Surface Areas And Volumes

Sum of volumes of balls that can be made from a large sphere will have same volume, as that of the sphere.

As we know that, volume of a sphere = . Given that, radius of large sphere, r₁ is 8 cm and that of the small spheres, r₂ is 1 cm.

Therefore, number of small balls that can be made from the big sphere =

spheres

Sum of volumes of spherical bullets that can be cast from a rectangular block of metal will have same volume, as that of the block.

As we know that, volume of the rectangular block of metal=lbh. Also, volume of spherical bullets =

Given that, dimension of rectangular block of metal is 11 dm × 1 m × 5 dm = 110 cm × 100 cm × 50 cm and diameter of spherical bullets = 5cm (radius = = cm)

Therefore, number of small bullets that can be made from the block =

Given that, radius of spherical ball = 3 cm

We know that, Volume of the sphere =

So, its volume, v =

The spherical ball of radius 3 cm is melted and recast into three spherical balls.

Let the radius of the third ball be r.

Then, volume of third spherical ball =

Volume of first ball =

Volume of second ball =

The volume of the original spherical ball is equal to that of the total volumes of the three balls.

⇒ r = 2.5 cm

Thus, diameter = 5 cm

Given: 2.2 dm³ of brass is to be drawn into a cylindrical wire 0.25 cm in diameter.

Diameter of cylindrical wire = 0.25cm

Radius of wire, r =

Let length of wire be h

Volume of the cylinder = πr²h

Volume of brass of 2.2dm³ is equal to volume of cylindrical wire

⇒ h = 448m

The solid cylinder should be to hollow cylinder

Given that, diameter of solid cylinder = 2 cm

Length of hollow cylinder = 16cm

External diameter = 20cm

Thickness = 2.5mm=0.25cm

Volume of solid cylinder = πr²h ------(i)

Radius of the cylinder = 1cm

So, volume of the solid cylinder = π(1)²h = π h

Let length of the solid cylinder be h

Volume of hollow cylinder = πh(R² – r²)

Thickness = R – r

⇒ r = 10 – 0.25

⇒ Internal radius = 9.75cm

So, volume of hollow cylinder = π × 16(100 – 95.0625)

= π × 16 × 4.9375

≈ π × 16 × 4.94 -------(2)

As we know that, volume of both solid and hollow cylinder must be the same, therefore

π h = π × 16 × 4.94

⇒ h ≈ 79cm

∴ Length of the solid cylinder = 79cm

Given: diameter of the cylindrical vessel = height of the vessel

∴ h = 2r

Given: water filled in the cylindrical vessel is poured in two cylindrical vessels

Volume of a cylindrical vessel = π r² h

So, volume = πr²(2r) = 2 π r³

Volume of vessel with diameter 42 cm =

= π (21)³ cm³

Since cylinders are filled by water completely, so volume of large cylinder = sum of these small cylinders

∴ 2 π r³ = 2 × π (21)³

⇒ r = 21cm

∴ diameter of the cylindrical vessel = 42cm

Diameter of each circular plate = 14 cm

∴ their radius, r = 7 cm

Thickness of each plate = 0.5 cm

50 plates are placed one above the other to form a right circular cylinder

∴ height of the cylinder formed, h = 50 × 0.5 cm

Or h = 25cm

Total surface area of the cylinder = 2πr(r + h)

= 1408 cm²

Radius of each circular plate = 10.5 cm

Thickness of each plate = 1.6 cm

25 plates are placed one above the other to form a solid circular cylinder

∴ height of the cylinder formed, h = 25 × 1.6 cm = 40cm

Total curved surface area of the cylinder = 2πrh

= 2640 cm²

Volume of cylinder = π r² h

= π (10.5)² × 40

= 13860 cm³

Diameter of circular pond is given as 40 cm

∴ Radius of the pond, r = 20cm

Also, thickness, t = R – r

⇒ 2 = R – r

⇒ 2 = R – 20

⇒ R = 22cm

Volume of a hollow cylinder = π (R² – r²)h

= 3.14 × (22² – 20²) × 0.2

≈ 52 .8

∴ 52.8 cubic metres of gravel are required to grave the path to a depth of 20 cm

Let us assume shape of well is like a solid right circular cylinder

Radius of cylinder, r = 3.5/2 = 1.75m

Depth of well, h = 16m

Volume of right circular cylinder, V’ = π r² h

---------(I)

Given that, length of platform, l = 27.5 cm

Breadth of the platform, b = 7cm

Let height of the platform be x m

Volume of the rectangle, V’’ = lbh

= 27.5 × 7 × x

= 192.5x -----------(II)

Since well is spread evenly to form the platform, so volume V’ = V’’

⇒ x = 0.8m

∴ Height of the platform = 80cm

Let us assume shape of well is like a solid circular cylinder

Radius of cylinder, r = 2/2 = 1m

Depth(height) of well, h = 14m

Volume of right circular cylinder, V’ = π r² h

---------(I)

Given that, length of embankment, l = 40 cm

Let width of the embankment be x m

Volume of the embankment, V’’ = π r² h

= π ((1+x)² – 1)² x 0.4 -----------(II)

Since well is spread evenly to form the embankment, so their volumes, V’ = V’’

⇒ π × 14 = π ((1+x)² – 1)² x 0.4

⇒ x = 5m

∴ Height of the embankment, x = 5cm

Given that, side of the cube = 9cm


Let radius of cone be x.

Since largest cone is curved from cube.

