Previous Year Questions: Surface Areas & Volumes

Ans: (d)
Given, radius of sphere= radius of cone= r 
Also, volume of sphere= volume of cone

Ans: Let a be side of one cube and b be side of another cube. 
Given, ratio of their volumes= 8:125

Ratio of their surface are 

Ans: Let r and R be the radius of cone and cylinder respectively and h and H be the height of cone and cylinder respectively. 
The volume of cylinder (V1) = πR²H
Volume of cone (V2) =

∴  Ratio of their volumes is given by V₁/V₂

∴ Ratio of their volumes= 9: 8

Ans: (c)
Also, total surface area of a sphere of radius r is 4πr². 
So, Assertion is true, but Reason is false. 

Ans: (c)

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

For the cone of greatest possible volume hollowed out from the cylinder:

• Radius of cone = r
• Height of cone = h

Step 1: Volume of the cylinder

V_cylinder = πr²h

Step 2: Volume of the cone

V_cone = (1/3)πr²h

Step 3: Volume of remaining wood

V_remaining = Volume of cylinder - Volume of cone

= πr²h - (1/3)πr²h

= (2/3)πr²h

Step 4: Required ratio

Volume of remaining wood : Volume of cone

= (2/3)πr²h : (1/3)πr²h

Cancel common terms:

= 2 : 1

Answer: The required ratio is 2 : 1.

Ans: Given, diameter of conical part = Diameter of hemispherical part= 4 cm 
Radius of conical part (r) = radius of hemispherical part (r) = 4/2 = 2 cm.

∴ Volume of toy= Volume of hemisphere+ Volume of cone =

∴ Surface area of toy= Curved surface area of cone+ Curved surface area of hemisphere
Curved Surface area of cone

Curved Surface area of hemisphere
Now,

Ans: 
Step 1: Volume of conical vessel (filled with water)
Given:

• $\hspace{0.17em}cmR\; =\; 5\; \backslash ,\; \backslash text\{cm\}$R = 5cm
• $\hspace{0.17em}cmh\; =\; 8\; \backslash ,\; \backslash text\{cm\}$h = 8cm

Step 2: Amount of water that flows out

One-fourth of the water flows out:
This volume equals the volume displaced by the spherical lead shots.
Radius of sphere: $\hspace{0.17em}cmr\; =\; 0.5\; \backslash ,\; \backslash text\{cm\}$r = 0.5cm

Ans: Height of largest cone= Side of cube 
H = 14 cm 
Diameter of largest cone= Side of cube 
∴ 2R = 14cm 
R=7cm 
∴ Volume of the largest cone = 1/3πR²H

Volume of remaining solid = (side)³ - Volume of cone 

Slant height of cone

L = 15.4 cm 
Surface area of remaining solid= 6(side)² - πR² + πRL 

Ans:(a)

Given:

• Radius r = 5 cm
• Height of cone h = diameter = 2r = 10 cm

Step 1: Volume of cone

V_cone = (1/3)πr²h

= (1/3) × π × (5)² × 10

= (1/3) × π × 25 × 10

= 250π / 3 cm³

Step 2: Volume of hemisphere

V_hemisphere = (2/3)πr³

= (2/3) × π × (5)³

= (2/3) × π × 125

= 250π / 3 cm³

Step 3: Total volume of toy

V = Volume of cone + Volume of hemisphere

= 250π/3 + 250π/3

= 500π / 3 cm³

Using π = 3.14,

V = (500 × 3.14) / 3

V = 523.6 cm³ (approx)

Answer: Volume of the toy = 523.6 cm³.

(b)

Given: Radius of hemisphere r = 3.5 cm

Step 1: Smallest edge of cube

For the hemisphere to lie completely on the cube, the side of the cube must be equal to the diameter of the hemisphere.

