Mensuration

Given : radius of cylinder = r = 14 cm

height of cylinder = h = 8 cm

To find : Curved surface area = CSA of cylinder = ?

Total surface area = TSA of cylinder = ?

Formula : CSA of cylinder = 2πrh

TSA of cylinder = 2πr (h + r)

Solution : CSA of cylinder = 2rh = 2 × × 14 × 8

= 44 × 16 = 704 cm²

TSA of cylinder = 2r (h + r) = 2 × × 14 (8 + 14)

= 88 × 22 = 1936 cm²

∴ The curved surface area and the total surface area of the right circular cylinder is 704 cm² and 1936 cm² respectively.

Given : Total surface area (TSA) of cylinder = 660 sq.cm.

Diameter of the base of the cylinder = d = 14 cm

To find : Height of cylinder = h = ?

Curved surface area (CSA) of cylinder = ?

Formula : d = 2r

TSA of cylinder = 2πr (h + r)

CSA of cylinder = 2πrh

Solution : diameter = d = 2r

⇒ Radius = r = = = 7 cm

TSA of cylinder = 2r (h + r)

⇒ 660 = 2 × × 7 (h + 7)

660 = 44 (h + 7)

44 (h + 7) = 660

⇒ h + 7 =

h + 7 = 15

⇒ h = 15 – 7

h = 8 cm

CSA of cylinder = 2rh = 2 × × 7 × 8

= 44 × 8 = 352 cm²

∴ The height of the right circular cylinder is 8 cm and its curved surface area is 352 cm².

Given : CSA of cylinder = 4400 sq.cm

Circumference of the base of cylinder = 110 cm

To find : Height of cylinder = h = ?

Diameter = d = ?

Formula : circumference = 2πr

CSA of cylinder = 2πrh

Solution : circumference of base = 2r

⇒ 110 = 2r (1)

CSA of cylinder = 2rh = 2r × h

4400 = 110 × h [from (1)]

⇒ h = = 40 cm

Circumference = d

⇒ 110 = × d

d = = 35 cm

∴ The height and diameter of the right circular cylinder are 40 cm and 35 cm respectively.

Given : Radius of cylindrical pillar = r = 50 cm = 0.5 m

Height of cylindrical pillar = h = 3.5 m

Number of cylindrical pillars = 12

Cost of painting per square metre = Rs 20

To find : The cost of painting the lateral surface of the pillars

Formula : CSA of cylinder = 2πrh

Solution : The lateral surface of a cylinder is the curved surface of the cylinder.

Area to be painted for one pillar = CSA of cylinder = 2πrh

= 2 × × 0.5 × 3.5

= 11 m²

Total area to be painted for 12 cylinders = 11 × 12 = 132 m²

Total cost of painting = Area to be painted × cost of painting per square meter

= 132 × 20 = Rs 2640

∴ The cost of painting 12 cylindrical pillars is Rs 2640.

Given : TSA of cylinder = 231 cm²

CSA of cylinder = TSA of cylinder = × 231 = 154 cm²

To find : Radius of cylinder = r = ?

Height of cylinder = h = ?

Formula : CSA of cylinder = 2πrh

TSA of cylinder = 2πr (h + r)

Solution : CSA of cylinder = 2πrh

⇒ 2 × × r × h = 154

r × h = = 24.5 (1)

TSA of cylinder = 2r (h + r)

⇒ 2rh + 2r² = 231

154 + 2r² = 231

⇒ 2r² = 231 – 154 = 77

2 × × r² = 77

⇒ r² = =

⇒ r = = 3.5 cm

From (1),

r × h = 24.5

⇒ 3.5 × h = 24.5

⇒ h =

h = 7 cm

∴ The radius and height of the right circular cylinder is 3.5 cm and 7 cm respectively.

Given : TSA of cylinder = 1540 cm²

To find : Height of cylinder = h = ?

Formula : TSA of cylinder = 2πr (h + r)

Solution : Let the radius and height of the cylinder be r and h respectively.

According to the given condition,

h = 4r

⇒ r =

TSA of cylinder = 2r (h + r)

⇒ 1540 = 2 × × (h + )

1540 =

h² = = 28 × 28

⇒ h = 28 cm

∴ The height of the right circular cylinder is 8 cm.

Given : Ratio of radii of two cylinders = r₁ : r₂ = 3 : 2

Ratio of heights of the two cylinders = h₁ : h₂ = 5 : 3

To find : Ratio of CSA of the two cylinders = C₁ : C₂ = ?

Formula : CSA of cylinder = 2rh

Solution : CSA of first cylinder = C₁ = 2r₁h₁

CSA of first cylinder = C₂ = 2r₂h₂

∴

⇒

∴ C₁ : C₂ = 5 : 2

∴ The ratio of the curved surface areas of the two cylinders is 5 : 2.

Given : External surface area of hollow cylinder = CSA_ext = 540 cm²

Internal diameter = d = 16 cm

Height of the hollow cylinder = h = 15 cm

To find : Total surface area = TSA of hollow cylinder = ?

