Surface Areas And Volumes

The Nip of a sharpened pencil is in conical shape and the rest of the part is cylindrical therefore pencil is a combination of cylinder and a cone.

The top part of surahi is in cylindrical shape and bottom part is in spherical shape therefore surahi is a combination of Sphere and a cylinder.

The upper part of plumbline is hemispherical, and the bottom part is conical therefore, it is a combination of hemisphere and cone.

We know that, the shape of frustum of a cone is

So, the shape of glass is a frustum [ an inverted frustum]

As the left and right part of a gilli are conical and the central part is cylindrical

Therefore,

It is a combination of a cylinder and two cones.

The cork of a shuttle is in hemispherical shapes and the upper part is in the shape of frustum of a cone. Therefore, it is a combination of frustum of a cone and a hemisphere.

When a cone is divided into two parts by a plane through any point on its axis parallel to its base, the upper and lower parts obtained are cone and a frustum respectively.

Given,

Internal edge of cube, a = 22 cm

As we know,

Volume of cube = a³,

where a = side of cube, we have

Volume of given hollow cube = a³

= (22)³ = 10648 cm³

As of the cube remains unfilled, only of cube remains filled

Volume of Filled cube = 7/8 of total volume = 7/8×10648 = 9317 cm³

Now,

Diameter of marble, D = 0.5 cm

Radius of marble, r = D/2 = 0.5/2 = 0.25 cm

Volume of one marble = volume of sphere of radius r

Volume of one marble,

[As volume of sphere

∴

No of total marbles

No. of marbles

= 142,244.275

= 142,244 (appx)

Volume of spherical shell = Volume of cone recast by melting [1]

For Spherical Shell,

Internal diameter, d₁ = 4 cm

Internal radius, r₁ = 2 cm

[ as radius = 1/2 diameter]

External diameter, d₂ = 8 cm

External radius, r₂ = 4 cm

Now,

As volume of spherical shell

where r₁ and r₂ are internal and external radii respectively.

volume of given shell

Using [1] we have

Volume of cone = 224π /3 cm³

For cone,

Base diameter = 8 cm

Base radius, r = 4 cm

[ as radius = 1/2 diameter]

Let Height of cone be 'h'.

As,

volume of cone =

where r = Base radius and h = height of cone

Volume of given cone

⇒ 16h = 224

h = 14 cm

So, Height of cone is 14 cm.

As we know,

Volume of cuboid = lbh

where, l = length, b = breadth and h = height

For given cuboid,

Length, l = 49 cm

Breadth, b = 33 cm

Height, h = 24 cm

Volume of cube = 49×(33) × (24) cm³

Now,

Let the radius of cube be r.

As volume of sphere

where r = radius of sphere

Also,

Volume of cuboid = volume of sphere moulded

⇒

⇒ πr³ = 29106

⇒

⇒ r = ∛ 9261 cm = 21 cm

Hence, radius of sphere is 21 cm

Volume of cuboid = lbh

where, l = length, b = breadth and h = height

For wall,

Length, l = 270 cm

Breadth, b = 300 cm

Height, h = 350 cm

Volume of wall = 270 × (300) × (350) = 28350000 cm³

As of this volume is covered by mortar, therefore of this volume is covered by bricks.

Volume of wall covered by bricks = 7/8 (volume of wall)

= 7/8 (2835000)

= 24806250 cm³

For one brick,

Length, l = 22.5 cm

Breadth, b = 11.25 cm

Height, h = 8.75 cm

Volume of one brick = lbh

= 22.5(11.25)(8.75) cm³

As we know,

Volume of cylinder = πr²h

where r = base radius

h = height of cylinder

For Given Solid cylinder,

Base diameter = 2 cm

Base radius, r = 1 cm [as radius = 1/2 diameter]

Height, h = 16 cm

Volume of cylinder, V = π(1)²(16) = 16π cm³

Now,

Let the radius of sphere be r

Volume of sphere

where r = radius

Volume of 12 spheres of radius r =

Now,

Volume of 12 spheres = volume of cylinder

[As volume remains same, when a metal body is melted and recast into another metal body]

⇒ 16πr³ = 16π

⇒ r³ = 1

⇒ r = ∛1 = 1 cm

Diameter = 2r = 2 cm

Hence, diameter of each sphere is 4 cm.

