Class 10 - Maths - Surface Areas Volumes

Exercise 13.1

Unless stated otherwise, take π = 22/7.

Question 1:

2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Answer:

Given that,

Class_10_Surface_Areas_&_2_Cuboids_Joined_Together

Volume of cubes = 64 cm

=> (Edge)3 = 64

=> (Edge)3 = 43

=> Edge = 4 cm

If cubes are joined end to end, the dimensions of the resulting cuboid will be 4 cm, 4 cm, 8 cm.

Now, surface area of cuboid = 2(lb + bh + lh)

                                                    = 2(4 * 4 + 4 * 8 + 4 * 8)

                                                    = 2(16 + 32 + 48)

                                                    = 2 * 80  

                                                    = 160 cm2

Question 2:

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder.

The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm.

Find the inner surface area of the vessel.

Answer:

Class_10_Surface_Areas_&_Vessel

It can be observed that radius (r) of the cylindrical part and the hemispherical part is the same

(i.e., 7 cm).

Height of hemispherical part = Radius = 7 cm

Height of cylindrical part (h) = 13 −7 = 6 cm

Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part

                                                         = 2πrh + 2πr2

                                                         = 2 * 22/7 * 7 * 6 + 2 * 22/7 * 7 * 7

                                                         = 44(6 + 7)

                                                         = 44 * 13

                                                         = 572 cm2                

Question 3:

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm.

Find the total surface area of the toy.

Answer:

It can be observed that the radius of the conical part and the hemispherical part is same      

(i.e., 3.5 cm).

Class_10_Surface_Areas_&_Toy_In_Form_Of_Toy

Height of hemispherical part = Radius (r) = 3.5 = 35/10 = 7/2 cm

Height of conical part (h) = 15.5 −3.5 = 12 cm

Slant height (l) of conical part = √(r2 + h2)

                                                      = √{(7/2)2 + 122)}

                                                      = √(49/4 + 144)

                                                      = √{(49 + 576)/4}

                                                      = √(625/4)

                                                      = 25/2                   

Total surface area of toy = CSA of conical part + CSA of hemispherical part

                                             = πrl + πr2

                                            = 22/7 * 7/2 * 25/2 + 2 * 22/7 * 7/2 * 7/2

                                            = 137.5 + 77

                                            = 214.5 cm2          

Question 4:

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have?

Find the surface area of the solid.

Answer:

From the figure, it can be observed that the greatest diameter possible for such hemisphere is

equal to the cube’s edge, i.e., 7 cm.

Radius (r) of hemispherical part = 7/2 = 3.5 cm

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part

                                                   − Area of base of hemispherical part

                                                = 6(Edge)2 + 2πr2 – πr2

                                                 = 6(Edge)2 + πr2

                                                 = 6 * 72 + 22/7 * 7/2 * 7/2

                                                 = 294 + 38.5     

                                                 = 332.5 cm2          

Question 5:

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube.

Determine the surface area of the remaining solid.

Answer:

Class_10_Surface_Areas_&_Volumes_WoodenBlock

Diameter of hemisphere = Edge of cube = l

Radius of hemisphere = l/2

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part− Area of

                                                    base of hemispherical part

                                                 = 6(Edge)2 + 2πr2 – πr2

                                                 = 6l2 + πr2

                                                 = 6l2 + π(l/2)2

                                                 = 6l2 + πl2/4

                                                 = l2(24 + π)/4 unit2

Question 6:

A medicine capsule is in the shape of a cylinder with  two hemispheres stuck to each of its ends (see Fig. 13.10).     

The length of the entire capsule is 14 mm and the diameter  of the capsule is 5 mm. Find its surface area.

Class_10_Surface_Areas_&_Volumes_MedicineCapsule1

Answer:

       Class_10_Surface_Areas_&_Volumes_MedicineCapsule                                     

It can be observed that

Radius (r) of cylindrical part = Radius (r) of hemispherical part

                                                   = Diameter of the capsule/2

                                                   = 5/2      

Length of cylindrical part (h) = Length of the entire capsule − 2 * r

                                                    = 14 – 2 * 5/2

                                                    = 14 - 5

                                                    = 9 cm

Surface area of capsule = 2 * CSA of hemispherical part + CSA of cylindrical part

                                         = 2 * 2πr2 + 2πrh

                                         = 4πr2 + 2πrh

                                         = 4π(5/2)2 + 2π * 5/2 * 9

                                         = 25π + 45π

                                         = 70π

                                         = 70 * 22/7

                                         = 10 * 22

                                         = 220 mm2                        

Question 7:

A tent is in the shape of a cylinder surmounted by a conical top.

If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m,

find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2.

