EXERCISE – 5 - livehindustan.com...A solid consisting of a right circular cone of height 120 cm and radius 60cm standing on a hemisphere of radius 60 cm is placed upright in a right

NCERT Textual Exercise (Solved)

143

Surface Areas and Volumes

EXERCISE – 5.1 1. Two cubes each of volume 64 cm3 are joined end to end. Find the surface

area of the resulting cuboid. 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.

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.

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.

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.

6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

14 mm

5 mm

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 `500 per m2. (Note that the base of the tent will not be covered with canvas.)

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.

9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure below. 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.


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10 mm

Test Yourself – SAV 1 1. Three equal cubes are placed adjacently in a row. Find the ratio of total

surface area of the new cuboid to that of the sum of the surface areas of the three cubes.

2. A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 40 m, calculate the total area of canvas used at `15 per m, if the width is 1.5 m?

3. The length of a hall is 20 m and width 16 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the hall.

4. A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. The radius of the common base is 3.5 cm and the height of the cylindrical and conical portion are 10 cm and 6 cm

respectively. Find the total surface area of the solid (use π = 227

).

5. A decorated cloth hanger is in the form of a cylinder having cones at both ends. The common radius is 8 cm. The height of the cone at one end is 6 cm and that at another end is 15 cm. Total height of the hanger is 35 cm. Find the area of the cloth required to make the hanger?

6. A metal container in the form of a cylinder is surmounted by a hemisphere of the same radius. The internal height of the cylinder is 7 m and the internal radius is 3.5 m. Calculate the total area of the internal surface excluding the base.

7. A solid toy is in the form of a hemisphere surmounted by a right circular cone. If the height of the cone is 4 cm and diameter of the base is 6 cm. Calculate the total surface of the toy (π = 3.14) .


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8. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 58 cm and the diameter of the cylinder is 28 cm.Find the total surface area of the solid.

9. Find the total surface area of the solid shown in the following figure :

EXERCISE – 5.2 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.

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.)

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.

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.


146


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.

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 8 g mass (use π = 3.14).

7. A solid consisting of a right circular cone of height 120 cm and radius 60cm 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.

8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter, and 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.

Test Yourself – SAV 2 1. A solid is in the form of a cylinder with hemispherical ends. The total height

of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and surface area of the solid.

2. A cone of height 15 cm and diameter 7 cm is mounted on a hemisphere of same radius. Find the volume of the solid thus formed.

3. A cylindrical copper rod 405 cm in length and 0.6 cm in radius is melted and drawn into a wire of length of 720 cm. Find the new radius.

4. A spherical ball of diameter 12 cm is melted and recast into three spherical balls. The radii of the two balls are 5 cm and 4 cm. Find the diameter of the third ball.


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5. 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.

6. A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised

by 6.75 cm. Find the radius of the ball (use π = 227 ).

7. From a cylinder of base radius 10 cm and height 7 cm, a cylindrical hole of radius 3 cm and of same height is cut out. Find the volume of remaining part of cylinder.

8. Water is flowing at the rate of 3 km/h through a circular pipe of 20 cm internal diameter into a circular cistern of diameter 10 m and depth 2 m. In how much time will the cistern be filled?

9. The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen is used up on writing 330 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre?

10. A sector of a circle of radius 12 cm has the angle 120°. It is rolled up so that two bounding radii are joined together to form a cone. Find the volume of the cone.

EXERCISE – 5.3

Take π = 227

, unless stated otherwise.

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.

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.

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.

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.


148


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.

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?

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.

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 min, if 8 cm of standing water is needed?

9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his/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?

Test Yourself – SAV 3 1. A cylinder, whose height is two-third of its diameter, has the same volume

as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.


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2. A solid sphere of radius 3 cm is melted and then cast into small spherical balls each of diameter 0.6 cm. Find the number of small balls thus obtained.