Diameter of base of cone = side of cube

⇒ 2x = 9

⇒ x = cm

Height of cone is:


h = 9 cm

Volume of largest cone =

V

V

V = 190.92 cm³

∴ volume of largest cone, v = 190.92 cm³

Volume of cylindrical bucket = π r²h

Here, r = 18cm and h = 32cm

So, volume of cylindrical bucket = π (18)²(32) cm²

When emptied on ground, volume of conical heap of sand = volume of cylindrical bucket

Now,

Volume of rainfall = lbh

= 6 × 4 × 0.01

= 0.24 m³

∵ Volume of water in cylindrical vessel = volume of rainfall

∴ πr²h = 0.24

⇒ h ≈ 1.91m = 191cm

Volume of conical flask = πr²h

∵ Volume of water in cylindrical flask = volume of conical flask

∴ π(rm)²h’ = πr²h

Volume of liquid in the new rectangular tank = volume of liquid filled in full cylindrical tank

lbh = π r² h’

Here, r = m, h’ = 5m, l = 15m and b = 11m, then

Number of bottles = (volume of cylindrical shaped small bottles/ Internal volume of hemispherical bowl)

Internal radius of the hollow spherical shell = 6/2 = 3cm

And external radius of that shell = 10/2 = 5cm

Radius of the cylinder = 14/2 = 7

Let height of the cylinder be x cm

According to the question,

Volume of the cylinder = volume of the spherical shell

∴ height of cylinder = cm

Given, internal radius of hollow sphere, r = (4/2) = 2cm

External radius = 8/2 = 4cm

Volume of hollow sphere =

--------(i)

Given, diameter of the cone = 8cm

∴ Radius of the cone = 4 cm

Let height of the cone be h

Volume of the cone =

--------(ii)

Since hollow sphere is melted into a cone so their volumes are equal,

∴ h = 12cm

Given, the radius of the cylindrical tube, r = 12cm

Level of water raised in tube, h = 6.75cm

Volume of cylinder = πr² h

= π (12)² × 6.75 cm³

----------(i)

let r be the radius of a spherical shell balls

volume of the sphere = -------(ii)

∵ volume of cylinder = volume of spherical ball

⇒ r = 9cm

∴ radius of spherical ball, r = 9cm

To find: the rise in the level of water in the tank, if the average displacement of water by a person is 0.04 m³ and 500 persons have to dip in a rectangular tank which is 80 m long and 50 m broad.

Solution:


Length of rectangular tank, l = 80m

Breadth, b = 50m

Total displacement of water in rectangular tank by 500 persons = 500 × 0.04 m³ = 20m³ -----(i)

Let depth of the rectangular tank be h


The amount of water displaced is equal to the volume of the tank.

Volume of rectangular tank = lbh

Volume of rectangular tank = 80 × 50 × h m³ -----------(ii)

Equating (i) and (ii), we get

20 = 80 × 50 × h

⇒ h = 0.005 m


As 1 m = 100 cm


So 0.005 m = 0.005 × 100 cm


⇒ h = 0.5 cm

∴ rise in level of water due to angular displacement = 0.5cm

Given that, radius of cylindrical jar, r = 6cm

Depth/height of the jar, h = 2cm

Volume of the jar, V’ = π r² h

⇒ V’ = π (6)² (2) cm³

Radius of the sphere = 1.5cm

So, volume of the sphere =

Volume of oil in cylindrical jar = volume of n spheres needed to raise the level by 2 cm

⇒ n = 16

∴ number of iron sphere needed = 16

Volume of hollow sphere before melting = volume of cone

Here, R = 4cm, r = 2cm, r’’ = 4cm, therefore

⇒ h’’ = 14cm = height of the cone

Now, slant height, l =

= 14.56 cm

Given that, internal diameter of hollow hemisphere, r = 10.5cm

External radius, R = 12.6cm

Total surface area = 2πR² +2πr² + π(R² - r²)

=2π(12.6)² +2π(10.5)² + π((12.6)² – (10.5)²)

=1843.38 cm²

Given that, cost of painting 1 cm² of surface = 10paise

Therefore, total cost to paint 1843.38 cm² =

Given that, radius of cylindrical tube = 12cm

Let height of cylindrical tube be h

Volume of the cylinder = πr₁²h = π (12)² h --------(i)

Also given that, spherical ball radius (r₂) = 9cm

Volume of sphere =

---------(ii)

Equating (i) and (ii), we get

⇒ h = 6.75cm

∴ level of water raised in tube = 6.75cm

Let radius of the third ball be r.

Volume of the larger ball =

Volume of larger ball = sum of volume of smaller balls

⇒ r³ = 27 – 3.375 – 8

⇒ r = 2.5cm

Curved surface area of sphere

Let ‘r’ be the radius of sphere

Height of cylinder=’2r’

Curved surface area of cylinder

⇒

⇒

∴ Surface area of a sphere is equal to the curved surface area of the circumscribed cylinder.

Let the radius of third sphere be r₃

Volume of big spherical ball=volume of all three balls

⇒ R³ = r₁³ + r₂³ + r₃³

⇒ 3 × 3 × 3 = (1.5 × 1.5 × 1.5) + (2 × 2 × 2) + r₃³

⇒ r₃³ = 27 - (3.375 + 8)

⇒ r₃³=15.625

⇒ r₃=

⇒ r₃=2.5 cm

Volume of the metallic sphere=volume of the wire(cylinder in shape)

