7 cross times 2.5 squared cross times 11 over denominator 4 over 3 cross times 22 over 7 cross times 0.25 cubed end fraction end cell row cell equals fraction numerator 25 squared cross times 11 cross times 100 cubed cross times 2 over denominator 4 cross times 100 cross times 25 cubed cross times 5 end fraction end cell row cell equals fraction numerator 11 cross times 100 squared cross times 2 over denominator 4 cross times 25 cross times 5 end fraction end cell row cell equals 440 end cell row cell therefore T h e text end text n u m b e r text end text o f text end text b a l l s text end text equals 440 end cell end table$

Solution 48

Solution 49

Solution 50

Solution 51

Solution 52

$table attributes columnalign left end attributes row cell R a d i u s text end text o f text end text t h e text end text b a s e equals 1.5 text end text m end cell row cell H e i g h t text end text o f text end text t h e text end text t a n k equals 3.5 text end text m end cell row cell V o l u m e text end text o f text end text h a l f text of end text t h e text end text t a n k equals pi cross times r squared cross times h over 2 end cell row cell equals fraction numerator pi left parenthesis 1.5 right parenthesis squared cross times 3.5 over denominator 2 end fraction end cell row cell equals 22 over 7 cross times 9 over 4 cross times fraction numerator 3.5 over denominator 2 end fraction end cell row cell equals fraction numerator 11 cross times 9 over denominator 8 end fraction m cubed end cell row cell equals 99 over 8 cross times 1000 text end text l i t r e s end cell row blank row cell T i m e text taken to empty half the tank end text end cell row cell a t text the rate 225 litres per end text m i n u t e equals 99 over 8 cross times 1000 over 225 end cell row cell equals 55 text end text m i n u t e s end cell end table$

Solution 53

Solution 54

Solution 55

Solution 56

Solution 57

Solution 58

Solution 59

Solution 60

$table attributes columnalign left end attributes row cell D i a m e t e r equals 14 text end text m end cell row cell R a d i u s comma text end text r equals 7 text end text m end cell row cell H e i g h t comma text end text h equals 24 text end text m end cell row cell S l a n t text end text h e i g h t comma l equals square root of h squared plus r squared end root equals square root of 7 squared plus 24 squared end root equals square root of 49 plus 576 end root equals square root of 625 equals 25 text end text m end cell row cell C u r v e d text end text s u r f a c e text end text a r e a text end text o f text end text t h e text end text t e n t equals pi r l end cell row cell equals 22 over 7 cross times 7 cross times 25 end cell row cell equals 550 text end text m squared end cell row cell L e n g t h text end text o f text end text t h e text end text c l o t h equals fraction numerator T o t a l text end text a r e a text end text o f text end text t h e text end text c l o t h over denominator W i d t h text end text o f text end text t h e text end text c l o t h end fraction end cell row cell equals 550 over 5 equals 110 m end cell row cell I t text end text i s text end text g i v e n text end text t h a t text end text t h e text end text c o s t text end text o f text end text c l o t h text end text i s text end text R s.25 text end text p e r text end text m e t r e. end cell row cell C o s t text end text o f text end text t h e text end text c l o t h equals 110 cross times 25 end cell row cell equals R s.2750 end cell end table$

Solution 61

Solution 62

Solution 63

Solution 64

Solution 65

$table attributes columnalign left end attributes row cell L e t text end text t h e text end text r a d i u s text end text o f text end text t h e text end text h e m i s p h e r e text end text b e apostrophe r apostrophe. end cell row cell T o t a l text end text s u r f a c e text end text a r e a text end text o f text end text t h e text end text h e m i s p h e r e equals 462 text end text c m squared end cell row cell T o t a l text end text s u r f a c e text end text a r e a text end text o f text end text t h e text end text h e m i s p h e r e equals 3 pi r squared end cell row cell therefore 462 equals 3 pi r squared end cell row cell rightwards double arrow 154 equals pi r squared end cell row cell rightwards double arrow 154 equals 22 over 7 cross times r squared end cell row cell rightwards double arrow r squared equals fraction numerator 154 cross times 7 over denominator 22 end fraction end cell row cell rightwards double arrow r squared equals 49 end cell row cell therefore r equals 7 text end text c m end cell row cell V o l u m e equals 2 over 3 pi r cubed end cell row cell equals 2 over 3 cross times 22 over 7 cross times left parenthesis 7 right parenthesis cubed end cell row cell equals 2 over 3 cross times 22 cross times 49 end cell row cell equals 718.67 text end text c m cubed end cell end table$

*Answer given in the book is incorrect.

Solution 66

Solution 67

Solution 68

Solution 69

Solution 70

Solution 71

Solution 72

Height of the conical vessel h = 24 cm

Radius of the conical vessel r =5 cm

Let h be the height of the cylindrical vessel which is filled by water of the conical vessel.

Radius of the cylindrical vessel =10 cm

Volume of the cylindrical vessel = volume of water

π(10)²h=150π

h = 150π¸ 100π

h = 1.5 cm

Thus, the height of the cylindrical vessel is 1.5 cm.