3. The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire (cylindrical shape) of uniform circular cross section. If the length of the wire is 36 cm, find its radius.

4. Rain water from a roof 22 m × 20 m drains into a conical vessel having diameter of base as 2 m and height 3.5 m. If the vessel is just full, find the rainfall in mm.

5. Water flows at the rate of 10 m per minute through a cylindrical pipe having its diameter as 5 mm. How much time will it take to fill a conical vessel whose diameter of base is 40 cm and depth 24 cm.

6. A solid rectangular block of metal 49 cm by 44 cm by 18 cm is melted and formed into a solid sphere. Calculate the radius of solid sphere.

7. Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Determine the time in which the level of water in the tank will rise by 21 cm.

8. Two solid right circular cones have same height. The radii of their bases are 4 cm and 3 cm. They are melted and recast into a right circular cylinder of same height. Find the radius of the base of the cylinder (use 1/ 3 = 0.577).

9. A well whose diameter is 7 m has been dug 22.5 m deep and the earth dug out is used to form an embankment 10.5 m wide around it. Find the height of the embankment.

10. A slab of iron whose dimensions are 60 cm × 20 cm × 28.26 cm is used to cast an iron pipe. The outer diameter of the pipe is 10 cm. If the wall of the pipe is 1 cm thick. Calculate the length of the pipe that can be cast from the slab (use π = 3.14).

EXERCISE – 5.4

Use π = 227 , unless stated otherwise.

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.

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.


150


3. A fez, the cap used by the Turks, is shaped like the frustum of a cone. 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.

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

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 116

cm, find the length of the wire.

Test Yourself – SAV 4 1. The diameters of the top of a bucket is 18 cm and that of the bottom is 6 cm

and its depth is 24 cm. Find the capacity of the bucket. 2. A bucket open at top is made up of 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 bucket if the cost of metal sheet used is `15 per 100 cm2 (take π = 3.14).

3. If the radii of the circular ends of a bucket, which is 45 cm high, are 28 cm and 7 cm, find the capacity of the bucket.

4. If the radii of the ends of a bucket 24 cm high are 5 cm and 15 cm respectively, find the curved surface area of the bucket.

5. The radii of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. Find its total surface area.

6. The lower portion of a hay stock is an inverted cone frustum and upper part is a cone. Find the volume of the hay stock.


151


7. A container open at top is 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 `15 per litre and the cost of the metal sheet used, if it costs `5 per 100 cm2 (use π = 3.14).

8. The height of a cone is 40 cm. A small cone is cut off at the top by a plane

parallel to the base. If the volume of the small cone be 164

th of the volume

of the given cone, at what height above the base is the section made? 9. The perimeters of the ends of a frustum are 48 cm and 36 cm. If the height

of the frustum be 11 cm, find the volume of the frustum (take π = 227

).

10. The slant height of the frustum of a cone is 4 cm. If the perimeters of its circular bases are 18 cm and 6 cm, find the curved surface of the frustum and also find the cost of painting its total surface at the rate of `12.50 per 100 cm2.

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NCERT Textual Exercises and Assignments

Exercise – 5.1 1. Volume of one cube = 64 cm³ \ a3 = (4)3

\ a = 4 cm \ After joining the cuboid, l = 4 + 4 = 8 cm \ Surface area = 2 (lb + bh + lh) = 2 (8 × 4 + 4 × 4 + 8 × 4) = 2 (32 + 16 + 32) = 2 × 80 \ Surface area of the cuboid = 160 cm2.

2. Radius = 142 = 7 cm

\ Surface area of cylinder (s1) = 2πrh = 2 × π × 7 × (13 – 7) = 2 × π × 7 × 6 = 84π cm² \ Surface area of hemisphere(s2) = 2πr2

= 2 × π × 7 × 7 = 98π cm² \ Inner surface area of the vessel = s1 + s2

= 84π + 98π = 182 × π

= 182 × 227

\ Inner surface area of the vessel = 572 cm2. 3. Height of the cone (h) = 15.5 – 3.5 = 12 cm.