(Note that the base of the tent will not be covered with canvas.)  [Use π = 22/7]

Answer:

Given that,

Class_10_Surface_Areas_&_Volumes_Tent

Height (h) of the cylindrical part = 2.1 m

Diameter of the cylindrical part = 4 m

Radius of the cylindrical part = 2 m

Slant height (l) of conical part = 2.8 m

Area of canvas used = CSA of conical part + CSA of cylindrical part

                                     = πrl + 2πrh

                                     = π * 2 * 2.8 + 2 π * 2 * 2.1

                                     = 2π(2.8 + 4.2)

                                     = 2π * 7

                                     = 2 * 22/7 * 7

                                     = 44 m2            

Cost of 1 m2 canvas = Rs 500

Cost of 44 m2 canvas = 44 * 500 = 22000

Therefore, it will cost Rs 22000 for making such a tent.

Question 8:

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out.

Find the total surface area of the remaining solid to the nearest cm2.

Answer:

Given that,

Class_10_Surface_Areas_&_Volumes_SolidCylinder

Height (h) of the conical part = Height (h) of the cylindrical part = 2.4 cm

Diameter of the cylindrical part = 1.4 cm

Therefore, radius (r) of the cylindrical part = 0.7 cm

Slant height (l) of conical part = √(r2 + h2)

                                                      = √{(0.7)2 + (2.4)2}

                                                      = √(0.49 + 5.76)

                                                      = √(6.25)

                                                      = 2.5

Total surface area of the remaining solid will be

= CSA of cylindrical part + CSA of conical part + Area of cylindrical base

= 2πrh + πrl + πr2

= 2 * 22/7 * 0.7 * 2.4 + 22/7 * 0.7 * 2.5 + 22/7 * 0.7 * 0.7

= 4.4 * 2.4 + 2.2 * 2.5 + 2.2 * 0.7

= 10.56 + 5.50 + 1.54

= 17.60 cm2

The total surface area of the remaining solid to the nearest cm2 is 18 cm2.

Question 9:

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 13.11.

If the height  of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Answer:

Given that,

Class_10_Surface_Areas_&_Volumes_WoodenArticle

Radius (r) of cylindrical part = Radius (r) of hemispherical part = 3.5 cm

Height of cylindrical part (h) = 10 cm

Surface area of article = CSA of cylindrical part + 2 * CSA of hemispherical part

= 2πrh + 2 * πr2

= 2π * 3.5 * 10 + 2 * 2π * 3.5 * 3.5

= 70π + 49π

= 119π

= 119 * 22/7

= 17 * 22

= 374 cm2

 

 

 

                                                                    Exercise 13.2

Unless stated otherwise, take π = 22/7.

Question 1:

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius.

Find the volume of the solid in terms of π.

Answer:

Given that,

Class_10_Surface_Areas_&_Volumes_SolidShape

Height (h) of conical part = Radius(r) of conical part = 1 cm

Radius(r) of hemispherical part = Radius of conical part (r) = 1 cm

Volume of solid = Volume of conical part + Volume of hemispherical part

                              = πr2 h/3 + 2πr3 /3

                              = (π(1)2 * 1)/3 +( 2π * 13 )/3

                              = π/3 + 2 π/3

                               = π cm3  

Question 2:

Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet.

The diameter of the model is 3 cm and its length is 12 cm.

If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made.

(Assume the outer and inner dimensions of the model to be nearly the same.)

Answer:

From the figure, it can be observed that

Height (h1) of each conical part = 2 cm

Height (h2) of cylindrical part = 12 − 2 * Height of conical part

                                                    = 12 − 2 * 2

                                                     = 8 cm

Class_10_Surface_Areas_&_Volumes_Model_In_Shape_Of_Cylinder

Radius (r) of cylindrical part = Radius of conical part = 3/2 cm

Volume of air present in the model = Volume of cylinder + 2 * Volume of cones

                                                                = πr2 h2 + 2 * (πr2 h1)/3

                                                                = π(3/2)2 * 8 + {2π * (3/2)2 * 2}/3

                                                                = π * 9/4 * 8 + 2π/3 * 9/4 * 2

                                                                = 18π + 3 π

                                                                = 21π

                                                                 = 21 * 22/7

                                                                 = 3 * 22

                                                                 = 66 cm2  

Question 3:

A gulab jamun, contains sugar syrup up to about 30%  of its volume. Find approximately how much syrup would 

 be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm   (see Fig. 13.15).

