Surface Area and Volume of Cylinder Class 10 Concise Exe-20A ICSE Maths Selina Solutions Ch-20 Cylinder, Cone and Sphere (Surface Area and Volume). In this article you would learn how to solve problems / questions on Surface Area and Volume of Cylinder.  Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

Surface Area and Volume of Cylinder Class 10 Concise Exe-20A ICSE Maths Selina Solutions Ch-20 Cylinder, Cone and Sphere (Surface Area and Volume)

Board	ICSE
Publications	Selina
Subject	Maths
Class	10th
Chapter-20	Cylinder, Cone and Sphere (Surface Area and Volume)
Writer	R.K. Bansal
Exe-20A	Surface Area and Volume of Cylinder
Edition	2025-2026

Practice Problems / Questions on Surface Area and Volume of Cylinder with Solutions / Answer

Class 10 Concise Exe-20A ICSE Maths Selina Solutions Ch-20 Cylinder, Cone and Sphere (Surface Area and Volume)

Que-1: The height of a circular cylinder is 20 cm and the radius of its base is 7 cm. Find : (i) the volume (ii) the total surface area.

Sol: (i) For circular cylinder,
Height = h = 20 cm
Radius of the base = r = 7 cm
Volume of a cylinder = πr²h
= (22/7) × 7 × 7 × 20 𝑐⁢𝑚³
= 3080 cm³

(ii) For a circular cylinder ,
Height = h = 20 cm
Radius of the base = r = 7 cm
Total surface area of a cylinder = 2πr (h + r)
= 2 × (22/7) ×7⁢(20+7) cm²
=2 × 22 × 27⁢ cm²
= 1188 cm²

Que-2: The inner radius of a pipe is 2.1 cm. How much water can 12 m of this pipe hold?

Sol: Inner radius of pipe = 2.1 cm
Length of the pipe = 12 m = 1200 cm
∴ Volume = πr²h
= (22/7) × 2.1 × 2.1 × 1200 cm³
= 16632 cm³

Que-3: A cylinder of circumference 8cm and length 21cm rolls without sliding for 4*(1/2) seconds at the rate of 9 complete rounds per second. Find : (i) distance travelled by the cylinder in 4*(1/2) seconds and (ii) the area covered by the cylinder in 4*(1/2) seconds

Sol: (i) Circumference of cylinder = 8 cm
Therefore, radius = 𝑐/2⁢𝜋 = (8×7)/(2×22) = 14/11⁢𝑐⁢𝑚
Length of the cylinder (h) = 21 cm
If distance covered in one revolution is 8 cm, then distance covered in 9 revolution = 9 × 8 72
Therefore, distance covered in 4*(1/2) seconds
= 72 × (9/2) ⁢𝑐⁢𝑚
= 324 cm

(ii) Curverd surface area = 2πrh
= 2 × (22/7) × (14/11) × 21
= 168 cm²
Area covered in one revolution = 168 cm²
Area covered in 9 revolution = 168 cm² × 9 = 1512 cm²
Therefore, area covered in 1 second = 1512 cm²
Hence, area covered in 4*(1/2) seconds = 1512 𝑐⁢𝑚² × (9/2)
= 6804 cm²

Que-4: How many cubic meters of earth must be dug out to make a well 28 m deep and 2.8 m in diameter? Also, find the cost of plastering its inner surface at Rs 4.50 per sq meter.

Sol: Radius of the well = 2.82 =1.4⁢m
Depth of the well = 28 m
Therefore, volume of earth dug out = πr²h
= (22/7) × 1.4 ×1.4 × 28
=17248100
= 172.48 m³
Area of curved surface = 2πrh
=2 × (22/7) × 1.4 × 28
= 246.4 m²
Cost of plastering at the rate of ₹ 4.50 per sq m.
= ₹ 246.40 × 4.50
= ₹ 1108.80

Que-5: What length of solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of external diameter 20 cm, 0.25 cm thick and 15 cm long?