\ Slant height (l) = r h2 2+

= 3.5 + 122 2

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= 12.25 144+

= 156.25

= 12.5 cm \ Surface area of cone (s1) = πrl = π × 3.5 × 12.5 = 43.75 π cm² \ Surface area of hemisphere(s2) = 2πr2

= 2π × (3.5)2 = 24.5 π cm² \ Total surface area of the toy = s1 + s2

= 43.75 π + 24.5 π = 68.25 π

= 68.25 × 227

\ Total surface area of the toy = 214.5 cm2. 4. Greatest diameter = side of a square = 7 cm. Greatest diameter of the hemisphere = 7 cm. SA of the cube = 6l2

= 6 × 7 × 7 = 6 × 49 = 294 cm2

SA of the solid = TSA of the cube – base area of hemisphere + CSA of hemisphere \ = 294 – πr² + 2πr² = 294 + πr²

= 294 + 227

11

× ×72

72

= 294 + 772

= 294 + 38.5 = 332.5 cm2

\ Total surface area of the solid = 332.5 cm2. 5. Edge of the cube = diameter of hemisphere

\ Radius of hemisphere = l2

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\ r = l2

Surface area of the remaining solid = [surface area of a cube] – [area of the top of the + [inner curved surface area of the hemispherical top] of the hemisphere]

= 6l² – πr² + 2πr² = 6l² + πr²

= 6l² + πl2

2

= 6l² + πl 2

4

= 24l l² ²

4

= 14

² sq.unit (24+ ) l π

Surface area of the remaining solid = 14

l2 (24 + π) sq. unit.

6. Radius = 52

= 2.5 mm

Length of the cylinder (h) = 14 – 2.5 – 2.5

14 mm

5 mm = 14 – 5 = 9 mm \ Surface area of the cylinder (s1) = 2πrh = 2 × π × 2.5 × 9 = 45 π mm² \ Surface area of 2 hemispheres (s2) = 2 × 2πr2

= 2 × 2 × π × (2.5)2

= 25π mm² \ Total surface area of the capsule = s1 + s2

= 45 π + 25 π = 70 π

= 70 × 227

10

\ Total surface area of the capsule = 220 mm2. 7. For the cone,

r = 42

= 2 m

l = 2.8 m \ Curved surface area of cone (s1) = πrl

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= π × 2 × 2.8

2.8 m

2.1 m4 m

= 5.6π m2

\ For the cylinder, r = 2 m h = 2.1 m \ Curved surface area of cylinder (s2) = 2πrh = 2 × π × 2 × 2.1 = 8.4 π m² \ Total surface area of the canvas = s1 + s2

= 5.6π + 8.4 π = 14 × π

= 14 × 227

Total surface area of the canvas = 44 m2

Total cost of the canvas = Total surface area × rate \ Total cost of the canvas = 500 × 44 \ Total cost of the canvas = ` 22,000. 8. For the cylinder,

2.4 cm

1.4 cm

h = 2.4 cm

r = 1.42 = 0.7 cm

\ Curved surface area of cylinder (s1) = 2πrh = 2 × π × 0.7 × 2.4 = 3.36 π cm2

\ Area of the base of cylinder = πr2

= π × (0.7)2

= 0.49 π cm2 Now, l2 = r2 + h2

\ l = 0 7 2 42 2. . \ l = 0.49 + 5.76 \ l = 6.25 \ l = 2.5 cm \ Curved surface area (s2) of cone = πrl = π × 0.7 × 2.5 = 1.75π cm²

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\ Total surface area of the remaining solid = s1 + s2 + area of the base = 3.36π + 0.49π + 1.75π = 5.6π

= 5.6 × 227

= 0.8 × 22 = 17.6 cm² \ Total surface area of the remaining solid ≈ 18 cm2. 9. Surface area of the cylinder (s1) = 2πrh

10 mm

= 2 × π × 3.5 × 10 = 70π cm2

\ Surface area of 2 hemispheres (s2) = 2 × 2πr2

= 2 × 2 × π × (3.5)2

= 49π cm2 \ Total surface area of the article = s1 + s2

= 70π + 49π = 119π

= 119 × 227

\ Total surface area of the article = 374 cm2.