Edge of cube a = 2r

a = 2 × 3.5 = 7 cm

Step 2: Surface area of cube

Surface area of cube = 6a²

= 6 × (7)²

= 6 × 49

= 294 cm²

Step 3: Area of circular base covered by hemisphere

Area = πr²

= (22/7) × (3.5)²

= (22/7) × 12.25

= 38.5 cm²

Step 4: Curved surface area of hemisphere

CSA = 2πr²

= 2 × 38.5

= 77 cm²

Step 5: Total surface area of solid

Total surface area = Surface area of cube - area covered + curved surface area of hemisphere

= 294 - 38.5 + 77

= 332.5 cm²

Ans:

The pole consists of two solid cylinders.

For the smaller cylinder:

• Radius r₁ = 7 cm
• Height h₁ = 50 cm

For the bigger cylinder:

• Diameter = 28 cm ⇒ Radius r₂ = 14 cm
• Height h₂ = 200 cm

Density of iron = 8 g/cm³

Step 1: Volume of smaller cylinder

V₁ = πr₁²h₁

V₁ = π × 7² × 50

V₁ = 2450π cm³

Step 2: Volume of bigger cylinder

V₂ = πr₂²h₂

V₂ = π × 14² × 200

V₂ = 39200π cm³

Step 3: Total volume of pole

V = V₁ + V₂

V = 2450π + 39200π

V = 41650π cm³

Using π = 22/7,

V = 41650 × 22 / 7

V = 130900 cm³

Step 4: Mass of the pole

Mass = Volume × Density

Mass = 130900 × 8

Mass = 1047200 g

Converting into kilograms:

Mass = 1047.2 kg

Therefore, Mass of the iron pole = 1047.2 kg

Ans: Here, two figures are combined, 2 hemisphere and cylinder.

Given:

• Total length = 14 mm
• Diameter = 4 mm ⇒ Radius r = 2 mm

Step 1: Height of cylindrical part

h = Total length - diameter

h = 14 - 4

h = 10 mm

(a) Surface Area of Capsule

Surface area = Curved surface area of cylinder + Surface area of sphere

Surface area = 2πrh + 4πr²

Substituting values:

= 2π × 2 × 10 + 4π × 2²

= 40π + 16π

= 56π

Using π = 22/7,

Surface area = 56 × 22 / 7

Surface area = 176 mm²

Surface Area = 176 mm²

(b) Volume of Capsule

Volume = Volume of cylinder + Volume of sphere

Volume = πr²h + (4/3)πr³

Substituting values:

= π × 2² × 10 + (4/3)π × 2³

= 40π + (32/3)π

= (120π + 32π) / 3

= 152π / 3

Using π = 22/7,

Volume = (152 / 3) × (22 / 7)

Volume = 159.24 mm³

Volume = 159.24 mm³

Ans: (c)
We have, the height of cone. h = 24 cm and radius, r = 7 cm.

We know that,
=
= √625

= 25
Now. curved surface area = πrl
= 22/7 x 7 x 25
= 550 cm²

Ans: (b)
Curved surface area of cylinder = 2πrh
⇒ 94.2 = 2 x 3 .14 x r x 5
⇒⇒ r = 3 cm

Ans: 

Radius of the cylinder (r) = 3.5 cm
Height of the cylinder (h) = 10 cm

Curved surface area = 2πrh

= ( 2 × 22/7 × 35/10 × 10 ) cm²

= 220 cm²

Curved surface area of a hemisphere = 2πr²

∴ Curved surface area of both hemispheres
2 × 2πr² = 4πr² = ( 4 × 22/7 × 35/10 × 35/10 ) cm²

= 154 cm²

Total surface area of the Article

= (220 + 154) cm²

= 374 cm²

Ans: 

Ans: Radius of cone, r = 3 cm 
Height of cone, h = 12 cm 
Let x be the volume of unfilled part of cone.