Radius of sphere cm

=4.5 cm

Volume of sphere = ³

³

Radius of cylinder = 1mm =0.1cm

Volume of cylinder r²h

⇒ πr²h ³

⇒ r²h R³

⇒ 0.1 × 0.1 × h = × 4.5 × 4.5 × 4.5

⇒ 0.1 × 0.1 × h =4 × 1.5 × 4.5 × 4.5

⇒ h = 12150 cm

Let the radius of bigger spherical be R

Volume of bigger spherical ball = ³

Radius of smaller spherical ballR

Volume of smaller ball ³

Let number of equal size spherical balls be n

Volume of n equal spherical ball =Volume of bigger spherical ball

⇒ n × (R/4)³ R³

⇒ n = 4³

⇒ n = 64 balls

Surface area of bigger spherical ball =4π R²

Surface area of smaller spherical ball ²

Ratio between the surface area of bigger and 64 smaller spherical ball

Height of tent =77dm

=7.7m

Radius of the cylinder =18 m

Height of cylinder =44dm =4.4m

Height of cone = (7.7-4.4) m

=3.3m

Curved Surface area of cylinder portion of the tent

= 2 × (22/7) × 18 × 4.4

= 497.83 m²

Curved surface area of conical tent =π r l

Slant height of cone

= 18.3m

Curved surface area of cone = (22/7) × 18 × 18.3

= 1035.26m²

Total surface area of the tent = 497.83m² + 1035.26m²

= 1533.09m²

Total cost of canvas = 3.50 × 1533.09

= Rs 5365.82

Volume of sphere ³

= 33.50 cm³

Volume of 16 spheres = 16 × 33.50

= 536cm³

Volume of rectangle = 16cm × 8cm × 8cm

= 1024cm³

Volume of this liquid = 1024cm³ – 536cm³

= 488cm³

Diameter of the sphere =Height of the cylinder

Diameter of the sphere =14 cm

Radius of the sphere cm

=7 cm

Volume of the sphere ³

= 1437.3cm³

Volume of the sphere ³

Volume of the cone

Volume of the cone = volume of the sphere

=

⇒ r² = 36 cm

⇒ r = 6 cm

Given that area of cuboid =160 cm²

Level of water is increased = 2 cm

Volume of vessel =160× 2 cm³ ……. (1)

Volume of each sphere ³

Volume of 3 sphere ³ ……. (2)

From eq 1 and eq 2

⇒ 3×(4/3) π R³ = 160× 2

⇒ R = 2.94 cm

Given: A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness.
To find: the thickness of the wire.
Solution:
Diameter of copper wire =1 cm

Radius cm

Height of copper rod =Length of copper rod =8 cm

Volume of rod

Let the radius of wire be ‘r’ cm

length of the wire=Height of the wire =18 m =1800 cm
As rod is converted into wire,
Therefore,

Volume of cylinder wire = Volume of cylindrical rod

As radius cannot be negative,

⇒ r = 0.033 cm
Hence thickness of the wire is 0.033 cm.

⇒ r²

⇒ r cm

⇒ r = 1/3 mm

∴ Diameter of cross section = (1/3) × 2 mm

= 0.67 mm

Diameter of internal surface =10 cm

∴ Radius of internal surface cm =5 cm

Diameter of external surface =6 cm

∴ Radius of external surface cm =3 cm

Volume of spherical shell hollow =

Height of solid cylinder cm

Let the radius of the solid cylinder be ‘r’ cm

Volume of the solid cylinder

Volume of the solid cylinder = Volume of spherical shell hollow

⇒ r² × (8/3) = (4/3) × 98

⇒ r = 7

Diameter of cylinder = 14 cm

Height of the cone (h₁)=3 cm

Radius of the cone (r₁) = 4 cm

Slant height of the cone (l₁) = 5 cm

Volume of cone (V₁)=

cm³

cm³

Again after rotating

Height of the cone (h₂)=4 cm

Radius of the cone (r₂)=3 cm

Slant height of the cone (l₂)=5 cm

Volume of cone (V₂₎

cm³

cm³

Difference between the volume two cones =16π – 12π

= 4π cm³

Curved surface area of first cone =π × r₁× l₁

= π × 4 × 5

= 20 cm²

Curved surface area of second cone = 15 cm²

Volume of cuboid = l × b × h = 11 × 10 × 7

Diameter of coin = 1.75 cm

Radius cm = 0.875 cm

Thickness of coin =2 mm = 0.2 cm

Volume of coin (cylinder)

= (22/7) × 0.875 × 0.875 × 0.2 cm³

Let number of coins be n

⇒ n × volume of coins = volume of cuboid

⇒ n × (22/7) × 0.875 × 0.875 × 0.2 = 11 × 10 × 7

⇒ n = 1600

Let the height of the embankment be ‘h’ m

Radius of well = 4 m

Height of the well = 14m

Volume of earth dug out (volume of cylinder)

= 704 m³

Outer Radius of the embankment(R) = 4 m + 3 m = 7 m

Inner radius of the embankment(r) = 4 m

Volume of embankment =Volume of earth dug out

Volume of embankment

= πh (R² – r²)

= πh (7² – 4² )

= 33πh

Volume of embankment = Volume of earth dug out

⇒ 33πh = 704

⇒ h = 6.78 m

Speed of flowing water =10 km/hr

In 30 minutes length of flowing water =10× km

= 5 km = 5000 m

Volume of flowing water in 30 minutes =5000 × width × depth

Width of canal = 1.5 m

Depth of canal = 6 m

Volume of canal =5000 × 1.5 × 6 m³

= 45000 m³

Irrigated area in 30 minutes if 8 cm of flowing water is required

= 562500 m²

Let the time-taken by the pipe to fill the tank be x hrs.

Since the water is flowing at a rate of 3 kmph, therefore

Length of water column in x hrs is

3 × x = 3x km = 3000x m

Therefore the length of the pipe is

Therefore, the volume of water flowing through the pipe in x hours, V’ = π r² h

= π (0.1)² (3000x) m³

Also, volume of water that falls into the tank in x hours = π r² h

= π (5)² (2)

The volume of water flowing through the pipe in x hours = volume of water that falls into the tank in x hours

∴ π (0.1)² (3000x) = π (5)² (2)

⇒ 30x = 50

= 1hr 40min

Let the height of the embankment be ‘h’ m

Radius of well

=1.5 m

Height of the well =14m

Volume of earth dug out (volume of cylinder)