Surface Areas and Volumes Exercise Ex. 14.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

$table attributes columnalign left end attributes row cell H e i g h t text end text o f text end text t h e text end text c y l i n d e r comma h equals 2.8 text end text c m end cell row cell R a d i u s text end text o f text end text t h e text end text c y l i n d e r comma r equals 2.1 text end text c m end cell row cell S l a n t text end text h e i g h t text end text o f text end text t h e text end text c o n e equals square root of 2.8 squared plus 2.1 squared end root end cell row cell equals square root of 7.84 plus 4.41 end root equals square root of 7.84 plus 4.41 end root equals square root of 12.25 end root equals 3.5 c m end cell row blank row cell T o t a l text end text s u r f a c e text end text a r e a text end text o f text end text t h e text end text r e m a i n i n g text end text s o l i d equals C u r v e d text end text s u r f a c e text end text a r e a text end text o f end cell row cell t h e text end text c y l i n d e r plus C u r v e d text end text s u r f a c e text end text a r e a text end text o f text end text t h e text end text c o n e plus T o p text end text c i r c u l a r text end text a r e a end cell row cell o f text end text t h e text end text c o n e end cell row cell equals 2 pi r h plus pi r l plus pi r squared end cell row cell equals 2 cross times 22 over 7 cross times 2.1 cross times 2.8 plus 22 over 7 cross times 2.1 cross times 3.5 plus 22 over 7 cross times 2.1 squared end cell row cell equals 22 over 7 cross times 2.1 cross times left parenthesis 5.6 plus 3.5 plus 2.1 right parenthesis end cell row cell equals 22 over 7 cross times 2.1 cross times 11.2 end cell row cell equals 73.92 text end text c m squared end cell end table$

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Total area of the canvas = curved surface area of the cone + curved surface area of a cylinder radius = 28 m height (cylinder) = 6 m

height (cone) = 21 m

l = slant height of cone

curved surface area of the cone = πrl

=π×28×35

=×28×35 = 3080 m²

curved surface area of the cylinder = 2πrh

=2××28×6

=1056

Total area of the canvas = 3080+1056 =4136 m²

Surface Areas and Volumes Exercise Ex. 14.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

$table attributes columnalign left end attributes row cell L e t text end text t h e text end text h e i g h t text end text o f text end text t h e text end text c o n e text end text b e text end text 2 h. end cell row cell S i n c e text end text t h e text end text p l a n e text end text d i v i d e s text end text t h e text end text c o n e text end text i n t o text end text h e i g h t s text end text o f text end text e q u a l text end text l e n g t h. end cell row cell B a s e text end text r a d i u s text end text o f text end text t h e text end text c o n e comma R equals 10 text end text c m end cell row cell L e t text end text t h e text end text r a d i u s text end text o f text end text t h e text end text u p p e r text end text p a r t text end text o f text end text t h e text end text f r u s t u m text end text b e text end text apostrophe r apostrophe. end cell row blank row cell I n text end text capital delta A D E comma end cell row cell fraction numerator A D over denominator D E end fraction equals fraction numerator A B over denominator B C end fraction end cell row cell rightwards double arrow h over r equals fraction numerator 2 h over denominator 10 end fraction end cell row cell therefore r equals 5 text end text c m end cell row cell V o l u m e text end text o f text end text t h e text end text s m a l l e r text end text c o n e comma V subscript c equals 1 third pi r squared h end cell row cell equals 1 third cross times 22 over 7 cross times 5 squared cross times h end cell row cell V o l u m e text end text o f text end text t h e text end text f r u s t u m text end text comma V subscript f equals 1 third pi h left parenthesis r squared plus R squared plus R r right parenthesis end cell row cell equals 1 third pi h left parenthesis 5 squared plus 10 squared plus 10 cross times 5 right parenthesis end cell row cell equals 1 third pi h cross times 175 end cell row blank row cell V subscript c over V subscript f equals fraction numerator 1 third cross times 22 over 7 cross times 5 squared cross times h over denominator 1 third pi h cross times 175 end fraction end cell row cell rightwards double arrow V subscript c over V subscript f equals fraction numerator 1 third pi cross times 5 squared cross times h over denominator 1 third pi h cross times 175 end fraction end cell row cell rightwards double arrow V subscript c over V subscript f equals 25 over 175 end cell row cell therefore V subscript c over V subscript f equals 1 over 7 end cell row cell R a t i o text end text o f text end text v o l u m e s equals 1 colon 7 end cell end table$

Solution 19

$table attributes columnalign left end attributes row cell H e i g h t text end text o f text end text t h e text end text c o n e comma h equals 24 text end text c m end cell row cell U p p e r text end text r a d i u s text end text o f text end text t h e text end text c o n e comma R equals 15 text end text c m end cell row cell L o w e r text end text r a d i u s text end text o f text end text t h e text end text c o n e comma r equals 5 text end text c m end cell row cell S l a n t text end text h e i g h t text end text o f text end text t h e text end text c o n e comma l equals square root of h squared plus left parenthesis R minus r right parenthesis squared end root end cell row cell equals square root of 24 squared plus left parenthesis 15 minus 5 right parenthesis squared end root end cell row cell equals square root of 24 squared plus 10 squared end root end cell row cell equals square root of 576 plus 100 end root end cell row cell equals square root of 676 equals 26 text end text c m end cell row cell T o t a l text end text s u r f a c e text end text a r e a text end text o f text end text t h e text end text b u c k e t equals pi left parenthesis R plus r right parenthesis l plus pi r squared end cell row cell equals pi left parenthesis 15 plus 5 right parenthesis 26 plus pi 5 squared end cell row cell equals pi left square bracket 520 plus 25 right square bracket end cell row cell equals pi cross times 545 end cell row cell equals 1711.3 c m squared end cell row blank row cell 100 text end text c m squared text end text o f text end text t h e text end text m e t a l text end text c o s t s text end text R s.10 end cell row cell C o s t text end text o f text end text t h e text end text s h e e t equals fraction numerator 1711.3 over denominator 100 end fraction cross times 10 end cell row cell equals R s.171.13 end cell row cell therefore C o s t text end text o f text end text t h e text end text m e t a l text end text s h e e t equals R s.171.13 end cell end table$

Solution 20

Solution 21

Let the height of the cone be H and the radius be R. This cone is divided into two equal parts.

AQ=1/2 AP

Also,

QP||PC

Therefore,ΔAQD~ΔAPC.

So,

Solution 22

A bucket, made of metal sheet, is in the form of a cone.