Class_10_Surface_Areas_&_Volumes_GulabJamunInSugarSyrup

Answer:

    Class_10_Surface_Areas_&_Volumes_HemisphereEnds_With_Cylinder                                                        

It can be observed that

Radius (r) of cylindrical part = Radius (r) of hemispherical part = 2.8/2 = 1.4 cm

Length of each hemispherical part = Radius of hemispherical part = 1.4 cm

Length (h) of cylindrical part = 5 − 2 * Length of hemispherical part

                                                    = 5 − 2 * 1.4

                                                    = 5 – 2.8 = 2.2 cm

Volume of one gulab jamun = Volume of cylindrical part + 2 * Volume of hemispherical part

                                                  = πr2 h + 2 * (2πr3)/3

                                                  = πr2 h + 4πr3/3

                                                   = π(1.4)2 * 2.2 + {4π * (1.4)3}/3

                                                   = 22/7 * 1.4 * 1.4 * 2.2 + (4 * 22/7 * 1.4 * 1.4 * 1.4)/3

                                                   = 13.552 + 11.498

                                                   = 25.05 cm3

Volume of 45 gulab jamuns = 45 * 25.05 = 1127.25 cm3

Volume of sugar syrup = 30% of volume

                                         = 30/100 * 1127.25

                                         = 338.17

                                         = 338 cm3   

Question 4:  

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. 

The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm.

The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm.

Find the volume of wood  in the entire stand (see Fig. 13.16).

Class_10_Surface_Areas_&_Volumes_PenStand

Answer:

Class_10_Surface_Areas_&_Volumes_InvertedCone1

Depth (h) of each conical depression = 1.4 cm

Radius (r) of each conical depression = 0.5 cm

Volume of wood = Volume of cuboid − 4 * Volume of cones

                               = lbh – 4 * πr2 h/3

                               = 15 * 10 * 3.5 – 4 * 1/3 * 22/7 * (1/2)2 * 1.4             

                               = 525 – 1.47

                               = 523.53 cm3

Question 5:

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm.

It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel,

one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Answer:

Height (h) of conical vessel = 8 cm

Radius (r1) of conical vessel = 5 cm

Radius (r2) of lead shots = 0.5 cm

Class_10_Surface_Areas_&_Volumes_InvertedCone

Let n number of lead shots were dropped in the vessel.

Volume of water spilled = Volume of dropped lead shots

=> 1/4 * Volume of cone = n * 4πr23/3

=> 1/4 * πr12 h/3 = n * 4πr23/3

=> r12 h = n * 16r23

=> 52 * 8 = n * 16(0.5)3

=> 25 * 8 = n * 16(1/2)3

=> 25 * 8 = n * 16 * 1/8

=> 25 * 8 = n * 2

=> 2n = 200

=> n = 200/2 = 100

Hence, the number of lead shots dropped in the vessel is 100.

Question 6:

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by

another cylinder of height 60 cm and radius 8 cm.

Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use π = 3.14)

Answer:

From the figure, it can be observed that

Class_10_Surface_Areas_&_Volumes_Solid_Iron_Pole

Height (h1) of larger cylinder = 220 cm

Radius (r1) of larger cylinder = = 12 cm

Height (h2) of smaller cylinder = 60 cm

Radius (r2) of smaller cylinder = 8 cm

Mass of 1 iron = 8 g

Total volume of pole = Volume of larger cylinder + volume of smaller cylinder

                                      = πr12 h1 + πr22 h2

                                      = π(12)2 * 220 + π(8)2 * 60

                                      = π[144 * 220 + 64 * 60]

                                      = π[31680 + 3840]

                                      = 3.14 * 35520

                                      = 111532.8 cm3             

Mass of 1 cm3 iron = 8 g

Mass of 111532.8 cm3 iron = 111532.8 * 8 = 892262.4 g = 892262.4/1000 kg                                           

                                                 = 892.262 kg

Question 7:

A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a

right circular cylinder full of water such that it touches the bottom.

Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Answer:

Class_10_Surface_Areas_&_Volumes_Figure_Having_Cone_Hemisphere_Cylinder

Radius (r) of hemispherical part = Radius (r) of conical part = 60 cm

Height (h2) of conical part of solid = 120 cm

Height (h1) of cylinder = 180 cm

Radius (r) of cylinder = 60 cm

Volume of water left = Volume of cylinder − Volume of solid

                                       = Volume of cylinder – (Volume of cone + Volume of Hemisphere)

                                       = πr2 h1 – (πr2 h2/3 +2 πr3/3)

                                       = π(60)2 * 180 – {π(60)2 * 120/3 +2 π(60)3/3}

                                       = π(60)2 [180 – (40 + 40)]

                                       = π(60)2 [180 – 80]

                                       = 3.14 * 3600 * 100

                                       = 1131428.57 cm3

                                       = 1131428.57/1000000 m3

                                        = 1.131 m3                 

Question 8:

A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm.

By measuring the amount of water it holds, a child finds its volume to be 345 cm3.

Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Answer:

Class_10_Surface_Areas_&_Volumes_SphericalGlassVessel

Height (h) of cylindrical part = 8 cm

Radius (r2) of cylindrical part = 2/2 = 1 cm

Radius (r1) spherical part = 8.5/2 = 4.25 cm

Volume of vessel = Volume of sphere + Volume of cylinder

                                = 4πr13/3 + πr22 h     

                                = {4π(4.25)3}/3 + π(1)2 * 8

                                = {4 * 3.14 * 76.765625}/3 + 3.14 * 8

                                = 321.392 + 25.12

                                = 346.512 cm3       

Hence, she is wrong.

                                            Exercise 13.3

Unless stated otherwise, take π = 22/7.

Question 1:

A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Answer:

Radius (r1) of hemisphere = 4.2 cm

Radius (r2) of cylinder = 6 cm

Let the height of the cylinder be h.

The object formed by recasting the hemisphere will be the same in volume.

Volume of sphere = Volume of cylinder

=> 4πr13/3 = πr22 h

=> 4π * (4.2)3/3 = π(6)2 h

=> 4(4.2)3/3 = 36h

=> h = {4(4.2)3/3}/36

=> h = {4 * 4.2 * 4.2 * 4.2}/(3 * 36)

=> h = 1.4 * 1.4 * 1.4

=> h = 2.74

Hence, the height of the cylinder so formed will be 2.74 cm.

Question 2:

Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Answer:

Radius (r1) of 1st sphere = 6 cm

Radius (r2) of 2nd sphere = 8 cm

Radius (r3) of 3rd sphere = 10 cm

Let the radius of the resulting sphere be r.

The object formed by recasting these spheres will be same in volume as the sum of the

volumes of these spheres.

      Volume of 3 spheres = Volume of resulting sphere

=> 4π[r13 + r23 + r33]/3 = 4πr3/3

=> r13 + r23 + r33 = r3

=> r3 = 63 + 83 + 103

=> r3 = 216 + 512 + 1000

=> r3 = 1728

=> r = 12 

Therefore, the radius of the sphere so formed will be 12 cm.

Question 3:

A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m.

Find the height of the platform.

Answer:

The shape of the well will be cylindrical.

Class_10_Surface_Areas_&_Volumes_Well

Depth (h) of well = 20 m

Radius (r) of circular end of well = 7/2 m

Area of platform = Length × Breadth = 22 * 14 m2

Let height of the platform = H

Volume of soil dug from the well will be equal to the volume of soil scattered on the platform.

Volume of soil from well = Volume of soil used to make such platform

=> πr2 h = Area of platform * Height of platform

=> π(7/2)2 * 20 = 22 * 14 * H

=> 22/7 * 7/2 * 7/2 * 20 = 22 * 14 * H

=> 22/7 * 7/2 * 7/2 * 20 = 22 * 14 * H

=> 7 * 5 = 14 * H

=> 5 = 2 * H

=> H = 5/2

=> H = 2.5

Therefore, the height of such platform will be 2.5 m.

Question 4:

A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a

circular ring of width 4 m to form an embankment. Find the height of the embankment.

Answer:

The shape of the well will be cylindrical.

Class_10_Surface_Areas_&_Volumes_Well_With_A_Circular_Ring

Depth (h1) of well = 14 m

Radius (r1) of the circular end of well = 3/2 m

Width of embankment = 4 m

From the figure, it can be observed that our embankment will be in a cylindrical shape having

outer radius (r2) as 4 + 3/2 = 11/2 m and inner radius (r1) as 3/2 m.

Let the height of embankment be h2.

      Volume of soil dug from well = Volume of earth used to form embankment

=> πr12 h1 = π(r12 - r12) * h2

=> π(3/2)2 * 14 = π{(11/2)2 – (3/2)2} * h

=> 9/4 * 14 = 112/4 * h

=> h = 9/8

=> h = 1.125

Therefore, the height of embankment will be 1.1.25 m.

Question 5:

A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream.

The ice cream is to be filled into 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.

Answer:

Height (h1) of cylindrical container = 15 cm

Radius (r1) of circular end of container =

Radius (r2) of circular end of ice-cream cone =

Height (h2) of conical part of ice-cream cone = 12 cm

Let n ice-cream cones be filled with ice-cream of the container.

Volume of ice-cream in cylinder = n * (volume of 1 ice-cream cone + volume of hemisphere

                                                               shape on the top)

=> πr12 h1 = n(πr22 h2 + 2πr23 /3)

=> n = (62 * 16)/{1/3 * 9 * 12 + 2/3 * (3)3}

=> n = (36 * 15 * 3)/(108 + 54)

=> n = 10

Therefore, 10 ice-cream cones can be filled with the ice-cream in the container.