=99 m³

Outer Radius of the embankment(R) =1.5 m + 4 m

=5.5 m

Inner radius of the embankment(r) =1.5 m

Volume of embankment =Volume of earth dug out

Volume of embankment

= 27.5πh

Volume of embankment =Volume of earth dug out

⇒ 27.5h = 99

h = 1.145

⇒ h = 9/8 = 1.125 m

Surface area of the sphere

⇒ 4πr² = 616

⇒ r² = 49

⇒ r = √49

⇒ r = 7 cm

Diameter of base = 7 × 2

= 14 cm

Height of hollow cylinder =14 cm

Let the radius of outer radius be ‘R’ cm

Let the radius of inner radius be ‘r’ cm

Difference between inner and outer Curved surface area= 88 cm

Curved surface area of hollow cylinder

⇒ 2π(R – r)h = 88

⇒ R – r = 1 ……… (1)

Volume of metal = 176

Height of hollow cylinder = 14 cm

Volume of hollow cylinder (metal) =

⇒ (R + r)(R – r) = 4

⇒ (R + r) = 4 ………… (2)

Solving (1) and (2) we get,

R = 2.5cm

r = 1.5cm

∴ Outer diameter cm

= 5 cm

Inner diameter cm

= 3 cm

Volume of hemi-sphere

⇒ r = 10.5 cm

Curved surface area

= 693 cm²

Height of the cylindrical bucket =32 cm

Radius of the cylindrical bucket = 18 cm

Volume of the bucket = πr² h

=π × 18² × 32 cm³

Height of the conical heap =24 cm

Let Radius of the conical heap =r cm

Volume of conical heap

Volume of conical heap=Volume of cylindrical bucket

⇒ r² = 1296 cm

⇒ r = 36 cm

Slant height

∴ slant height

Total surface area of solid hemisphere

⇒ 3πr² = 462

⇒3× (22/7) × r² = 462

⇒ r² =49

⇒ r=7 cm

Volume of solid hemisphere

cm³

= 718.67 cm³

Diameter of spherical marbles cm

=0.7 cm

Diameter of cylinder vessel

=3.5 cm

Volume of 150 spherical balls =150× ³

cm³

=215.6 cm³

Let the rise in level of water be h cm

Volume of rise in level of water (Volume of cylinder )

Volume of Rise in level of water in the vessel =volume of 150 spherical balls

⇒ (22/7) × 3.5 × 3.5 × h = 215.6

⇒ h = 5.6 cm

Diameter of the cylindrical base =3 m

∴ Radius of cylindrical tank m

= 1.5 m

Height of the tank =3.5 m

Volume of the tank

= 24.75 m³

Now, 1 m³ = 1000 liters

24.75 m³ =1000 × 24.75 liters

= 24750 liters

Full quantity of the water when it is full = 24750 m³

Quantity of water when it is half filled liters

= 12375 liters

Time taken by it to empty 225 liters 0f water = 1 minute

∴ Time taken by it empty 12375 litersof water = minutes

= 55 minutes

Volume of the sphere = ³

Let number of cones be n

⇒n × volume of cone = volume of sphere

⇒ n = 28

Height of the conical vessel, h’ = 11cm

It’s radius, r’ and r’’ resp.: r’ = 2.5cm r’’ = 0.25cm

Assume the number of spherical balls that were dropped in the vessel as ‘n’

Volume of water replaces by the spherical ball = volume of spherical balls dropped into the vessel

(2/5) × Volume of the cone = n × each spherical ball

⇒ (2.5)² ×11 = n × 10 × (0.25)³

⇒ n = 440

Outer and inner radius of the pipe: r’ = 8/2 = 4cm and r’’ = cm

Volume of cuboid = volume of pipe

L b h = π h[R² – r²]

⇒ h = 2698.18cm ≈ 27m

Speed of water = 2.52kmph = 2520 m/hr

Length, h’ = 2520 m

Let the radius of the pipe be r’ and radius of the tank, r’’ = 40 cm
= 0.4 m

Level of water in the tank in half and hour = 3.15 m

Level of water in the tank in an hour, h’’ = 6.30 m

Now, the volume of pipe = volume of the tank

⇒ π r’² h’ = π r’’² h’’

⇒ r’² (2520) = (0.4)² (6.3)

∴ diameter = 2r’ = 2 × 2
= 4 cm

Diameter= 24 m

r =24/2= 12

Height of cylinder, H =1cm

Height of cone, h = 16-11= 5 m

Slant height (l) =

= = 13 m

Surface area of tent = curved surface area of cylinder + curved surface area of cone

= 2rH + πrl

= 2 × (22/7) × 12 × 11 + (22/7) × 12 × 13

= (22/7) × 12 (22 +13)

= (22/7) × 12 × 35

= 1320 m²

Radius of the cylinder portion (r) = 2.5 m

Height of the cylinder portion (h) = 21 m

∴ Surface area of the cylindrical portion = 2πrh

= 2 × × 2.5 × 21 m

= 330 m ………(1)

Radius of the conical portion (r) = 2.5 m

Slant height of the conical portion (l) = 8 m

∴ Curved surface area of the conical portion = πrl

= (22/7) × 2.5 × 8 m

= 62.86 m ……..(2)

Area of the circular top

= (22/7) × 2.5 × 2.5 m²

= 19.64 m² ……………(3)

∴Total surface area of the rocket

= (330 + 62.86 + 19.64) m² [From (1) , (2), (3)]