Formula : r =

CSA of cylinder = 2rh

Solution : Internal radius of cylinder = r = = = 8 cm

Internal surface area of hollow cylinder = CSA_int = 2πrh

= 2π × 8 × 15

= 16π × 15 = 240π cm²

TSA of hollow cylinder = CSA_ext + CSA_int = 540 + 240

= 780 cm²

∴ The total surface area of the hollow cylinder is 780 cm².

Given : External diameter of pipe = d_ext = 25 cm

Thickness of pipe = 1 cm

Height of the hollow cylinder = h = 20 cm

To find : Total surface area of pipe = TSA of hollow cylinder = ?

Formula : r =

CSA of cylinder = 2rh

Solution : External radius of pipe = r_ext = = = 12.5 cm

External surface area of pipe = CSA_ext = 2r_exth

= 2 × 12.5 × 20

= 500 cm²

Thickness of pipe = 1 cm

∴ Internal radius of pipe = r_int = r_ext – 1 = 12.5 – 1 = 11.5 cm

Internal surface area of pipe = CSA_int = 2r_inth

= 2 × 11.5 × 20

= 460 cm²

TSA of hollow cylinder = CSA_ext + CSA_int = 500 + 460

= 960 cm²

∴ The total surface area of the hollow cylinder is 960 cm².

Given : Radius of cone = r = 7 cm

Height of cone = h = 24 cm

To find : CSA of cone = ?

TSA of cone = ?

Formula : l =

CSA of cone = rl

TSA of cone = r (r + l)

Solution : l = = = = = 25 cm

CSA of cone = rl = × 7 × 25 = 22 × 25 = 550 cm²

TSA of cone = r (r + l) = × 7 (25 + 7) = 22 × 32 = 704 cm²

∴ The curved surface area and total surface area of the right circular cone is 550 cm² and 704 cm² respectively.

Given : Radius of cone = r = 15 cm

Vertical angle = θ = 60°

To find : height of cone = h = ?

Slant height of cone = l = ?

Formula : l =

Solution :

Here, the vertical angle = ∠ A = 60°

AB = height

BC = radius = 15 cm

AC = slant height

In Δ ABC, AC is the hypotenuse.

tan θ =

⇒ tan 60° =

=

∴ h = = 5√3 cm

l = = = l = = 10√3 cm

∴ The height and slant height of the right circular cone are 5√3 cm and 10√3 cm respectively.

Given : Circumference of base of cone = 236 cm

Slant height of cone = l = 12 cm

To find : CSA of cone = ?

Formula : Circumference = 2r

CSA of cone = rl

Solution : Circumference = 2r

⇒ 2 × r = 236

r = = 118

CSA of cone = rl = 118 × 12 = 1416 cm²

∴ The curved surface area of the solid right circular cone is 1416 cm².

Given : Diameter of paddy = d = 4.2 m

Height of paddy = h = 2.8 m

To find : Area of canvas needed to cover paddy = ?

Formula : l =

CSA of cone = rl

Solution : radius of paddy = r = = = 2.1 m

l = = = = = 3.5 m

Area of canvas needed = CSA of cone = rl = × 2.1 × 3.5 = 23.1 m²

∴ Area of canvas needed to cover the paddy is 23.1 m².

Given : Radius of circular disc = r = 21 cm

Central angle of sector = θ = 180°

To find : Radius of cone = R = ?

Formula : Area of sector =

CSA of cone = rl

Solution : When the edges of the sector are joined to form a cone, the radius of the sector becomes the slant height of the cone.

The area of the sector = CSA of the cone

⇒

21 × 21 = × R × 21

= 21R

⇒ R = = = 10.5 cm

∴ The radius of the hollow cone formed by the sector of the circular disc is 10.5 cm.

Given : Ratio of radius and slant height of cone = r : l = 3 : 5

CSA of cone = 60 sq.cm

To find : TSA of cone = ?

Formula : CSA of cone = rl

TSA of cone = r (r + l)

Solution : Ratio of radius and slant height = 3 : 5

⇒

r =

CSA of cone = rl

60 = × × l

⇒ l² = = 100

⇒ l = 10 cm

∴ r = = = 6 cm

TSA of cone = r (r + l) = × 6 (6 + 10) = = = cm²

∴ The total surface area of the solid right circular cone is cm².

Given : CSA of sphere = 98.56 cm²

To find : Radius of sphere = r = ?

Formula : CSA of sphere = 4r²

Solution : CSA of sphere = 4r²

⇒ 4r² = 98.56

4 × × r² = 98.56

r² = = 7.84

⇒ r = 2.8 cm

∴ The radius of the solid sphere is 2.8 cm.

Given : CSA of hemisphere = 2772 sq.cm

To find : TSA of hemisphere = ?

Formula : CSA of hemisphere = 2r²

TSA of hemisphere = r²

Solution : CSA of hemisphere = 2r²

⇒ 2r² = 2772

2 × × r² = 2772

⇒ r² = = 441

⇒ r = 21 cm

TSA of hemisphere = r² = 3 × × 21 × 21 = 4158 cm²

∴ The total surface area of the solid hemisphere is 4158 cm².