Question 6:

How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm * 10 cm * 3.5 cm?

Answer:

Coins are cylindrical in shape.

Height (h1) of cylindrical coins = 2 mm = 2/10 mm = 0.2 cm

Radius (r) of circular end of coins = 1.75/2 = 0.875 cm

Let n coins be melted to form the required cuboids.

Class_10_Surface_Areas_&_Volumes_Coins

      Volume of n coins = Volume of cuboids

=> n * πr2 h1 = l * b * h

=> n * π(0.875)2  * 0.2  = 5.5 * 10 * 3.5

=> n = (5.5 * 10 * 3.5)/{π(0.875)2  * 0.2}

=> n = (5.5 * 10 * 3.5 * 7)/{22 * (0.875)2  * 0.2}

=> n = 400

Therefore, the number of coins melted to form such a cuboid is 400.

Question 7:

A cylindrical bucket, 32 cm high and with radius of base 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.

Answer:

Height (h1) of cylindrical bucket = 32 cm

Radius (r1) of circular end of bucket = 18 cm

Height (h2) of conical heap = 24 cm

Let the radius of the circular end of conical heap be r2.

The volume of sand in the cylindrical bucket will be equal to the volume of sand in the conical

heap.

Class_10_Surface_Areas_&_Volumes_ConicalHeapOfSand Class_10_Surface_Areas_&_Volumes_CylindricalBucket

Now, volume of sand in the cylindrical bucket = Volume of sand in conical heap

=> πr2 h1 = πr22 h2/3  

=> π* 182 * 32 = (πr22  * 24)/3

=> 182 * 32 = r22 * 8

=> r22 = (182 * 32)/8

=> r22 = 182 * 4

=> r2 = 18 * 2

=> r2 = 36

Now, slant height = √(362 + 242) = √{122 * (32 + 22)} = 12√(9 + 4) = 12√13

Hence, the radius and slant height of the conical heap are 36 cm and 12√13 cm respectively.

Question 8:

Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes,

if 8 cm of standing water is needed?

Answer:

            Class_10_Surface_Areas_&_Volumes_WaterFlowingInACanal Class_10_Surface_Areas_&_Volumes_WaterFlowingInACanal1

Consider an area of cross-section of canal as ABCD.

Area of cross-section = 6 * 1.5 = 9 m2

Speed of water = 10 km/h = 10000/60 metre/min

Volume of water that flows in 1 minute from canal = 9 * 10000/60 = 1500 m3

Volume of water that flows in 30 minutes from canal = 30 × 1500 = 45000 m3

Let the irrigated area be A. Volume of water irrigating the required area will be equal to the

volume of water that flowed in 30 minutes from the canal.

Volume of water flowing in 30 minutes from canal = Volume of water irrigating the area

=> 45000 = (A * 8)/100

=> A = (45000 * 100)/8

=> A = 5625 * 100

=> A = 562500 m2

Therefore, area irrigated in 30 minutes is 562500 m2.

Question 9:

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her 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?

Answer:

      Class_10_Surface_Areas_&_Volumes_Pipe                 Class_10_Surface_Areas_&_Volumes_CylindricalTank

Consider an area of cross-section of pipe as shown in the figure.

Radius (r1) of circular end of pipe = 20/200 = 0.1 m

Area of cross-section = π * r12 = π * (0.1)2 = 0.01π m2

Speed of water = 3 km/h = 3000/60 = 50 metre/min

Volume of water that flows in 1 minute from pipe = 50 * 0.01π = 0.5π m3

Volume of water that flows in t minutes from pipe = t * 0.5π m3

Radius (r2) of circular end of cylindrical tank = 10/2 = 5 m

Depth (h2) of cylindrical tank = 2 m

Let the tank be filled completely in t minutes.

Volume of water filled in tank in t minutes is equal to the volume of water flowed in t minutes

from the pipe.

Volume of water that flows in t minutes from pipe = Volume of water in tank

=> t * 0.5π = π *(r2)2 * h2

=> t * 0.5 = 52 * 2

=> t = 50/0.5

=> t = (50 * 10)/5

=> t = 100

Therefore, the cylindrical tank will be filled in 100 minutes.

                                                                    Exercise 13.4

Use π = 22/7 unless stated otherwise.

Question 1:

A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm.

Find the capacity of the glass.

Answer:

Radius (r1) of upper base of glass = 4/2 = 2 cm

Radius (r2) of lower base of glass = 2/2 = 1 cm

Class_10_Surface_Areas_&_Volumes_Frustum_Of_Cone3

Capacity of glass = Volume of frustum of cone

                               = πh[r12 + r22 + r1 * r2]/3

                               = πh[22 + 12 + 2 * 1]/3

                               = 1/3 * 22/7 * 14 * 7

                               = 308/3

                               = 102⅔ cm3

Therefore, the capacity of glass is 102⅔ cm3.