R = 15 cm, r = 6 cm and H=35 cm

Now, using the similarity concept, we can writ

Volume of the frustum is

The rate of milk is Rs. 40 per litre.

So, the cost of 51.48 litres is Rs. 2059.20.

Solution 23

(i)

Given:

Radius of lower end (r₁) = Diameter/2 = 5 cm

Radius of upper end (r₂) = Diameter/2 = 15 cm

Height of the bucket (h) = 24 cm

Area of metal sheet used in making the bucket

= CSA of bucket + Area of smaller circular base

Hence, area of the metal sheet used in making the bucket is 1711.3 cm².

(ii)

We should avoid the bucket made by ordinary plastic because it is less strength than metal bucket and also not ecofriendly.

Surface Areas and Volumes Exercise Rev. 14

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47

Solution 48

Solution 49

Solution 50

Solution 51

Solution 52

Solution 53

Solution 54

Solution 55

Solution 56

Solution 57

Solution 58

Solution 59

Solution 60

Solution 61

Solution 62

Solution 63

Solution 64

Solution 65

Solution 66

Solution 67

$table attributes columnalign left end attributes row cell H e i g h t text end text o f text end text t h e text end text c o n e comma text end text h equals 24 text end text c m end cell row cell R a d i u s text end text o f text end text t h e text end text u p p e r text end text e n d comma text end text r equals 8 text end text c m end cell row cell R a d i u s text end text o f text end text t h e text end text l o w e r text end text e n d comma R equals 20 text end text c m end cell row blank row cell V o l u m e text end text o f text end text t h e text end text f r u s t u m comma V equals 1 third pi h left parenthesis r squared plus R squared plus straight R r right parenthesis end cell row cell equals 1 third cross times 22 over 7 cross times 24 cross times left parenthesis 8 squared plus 20 squared plus 20 cross times 8 right parenthesis end cell row cell equals 1 third cross times 22 over 7 cross times 24 cross times 624 end cell row cell equals 15689.14 text end text c m cubed end cell row cell therefore C o s t text end text o f text end text m i l k equals fraction numerator 15689.14 over denominator 1000 end fraction cross times 21 end cell row cell equals R s. text end text 329.47 end cell end table$

Solution 68

Solution 69

$begin mathsize 12px style table attributes columnalign left end attributes row cell text In two similar triangles OAB and DCB, we have end text end cell row cell OA over CD equals OB over BD end cell row cell rightwards double arrow text end text 4 over straight r equals fraction numerator straight h over denominator bevelled straight h over 2 end fraction end cell row cell rightwards double arrow text end text straight r equals 2 end cell row cell Now comma text end text fraction numerator Volume text end text of text end text the text end text smaller text end text cone over denominator Volume text end text of text end text the text end text frustum text end text of text end text the text end text cone end fraction end cell row cell equals fraction numerator 1 third straight pi left parenthesis 2 right parenthesis squared cross times left parenthesis straight h over 2 right parenthesis over denominator 1 third straight pi cross times left parenthesis straight h over 2 right parenthesis left square bracket 4 squared plus 2 squared plus 4 cross times 2 right square bracket end fraction end cell row cell equals 4 over 28 end cell row cell equals text end text 1 over 7 end cell row cell Therefore comma text end text the text end text ratio text end text of text end text volume text end text of text end text the text end text smaller text end text cone text end text to text end text the text end text volume text end text end cell row cell text end text of text end text the text end text frustum text end text of text end text the text end text cone text end text is text end text 1 colon 7. end cell end table end style$

Solution 70

Solution 71

Solution 72

Solution 73

Solution 74

Solution 75

Solution 76

Surface Areas and Volumes Exercise 14.88

Solution 1

After melting a sphere and converting it into a wire, the volume remains same.

$begin mathsize 12px style radius space of space sphere space equals space space fraction numerator 6 space cm over denominator 2 end fraction equals 3 space cm radius space of space wire space equals space fraction numerator 2 space mm over denominator 2 end fraction equals space 1 space mm equals 0.1 space cm volume space of space sphere space equals space volume space of space wire rightwards double arrow 4 over 3 straight pi space straight r cubed space equals space straight pi space straight a squared straight h rightwards double arrow 4 over 3 straight pi space open parentheses 3 close parentheses cubed space equals space straight pi space open parentheses 0.1 close parentheses squared straight h rightwards double arrow fraction numerator 4 space cross times space 9 over denominator open parentheses 0.1 close parentheses squared end fraction equals straight h rightwards double arrow space straight h space equals space 3600 space cm space space space space space space space space space space equals space 36 space straight m end style$

So the correct option is (c).

Solution 2

$begin mathsize 12px style Let space number space of space cones space be space straight n rightwards double arrow volume space of space sphere space equals space straight n space straight x space volume space of space cone rightwards double arrow 4 over 3 straight pi space straight r cubed space equals straight n cross times 1 third straight pi space straight a squared straight h rightwards double arrow space 4 space cross times open parentheses 10.5 close parentheses cubed space equals space straight n space cross times space open parentheses 3.5 close parentheses squared space cross times space 3 straight n equals space fraction numerator 4 space straight x space open parentheses 10.5 close parentheses cubed over denominator 3 space straight x space open parentheses 3.5 close parentheses squared end fraction space space space space space equals space fraction numerator 4 space cross times 10.5 space cross times space 9 over denominator 3 end fraction space space space space space space equals space 4 space cross times space 10.5 space cross times space 3 space space space space space space space equals space 126 end style$

So, the correct option is (b).

