RD Sharma Solutions for CBSE Class 10 Mathematics chapter 14 - Surface Areas and Volumes

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Chapter 14 - Surface Areas and Volumes Excercise Ex. 14.1

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 vessel

Solution 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 14 - Surface Areas and Volumes Excercise Ex. 14.2

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 

×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 14 - Surface Areas and Volumes Excercise Ex. 14.3

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

Solution 21

Question 22

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 22

  

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

ThereforeAQD~ΔAPC.

So,

Question 23

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 23

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.

Chapter 14 - Surface Areas and Volumes Excercise Rev. 14

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

Solution 10

Question 11

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

Solution 18

Question 19

Solution 19

Question 20

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

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31



Question 32

Solution 32

Question 33

A right circular cylinder and a right circular cone have equal bases and equal heights. If their curved surfaces are in the ratio 8:5, determine the ratio of the radius of the base to the height of the either of them.

Solution 33

Question 34

[Use  = 22/7]
Solution 34

Question 35

Solution 35

Question 36

A solid metal sphere of 6 cm diameter is melted and a circular sheet of thickness 1 cm is prepared Determine the diameter of the sheet.

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

Solution 46



Question 47

Solution 47



Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50



Question 51

An iron pole consisting of a cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that 1 cm3 or iron has 8 gram mass approximately. (Use = 355/115)

Solution 51


Question 52

Solution 52



Question 53

Solution 53



Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Question 61

[Use  = 22/7]
Solution 61

Question 62

Solution 62

Question 63

Solution 63



Question 64

Solution 64



Question 65




Solution 65

Question 66

Solution 66

Question 67

A container open at the top, is in the form of a frustum of a cone of height 24 cm with radii of its lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill in the container at the rate of Rs. 21 per litre. (π = 22/7)

 

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


 


Question 68

A cone of maximum size is carved out from a cube of edge 14 cm. Find the surface area of the cone and of the remaining solid left out after the cone carved out.

Solution 68

Question 69

A cone of radius 4 cm is divided into two parts by drawing a plane through the midpoint of its axis and parallel to its base. Compare the volumes of two parts.

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 

 

 

Question 70

A wall 24 m, 0.4 m thick and 6 m high is constructed with the bricks each of dimensions 25 cm × 16 cm × 10 cm. If the mortar occupies   of the volume of the wall, then find the number of bricks used in constructing the wall.

Solution 70

Question 71

A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm respectively. Find the height of the bucket.

Solution 71

Question 72

Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.

Solution 72

Question 73

Two cones with same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape formed.

Solution 73

Question 74

From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the volume of the remaining solid.

Solution 74

Question 75

Two solids cones A and B are placed in a cylindrical tube as shown in figure. The ratio of their capacities are 2 : 1. Find the heights and capacities of the cones. Also, find the volume of the remaining portion of the cylinder.

 

 

 

 

Solution 75

 

 

 

Question 76

An ice cream cone full of ice cream having radius 5 cm and height 10 cm as shown in figure. Calculate the volume of ice cream, provided that its 1/6 part is left unfilled with ice cream.

 

 

 

Solution 76

Chapter 14 - Surface Areas and Volumes Excercise 14.88

Question 1

The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. The length of the wire is

(a) 12 m

(b) 18 m

(c) 36 m

(d) 66 m

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).

Question 2

A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. The number of such cones is

(a) 63

(b) 126

(c) 21

(d) 130 

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).

Question 3

A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of its radius and the height of its conical part is

(a) 1:3

(b) begin mathsize 12px style 1 colon square root of 3 end style

(c) 1:1

(d) begin mathsize 12px style square root of 3 colon 1 end style  

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).

Question 4

A solid sphere of radius r is melted and cast into the shape of solid cone of height r, the radius of the base of the cone is

(a) 2r

(b) 3r

(c) r

(d) 4r

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

Question 5

The material of a cone is converted into the shape of a cylinder of equal radius. If height of the cylinder is 5 cm, then height of the cone is

(a) 10 cm
(b) 15 cm
(c) 18 cm
(d) 24 cm

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).