Question 2:

The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm.

Find the curved surface area of the frustum.

Answer:

Perimeter of upper circular end of frustum = 18 cm

=> 2πr1 =18

=> r1 =18/π

Class_10_Surface_Areas_&_Volumes_Cap_In_TheShape_Of_Frustrum_Of_Cone1

Perimeter of lower end of frustum = 6 cm

=> 2πr2 = 6

=> r2 = 3/π

Slant height (l) of frustum = 4 cm

CSA of frustum = π(r1 + r2)l

                            = π(9/π + 3/π) * 4

                            = 12 * 4

                            = 48 cm2

Therefore, the curved surface area of the frustum is 48 cm2.

Question 3:  

A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig. 13.24). If its radius on the

open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material

used for making it.

Class_10_Surface_Areas_&_Volumes_Cap_In_TheShape_Of_Frustrum_Of_Cone1

Answer:

Radius (r2) at upper circular end = 4 cm

Radius (r1) at lower circular end = 10 cm

Slant height (l) of frustum = 15 cm

Class_10_Surface_Areas_&_Volumes_Cap_In_TheShape_Of_Frustrum_Of_Cone

Area of material used for making the fez = CSA of frustum + Area of upper circular end

                                                                         = π(r1 + r2)l + πr22

                                                                         = π (10 + 4)15 + π (4)2

                                                                         = π *14 * 15 + 16π

                                                                         = 210 π + 16 π

                                                                         = 226 π

                                                                        = (226 * 22)/7

                                                                        = 4972/7

                                                                        = 710 2/7 cm2

Hence, the area of material used for making it is 710 2/7 cm2.    

Question 4:

A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm

with radii of its lower and upper ends as 8 cm and 20 cm, respectively.

Find the cost of the milk which can completely fill the container, at the rate of Rs 20 per litre.

Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2.           (Take π = 3.14)

Answer:

Class_10_Surface_Areas_&_Volumes_Container_In_The_Shape_Of_Cone

Radius (r1) of upper end of container = 20 cm

Radius (r2) of lower end of container = 8 cm

Height (h) of container = 16 cm

Slant height (l) of frustum = √{(r1 – r2)2 + h2}

                                               = √{(20 - 8)2 + 162}

                                               = √(144 + 256)

                                               = √400

                                               = 20     

Capacity of container = Volume of frustum

                                       = πh[r12 + r22 + r1 * r2]/3

                                       = 1/3 * 3.14 * 16 * [202 + 82 + 20 * 8]

                                       = 1/3 * 3.14 * 16 * (400 + 64 + 160)

                                       = 1/3 * 3.14 * 16 * 624  

                                       = 3.14 * 16 * 208

                                      = 10449.92 cm3

                                      = 10449.92/1000 litres           [1000 cm3 = 1 litre]

                                      = 10.45 litres

Cost of 1 litre milk = Rs 20

Cost of 10.45 litres milk = 10.45 * 20 = Rs 209

Area of metal sheet used to make the container = π(r1 + r2)l + πr22

                                                                                       = π (20 + 8)20 + π (8)2

                                                                                       = π *28 * 20 + 64π

                                                                                       = 560 π + 64 π

                                                                                       = 624 π cm2

Cost of 100 cm2 metal sheet = Rs 8

Cost of 624 π cm2 metal sheet = (624 π * 8)/100

                                                       = (624 * 3.14 * 8)/100

                                                       = 15674.88/100

                                                       = Rs 156.75     

Therefore, the cost of the milk which can completely fill the container is Rs 209 and the cost of

metal sheet used to make the container is Rs 156.75

Question 5:

A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by

a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1 cm, 16 find the length of the wire.

Answer:

Class_10_Surface_Areas_&_Volumes_Right_Circular_Cone

In ΔAEG,

       EG/AG = tan 30°

=> EG/10 = 1/√3

=> EG = 10/√3

=> EG = (10 * √3)/(√3 * √3)

=> EG = 10√3/3

In ΔABD,

      BD/AD = tan 30°

=> EG/20 = 1/√3

=> EG = 20/√3

=> EG = (20 * √3)/(√3 * √3)

=> EG = 20√3/3

Radius (r1) of upper end of frustum = 10√3/3 cm

Radius (r2) of lower end of container = 20√3/3 cm

Height (h) of container = 10 cm

Volume of frustum = πh[r12 + r22 + r1 * r2]/3

                                   = 1/3 * π * 10 * [(10√3/3)2 + (10√3/3)2 + (10√3/3) * (10√3/3)]

                                   = 1/3 * π * 10 * [100/3 + 400/3 + 200/3]

                                   = 10/3 * 22/7 * 700/3 

                                   = 22000/9 cm3

Radius (r) of wire = 1/16 * 1/2 = 1/32 cm

Let the length of wire be l.