Clearly, the given bucket is in the form of frustum of a cone.

And we know

Curved surface area of frustum of a cone = πl(r₁ + r₂)

Where, l = slant height,

r₁ and r₂ are radii (r₁ > r₂)

For given bucket,

Radius of top, r₁ = 28 cm

Radius of bottom, r₂ = 7 cm

Slant height, l = 45 cm

Therefore,

Curved surface area of bucket = π(45)(28 + 7)

= 4950 cm²

So, the curved surface area of bucket is 4950 cm^2.

Capacity of capsule = Volume of 2 hemispherical part + volume of cylindrical part

Now,

Diameter of capsule = 0.5 cm

Radius of capsule, r = 0.25 cm

[as radius = 1/2 diameter]

Also,

Radius of hemispherical part = Radius of cylindrical part = r

Height of entire capsule = 2 cm

Let the height of cylindrical part be h.

Then

Height of entire capsule = radius of 2 hemispherical parts + height of cylindrical part

⇒ 2 = 2r + h

⇒ 2 = 2(0.25) + h

⇒ h = 1.5 cm

Volume of capsule

Volume of hemisphere of radius, r = (2/3) πr³

And volume of cylinder = πr²h,

where r = base radius,

h = height of cylinder

Volume of capsule

When two hemispheres are joined together along their bases, a sphere of same base radius is formed.

And Curved surface area of sphere is 4πr².

As the sphere just encloses in a cylinder,

The diameter of sphere will be equal to diameter of cylinder.

Diameter of base of cylinder = 2(radius of base) = 2r

So, Dance of sphere = 2r cm

When a solid is converted from one shape to another, the volume of new shape remains constant.

Clearly, the given bucket is in the form of frustum of a cone.

And we know

Where, h = height, r₁ and r₂ are radii (r₁ > r₂)

For given bucket,

Diameter of top = 44 cm

Radius of top, r₁ = 22 cm

[as radius = 1/2 diameter]

Diameter of bottom = 24 cm

Radius of bottom, r₂ = 12 cm

[as radius = 1/2 diameter]

height, h = 35 cm

Therefore,

= 32706.67 cm³

As,

32706.67 cm³ = 32.7 L [appx]

So, The volume of bucket is 32.7 L^.

As we know, The cross section when a plane is sliced through a cone is a circle.

Let the radius of spheres be r₁ and r₂ and we know that

Given,

Ratio of volumes is 64 : 27

…(1)

Now,

Surface area of sphere = 4πr²

where r = radius

So, ratio of surfaces areas is 16 : 9

False

When two hemispheres are joined together along their bases, a sphere of same base radius is formed.

And CSA of sphere is 4πr².

False

As one cylinder is placed over another, so the base of first cylinder and top of other cylinder will not be covered in total surface area.

So,

Total surface area of shape formed = 2(Total surface of single cylinder) - 2(Area of base of cylinder)

= 2(2πrh + 2πr²) - 2(πr²)

= 4πrh + 2πr²

[As Total surface area of cylinder = 2πrh + 2πr²h, where r = base radius and h = height]

False

When a solid cone is placed over a solid cylinder of same base radius, the base of cone and top of the cylinder will not be covered in total surface area.

As the height of cone and cylinder is same

Total surface area of shape formed = Total surface area of cone + Total Surface area of cylinder - 2(Area of base)

[ As Total surface area of cone = πrl + πr²,

where r = base radius and l = slant height and

Total surface area of cylinder = 2πrh + 2πr²h,

where r = base radius and h = height]

= πr(r + l) + (2πrh + 2πr²) - 2(πr²)

= πr² + πrl + 2πrh + 2πr² - 2πr²

= πr(r + l + h)

[ As slant height, ]

False

Let the radius of sphere be r

As the ball is exactly fitted into the cubical box.

Diameter of ball = Edge length of cube

2r = a

[As Diameter = 2(Radius)]

As we know,

where r is radius

False

As we know

Volume of frustum of a cone

Where, h = height, r₁ and r₂ are radii (r₁ > r₂)

True

The capacity of given shape = volume of cylinder - volume of hemispherical portion

And the base radius of cylinder will be equal to radius of hemisphere.