Test Yourself – SAV 1 1. 7 : 9 2. 7,920 m2, `79,200 3. 8.88 m 4. 337.45 cm2 5. 440 π cm2 6. 231 cm2

7. 103.62 cm2 8. 5104 cm2 9. 269.3 cm2

Exercise – 5.2 1. Now h = r = 1 cm

\ Volume of cone (V1) =13

πr2h

= 13

π × 1 × 1 × 1

= 13

π cm³

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\ Volume of hemisphere (V2) = 23 πr3

= 23 π × 1 × 1 × 1

= 23

π cm3

\ Volume of the solid = V1 + V2

= 13

π + 23

π

\ Volume of the solid = π cm3.

2. Radius = 32

= 1.5 cm

\ Height of the cylinder = 12 – 2 – 2 = 8 cm \ Volume of cylinder (V1) = π × r2 × h = π × 1.5 × 1.5 × 8 = 18π cm3

\ Volume of the two cones (V2) = 2 13

× πr h²

= 2 × 13 × π × 1.5 × 1.5 × 2

= 2 × 13

× × 32

× 32

× 2π

= 3π cm3

\ Volume of the model = V1 + 2V2

= 18π + 3π = 21π

= 21 × 227

\ Volume of the model = 66 cm3.

3. r = 2 82. = 1.4 cm

h = 5 – (1.4 + 1.4) = 5 – 2.8 = 2.2 cm \ Volume of cylinder = πr2h = π × 1.4 × 1.4 × 2.2 = 4.312π cm3

\ Volume of hemisphere = 23

πr3

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= 32 × π × 1.4 × 1.4 × 1.4 cm3

\ Volume of two hemispheres = 2 × 32

× π × (1.4)3

= 3.66π cm3 \ Volume of one gulab jamun = volume of cylinder + volume of two hemispheres = 4.312π + 3.66π = 7.972π cm3

\ Volume of 45 gulab jamun = 45 × 7.972π

= 45 × 7.97 × 227

= 7 890 3

7, .

= 1,127.14 cm3

\ 30% of the volume of 45 gulab jamun = 30

100 × 1,127.14 = 338.14 cm³

\ 30% of the volume of 45 gulab jamun = 338 cm3 (approx.). 4. ... Volume of the cuboid = 15 × 10 × 3.5 = 525 cm3

... Volume of the 4 cones = 4 × 13 πr2h

= 4 × 13

× 227

× 0.5 × 0.5 × 1.4

= 30.821

= 1.47 cm3

\ Volume of the wood in the entire stand = volume of cuboid – volume of 4 cones = 525 – 1.47 \ Volume of the wood in the entire stand = 523.53 cm3 .

5. Volume of water = volume of cone = 13 πr2h

= 13

× π × 5 × 5 × 8

= 2003

π cm3

\ Now, 14

of the volume of water = 14

2003

= 503π cm3

Lead is in the form of sphere, r = 0.5 cm

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\ Volume of one lead shot (sphere) = 43

3πr

= 43

× π × (0.5)3

=0.53

π cm3

\ Number of lead shots = Volume of waterVolume of one shot

=

53

0.53

0 π

π

= 503

× 305

10

1

10

\ Number of lead shots = 100

6. Radius of the bigger cylinder (r1) = 242

= 12 cm

Volume of the bigger cylinder (V1) = πr12 h

= π × 12 × 12 × 220 = 31,680π cm3

Radius of smaller cylinder (r2) = 8 cm \ Volume of the smaller cylinder (V2) = πr2

