Chapter 16 – Surface Areas and Volumes Exercise Ex. 16.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire.Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Find the number of metallic circular discs with 1.5 cm base diameter and of height 0.2 cm to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.Solution 9

Question 10

How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm × 42 cm × 21 cm.Solution 10

Question 11

How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm.Solution 11

Question 12

Three cubes of a metal whose edges are in the ratio 3 : 4: 5 are melted and converted into a single cube whose diagonal is   cm. Find the number of cones so formed.Solution 12

Question 13

A solid metallic sphere of radius 10.5 cm is melted and recast into a number of smaller cones, each of radius 3.5 cm and height 3 cm. Find the number of cones so formed.Solution 13

Question 14

Solution 14

Question 15

An iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls is 1/4 of the radius of the original ball, how many such balls are made? Compare the surface area, of all the smaller balls combined together with that of the original ball.Solution 15

Question 16

Solution 16

Question 17

A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.Solution 17

Question 18

Solution 18

Question 19

How many coins 1.75 cm in diameter and 2 mm thick must be melted to form a cuboid 11 cm  10 cm  7 cm?Solution 19

Question 20

Solution 20

Question 21

A cylindrical bucket, 32 cm high and with a radius of base 18 cm, is filled with sand. This bucket is emptied out on the ground and a conical heap of sand is formed. If the height of conical heap is 24 cm, find the radius and slant height of the heap.Solution 21

Question 22

A solid metallic sphere of radius 5.6 cm is melted and solid cones each of radius 2.8 cm and height 3.2 cm are made. Find the number of such cones formed.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

Question 23

A solid cuboid of iron with dimensions 53 cm x 40 cm x 15 cm is melted and recast into a cylindrical pipe. The outer and inner diameters of pipe are 8 cm and 7 cm respectively. Find the length of pipe.Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of the balls are 1.5 cm and 2 cm. Find the diameter of the third ball.Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Rain water, which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius 20 cm. What will be the height of water in the cylindrical vessel if a rainfall of 1 cm has fallen? [Use  = 22/7]Solution 35

Question 36

The rain water from a roof of dimensions 22 m × 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. if the rain water collected from the roof just fills the cylindrical vessel, then find the rain fall in cm.Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

150 spherical marbles, each of diameter 1.4 cm are dropped in cylindrical vessel of diameter 7 cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vesselSolution 46

*Answer given in the book is incorrect.Question 47

Sushant has a vessel, of the form of an inverted cone, open at the top, of height 11 cm and radius of top as 2.5 cm and is full of water. Metallic spherical balls each of diameter 0.5 cm are put in the vessel due to which 

of the water in the vessel flows out. Find how many balls were put in the vessel. Sushant made the arrangement so that the water that flows out irrigates the flower beds. What value has been shown by shushant ?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

Question 48

16 glass spheres each of radius 2 cm are picked into a cuboidal box of internal dimensions 16 cm × 8 cm × 8 cm and then the box is filled with water. Find the volume of water filled in the box.Solution 48

Question 49

Water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of 80 cm/sec in an empty cylindrical tank, the radius of whose base is 40 cm. What is the rise of water level in tank in half an hour?Solution 49

Question 50

Water in a canal 1.5 m wide and 6 m deep is flowing with a speed of 10 km/hr. How much area will it irrigate in 30 minutes if 8 cm of standing water is desired?Solution 50

Question 51

A farmer runs a pipe of internal diameter 20 cm from the canal into a cylindrical tank in his field which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?Solution 51

Question 52

A cylindrical tank full of water is emptied by a pipe at the rate of 225 liters per minute. How much time will it take to empty half the tank, if the diameter of its base is 3 m and its height is 3.5 m? (π = 22/7)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

Question 53

Water is flowing at the rate of 2.52 km/h through a cylindrical pipe into a cylindrical tank, the radius of the base is 40 cm. If the increase in the level of water in the tank, in half an hour is 3.15 m, find the internal diameter of the pipe.Solution 53

Question 54

Water flows at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will level of water in the pond rise by 21 cm?Solution 54

Question 55

A canal 300 cm wide and 120 cm deep. The water in the canal is flowing with a speed of 20 km/h. How much area will it irrigate in 20 minutes if 8 cm of standing water is desired?Solution 55

Question 56

The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 cm2, find the volume of cylinder.Solution 56

Question 57

Solution 57

Question 58

Solution 58

Question 59

Solution 59

Question 60

A 5 m wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. find the cost of cloth used at the rate of Rs. 25 per metre. (π = 22/7)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

Question 61

Solution 61

Question 62

The difference between the outer and inner curved surface areas of a hollow right circular cylinder 14 cm long is 88 cm2. If the volume of metal used in making the cylinder is 176 cm3, find the outer and inner diameters of the cylinder. (Use  = 22/7)Solution 62

Question 63

Solution 63

Question 64

Solution 64

Question 65

If the total surface area of a solid hemisphere is 462 cm2, find its volume.