Now, we know

Volume of a cylinder = πr²h,

where r = base radius and h = height of cylinder

And

Volume of hemisphere

Capacity of given shape

So, Given statement is true.

False

We know that,

Curved surface area of frustum = πl(r₁ + r₂)

Where, r₁ and r₂ are the radii of two ends (r₁ > r₂)

And l = slant height and

[ In the statement, formula for calculating slant height is incorrect]

True

As the resulting shape will be as in figure,

And the area of base of frustum, top of frustum and top of cylinder will not be counted, as no metal sheet is used there.

So required area will be = Curved surface area of frustum + area of base of cylinder + curved surface area of cylinder

We know that,

Volume of cube = a³,

where a = side of cube

Now,

Side of first cube, a₁ = 3 cm

Side of second cube, a₂ = 4 cm

Side of third cube, a₃ = 5 cm

Now, Let the side of cube recast from melting these cubes is 'a'.

As the volume remains same,

Volume of recast cube = (volume of 1^st + 2^nd + 3^rd cube)

⇒ a³ = a₁³ + a₂³ + a₃³

⇒ a³ = (3)³ + (4)³ + (5)³

⇒ a³ = 27 + 64 + 125 = 216

⇒ a = 6 cm

So, side of cube so formed is 6 cm.

Volume of cuboid = lbh

where, l = length, b = breadth and h = height

For cuboidal lead:

Length, l = 9 cm

Breadth, b = 11 cm

Height, h = 12 cm

Volume of lead = 9(11)(12) = 1188 cm³

Volume of sphere

where r = radius of sphere

For spherical shots,

Diameter = 3 cm

Radius, r = 1.5 cm

[As radius = diameter/2]

Volume of one shot

Now,

So, 84 bullets can be made from lead.

Clearly, Given bucket is in the form of frustum of a cone.

As we know

Volume of frustum of a cone

Where, h = height, r₁ and r₂ are radii of two ends (r₁ > r₂)

For given bucket,

Volume of bucket = 28.490 L

Volume of bucket = 28490 cm³

[As, 1 L = 1000 cm³ ]

Radius of top, r₁ = 28 cm

Radius of bottom, r₂ = 21 cm

Let the height be h.

Using these values,

We have.

Let ORN be the cone, when this is cone is divided into two parts by a plane through the mid-point of its axis parallel to its base, the upper and lower parts obtained are cone and a frustum respectively.

Given,

Height of cone = OM = 12 cm

As, the cone is divided from mid-point, let P be the mid-point of cone, then

OP = PM = 6 cm

Now, In △OPD and △OMN

∠POD = ∠POD [Common]

∠OPD = ∠OMN [Both 90°]

So, By Angle-Angle similarity criterion

△OPD ~ △OMN

And

[Similar triangles have corresponding sides in equal ratio]

[MN = 8 cm = radius of base of cone]

Now, For First part i.e. cone

Base Radius, r = PD = 4 cm

Height, h = OP = 6 cm

As we know, volume of cone for radius r and height is

Now, For second part, i.e. Frustum

Bottom radius, r₁ = MN = 8 cm

Top Radius, r₂ = PD = 4 cm

Height, h = PM = 6 cm

As we know

Volume of frustum of a cone

Where, h = height, r₁ and r₂ are radii

(r₁ > r₂)

So,

Volume of first part : Volume of second part = 32π : 224π = 1 : 7

Let 'a' be the side of one cube.

As two cubes are joined together, the surfaces that are joined together will not be included in the surface area of resulting cuboid.