Given : Ratio of radii of two hemispheres = r₁ : r₂ = 3 : 5

To find : Ratio of curved surface areas = C₁ : C₂ = ?

Ratio of total surface areas = T₁ : T₂ = ?

Formula : CSA of hemisphere = 2r²

TSA of hemisphere = r²

Solution : CSA of first hemisphere = C₁ = 2r₁²

CSA of second hemisphere = C₂ = 2r₂²

⇒

TSA of first hemisphere = T₁ = r₁²

TSA of second hemisphere = T₂ = r₂²

⇒

∴ The ratio of the curved surface areas and total surface areas of the two solid hemispheres is 9 : 25 and 9 : 25 respectively.

Given : Inner radius of hollow hemisphere = r_int = 2.1 cm

Outer radius of hollow hemisphere = r_ext = 4.2 cm

To find : CSA of hollow hemisphere = ?

TSA of hollow hemisphere = ?

Formula : CSA of hemisphere = 2r²

TSA of hemisphere = r²

Solution : Outer surface area = CSA_ext = 2r_ext² = 2 × 4.2 × 4.2 = 35.28 cm²

Inner surface area = CSA_int = 2r_int² = 2 × 2.1 × 2.1 = 8.82 cm²

Area of the edges = (r_ext² – r_int²) = (4.2² – 2.1²) = 13.23 cm²

Curved surface area of hollow hemisphere = CSA_ext + CSA_int = 35.28 + 8.82 = 44.1 cm²

Total surface area of hollow hemisphere = CSA + Area of edges = 44.1 + 13.23 = 57.33 cm²

∴ The curved surface area and total surface area of the hollow cylinder is 44.1 cm² and 57.33 cm² respectively.

Given : Circumference of base of dome = 17.6 m

Rate of painting = Rs 5 per sq.m

To find : Total cost of painting inside the dome = ?

Formula : Circumference = 2r

CSA of hemisphere = 2r²

Solution : Circumference of the base = 2r

⇒ 2 × × r = 17.6

r = = 2.8 m

The area to be painted = CSA of hemisphere = 2r² = 2 × × 2.8 × 2.8 = 49.28 m²

Rate of painting = Rs 5 per sq.m

The total cost of painting = 49.28 × 5 = Rs 246.40

∴ The cost of painting the hemispherical dome on the inside is Rs 246.40.

GIVEN : radius of solid cylinder “r” = 14cm

height of solid cylinder “h” = 30cm

we take π =

TO FIND : volume of a solid cylinder = ?

PROCEDURE :

As we know, that

Volume of a cylinder “V” = π r²h

So we now put the values of r and h in the above formula

So we get, V = π ×(14)²×(30)

= π ×196×30

= ×196×30

= 22 × 28 × 30

= 18480 cm³

∴ The volume of the given cylinder is 18480 cm³.

GIVEN : diameter of cylindrical bowl = 7cm

Radius of cylinder “r” = 3.5cm

Height upto which the soup is filled “h” = 4cm

No. of patients daily = 250

we take π =

TO FIND : quantity of soup to be prepared daily in the hospital to serve 250 patients = ?

PROCEDURE :

Volume of a cylindrical bowl “V” = π r²h

Putting the values in the above formula, we get

V = × (3.5)² × 4

V = ×3.5× 3.5× 4

= ×3.5×4

= 22× 3.5× 2

= 154cm³

∴ volume of 1 bowl = 154cm³

⇒ volume of 250 bowls = 250 × 154cm³ = 38500cm³

Quantity of soup in the 250 bowls = 38500 × litres

{∵ 1litre = 1000cm³)

Quantity of soup in the 250 bowls = litres = 38.5 litres.

∴ Quantity of soup in the 250 bowls is 38.5 litres.

GIVEN : radius of cylinder be “r” and its height be “h”

r + h = 37cm

TSA (total surface area) of the cylinder = 1628 sq.cm

we take π =

TO FIND : volume of a solid cylinder = ?

PROCEDURE :

As we know that, TSA of a cylinder = 2π r(r + h)

Now putting the values of “r + h” and “TSA” in the above formula,

We get,

1628 = 2π r(37)

= 2π r

44 = 2π r

r = = = = 44 × = 7cm

∴ r = 7cm

Now we have, r + h = 37

So, we have, 7 + h = 37

So, h = 37-7 = 30cm

∴ h = 30cm

Now that we have values of radius and height of the cylinder, we can find the volume of the cylinder.

Volume of the cylinder “V” = π r²h

V = π ×(7)²×(30)

= × 7 × 7 × 30

= 22 × 7 × 30

= 4620 cm³

∴ The volume of the given cylinder is 4620 cm³.

GIVEN : radius of solid cylinder = “r”

height of solid cylinder “h” = 4.5cm

volume of the cylinder “V” = 62.37cm²

we take π =

TO FIND : radius of a solid cylinder = ?

PROCEDURE :

As we know, that

Volume of a cylinder “V” = π r²h

So we now put the values of V and h in the above formula

So we get, 62.37 = π ×(r)²×(4.5)

= π ×(r)²

13.86 = ×(r)²

r² = = × 7 = 0.63 × 7 = 4.41cm²

∴ r = √ 4.41 = 2.1cm

∴ radius of the given cylinder = 2.1cm.