= 412.5 m²

Volume of the cylindrical portion h

= 412.5 m²

Given: height of the tent, H = 77dm = 7.7 m

Height of the cylindrical portion, h = 44dm = 4.4m

∴ Height of the conical portion, h’ = 7.7 – 4.4 = 3.3 m

Radius of the cylinder, r = 36/2 = 18m

Curved surface area of the cylindrical portion of the tent = 2 π r² h

= 2 × π × (18)² × 4.4

= 158.4 π m²

Slant height of the conical portion of the tent =

= 18.3 m

∴ Curved surface area of the conical portion = π × 18 × 18.3

= 329.4 π m²

∴ Total surface area of the tent = 158.4 π + 329.4 π

= 487.8 π

= 1533.09 m²

∴ Canvas required to make the tent = 1533.09 m²

Total cost of the canvas = 1533.09 × 3.50

= Rs5365.80

Height of the cone =4 cm

Radius of the cone = radius of hemisphere = 3 cm

Volume of toy = Volume of conical part + Volume of hemispherical part

Volume of cone

∴ Volume of conical part =

= 37.6 cm²

Volume of hemisphere =

Volume of toy = 37.6 +56.52 = 94.12 cm²

And total surface area of toy = Curved surface area of conical part + Curved surface area of hemispherical part

Curved surface area of cone , Where l =

= 3.14 × 3 ×

= 47.1 c

And curved surface area of hemisphere =2πr²

= 2 × 3.14 × 3 × 3

= 56.52 cm²

Then, total surface area of the toy = 47.1 + 56.52 = 103.62 cm²

Height of cylinder (h) = 10 cm

Height of conical part = 6 cm

Slant height of cone (l) =

⇒ l = 48.25 cm

Curved surface area of cone rl

= π (3.5) (48.25)

= 76.408 cm² …(1)

Curved surface area of cylinder = 2π rh

=2π (3.5) (10)

=220 cm² …(2)

Curved surface area of hemisphere = 2

= 2π (3.5)²

= 77 cm² ….(3)

∴ Total curved surface area = Curved surface area of(cone + cylinder + hemisphere)

= 76.408 + 220 + 77

= 373.408

∴ Total surface area of solid = 373.408 cm²

Total surface area of toy = C.S.A. of cylinder + C.S.A. of hemisphere + C.S.A. of cone

= (2× ×5 ×13) + (2× ×5×5) + ( ×5×13)

= ×5[(2×13) + (2×5)13]

= ×5[26+10+13]

= ×5×49

=770 cm²

Given radius of cylindrical tube(r) = 5 cm

Height of cylindrical tube (h) =9.8 cm

Volume of cylindrical =πr²h

V1 =π 5²(9.8)

= 770 cm

Given of hemisphere = 3.5 cm

Height of cone (h) = 5 cm

Volume of hemisphere =

= 89.79 cm³

Volume of cone =

= 64.14 cm³

Volume of cone + Volume of hemisphere (v2) = 89.79 + 64.14 = 154 cm³

Given radius of Cylindrical base = 20 m

Height of Cylindrical part (h) = 4.2

Volume of Cylindrical = ….(1)

V1 =π(20)² 4.2

= 5280 m³

Volume 0f Cone =

Height of conical part (h) = 2.1 m …..(2)

V2 (20)² (2.1)

= 880 m³

Volume of tent (v) = V1 + V2

V = 5280 + 880

= 6160 m³

Diameter =21 cm

∴ Radius, r =10.5 cm

Length of the cylindrical part, l =18 cm

Height of the conical part, h = 9 cm

Volume of the tank = Volume of the cylindrical part + 2 × Volume of the conical part

= (22/7) × 10.5 × 10.5 × 24

= 8316 cm³

Radius of cylinder(r) = 5 cm

Height of cylinder, (h) = 12 cm

Let, l be slant height of cone

Slant height of the cone = =13 cm

Volume of cylinder = π × × 12 = 300 cm³

Volume of the conical hole = × π × 5 × 12 = 100 cm³

Therefore, Volume of the remaining solid

= Volume of the cylinder – Volume of the removed conical part

= 300 π -100 π = 200 π cm³

Curved surface of the cylinder

= 2 × π r h = 2 × π × 5 × 12 = 120π cm

Curved surface of cone = π r l = π × 5 × 13 = 65π cm

Base area of cylinder = π × 5² = 25cm²

The whole surface area of the remaining solid includes the curved surface of the cylinder and cylinder and the cone and area of the base

Therefore, Whole surface area = 120 π + 65π + 25π =210π cm²

Given: A tent is in the form of a cylinder of diameter 20 m and height 2.5 m, surmounted by a cone of equal base and height 7.5 m.

To find: the capacity of the tent and the cost of the canvas at Rs. 100 per square metre.

Solution:



height of Cylinder (h) = 2.5 m

Volume of a cylinder (v)= πr²h

V₁ = π(10)²(2.5) cm …..(1)

Volume of cone (V₁) ₌


Volume of cone (V₁) ₌..... (2)

Volume of boiler = (1) + (2)

V = V₁ + V₂

Cost of 1 m² canva = Rs 100

cost of 550 m² canva = 550× 100
= Rs 55000

Given: radius of cylinder (r) = = 10 cm

Height of cylinder (h₁) = 2.5 m

Height of cone (h₂) = 7.5 m

Let l be slant height of cone

⇒ l = 12.5 m

Volume of cylinder (V₁) = πr²h

⇒ V₁ = π r(10)²(2.5) ……….. (1)

Volume of cone (V₂) = πr²h

= π(10)² (7.5) …………. (2)

Total capacity of tent = (1) + (2)

V = V₁ + V₂

⇒ V = π(10)²(2.5) + π(10)² (7.5)

⇒ V = 250π + 250π

⇒ V = 500 π m³

∴Total capacity of tent = 500 π m³

Radius of the hemispherical part of the hollow cylinder =

Hollow cylinder and the hemisphere have the same base

∴ Radius of the cylinder =

Depth of the cylinder, h =

Volume of the vessel = Volume of cylindrical part + Volume of the hemispherical part

≈ 56.15 m³

Internal surface area of the vessel = CSA of (cylinder + hemisphere)

= 2πrh + 2π r²

= 2π r(h+r)

= 70.58 m²

Radius of hemispherical ends, r = 7 cm

Height of body (h + 2r) = 104 cm

Curved surface area of cylinder = 2πrh

= 2π (7)h ……….(1)