So,

Surfaces area of resulting cuboid = 2(Total surface area of a cube) - 2(area of single surface)

⇒ Surfaces area of resulting cuboid = 2(6a²) - 2(a²) = 10a²

[As total surface area of cube = 6a² , where a = side of cube]

Also,

Volume of cube = a³

64 = a³

[Given volume of cube is 64 cm³ ]

a = 4 cm

Therefore,

Surface area of resulting cuboid = 10a² = 10(4)² = 160 cm²

Since the conical cavity is hollowed out from the cube,

Volume of remaining solid = volume of cube - volume of cone

Now,

For Cube

Side, a = 7 cm

Volume of cube = (7)³ = 343 cm³

[ As Volume of cube = a³,

where a = side of cube]

For cone

Radius, r = 3 cm

Height, h = 7 cm

Volume of cone

[As volume of cone

where r = radius and h = height

Volume of remaining solid = 343 - 66 = 277 cm³

The diagram is:

Diameter of Cylinder = 6 cm

Radius of cylinder = r = 3 cm

[As radius = diameter/2]

As both cones have equal radius

Radius of cone A = radius of cone B = r = 3 cm

Let the height of cone A be h₁ and Cone B be h₂

Given,

Ratio of volume of cones is 2 : 1

i.e.

As volume of cone =

where r = base radius and h = height

⇒

⇒ h₁ = 2h₂

Now,

Total height of cylinder is 21 cm

h₁ + h₂ = 21

2h₂ + h₂ = 21

3h₂ = 21

h₂ = 7 cm

h₁ = 2h₂ = 2(7) = 14 cm

We know,

Volume of cylinder = πr²h ,

where r = radius and h = height

Volume of given cylinder = π(3)²(21)

Volume of remaining solid = (Volume of cylinder) – (volume of cone A) – (volume of cone B)

= 594 - 132 - 66 = 396 cm³

As ice cream is a combination of hemisphere and cone

Radius of cone = radius of hemisphere = r = 5 cm

Height of cone, h = 10 cm

Volume of ice cream = volume of hemispherical part + volume of conical part

[ As Volume of cone = ,

where r is radius and h is height

And

volume of hemisphere ,

where r is radius]

Now, part of is left unfilled i.e. part is filled with ice-cream

Therefore,

Let x no of marbles are dropped, so that water level rises by 5.6 cm.

The increase in volume of water in beaker = Volume of x marbles.

Now,

Required raise in height, h = 5.6 cm

Diameter of beaker = 7 cm

Radius of beaker, r = 3.5 cm

[Radius = diameter/2]

Required increase in volume = volume of cylinder of above dimensions = πr²h

[As volume of cylinder = πr²h,

where r = Base radius and h = height]

Required increase in volume = π(3.5)²(5.6) cm³

Now, As diameter of marble is 1.4 cm

Radius of marble, r = 0.7 cm

[As radius = diameter/2]

So, we have,

Therefore, 150 marbles are required.

Volume of cuboid = lbh where, l = length, b = breadth and h = height

For cuboidal lead ;

Length, l = 66 cm

Breadth, b = 42 cm

Height, h = 21 cm

Volume of lead = 66(42)(21) = 58212 cm³

For spherical shots,

Diameter = 4.2 cm

Radius, r = 2.1 cm

Now,

So, 1500 bullets can be made from lead.

Volume of cube = a³ where, a = side of cube

For cubical lead ;

side, a = 44 cm

Volume of lead = 44(44)(44) cm³

For spherical shots,

Diameter = 4 cm

Radius, r = 2 cm

[As radius = diameter/2]

Now,

So, 2541 bullets can be made from lead.

Volume of cuboid = lbh,

where, l = length, b = breadth and h = height

For wall,

Length, l = 24 m

Breadth, b = 0.4 m

Height, h = 6 cm

Volume of wall = 24(0.4)(6) = 57.6 m³

As of this volume is covered by mortar, therefore of this volume is covered by bricks

For one brick,

Length, l = 25 cm = 0.25 m

[As 1 m = 100 cm]

Breadth, b = 16 cm = 0.16 m

Height, h = 10 cm = 0.10 m

Volume of one brick = lbh = 0.25(0.16)(0.10) m³

Hence Total No of bricks used in constructing the wall is 12960.

For Cone,

Base Diameter = 4.5 cm

Base radius, r = 2.25 cm

[as radius = diameter/2]

Height of cone, h = 10 cm

We know that,

Where, r is base radius and h is the height of the cone.