GIVEN :

For cylinder C1 : radius is r1

Height is h1

Volume is V1

For cylinder C2 : radius is r2

Height is h2

Volume is V2

So we have, =

=

we take π =

TO FIND : = ?

PROCEDURE :

As we know, that

Volume of a cylinder “V” = π r²h

So, = = ×

= ×

= ×

= ×

=

∴ the ratio between the 2 cylinders ie. =

ie. V₁ : V₂ = 20:27

GIVEN : radius of solid cylinder = “r”

height of solid cylinder = “h”

=

volume of the cylinder “V” = 4400cm²

we take π =

TO FIND : radius of a solid cylinder = ?

PROCEDURE :

As we know, that

Volume of a cylinder “V” = π r²h

Also, we have, =

⇒ = ⇒ h = r ×

So we now put the values of V and h in the above formula for V,

So we get, 4400 = π ×(r)²×( r × )

4400 = × r³×

4400 × × = r³

200 × 5 = r³

1000 = r³

So, r = ∛ 1000 = 10

∴ radius of the given cylinder = 10cm.

GIVEN : rectangular sheet of metal foil with dimension :

66 cm x 12cm ie. l × b

Height of the cylinder = 12cm

we take π =

TO FIND : volume of the cylinder = ?

PROCEDURE :

As the cylinder if formed from the rectangular foil, we can say that length of the rectangle is now the circumference of the base of the cylinder.

⇒ 66 = 2π r

So, r = = = = = = 10.5cm

∴ r = 10.5cm

Now, As we know, that

Volume of a cylinder “V” = π r²h

We put the values of r and h in the above formula, we get,

V = × (10.5)² × 12

= × 110.25 × 12

= 4158 cm³

∴ volume of the cylinder is 4158 cm³

GIVEN :

inner diameter of the cylinder of lead “r” = 1mm = 0.1cm

height of the pencil “h” = 28cm

outer radius of the pencil “R” = 3mm = 0.3cm

we take π =

TO FIND : volume of the wood used in the pencil “V” = ?

PROCEDURE :

Here, we will use the concept of hollow cylinder.

So, volume of the wood used in the pencil,

“V” = π × h× (R + r )× (R – r)

So, we put all the values in the above formula, and get

V = × 28 × (0.3 + 0.1)× (0.3 - 0.1)

= 22 × 4 × (0.4)× (0.2)

= 22 × 4 × 4 × 2 × 10^-2

= 704 × 10^-2

= 7.04 cm³

∴ volume of the wood used in the pencil is 7.04 cm³

GIVEN : radius of cone “r” = 20cm

Slant height of cone “l” = 29cm

we take π =

TO FIND : volume of the cylinder “V” = ?

PROCEDURE :

As we know that, volume of cone “V” = × π r²h

As we are not given the value of h, we have to find it.

As it is a right cone, l² = r² + h²

h² = l² – r²

= (29)² – (20)²

= 841 – 400

= 441

h² = 441

h = √ 441 = 21cm

So now, we put the values of r and h in the above formula

We get, V = × π r²h

= × π × (20)²× (21)

= × × 400 × 21

= 8800

∴ volume of the given cone = 8800cm³

GIVEN:

Height of the cone = 12m
circumference of the base = 44m

we take π =

TO FIND : volume of the cone = ?

PROCEDURE :

As we know the circumference of the cone = 2π r

2π r = 44

r = = = = = 7cm

now, as we know that, volume of cone “V” = × π r²h

putting all the values in the above formula, we get

V = × π × (7)²× (12)

= × × 49 × 12

= × 22 × 7 × 12

= 22 × 7× 4

= 616 cm³

∴ volume of the given cone is 616 cm³

GIVEN :

Volume of the frustum cone “V” =

Radius of another end = “r”

Radius (R) = 8 cm

height (h) = 14 cm

we take π =

TO FIND : Radius of another end = ?

PRODEDURE :

As we know, the volume of frustum of cone,

×π ×h×(R² + r² + R×r) =

××(14)×(8² + r² + 8×r) =

(8² + r² + 8×r) = × 3 × ×

(64 + r² + 8r) = 258× 7×

(64 + r² + 8r) =

r² + 8r + 64 = 129

r² + 8r + 64 -129 = 0

r² + 8r - 65 = 0

(r + 13) (r - 5) = 0

r + 13 = 0 or r - 5 = 0

r = -13 or r = 5 cm

as the radius cannot be negative,

∴ the required radius = 5cm.

GIVEN :

Perimeter of the upper end of frustum of a cone = 44cm

Perimeter of the lower end of frustum of a cone = 8.4cm

Height of the frustum of the cone is 14 cm

we take π =

TO FIND : volume of the frustum “V” = ?

PROCEDURE :

As we know, the volume of frustum of cone,

V = ×π ×h×(R² + r² + R×r)

Now first, for upper end,

2π R = 44

R = = = = 7cm

For the lower end,

2π r = 8.4π

2r = 8.4

r = 4.2cm

now that we have all the values, we will find the volumeof frustum.