Sol: External diameter of hollow cylinder = 20 cm
Therefore , radius = 10 cm
Thickness = 0.25 cm
Hence, Internal radius = (10-0.25) = 9.75 cm
Length of cylinder (h) = 15 cm
∴volume = 𝜋⁢ℎ⁡(𝑅²−𝑟²) = 𝜋 × 15⁢(10²−9.75²)
= 15⁢𝜋⁢(100−95.0625) ⁢𝑐⁢𝑚³
= 15⁢𝜋 × 4.9375⁢ 𝑐⁢𝑚³
Diameter = 2⁢𝑐⁢𝑚
Therefore, radius (r) = 1 cm
Let h be the length
then, volume = 𝜋⁢𝑟²⁢ℎ = 𝜋⁢(1×1)⁢ℎ = 𝜋⁢ℎ
Now, according to given condition:
𝜋⁢ℎ =15⁢𝜋 ×4.9375
⇒ℎ =15 ×4.9375
⇒ℎ =74.0625
Length of cylinder =74.0625⁢𝑐⁢𝑚

Que-6: A cylinder has a diameter of 20 cm. The area of curved surface is 100 sq cm. Find: (i) the height of the cylinder correct to one decimal place. (ii) the volume of the cylinder correct to one decimal place.

Sol: (i) Diameter of the cylinder = 20 cm
Hence, Radius (r) = 10 cm
Height = h cm
Curved surface area = 2πrh
∴ 2πrh = 100 cm²
 ⇒2 × (22/7) × 10 × ℎ = 100
⇒ ℎ = (100×7)/(22×10×2)
= 35/22
⇒ h = 1.6 cm

(ii) Diameter of the cylinder = 20 cm
Hence, Radius (r) = 10 cm
Height = h cm
Volume of the cylinder = πr²h
= (22/7) ×10 ×10 ×1.6
= (22×100×1.6)/7
= 502.85/7 cm³
= 502.9 cm³
or
= (22/7) ×10 ×10 × (35/22)
= 500 cm³

Que-7: A metal pipe has a bore (inner diameter) of 5 cm. The pipe is 5 mm thick all round. Find the weight, in kilogram, of 2 metres of the pipe if 1 cm3 of the metal weights 7.7 g.

Sol: Inner radius of the pipe = r
= 52 cm
= 2.5 cm
External radius of the pipe = R
= Inner radius of the pipe + Thickness of the pipes
= 2.5 cm + 0.5 cm
= 3 cm
Length of the pipe = h
= 2 m
= 200 cm
Volume of the pipe = External volume – Internal volume
= 𝜋⁢𝑅²⁢ℎ − 𝜋⁢𝑟²⁢⁢ℎ
= 𝜋⁢(𝑅²⁢−𝑟²⁢)⁢ℎ
= 22/7⁢(3²⁢−(5/2)²⁢) × 200
=  (22/7) × (9−(25/4)) × 200
= (22/7) × ((36−25)/4) × 200
= (22/7) × (11/4) × 200
= ((22/7) × 550)
= 1728.6 cm³
Since 1 cm³ of the metal weights 7.7 g,
∴ Weight of the pipe = (1728.6 × 7.7)g
= (1728 × (7.7/1000))⁢𝑘⁢𝑔
= 13.31 kg

Que-8: A cylindrical container with diameter of base 42 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22 cm x 14 cm 10.5 cm. Find the rise in level of the water when the solid is submerged.

Sol: Diameter of cylindrical container = 42 cm
Therefore, radius (r) = 21 cm
Dimensions of rectangular solid = 22 cm × 14 cm × 10.5 cm
Volume of solid = 22 × 14 × 10.5 cm³  …(1)
Let height of water = h
Therefore, volume of water in the container = πr²h
= (22/7) × 21 × 21 × ℎ 𝑐⁢𝑚³
= 22 × 63h cm³  …(2)
From (1) and (2)
22 × 63h = 22 × 14 × 10.5
⇒ℎ = (22×14×10.5)/(22×63)
⇒ℎ = 7/3
⇒ℎ = 2*(1/3) or 2.33 cm

Que-9: A cylindrical container with internal radius of its base 10 cm, contains water up to a height of 7 cm. Find the area of wetted surface of the cylinder.

Sol: Internal radius of the cylinderical container = 10 cm
Height of water = 7 cm
Therefore, surface area of the wet surface = 2πrh + πr²
= πr(2h + r)
= (22/7) × 10 × (2×7+10)
= (220/7) × 24
= 754.29 cm²

Que-10: Find the total surface area of an open pipe of length 50 cm, external diameter 20 cm and internal diameter 6 cm.