Question 6

A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 in and its slant height is 40 m, the total area of the canvas required in m2 is


(a) 1760
(b) 2640
(c) 3960
(d) 7920

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).

Question 7

The number of solid spheres, each of diameter 6 cm that could be molded to form a solid metal cylinder of height 45 cm and diameter 4 cm, is

(a) 3

(b) 4

(c) 5

(d) 6

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).

Question 8

A sphere of radius 6 cm is dropped into a cylindrical vessel partly filled with water. The radius of the vessel is 8 cm. If the sphere is submerged completely, then surface of the water rises by

(a) 4.5 cm
(b) 3 cm
(c) 4 cm
(d) 2 cm

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).

Question 9

If the radii of the circular ends of a bucket of height 40 cm are of lengths 35 cm and 14 cm, then the volume of the bucket in cubic centimeters, is

(a) 60060

(b) 80080 

(c) 70040

(d) 80160

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).

Question 10

If a cone is cut into two parts by a horizontal plane passing through the mid- point of its axis, the ratio of the volumes of the upper part and the cone is


(a) 1:2
(b) 1:4
(c) 1:6
(d) 1:8

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).

Question 11

The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be begin mathsize 12px style 1 over 27 end style of the volume of the given cone, then the height above the base at which the section has been made, is


(a) 10 cm
(b) 15 cm
(c) 20 cm
(d) 25 cm

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).

Question 12

A solid consists of a circular cylinder with an exact fitting right circular cone placed at the top. The height of the cone is h. If the total volume of the solid is 3 times the volume of the cone, then the height of the circular cylinder is

begin mathsize 12px style open parentheses straight a close parentheses space 2 straight h
open parentheses straight b close parentheses space fraction numerator 2 straight h over denominator 3 end fraction
open parentheses straight c close parentheses space fraction numerator 3 straight h over denominator 2 end fraction
open parentheses straight d close parentheses space 4 straight h end style

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).

Chapter 14 - Surface Areas and Volumes Excercise 14.89

Question 1

A reservoir is in the shape of a frustum of a right circular cone. It is 8 m across at the top and 4 m across at the bottom. If it is 6 m deep, then its capacity is

(a) 176 m3

(b) 196 m3

(c) 200 m3

(d) 110 m3

Solution 1

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).

Question 2

Water flows at the rate of 10 meter per minute from a cylindrical pipe 5 mm in diameter. How long will it take to fill up a conical vessel whose diameter at the base is 40 cm and depth 24 cm?

(a) 48 minutes 15 sec

(b) 51 minutes 12 sec

(c) 52 minutes 1 sec

(d) 55 minutes

Solution 2

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).

Question 3

A cylindrical vessel 32 cm high and 18 cm as the radius of the base, 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, the radius of its base is

(a) 12 cm

(b) 24 cm 

(c) 36 am

(d) 48 cm

Solution 3

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).

Question 4

The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is

begin mathsize 12px style open parentheses straight a close parentheses space 60 straight pi space cm squared
open parentheses straight b close parentheses space 68 straight pi space cm squared
open parentheses straight c close parentheses space 120 straight pi space cm squared
open parentheses straight d close parentheses space 136 straight pi space cm squared end style

Solution 4

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).

Question 5

A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the side of 3 cm to form a cone. The volume of the cone so formed is

begin mathsize 12px style open parentheses straight a close parentheses space 12 space straight pi space cm cubed
open parentheses straight b close parentheses space 15 space straight pi space cm cubed
open parentheses straight c close parentheses space 16 space straight pi space cm cubed
open parentheses straight d close parentheses space 20 space straight pi space cm cubed
end style

Solution 5

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).

Question 6

The curved surface area of a cylinder is 264 m2 and its volume is 924 m3. The ratio of its diameter to its height is

(a) 3:7

(b) 7:3

(c) 6:7

(d) 7:6

Solution 6

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).

Question 7

A cylinder with base radius of 8 cm and height of 2 cm is melted to form a cone of height 6 cm. The radius of the cone is

(a) 4 cm

(b) 5 cm

(c) 6 cm

(d) 8 cm

Solution 7

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).