Now, volume of cone, =

1/3 πr²h = 1/3 × 3.14 × (3)² × 12

Volume of filled part of cone = Volume of cone - Volume of unfilled part of cone

= 1/3 × 3.14 × (3)² × 12 - ((1/6) × 1/3 × 3.14 × (3)² × 12)

= 1/3 × 3.14 × (3)² × 12 × (1 - (1/6))

= 5/6 × 3.14 × (3)² × 12 = 94.2 cm³

Now, volume of ice-cream = volume of filled part of cone + volume of hemisphere

= 94.2 + (23 × 3.14 × (3)³)
= 150.72 cm³

Ans: Given ratio of radius and height of the right circular cylinder = 2:3
Let radius (r) of the base be 2x and height(h) be 3x.
Volume of cylinder, V = πr²h

1617 = 22/7 × (2x)² × 3x

⇒ 1617 = 22/7 × 4x² × 3x

x³ = 11319/264
x³ = 42.875
x = 3.5

Radius r = 2x = 2 × 3.5 = 7 cm

and height h = 3x = 3 × 3.5 = 10.5 cm

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

= 2 × 22 /7 × (10.5 + 7)

= 44 × 17.5 cm²

= 770 cm²

Ans: Paper required to make four caps is 2,200 sq.cm.
Weight of the cake for required dimensions is 1kg.
Step-by-step explanation:
(a) Given the base circumference of the cone, c = 44 cm
Height of a cone, h = 24 cm.
Base circumference of the cone, c = 2πr = 44 cm
Thus, the radius of the cone is

r = 44/2 × 22/7 = 7 cm

The curved surface area of the cone is given by

CSA = πr √ h² + r²

Substituting the values of h and r,

CSA = 227 × 7 × √ (24)² + 7²

= 22 × √ 576 + 49 = 22 × √ 625

= 22 × 25 = 550 sq. cm
Thus, to make one cap, 550 sq.cm of paper is required.
Then to make four caps, the required paper is
550 x 4 = 2200 sq. cm
Therefore, 2,200 sq.cm of paper is required to make four caps.
(b) Given the diameter of cylindrical shape cake, d = 24 cm
Height of cylindrical shape cake, h = 14 cm.
Radius of the cylindrical shape cake,

r = d2 = 242 = 12 cm

Volume of the cylinder is given by V = πr²h

Substituting the values of h and r,

V = 227 × (12)² × 14

V = 227 × 144 × 14

= 22 × 144 × 2 = 6,336 cm³

The required volume of the cylindrical shape cake is 6,336 cm³.

Given 650 cm³ equals 100 g of cake.

Then the required weight of the cake is

6336650 × 100 = 974.76 g

Given the bakery shop sells cakes by weight of 0.5 kg, 1 kg, 1.5 kg, etc.

Since, 974.76 g ≈ 1 kg, therefore, the cake of 1 kg should be ordered for required dimensions.

Ans: The dimension of the cuboids so formed are
length = 18 cm
breath = 6 cm and height = 6 cm.
Surface area of cuboids = 2 (l× b + b × h + l × h)
= 2 × (18 × 6 + 6 × 6+ 18 × 6)
= 504 cm²

Ans: According to given information, we have the following figure.

Clearly, the radius of conical part = radius of cylindrical part = 30/2 = 15 m = r  ...(say)
Let h and H be the height of conical and cylindrical part respectively.
Then h = 8 m and H = 9 m

= 17 m

(i) The area of the canvas used in making the tent
= Curved surface area of cone + Curved surface area of cylinder
= πrl + 2πrH
= πr(l + 2H)

= 1650 m²
(ii) Area of canvas bought for the tent
= (1650 + 30) m²
= 1680 m²
Now, this cost of the canvas height for the tent
= ₹ (1680 × 200)
= ₹ 3,36,000

Ans: Given diameter of circular pipe = 8 cm
So, radius of circular pipe = 4cm
Length of flow of water in one sec = 80 cm
length of flow of water in one hour =  80 x 60 x 60 cm=288000 cm=h
Volume of cylinderical pipe in one hour = πr²h

= 14482.28 litre [approx.]
14482.28 litres of water being pumped out through this pipe in 1 hr.