Given: h + (2r) = 104 cm

⇒ h = 104 – 2 (7)

⇒ h =90 cm

Substitute the value of h in eq. (1),

Curved surface area of cylinder = 2π (7) (90)

= 3958.40 cm …..(2)

Curved area of 2 hemisphere = 2(2πr²)

= 2(2 ×π ×7²)

= 615.75 cm ……….. (3)

Total curved surface area = 2+3

= 3958.40 + 615.75

= 4574.15 cm²

Cost of polishing for 1 dm² = Rs.10

Cost of polishing for 54.75 dm² = 45.74×10

= Rs.457.60

Given height of cylindrical vessel (h) = 42 cm

Inner radius of vessel (r’) = cm =7 cm

Outer radius of vessel (r’’) = =8 cm

Volume of cylinder = π (r’’² – r’²) h

= π (8² - 7²) 42

= 1980 cm³

Given that, internal radius of cylinder road roller (r1) = = 27cm

Thickness of road roller (t) =9 cm

Radius of cylinder road roller be R

t = R –r

t = R-27

R = 9 + 27 = 36 cm

Given height of cylindrical road roller (h) =1m

h = 100 cm

Volume of iron=πh(R²-r²)

= π(36² - 27²) × 100

= 1780.38 cm

Mass of 1 cm of iron = 7.8 gm

Mass of 1780.38 cm of iron = 1780.38 × 7.8

=1388696.4 gm

∴ Mass of roller (m) = 1388.7 kg

Given radius of hemisphere & cylindrical (r)

=7 cm

Given total height of vessel = 13 cm

∴ (h +r) = 13 cm

Inner surface area of vessel = 2π r (h +r)

= 2 ×π × 7 (13)

= 182π

=575 cm²

Given radius of cone (r) = 3.5 cm

Total height of toy (h) = 15.5 cm

height of cone (l) = 15.5 – 3.5

= 12 cm

The curved surface area of a cone,
S’ = πrl

= π (3.5) (12.5)

=137.5 cm²…….. (1)

The curved surface area of the hemisphere,
S’’ = 2πr²

= 2π(3.5)²

= 77 cm²………… (2)

∴ Total surface area of the toy = 137.5 + 77
= 214.5 cm²

Given: The difference between outside and inside surface areas of cylindrical metallic pipe 14 cm long is 44 m² and the pipe is made of 99 cm³ of metal.

To find: the outer and inner radii of the pipe.

Solution:




Let the inner radius of pipe = r1

And the radius of outer cylinder = r2

Length of cylinder (h) = 14 cm

Surface area of hollow cylinder =2πh (r2 –r1)

Given surface area of cylinder = 44 m².

⇒ 2π h(r2 –r1) = 44

⇒ 2π (14) (r2 –r1) = 44

⇒ (r2 –r1) =

⇒ (r2 –r1) = 1/2 ……. (1)

Given volume of a hollow cylinder = 99 cm³

Volume of a hollow cylinder = π h()

⇒ π h (r2² - r1²) = 99

⇒ 14π (r2² - r1²) = 99


Apply the formula a² - b² = ( a - b ) ( a + b ) in (r2² - r1²).

⇒ 14π (r2 + r1) (r2 – r1) = 99


Put the value of r2 - r1 from (1).

⇒ 14π (r2 + r1) = 99






⇒ 22 (r2+r1) = 99

⇒ (r2+r1) = ……………… (2)

Equating (1) & (2) equations we get

r2 = 5/2 cm

Substituting r2 value in (1)

⇒ r1 = 2 cm

∴Inner radius of pipe (r1) = 2 cm

Radius of outer cylinder (r2) = 5/2 cm

Given radius of cylinder (r’) = = 6 cm

Given radius of hemisphere (r’’) = =3 cm

Given height of cylinder (h) = 15 cm

Height of cone (l) = 12 cm

Volume of cylinder = πr²h

= πr(6)²(15) cm ………. (1)

Volume of each cone = Volume of cone + Volume of hemisphere

V = π(r)²(l) + π(r)³

V = π(3)²(12) + π(3)³ ………… (2)

Let number of cone be n

n (volume of each cone) = volume of cylinder

n = [( π(3)² (12) + π(3)³] = π(16)² 15

⇒ n = 10

Given radius of cylindrical part(r)= 12/2= 6 cm

Height of cylinder (h) = 110 cm

Length of cone (l)= 9 cm

Volume of cylinder=h

V1= π (6)² 110

Volume of cone=

V2=

=108π cm³

Volume of pole =v1 + v2

=+108

= 12785.14 108 cm³

Given mass of 1cm³ of iron = 8 gm

Mass of 1275.14 cm³ of iron = 12785.14 × 8

= 102281.12

=102.2 kg

∴ Mass of pole for 12785.14 cm³ of iron is 102.2 kg

Given radius of cone, cylinder & hemisphere (r) = 4/2 = 2 cm

Height of cone (l) = 2 cm

Height of cylinder (h) = 4 cm

Volume of cylinder = h = (4) cm …….. (1)

Volume of cone =

=

= × 4 × 2 cm ………………. (2)

Volume of hemisphere =

=

= …………….. (3)

So remaining volume of cylinder when toy is instead to it

= eq.(1) – eq.(2) + eq.(3)

= π(2)²(4) – ( × π × 8)

= 16π - π (4+8)

= 16π – 8π

= 8π cm³

∴ So remaining volume of cylinder when toy is instead to it = 8π cm³

Given: radius of circular cone (r) = 60 cm

Height of a cone (l) = 120 cm

Volume of a cone = πr² l

= π(60)²(120) cm …….. (1)

Volume of a hemisphere =

Given radius of hemisphere = 60 cm

……………. (2)

Radius of cylinder (r) = 60 cm

Height of cylinder (h) = 180 cm

Volume of cylinder = h

= π(60)²(180) cm³ ……… (3)