Question 8

The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is 

(a) 1 : 2

(b) 2 : 3

(c) 9 : 16

(d) 16 : 9

Solution 8

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).

Question 9

If three metallic sphere of radius 6 cm, 8 cm, 10 cm are melted to form a single sphere, the diameter of the sphere is

(a) 12 cm

(b) 24 cm

(c) 30 cm

(d) 36 cm

Solution 9

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).

Question 10

The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is

(a) 3 cm

(b) 4 cm

(c) 6 cm

(d) 12 cm

Solution 10

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).

Question 11

The volume of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is

begin mathsize 12px style open parentheses straight a close parentheses 4 over 3 straight pi
open parentheses straight b close parentheses 10 over 3 straight pi
open parentheses straight c close parentheses space 5 straight pi
open parentheses straight d close parentheses 20 over 3 straight pi end style

Solution 11

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).

Question 12

A cylindrical vessel of radius 4 cm contains water. A solid sphere of radius 3 cm is lowered into the water until it is completely immersed. The water level in the vessel will rise by

begin mathsize 12px style open parentheses straight a close parentheses space 2 over 9 cm
open parentheses straight b close parentheses space 4 over 9 cm
open parentheses straight c close parentheses space 9 over 4 cm
open parentheses straight d close parentheses space 9 over 2 cm end style

Solution 12

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).

Chapter 14 - Surface Areas and Volumes Excercise 14.90

Question 1

12 spheres of the same size are made from melting a solid cylinder of 16 cm diameter and 2 cm height. The diameter of each sphere is

begin mathsize 12px style open parentheses straight a close parentheses space square root of 3 cm
open parentheses straight b close parentheses space 3 space cm
open parentheses straight c close parentheses space 3 space cm
open parentheses straight d close parentheses space 4 space cm end style

Solution 1

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).

Question 2

A solid metallic spherical ball of diameter 6 cm is melted and recast into a cone with diameter of the base as 12 cm. The height of the cone is

(a) 2 cm

(b) 3 cm

(c) 4 cm

(d) 6 an

Solution 2

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).

Question 3

A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. The height of the cone is

(a) 12 cm

(b) 14 cm

(c) 15 cm

(d) 18 cm

Solution 3

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).

Question 4

A solid piece of iron of dimensions 49 x 33 x 24 cm is moulded into a sphere. The radius of the sphere is

(a) 21 cm

(b) 28 cm

(c) 35 cm

(d) None of these

Solution 4

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).

Question 5

The ratio of lateral surface area to the total surface area of a cylinder with base diameter 1.6 m and height 20 cm is

(a) 1 : 7

(b) 1 :5

(c) 7 : 1

(d) 5 : 1

Solution 5

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).

Question 6

A solid consists of a circular cylinder surmounted by a right circular cone. The height of the cone is h. If the total height of the solid is 3 times the volume of the cone, then the height of the cylinder is 

begin mathsize 12px style open parentheses straight a close parentheses space 2 straight h
open parentheses straight b close parentheses space fraction numerator 3 straight h over denominator 2 end fraction
open parentheses straight c close parentheses space straight h over 2
open parentheses straight d close parentheses space fraction numerator 2 straight h over denominator 3 end fraction end style

Solution 6

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).

Question 7

The maximum volume of a cone that can be carved out of a solid hemisphere of radius r is

begin mathsize 12px style open parentheses straight a close parentheses space space 3 πr squared
open parentheses straight b close parentheses space xr cubed over 3
open parentheses straight c close parentheses space xr squared over 3
open parentheses straight d close parentheses space space 3 πr cubed end style

Solution 7


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).

Question 8

The radii of two cylinders are in the ratio 3 : 5. If their heights are in the ratio 2 : 3, then the ratio of their curved surface areas is

(a) 2 : 5

(b) 5 : 2

(c) 2 : 3

(d) 3 : 5

Solution 8

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).

Question 9

A right circular cylinder of radius r and height h = 2r just encloses a sphere of diameter

(a) h

(b) r

(c) 2r

(d) 2h

Solution 9

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.