Ans: (d)

Diameter of sphere = Distance between opposite faces of cube = 2a
Radius of sphere = a
So, volume of spherical ball = 4/3 πr³
= 4/3 πr³

Ans: (c)
Let radius of the sphere be r.
According to the question,
⇒

Ans: Let height of one cone be h and height of another cone be 3h. Radius, of one cone is 3r and radius of another cone is r.
∴ Ratio o f their volumes =
= 3 : 1

Ans: Let n be the number of solid cubes of  2cm made from a solid cube of side 10 cm.

∴ n x Volume of one small cube = Volume of big cube
⇒ n x (2)³ = (10)³
⇒ 8n = 1000
⇒ n = 1000/8
= 125
Thus, the number of solid cubes formed of side 2 cm each is 125.

Ans: Let the radius and the height of the cylinder are r and h respectively.
So, radios of t he cone is r and height of the cone is 3h.
∴ Volume of the cylinder = πr2h
So Volume of cone =
So, require ratio =
= 1 : 1

Ans: Given that, the depth of the well is 14 m and the diameter is 3 m.
The width of the circular ring of the embankment is 4 m.
A figure is drawn below to visualize the shapes.

From the above figure, we can observe that the shape of the well will be cylindrical, and earth evenly spread out to form an embankment around the well in a circular ring will be cylindrical in shape (Hollow cylinder) having outer and inner radius.

Volume of the earth taken out from well = Volume of the earth used to form the embankment

Hence, Volume of the cylindrical well = Volume of the hollow cylindrical embankment

Let us find the volume of the hollow cylindrical embankment by subtracting volume of inner cylinder from volume of the outer cylinder.

Volume of the cylinder = πr²h where r and h are the radius and height of the cylinder respectively.

Depth of the cylindrical well, = h₁ = 14 m

Radius of the cylindrical well, = r = 3/2 m = 1.5 m

Width of embankment = 4 m

Inner radius of the embankment, r = 3/2 m = 1.5 m

Outer radius of the embankment, R = Inner radius + Width

R = 1.5 m + 4 m

= 5.5 m

Let the height of embankment be h

Volume of the cylindrical well = Volume of the hollow cylindrical embankment

πr²h₁ = πR²h - πr²h

πr²h₁ = πh (R² - r²)

r²h₁ = h (R - r )(R + r)

h = [(r²h₁)/(R - r)(R + r)]

h = [((1.5 m)² × 14 m)/(5.5 m - 1.5 m)(5.5 m + 1.5 m)]

= (2.25 m² × 14 m)/(4m × 7 m)

= 1.125 m

Therefore, the height of the embankment will be 1.125 m.

Ans: Given diameter of conical part = Diameter of hemispherical part = 4cm
∴ Radios of conical part (r) = Radius of hemispherical part (r) = 4/2 = 2 cm
Height of conical part (h) = 2 cm
∴ Volume of toy = Volume of hemisphere + volume of cone

= 3 .14 x 4(1.33 + 0.66)= 3.14 x 4 x 1.99 cm³
Volume of the toy = 24.99 cm³

Ans: Radius of the cone = Radius of the hemisphere = r = 7cm 
Height of the cone, h = 10 cm 
Now, volume of the toy = volume of hemisphere + volume of cone

23 πr³ + 13 πr² h = πr²3 (2r + h)

= 13 × 227 × 7 × 7 (2×7 + 10)

= 22 × 7 × 243

= 1232 cm³

Curved surface area of the toy = Curved surface area of cone + Curved surface area of hemisphere

= πrl + 2πr²

= πr √ h² + r² + 2πr²

= 227 × 7 × √ (10)² + 7² + 2πr²

= 22 × √ 100 + 49 + 14

= 22 × √ 149 + 14

= 22(12.2 + 14)
= 22 x 26.2
= 576.4 cm³

Ans: Radius of cylinder = Radius of cone = r = 6 cm. 
Height of cylinder = Height of cone = h = 14 cm
Volume of remaining solid = πr²h - 1/3πr²h
= 2/3πr²h
= 2/3 x 22/7 x 6 x 6 x 14
= 2 x 22 x 2 x 6 x 2
= 1056cm³
Hence, the volume of the remaining solid is 1056 cm³.