Solution 3

$begin mathsize 12px style Surface space area space of space hemisphere space equals space 2 space straight pi space straight r squared Surface space area space of space cone space equals space straight pi space rl Given comma space 2 πr squared space equals space πrl space space space space space space space space space space space rightwards double arrow 2 straight r squared space equals space straight l straight l squared space equals space straight h squared space plus space straight r squared rightwards double arrow open parentheses 2 straight r close parentheses squared space equals space straight h squared space plus space straight r squared rightwards double arrow 4 straight r squared space minus space straight r squared space equals space space straight h squared rightwards double arrow straight h space equals square root of 3 straight r rightwards double arrow space box enclose straight r over straight h equals fraction numerator 1 over denominator square root of 3 end fraction end enclose end style$

So the correct option is (b).

Solution 4

$begin mathsize 12px style Volume space of space sphere space equals space Volume space of space cone rightwards double arrow 4 over 3 straight pi space straight r cubed space equals space 1 third straight pi space straight R squared straight r rightwards double arrow 4 straight r squared space equals space straight R squared rightwards double arrow box enclose straight R space equals space 2 straight r end enclose So comma space the space correct space option space is space open parentheses straight a close parentheses. end style$

Solution 5

$begin mathsize 12px style Volume space of space cone space equals space volume space of space cylinder rightwards double arrow space 1 third straight pi space straight r squared straight h space equals space πr squared straight H rightwards double arrow 1 third straight pi space straight r squared straight h space equals space πr squared open parentheses 5 space cm close parentheses rightwards double arrow straight h space equals 15 space cm end style$

So, the correct option is (b).

Solution 6

$begin mathsize 12px style radius space equals space diameter over 2 space equals space 105 over 2 straight m slant space height space of space cylinder space equals space 40 space straight m Total space area space of space straight a space canvas equals curved space surface space area space of space the space cylinder space plus space curved space surface space area space of space the space cone equals 2 πrh plus πrl equals πr left parenthesis 2 straight h plus straight l right parenthesis equals 22 over 7 cross times 105 over 2 open parentheses 2 cross times 4 plus 40 close parentheses equals 22 over 7 cross times 105 over 2 cross times 48 equals 7920 space cm squared end style$

So, the correct option is (d).

Solution 7

$begin mathsize 12px style Let space straight n space solid space spheres space are space required volume space of space cylinder space equals space straight n space cross times space volume space of space solid space sphere πr squared straight h space equals space straight n space cross times space 4 over 3 straight pi open parentheses straight a close parentheses cubed rightwards double arrow space straight pi open parentheses 2 close parentheses squared 45 space equals straight n space cross times space 4 over 3 straight pi space open parentheses 3 close parentheses cubed rightwards double arrow straight n space equals space 5 end style$

So, the correct option is (c).

Solution 8

$begin mathsize 12px style rise space in space volume space of space vessel space equals space volume space of space sphere rightwards double arrow If space water space rises space by space straight h rightwards double arrow πr squared straight h space equals space 4 over 3 straight pi space straight a cubed rightwards double arrow straight pi open parentheses 8 close parentheses squared straight h space equals space fraction numerator 4 straight pi over denominator 3 end fraction open parentheses 6 close parentheses cubed rightwards double arrow straight h space equals space fraction numerator 4 space cross times space 6 space cross times space 6 cross times 6 over denominator 3 space cross times space 8 space cross times space 8 end fraction equals space 4.5 space cm end style$

So, the correct option is (a).

Solution 9

$begin mathsize 12px style Height space of space frustum space open parentheses straight h close parentheses space equals space 40 space cm space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight r subscript 1 space equals space 14 space cm space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight r subscript 2 space space equals 35 space cm Volume space of space frustum space equals space πh over 3 space open parentheses space straight r subscript 2 squared space plus space straight r subscript 2 space straight r subscript 1 plus space straight r subscript 1 squared close parentheses equals space fraction numerator straight pi space cross times space 40 over denominator 3 end fraction space open parentheses 35 to the power of 2 space end exponent plus 35 space cross times space 14 space plus 14 squared close parentheses equals space 80080 space cm cubed end style$

So, correct option is (b).

Solution 10

$begin mathsize 12px style Upper space part space is space cone. Volum space of space cone space equals space 1 third straight pi space open parentheses straight r over 2 close parentheses squared straight H over 2 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator πr squared straight H over denominator 24 end fraction volume space of space cone space equals space 1 third πr to the power of 2 space end exponent straight H fraction numerator volume space of space upper space part over denominator volume space of space cone end fraction equals fraction numerator space fraction numerator πr squared straight H over denominator 24 end fraction over denominator begin display style fraction numerator πr squared straight H over denominator 3 end fraction end style end fraction equals 1 over 8 end style$

So, the correct option (d).

Solution 11

$begin mathsize 12px style increment ADE space is space similar space to space increment ABC rightwards double arrow AD over AB equals DE over BC rightwards double arrow straight h over straight H equals straight r over straight R space space space space space space minus negative box enclose 1 straight H space equals space 30 space cm Volume space of space cone space equals space 1 third πR squared straight H Volume space of space Upper space cone space equals space 1 third πr squared straight h Given space 1 third πr squared straight H space equals space 1 over 27 space open parentheses 1 third straight pi straight R squared straight H close parentheses space space space space space space space space space space rightwards double arrow straight r squared straight H space equals space 1 over 27 space straight R squared straight H space space space space space space space space space rightwards double arrow open parentheses hR over straight H close parentheses squared straight h space equals space 1 over 27 straight R squared straight H space space space space space space space space space rightwards double arrow straight h cubed over straight H squared equals 1 over 27 straight H space space space space space space space space rightwards double arrow space space straight h cubed space equals space straight H cubed over 27 space space space space space space space space rightwards double arrow straight h space equals straight H over 3 equals space 10 space cm Height space above space base space at space which space section space is space made space equals space straight H space minus straight h space equals 20 space cm end style$

So, the correct option is (c).