V = ×π ×h×(R² + r² + R×r)

= ×× 14× (7² + (4.2)² + 7× 4.2)

= ×22 × 2 × (49 + 29.4 + 17.64)

= ×96.04

= 44 × 32.013

= 1408.57cm³

∴ volume of frustum is 1408.57cm³

GIVEN : Δ ABC with right angle at B

AC will be the hypotenuse = 13cm

AB = 12cm

BC = 5cm

we take π =

TO FIND : volume of the solid generated = ?

PROCEDURE :

The solid formed on rotating the ΔABC about AB is a cone with radius “r” = 5cm

And height “h” = 12cm

∴ volume of the cone “V” = × π r²h

= × × (5)² × 12

= × 25 × 4

= cm³

∴ volume of the solid generated ie. Cone is cm³

GIVEN : radius of cone = ”r”

Height of cone = “h”

=

Volume of the cone = 100.48 cu.cm

We take π = 3.14

TO FIND : slant height “l” = ?

PROCEDURE :

We know that volume of the cone “V” = × π r²h

Now =

⇒ r =

Now putting the values in the formula of volume, we get

V = × π r²h

100.48 = × 3.14 × ()² × h

100.48 = × 3.14 × × h³

h³ = 100.48 × 3 × ×

h³ = 32 × 3 × = 8× 27

h = ∛ 8× 27 = ∛ 2× 2× 2× 3× 3× 3

h = 2× 3 = 6cm

now, r = = = 4cm

now, slant height “l” can be found using :

l² = r² + h²

= 4² + 6² = 16 + 36 = 52

L = √ 52 = 2√13cm

∴ slant height of the given cone is 2√13cm.

GIVEN : volume of the cone “V” = 216cu.cm

Base radius of cone “r” = 9cm

we take π =

TO FIND : height of the cone “h” = ?

PROCEDURE :

We know that volume of the cone “V” = × π r²h

Now putting all the values in the above formula, we get

V = × π r²h

216 = × π × (9)² × h

h = 216 × 3 × = 216 × = 8cm

∴ height of the cone “h” is 8cm.

GIVEN : no. of steel balls “n” = 200

Radius of each ball “r” = 0.7cm

we take π =

Density of steel = 7.95 g/cm³

TO FIND : Mass of 200 balls = ?

PROCEDURE :

We know that, volume of sphere "V” = × π r³

Putting the values in the above formula, we get

V = × × (0.7)³

= × × 0.343

= 1.437cm³

This was the volume of one ball.

Now volume of 200 balls “V’ ” = n× V = 200 × 1.437cm³

V’ = 287.466 cm³

Now, it is given that, Density of steel = 7.95 g/cm³

So, mass of the balls = volume × density

Mass of 200 balls = volume of 200 balls × density

= 287.466 cm³ × 7.95 g/cm³

= 2285.46g

= 2.285 Kg

∴ Mass of 200 balls bearings is 2.285 Kg.

GIVEN : outer radius of hollow the sphere “R” = 12cm

inner radius of hollow the sphere “r” = 10cm

we take π =

TO FIND : volume of the hollow sphere = ?

PROCEDURE :

We know that volume of the hollow sphere “V” = × π × (R³ – r³)

Putting all the values in the above formula, we get

V = ×× (12³ – 10³)

= ××(1728-1000)

= ××728

= 3050.66 cm³

∴ volume of the hollow sphere is 3050.66 cm³

GIVEN : volume of a solid hemisphere “V” = 1152cu.cm

we take π =

TO FIND : curved surface area (CSA) = ?

PROCEDURE :

As we know that volume of a hemisphere “V” = × × π ×r³

So, putting all the values in the above formula, we get

V = × × π ×r³

1152 = × × π ×r³

1152 = × × r³

r³ = 1152 × 2 × = 1152 × = 576×3 = 1728

r = ∛ 1728 = 12cm

so, now that we have radius, we can find the curved surface area of the hemisphere.

Now, CSA = 2× π ×r²

= 2×π × (12)²

= 2×π ×144

= 288π cm²

∴ curved surface area of the hemisphere is 288π cm²

GIVEN : side of cube ”s” = 14cm

we take π =

TO FIND : volume of the largest right circular cone that can be cut out of a cube = ?

PROCEDURE :

For the cone to largest, its diameter should be equal to side of the cube ie. d = 14cm

And, height of the cone should also be equal to the side of the cube ie. h = 14cm

Now, radius of the cone “r” = 7cm

Now, volume of cone “V” = × π r²h

Now putting all the values in the above formula, we get

V = × π × (7)² × 14

= × × 7 × 7 × 14

= × 22 × 7× 14

= 718.666cm³

∴ volume of the largest right circular cone that can be cut out of a cube is 718.666cm³.

GIVEN : radius of balloon before the increase “r” = 7cm

radius of balloon after the increase “R” = 14cm

TO FIND : ratio of volumes of the balloon in both cases

PROCEDURE :

Volume of balloon before the increase “V1” = ×π ×r³

Volume of balloon after the increase “V2” = ×π ×R³

Ratio of the volumes in the 2 cases =

= = = = = =

∴ the ratio of volumes of the balloon in the two cases is = ie. 1 : 8.