Ans: Radius of cylinderical part (r) = Radius of each spherical part(r) = 7/2 cm
Height of cylinderical part (h) = 20 - 7/2 - 7/2 = 13 cm

Now. Volume of the solid = Volume of cylinderical part + Volume of two hemispherical endsVolume of the solid =

Volume of the solid = 680.17 cm³.

Ans: Base radius = 5/2 = 2.5 cm
Apparent capacity of glass = Volume of cylindrical portion
= πr²h
= 3.14 x (2.5)² x 10
= 196.25 cm³
Actual capacity of the glass = Volume of cylinder - Volume of hemisphere

= 196.25 - 32.71
= 163.54 cm³   

Ans: r = 12 cm, R = 20 cm, V = 12308.8 cm³ 

Volume of frustum = 13 πh (r² + R² + rR)

12308810 = 13 × 314100 × (144 + 400 + 240) × h

h = 123088 × 3 × 10314 × 784

h = 15 cm

l = √(h² + r²) = √(225 + 64) = √289 = 17 cm
Area of metal sheet used = π(R + r)l + πr² 
= π(20 + 12) x 17 + π x 144

Hence, the height of the bucket is 15 cm and the area of the metal sheet used is 2160.32 cm².

Ans: The total height of the bucket = 40 cm, which includes the height of the base. So, the height of the frustum of the cone = (40 - 6) cm = 34 cm.
Therefore, the slant height of the frustum,

l = √(R² + (r₁ - r₂)²)

where, r₁ = 452 cm = 22.5 cm

r₂ = 252 cm = 12.5 cm and h = 34 cm

So, l = √(34² + (22.5 - 12.5)²) = √(34² + 10²) = 35.44 cm

Area of the metallic sheet used

Curved surface area of frustum of cone + Area of circular base + Curved surface area of cylinder

= [π × 35.44 (22.5 + 12.5) + π × (12.5)² + 2π × 12.5 × 6] cm²

= 227 × (1240.4 + 156.25 + 150) cm²

= 4860.9 cm²

Now, the volume of water that the bucket can hold (also, known as the capacity of the bucket)

Volume of frustum:

= 13 π × h (r₁² + r₂² + r₁r₂)

= 13 × 227 × 34 × (22.5² + 12.5² + 22.5 × 12.5) cm³

= 13 × 227 × 34 × 943.75 = 33615.48 cm³

= 33.62 litres (approx.)

Ans: We have,
Radius of cylindrical bucket =18 cm
Height of cylindrical bucket = 32 cm
And the height of conical heap = 24 cm
Let the radius of the conical heap be r cm
Volume of the sand = volume of the cylindrical bucket
= πr²h = π x (18)² x 32
Now, volume of conical heap
Here, volume of the conical heap will be equal to the volume of sand
∴ 8πr² = π x (18)² x 32
⇒ r² = 18 x 18 x 4 = (18)² x (2)²
⇒ r² = (36)² or r = 36 cm

Ans: 

For bucket,

Upper diameter (D) = 30 cm

∴ Radius (R) = 302 = 15 cm

Lower diameter (d) = 10 cm

Radius (r) = 102 = 5 cm

Height of bucket (h) = 24 cm

The area of the metal sheet used = CSA of the frustum (bucket) + Area of bottom part (base)