So,

Now, Disk is in shape of a cylinder

For a single disk

Base diameter = 1.5 cm

Base radius, r = 0.75 cm

[as radius = diameter/2]

Height, h = 0.2 cm

As we know,

Volume of a cylinder = πr²h

Where r is base radius and h is the height of cylinder

Volume of a disk = π(0.75)²(0.2) = (0.75 × 0.75 × 0.2) π cm³

Now,

So, 150 disks can be made by melting the cone.

When two identical cones are joined along their base, the shape so formed will be as shown in figure

So,

Total surface area of shape formed = Curved area of first cone + Curved surface area of second cone

= 2(Surface area of cone)

[since, both cones are identical]

Now, As we know,

Surface area of cone = πrl,

where r = radius and l = slant height

Total Surface area of shape so formed = 2πrl

Given,

Radius, r = 8 cm

Height, h = 15 cm

So,

Area = 2(3.14)(8)(15) = 753.6 cm²

For hemisphere,

Radius, r = 8 cm

As we know,

Volume of hemisphere

Where, r = radius of hemisphere

Volume of given hemisphere

For cone that is recast from hemisphere,

Base radius, r = 6 cm

We know that,

volume of cone

Where, r is base radius and h is the height of the cone.

Now,

Volume of cone

As the volume remains same, when a body is reformed to another body

Volume of cylinder = Volume of cone

12πh = 1024π /3

h = 28.44 cm

Volume of water in tank = volume of cuboidal tank up to a height of 5 m

For cuboidal tank

Length, l = 11 m

Breadth, b = 6 m

Height, h = 5m

We know that,

Volume of tank = lbh

Where, l, b and h are the length, breadth and height of tank respectively

Volume of water = 11(6)(5) = 330 m³

Also,

Let the cylindrical tank is filled up to a height of h m

Base radius of cylindrical tank, r = 3.5 m

As we know,

Volume of a cylinder = πr²h

Where r is base radius and h is the height of cylinder

Volume of water in cylindrical tank = π(3.5)²h

h = 8.57 m

So, cylindrical tank is filled up to a height of 8.57 m

Let the length breath and height be the external dimension of an open box and thickness be

The volume of metal used in box = Volume of external box - Volume of internal box

For external box,

Length, l = 36 cm

Breadth, b = 25 cm

Height, h = 16.5 cm

We know that,

Volume of cuboid = lbh

Where, l, b and h are the length, breadth and height of tank respectively

Volume of external box = 36(25)(16.5) = 14850 cm³

Now,

As the box is open from top,

For internal box,

Length, l' = Length of external box - 2(thickness of box) = 36 - 2(1.5) = 33 cm

[i.e. the thickness of two sides is deduced]

Breadth, b' = Breadth of external box - 2(thickness of box) = 25 - 2(1.5) = 22 cm

[i.e. the thickness of two sides is deduced]

Height, h' = Height of external box - thickness of box = 16.5 - 1.5 = 15 cm

[i.e. the thickness of bottom is reduced]

Volume of internal box = 33(22)(15) = 10890

So,

Volume of metal in box = 14850 - 10890 = 3960 cm³

Also,

1 cm³ weighs 7.5 g

3960 cm³ weighs 3960(7.5) = 29,700 g

So,

The weight of box is 29,700 g i.e. 29.7 kg

Let us first calculate the volume of barrel of pen that is of cylindrical shape

For barrel,

Base diameter = 5 mm = 0.5 cm [As 1 cm = 10 mm]

Base radius, r = 0.25 cm

[as radius = diameter/2]

Height, h = 7 cm

As we know,

Volume of a cylinder = πr²h

Where r is base radius and h is the height of cylinder

volume of barrel

⇒ Volume of barrel

So,

1.375 cm³ of ink can write 3300 words

No of words that can be written by 1 cm^{^3}

[ Also of a liter i.e. 0.2 L and we know

1 L = 1000 cm³

0.2 L = 200 cm³ ]

No of words that can be written by 200 cm³ = 2400(200) = 480000 words

So,

One-fifth of a liter ink can write 480000 words on an average.

Let the time taken by pipe to fill vessel is t minutes

As water flows 10 m in 1 minute, it will flow 10t meters in t minutes.