2h = π × 8 × 8 × 60 = 3,840π cm3

\ Volume of the pole = V1 + V2

= 31,680π + 3,840π = 35,520π cm3

\ Mass of 1cm3 iron = 8 gm \ Mass of the pole = 35,520 × 3.14 × 8 = 8,92,262.4 g = 892.2624 kg \ Mass of the pole = 892.26 kg. 7. Dimensions of the cylinder EFGH Radius = 60 cm and height = 120 cm + 60 cm = 180 cm \ Volume of the vessel = πr2h = π × (60)2 × 180

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= 6,48,000π cm3. Volume of the solid inserted in the vessel = Volume of conical part + volume of hemispherical part

= 13 πr2h +

23

πr3

= 13 πr2 (h + 2r)

= 13 × π × (60)2 (120 + 2 × 60)

= 3 600 2403

, × π

= 2,88,000π cm3. \ Volume of water left in the cylinder = 6,48,000π – 2,88,000π = 3,60,000π cm3

= 3,60,000(100)3 × 22

7 × 7 92

7.

= 7 92

7.

\ Volume of water left in the cylinder = 1.131 m3 (approx.).

8. Radius of cylinder = 22

= 1 cm

\ Volume of cylinder (V1) = πr2h = π × (1)2 × 8 = 8π cm3

Radius of sphere = 8.52

cm

\ Volume of sphere (V2) =43 πr3

= 43 × π ×

8.52

3

= 43 × π ×

614 1258.

= 2456 524

. π

= 102.354 π cm3 \ Volume of the water it can hold = V1 + V2

= 8π + 102.354π

= 110.354π

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= 110.354 × 3.14 = 346.51 cm3

\ The volume found by the child, that is 345 cm3 is incorrect.

Test Yourself – SAV 2 1. 641.67 cm2, 418 cm2 2. 282.33 cm3

3. 9/20 cm 4. 6 cm 5. 150 6. 9 cm 7. 2,002 cm3 8. 1h 40 min 9. 48,000 10. 189.57 cm3

Exercise – 5.3 Take π = 22

7, unless stated otherwise.

1. Radius of sphere (r1) = 4.2 cm

\ Volume of sphere = 43

πr13

= 43

× π × (4.2)3 cm3

Radius of cylinder (r2) = 6 cm \ Volume of the cylinder = πr2

2h = π × 6 × 6 × h cm3

But volume of the sphere = volume of the cylinder

\ 43 × π × (4.2)3 = π × 6 × 6 × h

\ 4× 4.23 × 6 × 6

3 = h

h = 296 352

108.

\ h = 2.74 m

\ Height of the cylinder = 2.74 cm. 2. For metallic spheres r1 = 6 cm r2 = 8 cm r3 = 10 cm \ Volume of the three spheres = V1 + V2 + V3

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= 43 π(6)3 +

43 π(8)3 +

43 π(10)3

= 43 π (216 + 512 + 1,000)

= 43

π × 1,728 cm³

Let the radius of new sphere be ‘R’ cm But, volume of the bigger sphere = volume of the 3 spheres

\ 43π R3 =

43π × 1,728

\ R = 12 cm \ Radius of the resulting sphere is 12 cm.

3. Let the height of the platform be h. \ Volume of cylinder = volume of platform

\ 72

72

20 = 22 × 14 × h

\ 22 20 1

221

14772

72

5× × × × ×

2= h

\ h = 52

m = 2.5 m

\ Height of the platform is 2.5 m.

4. For well, diameter = 3m \ r = 32

m

For embankment, diameter = 3 + 4 + 4 = 11 m

\ R = 112 m

Volume of the earth taken out = πr2h

= 22 3

21

74

2

Also volume of earth taken = πR2h – πr2h = π (R2 – r2 ) × h

= 227

112

32

2 2

h

But π (R2 – r2 ) × h = 22 3

21

74

2

166


h =

22 32

32

1

227

112

32

2 27

4

=× ×

−

32

32

1214

94

147

632

4112

9 2

168

= 9

8 h = 1.125 m \ Height of the embankment is 1.125 m.