= π(R + r)l + πr²

= π(R + r) × √(h² + (R - r)²) + πr²

[∵ Slant height (l) = √(h² + (R - r)²)]

= 3.14 × (15 + 5) × √(24² + (15 - 5)²) + 3.14 × (5)²

= 3.14 × 20 × √(576 + 100) + 3.14 × 25

= 62.8 × √676 + 78.5

= 62.8 × 26 + 78.5

= 1632.8 + 78.5 = 1711.3 cm²

Ans: Let VAB be the metallic right circular cone of height 20 cm. Suppose this cone is cut by a plane parallel to its base at a point O' such that VO' = O' O i.e., O' is the midpoint of VO.
Let r₁ and r₂ be the radii of circular ends of the frustum ABB' A'.
Now, in ΔVOA and VO' A', we have

⇒ tan 30° = OAVO and tan 30° = O'A'V'O'

⇒ 1√3 = r₁20 and 1√3 = r₂10

⇒ r₁ = 20√3 and r₂ = 10√3

∴ Volume of the frustum = 13 πh (r₁² + r₂² + r₁r₂)

= 13 × π × 10 × [ (20/√3)² + (10/√3)² + (20/√3 × 10/√3) ]

= 10π3 × [400/3 + 100/3 + 200/3]

= 10π3 × 7003 cm³

Let the length of wire of diameter 1/16 cm be l cm. Then,

Volume of metal used in wire

= π × (1/32)²× l = πl/1024 cm³

Since the frustum is recast into a wire of length l cm and diameter 1/16 cm,

∴ Volume of the metal used in wire = volume of the frustum

⇒ πl1024 = 7000π9

⇒ l = 7000 × 10249

= 796144.4 cm = 7964.444 m

Ans:  Diameter of metallic cylinder = 12 cm
∴ Radius of metallic cylinder (r) = 6 cm
Height (h) = 15 cm
Volume of cylinder = πr²h = π(6)² x 15 cm³
Radius of cone = 3 cm
Height of cone = 9 cm
Volume of cone = 1/3 π(3)² x 9 = 3 x 9π cm³
Number of toys so formed = 
If the question is "A tent is in the form of a cylinder surmounted by a cone. Find the capacity of the tent and the cost of canvas for making the tent at Rs 100 per sq.m.", then the solution is given as "Tent is a combination of a cylinder and a cone. For capacity
∴ Volume (capacity) of the tent = Volume of the cylindrical part + Volume of the conical part
For the cost of the canvas, we find the total surface area.
Total surface area = Curved area of the cylindrical part + Curved surface area of the conical part
Now, proceed to find its cost."
Note: Don't solve as "Total surface area of canvas = Total surface area of cylinder + Total surface area of a cone and proceed further"
This is the wrong solution.

Ans: Radius of conical vessel (r) = 5 cm
Height of conical vessel (h) = 24 cm
Radius of cylindrical vessel (R) = 10 cm

Let H be the height of water in the cylindrical vessel.

Now, the total volume of the conical vessel

= 13 πr²h = 13 × 227 × (5)² × 24 cm³

= 22 × 25 × 247 × 3 cm³

According to the question,

3/4 of the volume of water from the conical vessel is emptied into the cylindrical vessel.

⇒ 34 × Volume of conical vessel = Volume of water in the cylindrical vessel

⇒ 34 × 22 × 25 × 247 × 3 = πR²H

⇒ 3 × 22 × 25 × 244 × 7 × 3 = 227 × (10)² × H

⇒ 25 × 6 = 10 × 10 × H

⇒ H1 = 25 × 610 × 10

⇒ H = 1.5 cm

∴ Height of water in the cylindrical vessel = 1.5 cm

Ans: Length of roof (l) = 22 m
Breadth of roof (b) = 20 m
Let the height of water column collected on roof= hm
∴ Volume of standing water on rooftop = lbh
= (22 x 20 x h) m³ 
This water is taken into a cylindrical tank of diameter of base 2 m and height 3.5 m.
Tank gets completely filled with this amount of water.
⇒ Volume of water from roof-top = Volume of cylinder
Diameter of tank = 2 m
∴ Radius of tank = 2/2 = 1 m

⇒ Volume of tank = πr²h

= 227 × (1)² × 3.5 m³

A.T.Q.