A play-top is made up of a hemisphere (at the top) and cone(at the bottom).

Given, Diameter of the hemisphere = 3.6 cm

Radius of the hemisphere =

= 1.8 cm

Total height of the play-top = 4.2 cm

Height of cone = (Total height of the play-top)-( Radius of the hemisphere)

⇒ Height of cone = 4.2-1.8

= 2.4 cm

Now, Total surface area of play-top = Lateral surface area of hemisphere+ Lateral surface area of cone

Lateral surface area of hemisphere = 2r²

= 21.81.8

= 6.48

Lateral surface area of cone = rl ,

where l =

⇒ l =

⇒ l =

⇒ l =

⇒ l = 3

So, Lateral surface area of cone = 1.83

= 5.4

Total surface area of play-top = 6.48+5.4

= 11.88 cm²

Total surface area is 11.88 cm².

Given, Diameter = 21 cm

Total height of the solid = 25.5 cm

Radius =

= 10.5 cm

Height of cylinder = 25.5-10.5

= 15 cm

Volume of cylinder = r²h

= 10.510.515

= 1653.75

Volume of hemisphere = r³

=

= 771.75

Volume of solid = Volume of cylinder+ Volume of hemisphere

= 1653.75 +771.75

= 2425.5

= 7616.07 cm³

Volume of solid is 7616.07 cm³

Given, Diameter of the capsule = 5 mm

Length of the entire capsule = 14 mm

Radius of the capsule = mm

= 2.5 mm

Height of cylinder = 14-5

= 9 mm

Lateral surface area of cylinder = 2rh

= 22.59

= 45 mm²

Lateral surface area of hemisphere = 2r²

= 22.52.5

= 12.5 mm²

Surface area of the solid = 12.5 +45 +12.5

= 70

= 70

= 220 mm²

Surface area of capsule is 220 mm².

Given, Total height = 13.5 m

Diameter of the base = 28 m

Height of the cylindrical portion = 3 m

Height of cone = 13.5-3

= 10.5 m

Radius of the base =

= 14 m

Lateral surface area of cylinder = 2rh

= 2143

= 84 m²

Lateral surface area of cone = rl ,

where l =

⇒ l =

⇒ l =

⇒ l =

⇒ l = 17.5

So, Lateral surface area of cone = 1417.5

= 245 m²

Total surface area = Lateral surface area of cylinder+ Lateral surface area of cone

Total surface area = 84 +245

= 329

= 329

= 1034 m²

Total surface area of the tent is 1034 m².

Given, height of cone(h) = 48 cm

Base radius(r) = 12 cm

Volume of cone = r²h

= 121248

= 2304

We know,

Volume of sphere = r³

To find radius of sphere,

r³ = 2304

⇒ r³ = 2304

⇒ r³ = 5763

⇒ r³ = 1728

⇒ r = 12 cm

Radius of the sphere is 12 cm.

Given, Radius of a solid sphere(r) = 24 cm

Radius of wire (R) = 1.2 mm = 0.12 cm

length of the wire(l) = ?

Solid sphere is melted and drawn into a long wire of uniform cross section. Thus,

Volume of sphere = Volume of wire(cylinder)

⇒ r³ = r²l

⇒ (24)³ = (0.12)²l

⇒ (24)³ = (0.12)²l

⇒ l =

⇒ l =

⇒ l = 128100100

⇒ l = 12.8 km

The length of the wire is 12.8 km.

Given, Radius of right circular conical vessel(r) = 5 cm

Height of right circular conical vessel(h) = 24 cm

Radius of cylinder(R) = 10 cm

Height of cylinder(H) = ?

As water is emptied into an empty cylindrical vessel,

Volume of right circular cone = Volume of cylinder

⇒ r²h = R²H

⇒ (5)²24 = (10)²H

⇒ H =

⇒ H = 2 cm

Height of the water level in the cylindrical vessel is 2 cm.

Given, Diameter of the sphere(d) = 6cm

Radius of the sphere (r) = = 3cm

Diameter of the cylindrical vessel(D) = 12 cm

radius of the cylindrical vessel (R) = = 6 cm

Let level of water raised be h cm.

Volume of the water raised in the vessel = Volume of the sphere

⇒ R²h = r³

⇒ h =

⇒ h =

⇒ h = 1 cm

Level of water raised in the vessel is 1 cm

We need to find volume of water (in liters) discharged through the pipe in half an hour.

Half an hour = 30 minutes

Let us change minutes into seconds. For that we have to multiply 30 by 60. Because 60 seconds = 1 minute.

⇒ 30 60 = 1800 seconds.

Volume of water discharged in half an hour = r² 5 1800

= (7)² 5 1800

= 7 7 5 1800

= 22 7 5 1800

= 1386000 cubic cm ()

= (1000 cm³ = 1 liter)

= 1386 liters

Volume of water discharged through the pipe in half an hour is 1386 liters.