Question 10

The radii of the circular ends of a frustum are 6 cm and 14 cm. If its slant height is 10 cm, then its vertical height is

(a) 6 cm

(b) 8 cm

(c) 4 cm

(d) 7 cm

Solution 10

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).

Question 11

The height and radius of the cone of which the frustum is a part are h1 and r3 respectively. If h2 and r2 are the heights and radius of the smaller base of the frustum respectively and h2 : h1 = 1: 2 then r2 : r1 is equal to

(a) 1 : 3

(b) 1 : 2

(c) 2 : 1

(d) 3 : 1

Solution 11

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).

Question 12

The diameters of the ends of a frustum of a cone are 32 cm and 20 cm. If its slant height is 10 cm, then its lateral surface area is 

begin mathsize 12px style open parentheses straight a close parentheses space 321 space straight pi space cm squared
open parentheses straight b close parentheses space 300 space straight pi space cm squared
open parentheses straight c close parentheses space 260 space straight pi space cm squared
open parentheses straight d close parentheses space 250 space straight pi space cm squared end style

Solution 12

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).


Question 13

A solid frustum is of height 8 cm. If the radii of its lower and upper ends are 3 an and 9 cm respectively, then its slant height is

(a) 15 an

(b) 12 an

(c) 10 cm

(d) 17 cm 

Solution 13

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).

Chapter 14 - Surface Areas and Volumes Excercise 14.91

Question 1

The radii of the ends of a bucket 16 cm high are 20 cm and 8 cm. The curved surface area of the bucket is 

(a) 1760 cm2

(b) 2240 cm2

(c) 880 cm2

(d) 3120 cm2

Solution 1

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).

Question 2

The diameters of the top and the bottom portions of a bucket are 42 cm and 28 cm respectively. If the height of the bucket is 24 cm, then the cost of painting its outer surface at the rate of 50 paice/cm2 is

(a) Rs. 1582.50

(b) Rs. 1724.50

(c) RS. 1683

(d) Rs. 1642

Solution 2

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)

Question 3

If four times the sum of the areas of two circular faces of a cylinder of height 8 cm is equal to twice the curve surface area, then diameter of the cylinder is

(a) 4 cm

(b) 8 cm

(c) 2 crn

(d) 6 cm

Solution 3

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).

Question 4

If the radius of the base of a right circular cylinder is halved, keeping the height the same, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is 

(a) 1 : 2
(b) 2 :1
(c) 1 : 4
(d) 4 : I

Solution 4

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).

Question 5

A metallic solid cone is melted to form a solid cylinder of equal radius. If the height of the cylinder is 6 cm, then the height of the cone was 

(a) 10 cm

(b) 12 cm

(c) 18 cm

(d) 24 cm

Solution 5

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).

Question 6

A rectangular sheet of paper 40 cm x 22 cm, is rolled to form a hollow cylinder of height 40 cm. The radius of the cylinder (in cm) is

(a) 3.5

(b) 7

(c) 80/7

(d) 5

Solution 6

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).

Question 7

The number of solid spheres, each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm is

(a) 3 

(b) 5

(c) 4

(d) 6

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 (b).

Question 8

Volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is

 

a. 3 : 4

b. 4 : 3

c. 9 : 16

d. 16 : 9

Solution 8

Question 9

A right circular cylinder of radius r and height h (h > 2r) just encloses a sphere of diameter

 

a. r

b. 2r

c. h

d. 2h

Solution 9

 

Correct option: (b)

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

Question 10

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

 

a. circle

b. frustum of a cone

c. sphere

d. hemisphere

Solution 10

 

 

Correct option: (a)

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

Question 11

If two solid-hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is

 

a. 4πr2

b. 6πr2

c. 3πr2

d. 8πr2

Solution 11

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

Question 12

The diameters of two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is

 

a. 32.7 litres

b. 33.7 litres

c. 34.7 litres

d. 31.7 litres

Solution 12

Question 13

A spherical ball of radius r is melted to make 8 new identical balls each of radius r1. Then r : r1 =

 

a. 2 : 1

b. 1 : 2

c. 4 : 1

d. 1 : 4

Solution 13