(22 × 20 × h) m³ = 227 × 1 × 3.5 m³

⇒ h = 22 × 3.57 × 22 × 20

⇒ h = 777 × 22 × 30

= 0.025 m = 2.5 cm
Hence rainfall is 2.5 cm.

Ans: We have,
Radius of the cylinder = Height of the cylinder = 2.4 cm
Also, radius of the cone = 0.7 cm and height of the cone = 2.4 cm
Now, slant height of the cone =

∴ Total surface area of the remaining solid
= curved surface area of cylinder + curved surface area of the cone + area of upper circular base of the cylinder.

Ans: 

Total height of the tent above the ground = 27 m

Height of the cylindrical part,

h₁ = 6 m

Height of the conical part,

h₂ = 21 m

Diameter = 56 m

Radius = 28 m

Curved surface area of the cylinder, CSA₁

= 2πrh = 2π × 28 × 6 = 336π

Curved surface area of the cone, CSA₂

= πrl = π × 28 × √ h₂² + r²

= π × 28 × √ 21² + 28²

= 28π × √ 441 + 784

= 28π × 35

= 980π

Total curved surface area = CSA of cylinder + CSA of cone

= CSA₁ + CSA₂

= 336π + 980π

= 1316π

= 4136 m²

Ans: Volume of sphere = 4/3 π(6)³ cm³

Volume of water rise in cylinder

= πr² 329 cm³

∴ πr² 329 = 4/3 π(6)³

⇒ r² = 4 × 2 × 36 × 932 = 81

⇒ r = 9 cm

Ans: Volume of the smaller sphere

= 4/3 πr³ = 4/3 π(3)³ = 4/3 π(27) = 36π

Volume of smaller sphere × density = mass

∴ 36π (density of metal) = 1

Density of metal = 1/36π

∴ Volume of bigger sphere × density = mass

= 4/3 π (R)³ × (1/36π) = 7

R³ = (7 × 36 × 3)/4 = 7 × 9 × 3

(i) Volume of new sphere = volume of smaller sphere + volume of bigger sphere

4/3 π(R')³ = 4/3 πr³ + 4/3 πR³ (where R' is the radius of the new sphere)

= 4/3 π(3)³ + 4/3 π(7 × 9 × 3) [using (i)]

= 4/3 π [3³ + 7 × 9 × 3]

(R')³ = [3³ + 7 x 3³]
(R')³ = 3³(l + 7)
(R')³ = 3³ x 8
(R')³ =  3³ x 2³
R' =3 x 2
R' = 6 cm
∴ Diameter of new sphere =12 cm.

Ans: Radius of hemispherical bowl, R = 36/2 = 18 cm
Radius of cylindrical bottle, r = 6/2 = 3 cm
Let height of cylindrical bottle = h
Since 10% liquid is wasted, therefore only 90% liquid is filled into 72 cylindrical bottles.
∴ The volume of 72 cylindrical bottles = 90% of the volume in the bowl

⇒ 72 × πr²h = 90% of 23 πR³

72 × π × 3 × 3 × h = 90100 × 23 × π × 18 × 18 × 18

h = 90 × 2 × π × 18 × 18 × 18100 × 3 × π × 72 × 3 × 3

h = 275 = 5.4 cm

Ans: Let.BC = r cm, DE = 10 cm
Since B is the mid-point of AD and BC is parallel to DE, therefore C is the mid-point of AE.
∴ AC = CE
Also, ΔABC ~ ΔADE




∴ The required ratio = 1 : 7.