Sol: Length of an open pipe = 50 cm
External diameter = 20 cm ⇒ External radius (R) = 10 cm
Internal diameter = 6 cm ⇒ Internal radius (r) = 3 cm
Surface area of pipe open from both sides = 2πRh + 2πrh
= 2πh(R + r)
= 2 × (22/7) × 50 × (10+3)
= 4085.71 cm²
Area of upper and lower part = 2π(R² – r²)
= 2 × (22/7) ×(10²−3²)
= 2 × (22/7) × (100−9)
= 2 × (22/7) × 91
= 572 cm²
Total surface area = 4085.714 + 572
= 4657.71 cm²

Que-11: The curved surface area of a cone is 12320 cm². If the radius of its base is 56 cm. Find its height.

Sol:  Curved surface area = 12320 cm²
Radius of base (r) = 56 cm
Let slant height = l
∴ πrl = 12320
⇒ (22/7) × 56 × 𝑙 = 12320
⇒ 𝑙 = (12320×7)/(56×22)
⇒ l = 70 cm
Height of the cone
= √(𝑙²−𝑟²)
= √{(70)²−(56)²}
= √(4900−3136)
= √1764
= 42 cm

Que-12: The radius of a solid right circular cylinder increases by 20% and its height decreases by 20%. Find the percentage change in its volume.

Sol:  Let the radius of a solid right circular cylinder be r = 100 cm
And let the height of a solid right circular cylinder be h = 100 cm
∴ Volume (original) of a solid right circular cylinder
= πr²h
= π × (100)² × 100
= 1000000π cm³
New radius = r’ = 120 cm
New height = h’ = 80 cm
∴ Volume (New) of a solid right circular cylinder = πr’²h’
= π × (120)² × 80
= 1152000π cm³
∴ Increase in volume = New volume – Original volume
= 1152000π cm³ – 1000000π cm³
= 152000π cm³
Thus, Percentage change in volume = [Increase in volume / Original volume] × 100%
= (152000⁢𝜋)/(1000000⁢𝜋) × 100%
= 15.2%

Que-13: The radius of a solid right circular cylinder decreases by 20% and its height increases by 10%. Find the percentage change in its : (i) volume  (ii) curved surface area

Sol: (i)  Let the radius of a solid right circular cylinder be r = 100 cm
And let the height of a solid right circular be h = 100 cm
Volume (original) of a solid right circular cylinder = πr²h
= π × (100)² × 100
= 1000000π cm³
New radius = r’ = 80 cm
New height = h’ = 110 cm
∴ Volume (New) of a solid right circular cylinder = πr’²h’
= π × (80)² × 110
= 704000π cm³
∴ Decrease in volume = Original volume – New volume
= 1000000π cm³ – 704000π cm³
= 296000π cm³
Percentage change in volume = (Decrease in volume)/(Original volume) × 100%
= (296000⁢𝜋)/(1000000⁢𝜋) × 100%
= 29.6%

(ii)  Curved surface area (Original) of a solid right circular cylinder = 2πrh
= 2π × 100 × 100
= 20000π cm²
Curved surface area (New) of a solid right circular cylinder
= 2πr’h’
= 2π × 80 × 110
= 17600π cm²
Decrease in curved area
= Original CSA – New CSA
= (20000π – 17600π) cm²
= 2400π cm²
Percentage change in curved surface area = (Decrease in curved surface area)/(Original curved surface area) × 100%
= (2400⁢𝜋)/(20000⁢𝜋) × 100%
= 12%

Que-14: Find the minimum length in cm and correct to nearest whole number of the thin metal sheet required to make a hollow and closed cylindrical box of diameter 20 cm and height 35 cm. Given that the width of the metal sheet is 1 m. Also, find the cost of the sheet at the rate of Rs. 56 per m. Find the area of metal sheet required, if 10% of it is wasted in cutting, overlapping, etc.

Sol:  Height of the cylindrical box = h = 35 cm
Base radius of the cylindrical box = r = 10 cm
Width of metal sheet = 1 m = 100 cm
Area of metal sheet required = Total surface area of the box
⇒ Length × width = 2πr(r + h)
⇒ Length × 100 = {2 × (22/7) × 10⁢ (10+35)}
⇒ Length × 100 = 2 × (22/7) ×10 ×45
⇒ Length = (2×22×10×45)/(100×7) = 28.28 cm = 28 cm
∴ Area of metal sheet = Length × Width
= 28 × 100
= 2800 cm²
= 0.28 m²
∴ Cost of the sheet at the rate of Rs. 56 per m²
= Rs. (56 × 0.28)
= Rs. 15.68
Let the total sheet required be x.
Then, x – 10% of x = 2800 cm²
⇒𝑥 − (10/100) × 𝑥 = 2800
⇒ (10⁢𝑥−𝑥)/10 = 2800
⇒ 9⁢𝑥/10 = 2800
⇒𝑥 = 2800 × (10/9)
⇒ x = 3111 cm²

Que-15: 3080 cm3 of water is required to fill a cylindrical vessel completely and 2310 cm³ of water is required to fill it upto 5 cm below the top. Find :
(i) radius of the vessel.
(ii) height of the vessel.
(iii) wetted surface area of the vessel when it is half-filled with water.