Solution 12

$begin mathsize 12px style Volume space of space cone space equals 1 third πr squared straight h Volume space of space cylinder space equals space straight pi straight r squared straight H Total space volume space equals space 3 space open parentheses volume space of space cone close parentheses rightwards double arrow 1 third πr squared straight h space space plus space πr squared straight H space equals space 3 space open parentheses 1 third πr squared straight h close parentheses rightwards double arrow space space 1 third πr squared straight h space space plus space πr squared straight H space equals space πr squared straight h rightwards double arrow space πr squared straight H space equals space 2 over 3 πr squared straight H rightwards double arrow straight H equals 2 over 3 straight h end style$

So, the correct option is (b).

Surface Areas and Volumes Exercise 14.89

Solution 13

$begin mathsize 12px style Upper space radius space equals space 4 space straight m lower space radius space equals space 2 space straight m height space equals space 6 space straight m volume space equals space fraction numerator straight pi space cross times 6 over denominator 3 end fraction open parentheses 4 squared plus 4 space cross times 2 space plus 2 squared close parentheses space space space space space space space space space space space space space space space space equals space 2 straight pi space open parentheses 28 close parentheses space space space space space space space space space space space space space space space space equals space 2 space cross times space 22 over 7 cross times 28 space equals 176 space straight m cubed end style$

So, the correct option is (a).

Solution 14

$begin mathsize 12px style Volume space of space water space that space flows space out space of space the space pipe space in space 1 space min equals space πr squared straight h equals space straight pi open parentheses 5 over 2 cross times 1 over 10 close parentheses squared open parentheses 10 space cross times 100 close parentheses.... left parenthesis Since space 1 space cm equals 10 space mm right parenthesis equals space fraction numerator straight pi space cross times 1000 over denominator 16 end fraction equals space fraction numerator 125 space straight pi over denominator 2 end fraction space cm cubed Volume space of space conical space vessel space equals space 1 third straight pi space open parentheses 20 close parentheses squared space cross times space 24 equals space straight pi space cross times 400 space cross times 8 equals 3200 straight pi space space cm cubed Time space required space to space fill space the space vessel space equals space fraction numerator volume space of space vessel over denominator volume space of space water space flow space in space 1 space min end fraction equals fraction numerator 3200 straight pi over denominator fraction numerator 125 space straight pi over denominator 2 end fraction end fraction equals fraction numerator 3200 cross times 2 over denominator 125 end fraction equals space 51.2 space min 1 space min space equals space 60 space sec 0.2 space min space equals space 12 space sec Hence comma space 51.2 space min space equals space 51 space min space 12 space sec end style$

So, the correct option is (b).

Solution 15

$begin mathsize 12px style Volume space of space cylindrical space vessel space equals space volume space of space conical space heap rightwards double arrow straight pi space open parentheses 18 close parentheses squared space cross times space 32 space equals space 1 third straight pi space open parentheses straight r squared close parentheses space 24 rightwards double arrow space fraction numerator 18 space cross times 18 space cross times 32 space cross times 3 over denominator 24 end fraction equals straight r squared rightwards double arrow space straight r squared space equals space 18 space cross times 18 space cross times 4 rightwards double arrow straight r space equals space 18 space cross times space 2 equals space 36 space cm space end style$

So, the correct option is (c).

Solution 16

$begin mathsize 12px style Curved space surface space area space equals space straight pi space rl straight l rightwards double arrow slant space height straight l squared space equals space 15 squared space plus 8 squared straight l space equals space 17 curved space surface space area space equals space straight pi space cross times 8 space cross times 17 equals space 136 space straight pi space cm squared end style$

So, the correct option is (d).

Solution 17

$begin mathsize 12px style height space of space cone space equals space 4 space cm radius space of space cone space equals space 3 space cm volume space equals space 1 third πr squared straight h equals space 1 third straight pi space open parentheses 3 close parentheses squared space open parentheses 4 close parentheses equals space 12 straight pi space cm cubed end style$

So, the correct option is (a).

Solution 18

$begin mathsize 12px style Curved space surface space area space of space cylinder space equals space 2 πrh volume space of space cylinder space equals space πr squared straight h rightwards double arrow 2 πrh space equals space 264 space space space space space space space space space space and space space space πr squared straight h space equals space 924 rightwards double arrow fraction numerator πr squared straight h over denominator 2 πrh end fraction equals 924 over 264 rightwards double arrow straight r over 2 equals 3.5 rightwards double arrow straight r space equals space 7 space straight m Also space space space straight h space equals space fraction numerator 264 over denominator 2 space cross times begin display style 22 over 7 end style cross times 7 end fraction equals 6 rightwards double arrow space diameter over height equals fraction numerator 2 straight r over denominator 7 end fraction space equals 7 over 3 end style$

So, the correct option is (b).

Solution 19

$begin mathsize 12px style Volume space of space cylinder space equals space volume space of space cone rightwards double arrow πr squared straight h space equals space 1 third πr squared straight h rightwards double arrow straight pi space open parentheses 8 close parentheses squared space cross times space 2 space equals space 1 third πr squared space cross times space 6 rightwards double arrow straight r space equals space 8 space cm end style$

So, the correct option is (d).

Solution 20

$begin mathsize 12px style fraction numerator begin display style 4 over 3 end style πr subscript 1 cubed over denominator 4 over 3 πr subscript 2 cubed end fraction equals 64 over 27 rightwards double arrow open parentheses straight r subscript 1 over straight r subscript 2 close parentheses cubed equals 64 over 27 rightwards double arrow straight r subscript 1 over straight r subscript 2 equals 4 over 3 rightwards double arrow open parentheses straight r subscript 1 over straight r subscript 2 close parentheses squared equals open parentheses 4 over 3 close parentheses squared rightwards double arrow fraction numerator 4 πr subscript 1 squared over denominator 4 πr subscript 2 squared end fraction equals open parentheses straight r subscript 1 over straight r subscript 2 close parentheses squared equals open parentheses 4 over 3 close parentheses squared rightwards double arrow fraction numerator 4 πr subscript 1 squared over denominator 4 πr subscript 2 squared end fraction equals 16 over 9 end style$

So, the correct option is (d).