5. Radius of the cylinder (R) = 122

= 6 cm Height of the cylinder = 15 cm \ Volume of the cylinder = πr2h = π × 6 × 6 × 15 = 540 π cm³

Radius of the cone = 62

= 3 cm Radius of hemisphere = 3 cm

\ Volume of the cone = 13

× 2πr h + 23

πr3

= 13

πr2 (h + 2r)

= 13 × π × 32 × (12 + 2 × 3)

= 13

93

18

= 54π cm³

167


\ Number of cones = Volume of cylinder

Volume of cone

= 54054

ππ

= 10 cones \ Tenconescanbefilledwithicecream. 6. Let the number of coins be x

Height of coin (h) = 2mm = 210

cm

Radius of coin (r) = 1 75

2. cm

\ Volume of 1 coin (cylinder) = πr2h \ Volume of x coins= x × π × r2h

= × 227

× 1.752

× 210

cm³2

\ Volume of cuboid = l × b × h = 5.5 × 10 × 3.5 cm³ But volume of x coins = volume of cuboid \ x × πr2h = l × b × h

\ x = 5.5 ×10 × 3.5

227

× 1.752

× 1.752

× 210

\ x = 400 \ 400 coins must be melted. 7. For the cylindrical bucket, h = 32 cm r = 18 cm \ Volume of the cylindrical bucket = πr2h = π × 182 × 32 cm³ For the cone, h = 24 cm

\ Volume of the cone = 13

πr2h

= 13 π r2 × 24 cm³

But volume of the cylinder = volume of the cone

\ π × 18 × 18 × 32 = 13p × r2 × 24

168


\ 183

× 18 × 328

× 32441

= r2

\ r² = 9 × 18 × 8 \ r2 = (3 × 3) × (3 × 3 × 2) × (2 × 4) \ r = 3 × 3 × 2 × 2 \ r = 36 cm \ Radius of the heap = 36 cm Now l² = r2 + h2 \ l² = 362 + 242 \ l² = 1,296 + 576 \ l² = 1872 \ l = 12 13 cm

\ Slant height of the heap = 12 13 cm.

8. Height of the water flowing in 30 min = 102

km

= 5 km = 5,000 m \ Volume of water in the canal (cuboid) = l × b × h = 5,000 × 6 × 1.5 m³ Standing water required = 8 cm

= 8

100 m

Volume = area × height

5,000 × 6 × 1.5 = area × 8100

\ Area = 5,000 × 6 × 1.5 × 1008

= 45 00 0008

, ,

= 5,62,500 m2

= 56.25 hectare ...(1 hectare = 10,000 m²) \ Area irrigated will be 5,62,500m² or 56.25 hectares. 9. Let the time taken to fill the tank be ‘x’ h. For pipe, the length of the water (h1) = 3 × x

169


= 3x km (h1) = 3,000 x m

Radius of pipe (r1) = 202

cm = 10 cm

= 10100

m = 1

10m

But radius of cylindrical tank (r2) =102

= 5 m height of cylindrical tank (h2) = 2 m But volume of the pipe = volume of the tank \ πr h1

21 = πr h2

22

\ 1

10

2

× 3,000x = 5² × 2

\ x = 5 × 5 × 2 ×10 ×10

30 0 06

\ x = 106 h

\ x = 106

× 6010

min ...... (... 1h = 60 min)

\ x = 100 min \ Thetankwillbefilledupin100min.

Test Yourself – SAV 3 1. 4 cm 2. 1,000

3. 1 cm 4. 813

mm

5. 51.2 min 6. 21 cm 7. 120 min 8. 2.885 cm 9. 1.5 m 10. 12 m

170


Exercise – 5.4 Use π = 22

7, unless stated otherwise.