Volume of water left in cylinder = eq.(3) – eq.(1) + eq.(2)

= 113.1 cm³= 1.131m³

Given internal radius (r1) = 10/2 = 5 cm

Height of cylindrical vessel (h) = 10.5 cm

Outer radius of cylindrical vessel (r2) = 7/2 = 3.5 cm

Length of cone (l) = 6 cm

(i)Volume of water displaced = Volume of cone

Volume of cone =

=

= 76.9 cm³

≈ 77 cm³

Volume of water displayed = 77 cm³

Volume of cylinder = πr²h

= π(5)²10.5

= 824.6

≈ 825

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

= 825 -77 = 748 cm³

Edge of wooden block (a) = 21 cm

Diameter of hemisphere = edge of cube

Radius = 21/2 = 10.5 cm

Volume of remaining block = Volume of box – Volume of hemisphere

= 6835.5 cm³

Surface area of box = 6a² ………… (1)

Curved surface area of hemisphere = 2πr³ ……….. (2)

Area of base of hemisphere ……………….. (3)

So remaining surface area of box = (1) – (2) + (3)

= 2992.5 cm²

Remaining surface area of box=2992.5 cm²

Volume of remaining block=6835.5 cm³

Radius of base = 21cm and its volume = 2/3 × volume of the hemisphere

Hence surface area of the toy = Surface area of the cone + Surface area of hemisphere

= π r () + 2 π r²

= π (21)[35 + 2 × 21]

= 5082 cm²

Height of the cylinder = height of solid – radius of the cone

= 9.5 – 3.5 = 6cm

Volume of the solid = volume of the cone + volume of the cone

= 166.83 cm³

the height of the cylinder = 10 cm, and its base is of radius = 3.5 cm

Volume of wood in the toy = volume of cylinder – 2 × volume of hemisphere

= 205.33 cm³

To be the largest, diameter of the sphere = edge of the cube

∴ its radius = 1/2 × edge

Volume of wood left = volume of cube – volume of sphere

= 343 – 179.67

= 163.33 cm³

Slant height of the cone, l =

Remaining area of the cylinder = Curved surface area of the cylindrical part + Area of the conical part + Area of cylindrical base

= 2πrh + πrl + πr²

= 2π × (2.1)× (2.8) + π × 2.1 × (3.5) + π (2.1)²

= 36.96 + 23.1 + 13.86

= 73.92 cm²

Largest cone is the right circular cone with base diameter and height = edge of the cube

∴ radius of the cone = (21/2) cm

and
where 'r' is radius and 'h' is height of the cone.


Volume of the remaining cube = volume of the cube – volume of cone cone)

Let h be the height of the cone and r be radius of its base

Now, volume of the wooden toy =

According to the question,

⇒ h = 6cm

The height of the wooden toy = 6cm + 3.5 cm

= 9.5cm

Now, curved surface area of the hemispherical part = 2 π r²

= 2π × (3.5)²

=77cm²

Hence the cost of painting the hemispherical part of the toy at the rate of ` 10 per cm² = 77 × 10 = Rs.770

Here, dimension of cuboid = 15 cm × 10 cm × 5 cm

Radius of cylindrical hole = cm

Surface area of the remaining block = surface area of cuboidal solid metallic block – 2 × base area of cylindrical hole

= 2(15 × 10 + 10 × 5 + 5 × 15) - 2 ×

= 583 cm²

Radius of top of the bucket, r’ = 40/2 = 20 cm

Radius of bottom of the bucket, r’’ = 20/2 = 10 cm

Depth, h = 12cm

Volume of the bucket =

= 8800 cub. cm

Let l be slant height of the bucket

⇒ l = 15.62 cm

Total surface area of bucket = π (r’ + r’’) × l + π r’’²

= π (20 + 10) × 15.620 + π (10)²

= 17.81 dm²

Cost of tin sheet = 1.20 × 17.87 = Rs21.40

For frustum:

Base radius, r’ = 20/2 = 10cm

Top radius, r’’ = 12/2 = 6cm

Height, h = 3cm

Volume =

= 616 cub. cm

Let l be the slant height of the cone, then

⇒ l = 5cm

Total surface area of the frustum = π (r’ + r’’) × l + π r’² + π r’’²

= π (20 + 10) × 15.620 + π (10)² + π (6)²

= 678.85 cm²

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

To find: the curved surface of the frustum.

Solution:


For the given frustum:

Slant height, l = 4 cm

Perimeter of its circular ends = 18 cm and 6 cm

Let radius of larger and shorter ends be 'r' and 'R' resp., then

Perimeter of larger circular end = 2 π r = 18 cm

⇒ π r = 9

Similarly, 2π R = 6

⇒ π R = 3

Curved surface of the frustum = (π r +π R) × l

= (9 + 3) × 4

= 12 × 4

= 48 cm²

For the frustum,

Perimeter of the larger end = 44cm

⇒ 2πr’ = 44

Perimeter of the smaller end = 33cm

⇒ 2πr’’ = 33

Height = 16cm

Its volume =

≈ 1900 cub. cm

Let l be the slant height of the cone, then

⇒ l = 16.1cm

Curved surface area of frustum = π (r’ + r’’) × l

= 3.14(7 + 5.25) × 16.1

= 619.65 cm²

Total surface area of the frustum = π (r’ + r’’) × l + π r’² + π r’’²

= 619.65 + 3.14 × (7² + 5.25²)

= 860.275 cm²

For the bucket:

Height = 45cm

Radius of top = 28cm

Radius of bottom = 7cm

Volume of the bucket =

= 48510 cub. cm

Height of the cone = 20cm

Let the small cone was cut at the height x from the top then height of the small cone = x cm

From cones ABC and AEF,

(h’, r’ and h’’, r’’ are heights and radius of larger and smaller cones resp.)