Solution 21

Volume of the sphere = Sum of the volume of the three spheres

$begin mathsize 12px style rightwards double arrow 4 over 3 πr cubed space equals space 4 over 3 straight pi space open parentheses 6 close parentheses cubed space plus space 4 over 3 straight pi open parentheses 8 close parentheses cubed space plus space 4 over 3 straight pi open parentheses 10 close parentheses cubed rightwards double arrow 4 over 3 πr cubed space equals 4 over 3 straight pi space open parentheses 6 to the power of 3 space end exponent plus space 8 to the power of 3 space end exponent plus space 10 cubed close parentheses rightwards double arrow space straight r cubed space equals space open parentheses 1728 close parentheses space straight r space equals space 12 space cm diameter space equals space 2 straight r space equals space 24 space cm end style$

So, correct option is (b).

Solution 22

$begin mathsize 12px style Surface space area space of space sphere space equals space 4 space πr squared Curved space surface space area space of space right space circular space cylinder space equals space 2 πr subscript 1 straight h subscript 1 height space of space cylinder space equals space 12 space cm radius space of space cylinder space open parentheses straight r subscript 1 close parentheses space equals space 6 space cm rightwards double arrow 4 πr squared space equals space 2 πr subscript 1 straight h subscript 1 rightwards double arrow 2 space straight r squared space equals space 6 space cross times 12 rightwards double arrow space straight r squared space equals space 36 space space space space space rightwards double arrow straight r space equals space 6 space cm end style$

So, correct option is (c).

Solution 23

$begin mathsize 12px style Largest space possible space diameter space of space shere space equals space diameter space of space cylinder Hence space radius space of space sphere space equals space 1 space cm Volume space of space the space greatest space sphere space equals space 4 over 3 straight pi space open parentheses 1 close parentheses cubed space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4 over 3 straight pi end style$

So, correct option is (a).

Solution 24

$begin mathsize 12px style Increase space in space volume space level space equals space volume space of space sphere rightwards double arrow If space rise space in space height space be space straight h rightwards double arrow πr squared straight h space equals space 4 over 3 straight pi space open parentheses 3 close parentheses cubed rightwards double arrow straight pi open parentheses 4 close parentheses squared straight h space equals space 4 over 3 straight pi space open parentheses 3 close parentheses cubed rightwards double arrow straight h space equals space fraction numerator 4 straight pi space cross times 9 over denominator 16 space straight pi end fraction straight h space equals space 9 over 4 cm end style$

So, the correct option is (c).

Surface Areas and Volumes Exercise 14.90

Solution 25

$begin mathsize 12px style Volume space of space the space cylinder equals 12 cross times Volume space of space the space sphere rightwards double arrow straight pi space cross times space open parentheses 8 close parentheses squared space cross times space 2 space equals space 12 space cross times space 4 over 3 πr cubed rightwards double arrow space straight pi space cross times space 128 space equals space 16 space πr cubed rightwards double arrow straight r cubed space equals space 8 rightwards double arrow straight r space equals 2 space cm diameter space equals space 4 space cm end style$

So, correct option is (d).

Solution 26

$begin mathsize 12px style Volume space of space spherical space ball space equals space Volume space of space cone rightwards double arrow space 4 over 3 straight pi space open parentheses 3 close parentheses cubed space equals space 1 third straight pi space open parentheses 6 close parentheses squared space cross times space straight h rightwards double arrow fraction numerator 4 space cross times space 3 space cross times space 3 cross times 3 over denominator 6 space cross times space 6 end fraction space equals space straight h rightwards double arrow space straight h space equals space 3 space cm end style$

So, the correct option is (b).

Solution 27

$begin mathsize 12px style volume space of space sphere space equals space volume space of space cone rightwards double arrow 4 over 3 straight pi space open parentheses straight r subscript 2 to the power of 3 space end exponent minus space straight r subscript 1 cubed close parentheses space equals space 1 third space πr squared straight h rightwards double arrow 4 space open parentheses 4 cubed minus space 2 cubed close parentheses space equals space open parentheses 4 close parentheses squared straight h rightwards double arrow 56 over 4 space equals space straight h rightwards double arrow straight h space equals space 14 space cm end style$

So, the correct option is (b).

Solution 28

$begin mathsize 12px style Volume space of space sphere space equals space volume space ofiron space piece rightwards double arrow 4 over 3 space πr cubed space equals space 49 cross times 33 cross times 24 rightwards double arrow straight r cubed equals fraction numerator 49 cross times 33 cross times 24 over denominator 4 cross times begin display style 22 over 7 end style end fraction rightwards double arrow straight r cubed equals fraction numerator 49 cross times 7 cross times 3 cross times 6 cross times 3 over denominator 2 end fraction rightwards double arrow space straight r cubed space equals 49 cross times 7 cross times 3 cross times 3 cross times 3 rightwards double arrow straight r equals 7 cross times 3 equals 21 space cm end style$

So, the correct option is (a).

Solution 29

Diameter = 1.6 m = 160 cm

So, radius = 80 cm

$begin mathsize 12px style Lateral space space surface space area space equals space 2 πrh Total space surface space area space equals space space 2 πrh space plus space space 2 πr squared rightwards double arrow fraction numerator space lateral space surface space area space over denominator total space area end fraction equals fraction numerator space 2 πrh over denominator space 2 πr open parentheses straight r space plus straight h close parentheses end fraction equals space fraction numerator straight h over denominator straight r space plus space straight h end fraction space equals fraction numerator 20 over denominator 20 space plus 80 end fraction equals space 1 space colon space 5 end style$

So, the correct option is (b).