Also,

Volume of conical vessel = Volume of water that passes through pipe in t minutes

Now, For conical pope

Base Diameter = 40 cm

Base radius, r = 20 cm

[as radius = diameter/2]

Height, h = 24 cm

We know that,

Where, r is base radius and h is the height of the cone.

Volume of conical vessel

For cylindrical pipe

Base diameter = 5 mm = 0.5 cm

[As 1 cm = 10 mm]

Base radius, r = 0.25 cm

[as radius = diameter/2]

Height, h = 10t m = 1000t cm

[∵ 1 m = 100cm]

[As water covers 10t m distance in pipe]

As we know,

Volume of a cylinder = πr²h

Where r is base radius and h is the height of cylinder

Volume of water passed in pipe = π(0.25)²(1000t) = 62.5tπ cm³

So, we have

62.5tπ = 3200

62.5t = 3200

t = 51.2 minutes

t = 51 minutes 12 seconds

[ as 0.2 minutes = 0.2(60) seconds = 12 seconds]

Given for conical heap,

Base Diameter = 9 cm

Base radius, r = 4.5 cm

[as radius = diameter/2]

Height, h = 3.5 cm

Slant height,

We know that,

Volume of cone

Where, r is base radius and h is the height of the cone.

Volume of rice = Volume of conical heap

Volume of rice

Also,

Canvas requires to just cover heap = Curved surface area of conical heap

And we know,

Curved surface area of a cone = πrl

Where r is base radius and l is slant height.

Canvas required = π(4.5)(5.7) = 80.61 cm² [appx]

The pencil is in shape of cylinder.

Let the radius of base be r cm

Circumference of base = 1.5 cm

2πr = 1.5 cm

[As Circumference of circle is 2πr]

Also, Height, h = 25 cm

As we know,

Curved surface area of cylinder = 2πrh

Where r is base radius and h is height

1 cm = 0.1 dm

1 cm² = 0.01 dm²

37.5 cm² = 0.375 dm²

Cost for coloring 1 dm² = Rs. 0.05

Cost for coloring 0.375 dm² (i.e. 1 pencil) = Rs. 0.01875

Cost for coloring 120000 pencils = 120000 ×0.01875 = Rs. 2250

Let the time taken by pipe to fill pond is t hours

As water flows 15 km in 1 hour, it will flow 15t meters in t hours.

Also,

Volume of cuboidal pond up to height 21 cm = Volume of water that passes through pipe in “t” hours

Now, For cuboidal pond

Length, l = 50 m

Breadth, b = 44 m

Height, h = 21 cm = 0.21 m

We know that,

Volume of tank = lbh

Where, l, b and h are the length, breadth and height of tank respectively

Volume of water = 50(44)(0.21) = 462 m³

For cylindrical pipe

Base diameter = 14 cm

Base radius, r = 7 cm = 0.07 m

[as radius = diameter/2]

Height, h = 15t km = 15000t m

[1 km = 1000 m]

As we know,

Volume of a cylinder = πr²h

Where r is base radius and h is the height of cylinder

Volume of water passed in pipe = π(0.07)²(15000t)

= 231t cm³

So, we have

231t = 462

t = 2 hours

Time required to fill tank up to a height of 25 cm is 2 hours.

For cuboidal block

Length, l = 4 m

Breadth, b = 2.6 m

Height, h = 1 m

We know that,

Volume of tank = lbh

Where, l, b and h are the length, breadth and height of tank respectively

Volume of cuboid = 4.4(2.6)(1) = 11.44 m³

Also,

As the volume remains same when a body is recast to another body.

We have

Volume of cylindrical pipe = 11.44 m³

Now, For pipe,[i.e. hollow cylinder]

Internal radius, r₂ = 30 cm = 0.3 m

Thickness = 5 cm

External radius, r₁ = Internal radius + thickness = 30 + 5 = 35 cm = 0.35 m

Let the length of pipe be h

Also, we know

Volume of a hollow cylinder = πh(r₁² - r₂²), as shown below:

Where h is height and r₁ and r₂ are external and internal diameters respectively.

So, we have

Volume of pipe = πh((0.35)² - (0.3)²)

So, the length of pipe is 112 m.