Given,

Diameter of the cylindrical tank = 4 m

Radius of the cylindrical tank = 2 m

Height of the tank = 10 m

Diameter of the cylindrical pipe = 10 cm

Radius of the cylindrical pipe = = 5 cm

= 0.05 m

speed of water = 2.5 km/hr

= 2.5 x 1000 (1000 m = 1 km)

= 2500 m/hr

⇒ Volume of water discharged from the cylindrical pipe = Volume of cylindrical tank

⇒ Area of cross section time speed = r² h

⇒ r² time speed = r² h

⇒ ()² time 2500 = (2)² (10)

⇒ () () time 2500 = (2) 2 (10)

⇒ Time = 2 10 () () ()

⇒ Time =

= 3.2 hour

= 3 hours 0.260 mins(1 hour = 60 minute)

= 3 hour 12 minute

Time taken to empty the half of the tank is 3 hour 12 minutes.

Volume of sphere = r³

⇒ (18)³ = (2)³ + (12)³ + r³

⇒ (18)³ = [(2)³ + (12)³ + r³ ]

⇒ (18)³ = [(2)³ + (12)³ + r³ ]

⇒ 5832 = 8 + 1728 + r³

⇒ 5832 = 1736 + r³

⇒ 5832-1736 = r³

⇒ 4096 = r³

⇒ r =

⇒ r = 16 cm

Radius of the third sphere is 16 cm.

Given, Length of cylindrical pipe (h) = 40 cm

Internal radius of the pipe (r) = 4 cm

External radius of the pipe (R) = 12 cm

Height of cylinder = 20 cm

Volume of hollow cylindrical pipe = Volume of cylinder

⇒ h (R² - r²) = r² h

⇒ h (R² - r²) = r² h

⇒ (40) (12² - 4²) = r² (20)

⇒ (40) (144 - 16) = r² (20)

⇒ (40) = r²

⇒ r² = (40)

⇒ r² = 2 128

⇒ r² = 256

⇒ r =

⇒ r = 16 cm

Therefore radius of cylinder = 16 cm

Let the number of lead shots be n.

Given,

Diameter of right circular cone = 8 cm

Radius of right circular cone (r) = 4 cm

Height of right circular cone (h) = 12 cm

Radius of spherical lead shot (r) = 4 mm

= cm (10 mm = 1 cm)

Volume of right circular cone = nNumber of spherical lead shots

n =

⇒ n =

⇒ n = (4)² (12) ()³

⇒ n = 4 4 12

⇒ n = 3 5 5 10

⇒ n = 750 lead shots

Therefore 750 lead shots can be made.

Given, Diameter of right circular cylinder = 12 cm

Radius of right circular cylinder (r₁) =

= 6 cm

Height of right circular cylinder (h₁) = 15 cm

Volume of Cylindrical ice-cream container = r₁²h₁

= 6 6 15

= cm³

Diameter of cone = 6 cm

Radius of cone (r₂) =

= 3 cm

Height of cone (h₂) = 12 cm

Radius of hemisphere = radius of cone = 3 cm

⇒ Volume of cone full of ice-cream = volume of cone + volume of hemisphere

= r₂²h₂ + r₂³

= ( r₂²h₂ + 2r₂³)

= (3²× 12 + 2× 3³)

= (9 ×12 + 2 × 27)

= (108 +54)

= 162

= cm³

Let n be the number of cones full of ice cream.

⇒ Volume of Cylindrical ice-cream container = nVolume of one cone full with ice cream

= n

⇒ N =

⇒ n = 10

Hence, the required number of cones = 10

Given,

Length of container(l) = 4.4 m

Breadth of container(b) = 2 m

Height of container(h) = 4 m

Radius of cylindrical vessel(r) = 40 cm

⇒ Volume of container = Volume of cylindrical vessel

⇒ lbh = r²h

⇒ 4.424 = (40)²h

⇒ H =

⇒ H = 70 cm

The height of the water level in the cylinder is 70 cm.

Given, Height of cylindrical bucket(h) = 32 cm

Radius of cylindrical bucket(r) = 18 cm

Height of conical heap (H) = 24 cm

Radius of conical heap (R) = ?

⇒ Volume of cylindrical bucket = Volume of conical heap

⇒ r²h = R²H

⇒ 181832 = R²24

⇒ 181832 = R²8

⇒ R² =

⇒ R² = 18184

⇒ R =

⇒ R = 182

⇒ R = 36 cm

Slant height (L) =

L =

L =

L =

L = 12 cm

Slant height = 12 cm

Radius of conical heap = 36 cm

Given, Depth of well = Height of well = 20 m

Diameter of well = 14 m

Radius of well =

= 7 m

Volume of well = r²h

= 7720

= 3080 m³

⇒ Volume of platform = Volume of well

⇒ lbh = 3080

⇒ 2014h = 3080

⇒ H =

⇒ H = 11 m

The height of the platform is 11 m.