1. r1 = 42

= 2 cm

r2 =22

= 1 cm h = 14 cm

\ Volume of frustum = 13

πh (r12 + r2

2 + r1r2)

= 13× 22

7 × 14 (22 + 12 + 2 × 1)

= 13

× × ×227

14 72

= 3083

= 102 23 cm3

\ Capacity of the glass =102 23

cm3.

2. Let the radius of the upper part be r1 and of lower part be r2. \ 2πr1 = 18

\ r1 = 182π

\ r1 = 9π

cm

and 2πr2 = 6

\ r2 = 62π

\ r2 = 3π

cm

Curved surface area of the frustum = π (r1 + r2) l

= π 9 3

+

× 4

\ Curved surface area of the frustum = 48 cm²

171


3. r1 = 10 cm r2 = 4 cm l = 15 cm One part of the cap is open, so cloth is not needed there. \ Area of material = πl (r1 + r2) + πr2

2

=227

× 15 10 + 4 + 227

× 4²( ) (\ The cap is open from the base)

= 227 × 15 × 14 +

227 × 16

= 660 + 3527

= 4972

7

\ Area of material used for making it = 71027cm2 .

4. Slant height

\ l = h r r21 2

2+ = 16 + 20 82 2 = 256 + 144

= 400

= 20 cm.

\ Volume of the frustum = 13

πh (r12 + r2

2 + r1r2)

= 13

× 3.14 × 16 (82 + 202 + 8 × 20)

= 13

× 3.14 × 16 (64 + 400 + 160)

= 3.14 × 16 × 6243

= 31,349.76

3

= 10,449.92 cm³ = 10.44992 l ..... (... 1 l = 1000 cm³) = 10.45 l \ Cost of milk = 10.45 × 20 \ Cost of milk = `209

172


\ Surface area of the frustum = πl (r1 + r2) + πr12+ πr²

= π × 20 (20 + 8) + π × 8 × 8 + 0 .... (... the container is open from the top) = π × 560 + π × 64 = 3.14 (560 + 64) = 3.14 × 624 = 1,959.36 cm2

\ Cost of metal sheet = ` 8100

× 1,959.36

\ Cost of metal sheet = ` 156.75.

5.

In ∆ ABC, In ∆ AED, ∠DAE = 30°

∠BAC =602

= 30° \ tan 30° = DEAE

tan 30° =BCAC

\ 13

2

20= r

\ 13=

r110 \ r2 =

203

cm

\ r1 =10

3 cm

\ Volume of the metal in the frustum = 13

πh (r12 + r2

2 + r1r2)

=13

× π × 10 10

3

2 2

+ 203

+ 103

203

173


= 13

× π × 10 1003

+ 4003

+ 2003

=13

× π × 10 × 7003

= 7,0009

π cm3

Let the length of the wire = h cm, radius of the wire =

1162

132

= cm

\ Volume of wire = πr²h = π × 132

×132

× h cm3

\ Volume of frustum = volume of wire

\ 7 0009

, π = π × 132

132

× × h

\ h = 7 000 32 329

, × × cm =7 000 32 32

9 100, × ×

× m

\ h = 71 680

9,

m \ h = 7,964.4 m

\ Length of the wire = 7964.4 m.

Test Yourself – SAV 4 1. 2941.71 cm3. 2. `294.17 3. 48510 cm3. 4. 520 π cm2.

5. 2418π cm2. 6. 135 23

cm3.

7. `156.75, `97.97 (apporox.) 8. 30 cm 9. 1,554 cm3 10. 48 cm2; ` 9.58

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EXERCISE – 5 - livehindustan.com...A solid consisting of a right circular cone of height 120 cm and radius 60cm standing on a hemisphere of radius 60 cm is placed upright in a right