Let the rise of water level in the pond be h meters, when 500 persons are taking a dip into a cuboidal pond.

Given that,

Average displacement by a person = 0.04 m³

Average displacement by 500 persons = 500 ×0.04 = 20 m³

So, the volume of water raised in pond is 20 m³

Also,

Length of pond, l = 80 m

Breadth of pond , b = 50 m

Height is h

And we know,

Where, l, b and h are the length, breadth and height of tank respectively

Volume of water raised in pond = 80(50)(h)

20 m³ = 4000h

h = 0.005 m = 0.5 cm

[As 1 m = 100 cm]

So, Raise in water height is 0.5 cm.

For one glass sphere

Radius , r = 2 cm

As we know

Where, r is radius of Hemisphere

So ,

For cuboidal box,

Length, l = 16 cm

Breadth, b = 8 cm

Height, h = 8 cm

As,

Volume of cuboid = lbh where l, b and h are length, breadth and height respectively.

Volume of cuboid box = 16(8)(8) = 1024 cm³

Volume of water = Volume of Cuboid - Volume of 16 spheres

Volume of water = 1024 - 16(33.52) = 487.6 cm³

So, Volume of water in bucket is 487.6 cm³

Clearly, Given container is in the form of frustum of a cone.

As we know

Volume of frustum of a cone

Where, h = height, r₁ and r₂ are radii of two ends (r₁ > r₂)

For given bucket,

Radius of lower end, r₂ = 8 cm

Radius of upper end, r₁ = 20 cm

Height of container, h = 16 cm

Using these values,

We have.

⇒Volume = 1/3 × 22/7 × 16 × (64 + 400 + 160)

⇒Volume = 10459.42 cm³

As 1000 cm³ = 1 L

10459.42 cm³ = 10.45942 L = 10.46 L [Appx]

Also,

Cost for 1 L milk = 22 Rs

Cost for 10.46 L milk = 22(10.46) = 230.12 Rs

For cylindrical bucket,

Radius, r = 18 cm

Height, h = 32 cm

As we know,

Volume of cylinder = πr²h

Where r is base radius and h is height of cylinder.

Volume of sand in bucket = π(18)²(32) cm³

Also, For conical heap

Let the radius be r and height, h = 24 cm is given,

As we know,

Where r is base radius and h is height of cone

As the volume of sand is constant

Volume of sand in bucket = Volume of conical heap

π (18)²(32) = 8πr²

(18)(18)(4) = r²

r = 18(2) = 36 cm

Also, we know

l² = h² + r², where h , r and l are height radius respectively.

l² = (24)² + (36)² = 576 + 1296 = 1876

l = 43.267 cm

So, radius and slant height of heap are 36 cm and 43.267 cm respectively.

The diagram is given as:

For upper conical part,

Radius of base, r = 3 cm

Slant height, l = 5 cm

As,

l² = h² + r², where h , r and l are height radius respectively.

h² = l² - r²

⇒ h² = (5)² - (3)²

h²= 25 - 9 = 16

h = 4 cm

Also,

volume of cone

Curved surface area of cone = πrl = π(3)(5) = 15π cm²

For cylindrical part,

Radius of base = Radius of base of conical part = r = 3 cm

Height, h = 12 cm

Also,

Volume of cylinder = πr²h = π(3)²(12) = 108π cm³

Curved surface area of cylinder = 2πrh = 2π(3)(12) = 72π cm²

Volume of rocket = volume of conical part + volume of cylindrical part

Volume of rocket = 12π + 108π = 120π

Also,

Surface area of rocket = Curved surface area of conical part + Curved surface area of Cylindrical part + Surface area of base of rocket

Surface area of base of rocket = πr² = π(3)² = 9π cm²

Therefore,

Surface area of rocket = 15π + 72π + 9π = 94π cm²

Let the radius of surmounted hemispherical dome = r

Diameter of hemispherical dome = 2r

Given,

Total height of dome = 2r

Height of hemispherical part = Radius of hemispherical part = r

Height of cylindrical part = r

As we know,

Volume of cylinder = πr²h

Where r is base radius and h is height of cylinder.