Radius (R) = 1 cm

Height (H) = 1 cm

Curved Surface Area = (2πRH) = 2π cm²

Out of all the four Options,

Correct Option: B

Total Surface Area = 2πrh + 2πr²

where r is the radius and h is the height

According to the problem r

Total surface area

⇒ Total surface area

⇒ Total surface area

Out of all the four Options,

Correct Option : C

Base Area (A) = 80 cm²

Height (H) = 5 cm

Volume = (A×H) = 400 cm³

Out of all the four Options,

Correct Option : A

Total Surface Area = 2πrh + 2πr²

where r is the radius and h is the height

Total Surface Area = 200π cm² (Given)

Radius (r) = 5 cm

Equating the Total Surface Area we can say

2πrh + 2πr² = 200π

⇒ rh + r² = 100

⇒ r(r + h) = 100

⇒ r + h = 20 cm

Out of all the four Options,

Correct Option : A

Radius (R) = a units

Height (H) = b units

Curved Surface Area = (2πRH) = 2πab sq. units

Out of all the four Options,

Correct Option: B

Volume of Cylinder = 3 × Volume of Cone

Volume of the cylinder = 120 cm³ (Given)

⇒ Volume of Cone

⇒ Volume of Cone = 40 cm³

Out of all the four Options,

Correct Option : C

Diameter = 12 cm.

⇒ Radius (R) = 6 cm.

Height (H) = 8 cm

Slant height (l) = √(R² + H²) = √(6² + 8²)

⇒ Slant height (l) = √100 = 10 cm

Out of all the four Options,

Correct Option : A

Base Circumference = 120π cm

Let radius be r

Base Circumference = 2πr

2πr = 120π

⇒ r = 60 cm

Slant Height (l) = 10 cm

Curved Surface Area = π×r×l

Curved Surface Area = 600π cm²

Out of all the four Options,

Correct Option :B

Volume of cone = 48π cm³

Base Area = 12π cm²

Volume of cone

Height

⇒ Height

⇒ Height = 12 cm.

Out of all the four Options,

Correct Option :D

Base Area (A) = 48 cm²

Height (H) = 5 cm

Volume

⇒ Volume

⇒ Volume = 80 cm³

Out of all the four Options,

Correct Option : C

Let R₁,R₂ be the radius and H₁,H₂ be the height V₁,V₂ of the two cylinders

Volume Of Cylinder = πR²H

So we can say,

Out of all the four Options,

Correct Option : C

Radius of sphere (r) = 2 cm

Surface Area of Sphere = 4πr²

⇒ Surface Area of Sphere = 4π × 2²

⇒ Surface Area of Sphere = 16π cm²

Out of all the four Options,

Correct Option :D

Diameter = 2 cm

Radius (r) = 1 cm

Total surface Area = Curved surface Area + Base Area

⇒ Total surface Area = 2πr² + πr² = 3πr²

⇒ Total surface Area = 3π × 1² = 3π cm²

Out of all the four Options,

Correct Option : D

Let radius of sphere be r cm

Volume of Sphere

Volume of Sphere

So we can say that

Out of all the four Options,

Correct Option : B

Let R₁,R₂ be the radius of the two spheres and V₁,V₂ be the Volume of this two spheres, A₁,A₂ be the area of two spheres

Surface Area of Sphere = 4πr²

So we can say,

…Equation (i)

Volume of Sphere

Putting the values from Equation (i) we get

Out of all the four Options,

Correct Option : D

Radius = a units

Total surface Area = Curved surface Area + Base Area

⇒ Total surface Area = 2πa² + πa² = 3πa² sq. units

Out of all the four Options,

Correct Option : B

Let radius of sphere be r cm

Surface Area of Sphere = 4πr²

Surface Area of Sphere = 100π cm²

So we can say,

4πr² = 100π

⇒ r² = 25

⇒ r = 5 cm

Out of all the four Options,

Correct Option: C

Let radius of sphere be r cm

Surface Area of Sphere = 4πr²

Surface Area of Sphere = 36π cm²

So we can say,

4πr² = 36π

⇒ r² = 9

⇒ r = 3 cm

Volume of Sphere

⇒ Volume of Sphere

⇒ Volume of Sphere = 4π × 3²

⇒ Volume of Sphere = 36π cm³

Out of all the four Options,

Correct Option : B

Total surface Area = Curved surface Area + Base Area

⇒ Total surface Area = 2πr² + πr² = 3πr²

Total surface area of a solid hemisphere = 12π cm²(Given)

3πr² = 12π

⇒ r² = 4

⇒ r = 2cm

Curved surface Area = 2πr²

Curved surface Area = 2π×2²

Curved surface Area = 8π cm³

Out of all the four Options,

Correct Option : D

Let R₁,R₂ be the radius of the two spheres and V₁,V₂ be the Volume of this two spheres

Volume of Sphere

R₂ = 2R₁

So we can say,

Out of all the four Options,

Correct Option :A

Curved Surface Area = 24 cm²

Let radius be r

Curved Surface Area of solid sphere = 4πr²

4πr² = 24

Total surface Area of hemisphere = 2πr² + πr² = 3πr²

Total surface Area of hemisphere

Total surface Area of hemisphere = 18 cm²

Out of all the four Options,

Correct Option :D

Curved Surface Area = πRL

where R is the radius and L is the slant height

Let L₁,L₂ be the slant heights of the two cone and A₁,A₂ be the area of this two cones

So we can say,

Out of all the four Options,

Correct Option : C