Volume of wire = Area of cross-section * Length

                             = πr2 * l

                             = π(1/32)2 * l 

                            = 22/7 * (1/32)2 * l

Volume of frustum = Volume of wire

=> 22000/9 = 22/7 * (1/32)2 * l

=> l = 7000/9 * 1024

=> l = 796444.44 cm

=> l = 796444.44/100 m

=> l = 7964.44 m

                                     Exercise 13.5 (Optional)*

Question 1:

A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm,

so as to cover the curved surface of the cylinder.

Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.

Answer:

It can be observed that 1 round of wire will cover 3 mm height of cylinder.

Class_10_Surface_Areas_&_Volumes_Cylinder

Number of rounds = Height of cylinder/Diameter of wire

                                  = 12/0.3           [3 mm = 3/10 cm = 0.3 cm]

                                  = (12 * 10)/3

                                  = 120/3

                                  = 40 rounds

Length of wire required in 1 round = Circumference of base of cylinder

                                                               = 2πr = 2π * 5 = 10π

Length of wire in 40 rounds = 40 * 10π

                                                   = 40 * 10 * 22/7

                                                   = 8800/7          

                                                   = 1257.14 cm

                                                   = 12.57 m

Radius of wire = 0.3/2 = 0.15 cm

Volume of wire = Area of cross-section of wire * Length of wire

                             = π(0.15)2 * 1257.14

                            = 88.898 cm

Mass = Volume * Density

           = 88.898 * 8.88

           = 789.41 gm  

Question 2:

A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse.

Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate.)

Answer:

The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse as

shown in the figure.

Class_10_Surface_Areas_&_Volumes_Double_Cone

Hypotenuse AC = √(32 + 42) = √(9 + 16) = √25 = 5 cm

Area of ΔABC = 1/2 * AB * AC

=> 1/2 * AC * OB = 1/2 * 4 * 3

=> 5 * OB = 4 * 3

=> OB = 12/5

=> OB = 2.4 cm

Volume of double cone = Volume of cone 1 + Volume of cone 2

                                           = πr2 h1/3 + πr2 h2/3

                                           = πr2 (h1 + h2)/3

                                           = πr2 (OA + OC)/3   

                                           = {3.14 * (2.4)2 * 5}/3

                                           = 30.14 cm3

Surface area of double cone = Surface area of cone 1 + Surface area of cone 2

                                                   = πrl1 + πrl2

                                                   = πr(4 + 3)

                                                 = 3.14 * 2.4 * 7

                                                 = 52.75 cm2

Question 3:

A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it.

Porous bricks are placed in the water until the cistern is full to the brim.

Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water,

each brick being 22.5 cm × 7.5 cm × 6.5 cm?

Answer:

Volume of cistern = 150 * 120 * 110 = 1980000 cm3

Volume to be filled in cistern = 1980000 – 129600 = 1850400 cm3

Let n numbers of porous bricks were placed in the cistern.

Volume of n bricks = n * 22.5 * 7.5 * 6.5 = 1096.875n

As each brick absorbs one-seventeenth of its volume, therefore, volume absorbed by these

bricks = n(1096.875)/17

Now, 1850400 + n(1096.875)/17 = 1096.875 * n

=> 1096.875 * n - n(1096.875)/17 = 1850400

=> 1096.875(n - n/17) = 1850400

=> 1096.875 * 16n/17 = 1850400

=> n = (1850400 * 17)/( 1096.875 * 16)

=> n = 1792.41

Therefore, 1792 bricks were placed in the cistern.

Question 4:

In one fortnight of a given month, there was a rainfall of 10 cm in a river valley.

If the area of the valley is 97280 km2, show that the total rainfall was approximately equivalent to the addition to the

normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Answer:

Area of the valley = 7280 km

If there was a rainfall of 10 cm in the valley then amount of rainfall in the valley

= Area of the valley * 10 cm

Amount of rainfall in the valley = 7280 km2 * 10 cm

                                                        = 7280 * (1000 m)2 * 10/100 m

                                                        = 7280 * 105 m3      

                                                        = 7.28 * 108 m3

Length of each river, l = 1072 km = 1072 * 1000 m = 1072000 m

Breadth of each river, b = 75 m

Depth of each river, h = 3 m

Volume of each river = l * b * h

                                      = 1072000 * 75 * 3

                                      = 2.412 * 108 m3

Volume of three such rivers = 3 * Volume of each river

                                                  = 3 * 2.412 * 108 m3

                                                  = 7.236 * 108 m3

Hence, total rainfall is approximately same as the volume of the three rivers.