Solution 30

$begin mathsize 12px style Volume space of space cone space equals space 1 third πr squared straight h Total space Volume space space equals space 1 third πr squared straight h plus πr squared straight H Total space Volume space equals space 3 space open parentheses Volume space of space cone close parentheses 1 third πr squared straight h space plus πr squared straight H equals πr squared straight h rightwards double arrow πr squared straight H space equals space 2 over 3 πr squared straight h rightwards double arrow straight H space equals space 2 over 3 straight h end style$

So, the correct option is (d).

Solution 31

$begin mathsize 12px style maximum space base space area space of space cone space equals base space area space of space hemisphere Hence comma space radius space of space cone space equals space straight r Also comma space maximum space height space of space the space cone equals straight r Maximum space possilbe space volume space of space the space cone space equals space 1 third πr squared straight h space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 third πr squared space cross times space straight r space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals πr cubed over 3 space space space space space space end style$

So, correct option is (b).

Solution 32

$begin mathsize 12px style straight r subscript 1 over straight r subscript 2 space equals space 3 over 5 straight h subscript 1 over straight h subscript 2 space equals space 2 over 3 Curved space surface space area space of space cylinder space equals space 2 πrh Hence space ratio space of space areas space equals space fraction numerator 2 πr subscript 1 straight h subscript 1 over denominator 2 πr subscript 2 straight h subscript 2 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 3 over 5 cross times 2 over 3 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2 over 5 end style$

So, the correct option is (a).

Solution 33

The cylinder completely encloses the sphere.

Hence, diameter of the sphere = diameter of the cylinder = 2r

Now, h is also given to be 2r.

So, the correct option is (a) or (c).

Note: Both can be the answer since h = 2r.

Solution 34

$begin mathsize 12px style AB space equals space 14 space minus 6 space space space space space space space space equals space 8 space cm OB space equals space 10 space cm OB squared space equals space AB squared space equals space OA squared rightwards double arrow 10 squared space equals space 8 squared space plus space OA squared rightwards double arrow space OA squared space equals space 100 space minus space 64 rightwards double arrow space OA squared space equals 36 rightwards double arrow OA space equals space 6 cm end style$

So, the correct option is (a).

Solution 35

$begin mathsize 12px style rightwards double arrow AD over AB space equals space DE over BC rightwards double arrow straight h subscript 2 over straight h subscript 1 equals straight r subscript 2 over straight r subscript 1 rightwards double arrow straight r subscript 2 over straight r subscript 1 equals 1 half end style$

So, the correct option is (b).

Solution 36

$begin mathsize 12px style Slant space height space of space the space complete space cone space equals space straight l rightwards double arrow straight r subscript 1 over straight r subscript 2 equals fraction numerator straight l subscript 1 over denominator straight l subscript 1 plus 10 end fraction rightwards double arrow 10 over 16 equals fraction numerator straight l subscript 1 over denominator straight l subscript 1 plus 10 end fraction rightwards double arrow 10 straight l subscript 1 space plus 100 space equals space 16 straight l subscript 1 rightwards double arrow 6 straight l subscript 1 space equals space 100 rightwards double arrow straight l subscript 1 space equals space 100 over 6 equals 50 over 3 cm straight l space equals space 10 space plus space straight l 1 space equals space 10 plus 50 over 3 space equals space 80 over 3 space cm lateral space surface space area space of space frustum space equals space lateral space surface space area space of space large space cone space minus space lateral space surface space area space of space small space cone equals space πr subscript 2 straight l space minus space πr subscript 1 straight l subscript 1 equals space straight pi space open parentheses 16 space cross times 80 over 3 minus 10 space cross times space 50 over 3 close parentheses equals space straight pi space cross times 780 over 3 equals space 260 space straight pi space cm squared end style$

So, the correct option is (c).

Solution 37

$begin mathsize 12px style AB space equals space 9 space minus space 3 space space space space space space space equals space 6 space cm OB squared equals AB squared plus OA squared space space space space space space space space space equals 6 squared space plus space 8 squared space space space space space space space space equals 36 space plus space 64 OB squared equals 100 OB space equals space 10 space cm end style$

So, the correct option is (c).

Surface Areas and Volumes Exercise 14.91

Solution 38

$begin mathsize 12px style straight r subscript 1 space equals space 8 space cm straight r subscript 2 space equals space 20 space cm straight r subscript 1 over straight r subscript 2 space equals space straight h subscript 1 over straight h subscript 2 rightwards double arrow straight h subscript 1 over straight h subscript 2 equals 8 over 20 rightwards double arrow straight h subscript 2 over straight h subscript 1 equals 20 over 8 equals 5 over 2 rightwards double arrow straight h subscript 2 over straight h subscript 1 minus 1 equals 5 over 2 minus 1 rightwards double arrow fraction numerator straight h subscript 2 space minus straight h subscript 1 over denominator straight h subscript 1 end fraction space equals 3 over 2 straight h subscript 1 equals 2 over 3 cross times 16 space equals 32 over 3 space cm straight h subscript 2 equals straight h subscript 1 space plus space 16 space space space space space equals 16 space plus 32 over 3 box enclose straight h subscript 2 equals 80 over 3 cm end enclose straight l subscript 1 squared space equals space straight r subscript 1 squared space plus straight h subscript 1 to the power of 2 space space space space space space space space space space space end exponent and space space space space straight l subscript 2 squared space equals space straight r subscript 2 squared space plus straight h subscript 2 to the power of 2 space space space space space space space space space space space end exponent equals space 8 squared space plus open parentheses 32 over 3 close parentheses squared space space space space space space space space space space space space space space space space space space space space equals space 20 squared space plus open parentheses 80 over 3 close parentheses squared equals space 1600 over 9 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 10000 over 9 straight l subscript 1 space space end subscript equals 40 over 3 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 straight l subscript 2 space equals space 100 over 3 Curved space surface space area space of space frustum space equals space πr subscript 2 straight l subscript 2 space minus straight pi straight r subscript 1 straight l subscript 1 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 space straight pi open parentheses 20 space cross times 100 over 3 space minus space 8 cross times 40 over 3 close parentheses 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 space straight pi open parentheses 2000 over 3 space minus space 320 over 3 close parentheses 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 space 560 space straight pi 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 1760 space cm squared end style$

So, the correct option is (a).