Volume of cylindrical part = πr²r = πr³ cm³

Also,

, where r is radius of hemisphere.

Volume of conical part

Volume of building = Volume of cylindrical part + volume of conical part

Volume of building

Volume of air in building = volume of building

r³ = 8

r = 2 m

Height of building = 2r = 2(2) = 4 meter

Let n bottles are needed to empty the bowl,

Therefore,

Volume of bowl = Volume of 'n' bottles

For hemispherical bowl,

Radius, r = 9 cm

As we know,

Volume of a hemispherical bowl

Where r is the radius of hemisphere

Volume of bowl

For single cylindrical bottle,

Base radius, r = 1.5 cm

Height, h = 4 cm

As we know,

Volume of cylinder = πr²h

Where r is base radius,

h is height of cylinder.

Volume of one bottle = π(1.5)²(4) = 9π cm³

As,

Volume of bowl = volume of n bottles

486π = (9π)n

Hence, 54 bottles are needed to empty the bowl.

Whenever we placed a solid right circular cone in a right circular cylinder with full of water, then volume of a solid right circular cone is equal to the volume of water felled from the cylinder and Total volume of water in a cylinder is equal to the volume of the cylinder.

Therefore, we have,

Volume of water left in the cylinder = (Volume of cylinder) – (Volume of cone)

For cylinder,

Base radius, r = radius of cone = 60 cm

Height, h = 180 cm

As we know,

Volume of cylinder = πr²h

Where r is base radius and h is height of cylinder.

Volume of cylinder = π(60)²(180)

For cone

Base radius, r = 60 cm

Height, h = 120 cm

As we know,

Where r is base radius and h is height of cylinder.

So,

Volume of water left in cylinder = 2036571.43 - 452571.43 = 1584000 cm³

Water flows in 1 sec = 80 cm

Water flows in 1/2 hour

[As 1 hour = 3600 seconds]

For cylindrical pipe,

Base radius, r = 1 cm

Height, h = water flowed in half hour = 144000 cm

As we know,

Volume of cylinder = πr²h

Where r is base radius and h is height of cylinder.

Volume of water flowed through pipe = π(1)²(144000) = 144000π cm³

For cylindrical tank,

Base radius, r = 40 cm

Let the height of water raised be h cm

As we know,

Volume of cylinder = πr²h

Where r is base radius and h is height of cylinder.

Volume of water in tank = π(40)² h

144000π = 1600πh

h = 90 cm

Height of water raised in tank is 90 cm.

Let the rainfall be 'x' cm [i.e. 0.01x meters because 1 m = 100 cm ] .

For cylindrical vessel,

Diameter of base = 2 m

Base Radius = 1 m

[As radius = diameter/2]

Height, h = 3.5 m

As we know,

Volume of cylinder = πr²h

Where r is base radius and h is height of cylinder.

So,

Volume of cuboidal vessel = π(1)²(3.5)

Also,

For roof

Length, l = 22 m

Breadth, b = 20 m

Height, h = height of rainfall = 0.01x m

As we know,

Volume of cuboid = lbh

Where l, b and h are length, breadth and height of the cuboid respectively.

Volume of water on roof = 22(20)(0.01x) = 4.4x m³

Given,

Volume of water on roof = volume of cuboidal vessel

4.4x = 11

x = 2.5 cm

Height of rainfall is 2.5 cm.

Given,

For cuboidal stand,

Length, l = 10 cm

Breadth, b = 5 cm

Height, h = 4 cm

We know that

Volume of a cuboid = lbh

Where l, b and h are length, breadth and height respectively.

So,

Volume of cuboidal stand = 10(5)(4) = 200 cm³

For one conical depression,

Radius, r = 0.5 cm

Height, i.e. depth, h = 2.1 cm

We know that

Where r is base radius and h is the height of the cone

For Cubical depression,

Side, a = 3 cm

We know that

Volume of cube = a³, where a is the side of the cube.

Volume of cubical depression = (3)³ = 27 cm³

Volume of wood in the entire stand = volume of cuboidal stand - volume of 4 conical depression - volume of one cubical depression.

Volume of wood = 200 - 4(5.5) - 27 = 170.8 cm³