-------(i)

Given: Volume of the small cone = × volume of the large cone

And Volume of a cone =

⇒ x = 4cm

Height at which section is made = 20 – 4 = 16cm

Given: for the bucket,

Height = 45cm

r’ = 5cm

r’’ = 15cm

Curved surface area = π(r’ + r’’)l + π r’’²

= π(5 + 15)(26) + π (15)²

= 745 π cm²

Curved surface area of bucket = 745π cm²

Given: radius of bases of frustum = 12cm and 3cm and its height = 12cm

Let l be the slant height of the cone, then

⇒ l = 15cm

Total surface area of the frustum = π (r’ + r’’) × l + π r’² + π r’’²

= π (12 + 3) + π × (12² + 3²)

= 378 cm²

Volume =

= 756 π cm³

Height of the frustum, h = 8m

Radii of frustum are: r’ = 13m and r’’ = 7m

Let l be the slant height of the frustum

⇒ l = 10cm

Curved surface area of frustum, A’ = π (r’ + r’’) × l

= π (13 + 7) × 10

= 628.57 cm²

Slant height of the conical cap = 12m

Base radius of the cap = 7m

Curved surface of the cap, A’’ = πrl

= π × 7 × 12

= 264m²

Total canvas required = A’ + A’’

= 628.57 + 264

= 892.57 m²

∴ Total canvas = 892.57 m²

Let height and slant height of the frustum be h and l.

Radius, r’ = 100m, r’’ = 50m

Volume of frustum = 44 × 10⁷ litres = 44 × 10⁴ m³

Slant height, l =

⇒ l = 55.46m

Lateral surface area of the frustum = π l (r’ + r’’)

= 26145.4 m²

let ABC be the given cone.

Here, the height of the metallic cone AO = 20cm

Cone is cut into two pieces at the middle point of the axis.

hence, the height of the frustum cone AD = 10cm

Since ∠A is right-angled, so ∠B = ∠C = 45°

From triangle ADE,

⇒ r'=10cm

Similarly, from triangle AOB,

⇒ r''=20cm
The volume of the frustum of a cone

Volume of the frustum = = 12308.8

⇒ h = 15cm

Let l be the slant height of the bucket

⇒ Slant height, l =

⇒ l = 17cm

∴ length of the bucket, l = 17cm

Curved surface area = π(r’ + r’’)l + π r’’²

= π(20 + 12)(17) + π (12)²

= 2160.32 cm²

Slant height, l =

⇒ l = 25cm

∴ Slant height of the bucket, l = 25cm

Curved surface area = π(r’ + r’’)l + π r’’²

= π(25 + 10)(25) + π (10)²

=3061.5 cm²

Cost of making bucket per 100 cm² = Rs70

Cost of making bucket per 3061.5 cm² =

∴ Total cost = Rs2143.05

Slant height of frustum cone, l = 10cm

Radii of circular ends of frustum cone: r’ = 33cm and r’’ = 27cm

Total surface area = π(r’ + r’’)l + π r’² + π r’’²

= π(33 + 27)10 + π (33)² + π (27)²

= 7599.42cm²

Given:

Height of frustum, h = 16cm

Radius of lower and upper end of bucket, r’ = 8cm and r’’ = 20cm

Slant height, l =

⇒ l = 20cm

Volume =

= 10449.92 cm³

Curved surface area = π(r’ + r’’)l + π r’’²

= π(20 + 8)(20) + π (8)²

= 624π cm²

Cost of making bucket per 100 cm² = Rs20

Cost of making bucket per 3061.5 cm² =

∴ Total cost = Rs.391.9

Given: height of the frustum = 9cm

Radius of its lower and upper ends: r’ = 30cm and r’’ = 18cm

Let slant height be l

Slant height, l =

⇒ l = 15cm

Volume =

= 5292 π cm³

Total surface area = π(r’ + r’’)l + π r’² + π r’’²

= π(30 + 18)15 + π (30)² + π (18)²

= 1944π cm²

Given: volume of the frustum = 10459 cm³

Radii of lower and upper ends resp.: r’ = 8cm and r’’ = 20cm

Volume =

⇒ h = 16cm

Let slant height be l

Slant height, l =

⇒ l = 20cm

Total surface area of the frustum = π(r’ + r’’)l + π r’² + π r’’²

= π(20 + 8)20 + π (20)² + π (8)²

= 3218.29 cm²

cost of metal sheet used in making the container at the rate of Rs. 1.40 per = 3218.28 × 1.40

= Rs4506

Given: let the height of the cone be H and its base radius be R

This cone is divided into two parts through the mid-point of the height of the cone such that

ED||BC

Therefore triangle AED is similar to triangle ABC

By the condition of similarity,

Volume of a cone = 1/3 πr²h

Volume of the frustum = Volume of the cone ABC – Volume of the cone AED

Given, h = 24cm, r’ = 15cm and r’’ = 5cm

Let slant height be l

Slant height, l =

⇒ l = 26cm

Total surface area of the frustum = π(r’ + r’’)l+ π r’’²

= π(15 + 5)(26) + π (5)²

= 1711.3 cm²

Cost of metal sheet used at the rate of Rs. 10 per 100=

= Rs171

In triangle ABC and ADE,

DE||BC

So, ∆ABC ~ ∆ADE

⇒ OE = 2cm

let slant height be l

Slant height, l =

⇒ l = 13.42cm

Total surface area of the frustum = πr(r + l)

= π × 6(6 + 13.42)

= 366.21 cm²

Now, for smaller cone:

Slant height, l’ =

⇒ l = 4.472 cm

So, curved surface area of smaller cone = πrl

= π × 2 × 4.472 = 28.11 cm²

Now, total surface area of the remaining frustum = total surface area of the bigger cone – curved surface area of smaller cone + area of base of smaller cone

= 366.21 – 28.11 + π (2)²

= 350.52 cm²