(π = 22/7)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.Question 66

Water flows at the rate of 10m/minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40 cm and depth 24 cm?Solution 66

Question 67

A solid right circular cone of height 120 cm and radius 60 cm is placed in a right circular cylindrical full of water of height 180 cm such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is equal to the radius of the cone.Solution 67

Question 68

A heap of rice in the form of a cone of diameter 9 m and height 3.5 m. Find the volume of rice. How much canvas cloth is required to cover the heap?Solution 68

Question 69

A cylindrical bucket of height 32 cm and base radius 18 cm is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.Solution 69

Question 70

A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped each of radius 1.5 cm and height 4 cm. How many bottles are needed to empty the bowl?Solution 70

Question 71

A factory manufactures 120,000 pencils daily The pencils are cylindrical in shape each of length 25 cm and circumference of base as 1.5 cm. Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at Rs. 0.05 per dm2.Solution 71

Question 72

πThe   part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical vessel.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.

Chapter 16 – Surface Areas and Volumes Exercise Ex. 16.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub. If the radius of the hemi-sphere is 3.5 cm and height of the cone outside the hemisphere is 5 cm, find the volume of the water left in the tub. (Take  = 22/7)Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

A cylinderical road roller made of iron is 1 m long. Its internal diameter is 54 cm and the thickness of the iron sheet used in making the roller is 9 cm. Find the mass of the roller, if 1 cm3 of iron has 7.8 gm mass. (Use  = 3.14)Solution 16

Question 17

A vessel in the form of a hollow hemisphere mounted by a hollow cylinder. The dijameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

A right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream. The ice-cream is to be filled in cones of height 12 cm and diameter 6 cm having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream.Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

A wooden toy is made by scooping out a hemisphere of same radius from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the volume of wood in the toy.

(π = 22/7)Solution 28

Question 29

The largest possible sphere is carved out of a wooden solid cube of side 7 cm. find the volume of wood left.(Use begin mathsize 12px style straight pi end style = 22/7)Solution 29

Question 30

From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid. (π = 22/7)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

Question 31

The largest cone is curved out from one face of solid cube of side 21 cm. Find the volume of the remaining solid.Solution 31

Question 32

A solid wooden toy is in the form of a hemisphere surmounted by a cone of same radius. The radius of hemisphere is 3.5 cm and the total wood used in the making of toy is 166 . Find the height of the toy. Also, find the cost of painting the 6 hemispherical part of the toy at the rate of Rs. 10 per cm2. (Take π = 22/7).Solution 32

Question 33

In Fig. 16.57, from a cuboidal solid metalic block, of dimensions 15 cm × 10 cm × 5 cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block. (Take π = 22/7).

Solution 33

Question 34

A building is in the form of a cylinder surmounted by a hemi-spherical vaulted done and contains   of air. If the internal diameter of done is equal to its total height above the floor, find the height of the building?Solution 34

Question 35

A pen stand made of wood is in the shape of a cuboid four conical depressions and a cubical depression to hold the pens and pins, respectively. The dimension of the cuboid are 10 cm × 5 cm × 4 cm. The radius of each of the conical depression is 0.5 cm and the depth is 2.1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand.Solution 35

Question 36

A building is in the form of a cylinder surmounted by a hemispherical dome. The base diameter of the dome is equal to   of the total height of the building. Find the height of the building, if it contains   of air.Solution 36

Question 37

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of cone is 4 cm and the diameter of the base is 8 cm. Determine the volume of the toy. If a cube circumscribes the toy, then find the difference of the volumes of cube and the toy. Also, find the total surface area of the toy.Solution 37

Question 38

A circus tent is in the shape of a cylinder surmounted by a conical top of same diameter. If their common diameter is 56m, the height of the cylindrical part is 6 m and the total height of the tent above the ground is 27 m, find the area of the canvas used in making the tent.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

Chapter 16 – Surface Areas and Volumes Exercise Ex. 16.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

A milk container of height 16 cm is made of metal sheet in the form of frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of Rs.44 per litre which the container can hold.Solution 10

Question 11

A bucket is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find the capacity and surface area of the bucket. Also, find the cost of milk which can completely fill the container, at the rate of Rs.25 per litre.Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

A solid cone of base radius 10 cm is cut into two parts through the mid-points of its height, by a plane parallel to its base. Find the ratio in the volumes of two parts of the cone.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

Question 19

A bucket open at the top, and made up of a metal sheet is in the form of a frustum of a cone. The depth of the bucket is 24 cm and the diameters of its upper and lower circular ends are 30 cm and 10 cm respectively. Find the cost of metal sheet used in it at the rate of Rs. 10 per 100 cm2. (π = 22/7)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

Question 20

In Fig. 14.75, from the top of a solid cone of height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid.

).

Solution 20

Question 21

The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the volumes of two parts.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,

Question 22

A bucket, made of metal sheet, is in the form of a cone whose height is 35 cm and radii of circular ends are 30 cm and 12 cm. How many liters of milk it contains if it is full to the brim? If the milk is sold at 40 per litre, find the amount received by the person.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.Question 23

The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm,

(i) Find the area of the metal sheet used to make the bucket.

(ii) Why we should avoid the bucket made by ordinary plastic? (use π = 3.14)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.

Discover more from EduGrown School

Subscribe to get the latest posts sent to your email.