Solution 39

$begin mathsize 12px style Given space space space space space straight r subscript 1 equals space 14 space cm space space space space space space space space space space space space space space space space straight r subscript 2 equals space 21 space cm space straight h subscript 2 space space minus space end subscript straight h subscript 1 equals space 24 space cm space space minus negative negative negative negative box enclose 1 fraction numerator space straight r subscript 1 over denominator space straight r subscript 2 end fraction space equals space straight h subscript 1 over straight h subscript 2 equals space straight h subscript 1 over straight h subscript 2 space equals space 14 over 21 space equals space 2 over 3 space space space minus negative negative negative negative negative box enclose 2 from space box enclose 1 space space & space box enclose 2 straight h subscript 2 space minus space 2 over 3 straight h subscript 2 space equals space 24 rightwards double arrow straight h subscript 2 space equals space 72 space cm rightwards double arrow straight h subscript 1 space equals space 48 space cm straight l subscript 1 squared space equals space straight r subscript 1 squared space plus straight h subscript 1 squared space space space space space space space space space space space space space and space space space space space space straight l subscript 2 squared space equals space straight r subscript 2 squared plus straight h subscript 2 squared space space space equals space 14 squared space plus 48 squared 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 21 squared space plus 72 squared equals space 2500 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 5625 space space space space space straight l subscript 1 space equals space 50 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 straight l subscript 2 space space equals space 75 space space space space space space space space space space space space space space curved space surface space area space of space bucket space equals space space πr subscript 2 straight l subscript 2 space space minus space space πr subscript 1 straight l subscript 1 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 space straight pi space open parentheses 21 space cross times space 75 space minus space 14 cross times 50 close parentheses 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 space 875 straight pi 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 2750 space cm squared space space space cost space of space painting space equals space 50 space paise divided by cm squared total space cost space equals space Rs. space 1375 But space we space have space to space paint space the space base space of space bucket space too space space space space space space space space space space space space space space space area space equals space πr subscript 1 squared space space space space space space space space space space space space space space space space cost space equals space 0.5 space cross times space straight pi straight r subscript 1 squared 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 space 0.5 space cross times 22 over 7 space cross times space open parentheses 14 close parentheses squared space space space equals space 308 rightwards double arrow total space cost space of space painting space the space outer space surface space of space the space bucket equals 308 space plus space 1375 space equals space Rs. space 1683 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 end style$

So, the correct option is (c)

Solution 40

$begin mathsize 12px style area space of space circular space face space equals space πr squared curved space surface space area space of space cylinder space equals space 2 πrh Given comma space 4 open parentheses πr squared space plus πr squared close parentheses space equals space 2 open parentheses πrh close parentheses space space space space space space space space space space space space space rightwards double arrow space 4 open parentheses 2 πr squared close parentheses space equals space 4 πrh space space space space space space space space space space space space space rightwards double arrow 2 straight r space equals space straight h space space space space space space space space space space space space rightwards double arrow space diameter space equals space straight h space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 8 space cm end style$

So, the correct option is (b).

Solution 41

$begin mathsize 12px style Initial space volume space equals space πr squared straight h After space radius space of space base space is space halved comma volume space of space cylinder space equals space straight pi open parentheses straight r over 2 close parentheses squared straight h space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator πr squared straight h over denominator 4 end fraction fraction numerator Final space Cylinder space Volume over denominator Original space Cylinder space Volume end fraction space equals space fraction numerator begin display style fraction numerator πr squared straight h over denominator 4 end fraction end style over denominator πr squared straight h end fraction space equals space 1 fourth end style$

So, the correct option is (c).

Solution 42

$begin mathsize 12px style Volume space of space cone space equals space Volume space of space cylinder rightwards double arrow 1 third space πr squared straight h space equals space space πr squared space open parentheses 6 close parentheses rightwards double arrow straight h over 3 equals 6 rightwards double arrow straight h space equals space 18 space cm end style$

So, the correct option is (c).

Solution 43

$begin mathsize 12px style If space this space sheet space of space paper space is space rolled space to space form space straight a space cylinder space of space height space 40 space cm. Then space circumference space of space cylinder space is space 22 space cm rightwards double arrow 2 πr space equals space 22 space cm rightwards double arrow straight r space equals space fraction numerator 22 over denominator 2 space cross times space 22 end fraction cross times space 7 space equals 3.5 space cm end style$

So, the correct option is (a).

Solution 44

So, the correct option is (b).

Solution 45

Solution 46

Correct option: (b)

From the figure, it is clear that diameter of sphere is 2r.

Solution 47

Correct option: (a)

In a right circular cone, the cross-section made by a plane parallel to the base is a circle.

Solution 48

Correct option: (a)

When two solid-hemispheres of same base radius r are joined together along their bases, it forms a sphere.

And, CSA of sphere = 4πr²

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Class 10 RD SHARMA Solutions Maths Chapter 14 - Surface Areas and Volumes

Surface Areas and Volumes Exercise Ex. 14.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

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Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

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Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47

Solution 48

Solution 49

Solution 50

Solution 51

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Solution 53

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Solution 72

Surface Areas and Volumes Exercise Ex. 14.2

Solution 1

Solution 2

Solution 3

Solution 4