Question 5:  

An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone.

If the total height is 22 cm, diameter of the cylindrical portion 

is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel   (see Fig. 13.25).

Class_10_Surface_Areas_&_Volumes_Funnel1

Answer:

Class_10_Surface_Areas_&_Volumes_Funnel

Radius (r1) of upper circular end of frustum part = 18/2 = 9 cm

Radius (r2) of lower circular end of frustum part = Radius of circular end of cylindrical part

                                                                                       = 8/2 = 4 cm 

Height (h1) of frustum part = 22 − 10 = 12 cm

Height (h2) of cylindrical part = 10 cm

Slant height (l) of frustum part = √{(r1 – r2)2 + h2}

                                                       = √{(9 - 4)2 + 122}

                                                       = √(25 + 144)

                                                       = √169

                                                       = 13

Area of tin sheet required = CSA of frustum part + CSA of cylindrical part

                                               = π(r1 + r2)l + 2πr2 h2         

                                               = 22/7 * (9 + 4) * 13 + 2 * 22/7 * 4 * 10

                                               = 22(169 + 80)/7

                                               = (22 * 249)/7

                                               = 5478/7

                                               = 782 4/7 cm2      

Question 6:

Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5,

using the symbols as explained.

Answer:

Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base. Let r1 and r2 be the

radii of the ends of the frustum of the cone and h be the height of the frustum of the cone.

Class_10_Surface_Areas_&_Volumes_Frustum_Of_Cone1

In ΔABG and ΔADF, DF||BG  

So, ΔABG ∼ ΔADF

       DF/BG = AF/AG = AD/AB

=> r2/r1 = (h1 – h)/h1 = (l1 - l)/l1

=> r2/r1 = 1 – h/h1 = 1 - l/l1

=> r2/r1 = 1 - l/l1

=> l/l1 = 1 - r2/r1 

=> l/l1 = (r1 - r2)/r1

=> l1/l = r1/(r1 - r2)

=> l1 = l * r1/(r1 - r2)

CSA of frustum DECB = CSA of cone ABC − CSA cone ADE

                                       = πr1 l1 – πr2(l1 – l)     

                                       = πr1 { l * r1/(r1 - r2)} – πr2{ l * r1/(r1 - r2) – l}

                                       = πr12 l/(r1 - r2)} – πr2 {(r1 l – r1 l + r2 l )/(r1 - r2)}     

                                       = πr12 l/(r1 - r2)} - πr22 l/(r1 - r2)} 

                                       = πl(r12 - r2 2)/(r1 - r2)   

CSA of frustum = π(r1 + r2)l

Total surface area of frustum = CSA of frustum + Area of upper circular end + Area of upper

                                                                                                                                           circular end

                                                     = π(r1 + r2)l + πr22 + πr12

                                                     = π[(r1 + r2)l + r22 + r12]          

Question 7:

Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Answer:

Let ABC be a cone. A frustum DECB is cut by a plane parallel to its base. Let r1 and r2 be the

radii of the ends of the frustum of the cone and h be the height of the frustum of the cone.

Class_10_Surface_Areas_&_Volumes_Frustum_Of_Cone

In ΔABG and ΔADF, DF||BG 

So, ΔABG ∼ ΔADF

       DF/BG = AF/AG = AD/AB

=> r2/r1 = (h1 – h)/h1 = (l1 - l)/l1

=> r2/r1 = 1 – h/h1 = 1 - l/l1

=> r2/r1 = 1 - h/h1

=> h/h1 = 1 - r2/r1 

=> l/l1 = (r1 - r2)/r1

=> h1/h = r1/(r1 - r2)

=> h1 = h * r1/(r1 - r2)

Volume of frustum of cone = Volume of cone ABC − Volume of cone ADE

                                                  = πr12 h1/3 – π r22 (h1 – h)/3

                                                  = 1/3 * π[r12 h1 – r22 (h1 – h)]

                                                  = 1/3 * π[r12 * h * r1/(r1 - r2) – r22 { h * r1/(r1 - r2) – h}]

                                                  = 1/3 * π[hr13/(r1 - r2) – r22 {(hr1 - hr1 + hr2)/ (r1 - r2)]

                                                  = 1/3 * π[hr13/(r1 - r2) – hr23/ (r1 - r2)]

                                                  = h/3 * π[(r13 - r23)/(r1 - r2)]

                                                  = h/3 * π[{(r1 - r2) r12 + r22 + r1 r2)}/(r1 - r2)]    

                                                  = πh[r12 + r22 + r1 r2]/3     

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