Sol:  Let r be the radius of the cylindrical vessel and h be its height
Now, volume of cylindrical vessel = volume of water filled in it
⇒ πr²h = 3080
⇒ (22/7) × 𝑟² × ℎ = 3080
⇒ r² × h = 980  …(i)
Volume of cylindrical vessel of height 5 cm = (3080 – 2310) cm³
⇒ πr² × 5 = 770
⇒ (22/7) × 𝑟² × 5 = 770
⇒ r² = 49
⇒ r = 7 cm
Substituting r² = 49 in (i), we get
49 × h = 980
⇒ h = 20 cm
Wetted surface area of the vessel when it is half-filled with water
= 2πrh + πr²
= πr(2h + r)
= (22/7) × 7⁢(2×10+7)   …[Half – filled ⇒ Height = 20/2 = 10 𝑐⁢𝑚]
= 22 × 27
= 594 cm²

Que-16: Find the volume of the largest cylinder formed when a rectangular piece of paper 44 cm by 33 cm is rolled along it : (i) shorter side.  (ii) longer side.

Sol: (i) Length of a rectangular paper = 44 cm
Breadth of a rectangular paper = 33 cm
When the paper is rolled along its shorter side, i.e. breadth, we have
Height of cylinder = h = 44 cm
Circumference of cross-section = 2πr = 33 cm
⇒ 2 × (22/7) × 𝑟 = 33
⇒𝑟 = (33×7)/(2×22) = 5.25 𝑐⁢𝑚
∴ Volume of cylinder = πr²h
= (22/7) × 5.25 × 5.25 × 44
= 3811.5 cm³

(ii) Length of a rectangular paper = 44 cm
Breadth of a rectangular paper = 33 cm
When the paper is rolled along its longer side , i.e. length, we have
Height of cylinder = h = 33 cm
Circumference of cross section = 2 π r = 44 cm
⇒2 × (22/7) × r =44
⇒r = (44×7)/(2×22) = 7 cm
∴Volume of cylinder = 𝜋⁢r²⁢ℎ
= (22/7) × 7 × 7 × 33 = 5082 ⁢cm³

Que-17: A metal cube of side 11 cm is completely submerged in water contained in a cylindrical vessel with diameter 28 cm. Find the rise in the level of water.

Sol: Given, side of a cube = 11 cm
∴ Volume of cube= (Side)³
= 11 × 11 × 11 cm³
= 1331 cm³
Given, diameter of cylinder = 28 cm
∴ Radius (r) of cylinder = 28/2 = 14 cm
Volume = 1331 cm³
∴ Rise in water level = Volume / 𝜋⁢𝑟²
= (1331×7)/(22×14×14) ⁢𝑐⁢𝑚
= 121/56 ⁢𝑐⁢𝑚
= 2.16 cm

Que-18: A circular tank of diameter 2 m is dug and the earth removed is spread uniformly all around the tank to form an embankment 2 m in width and 1.6 m in height. Find the depth of the circular tank.

Sol:  Given, diameter of circular tank = 2 m
And width of embankment = 2 m
Height = 1.6 m
Radius of tank = 22 = 1 m
Outer radius of embankment = 1 + 2 = 3 m
∴ Volume of earth = π(R² – r²) × h
= (22/7)⁢(3²−1²) × 1.6 𝑚³
= (22/7) × (9−1) × 1.6 𝑚³
= (22/7) × 8 × 1.6 𝑚³
= (22/7) × 12.8 𝑚³
Now volume of the earth dug out from the tank
= (22/7) × 12.8 𝑚³
Radius of tank = 1 m
∴ Depth of circular
= Volume/𝜋⁢𝑟²
= (22×12.8×7)/(7×22×1×1)
= 12.8 m

Que-19: The sum of the inner and the outer curved surfaces of a hollow metallic cylinder is 1056 cm² and the volume of material in it is 1056 cm³. Find its internal and external radii. Given that the height of the cylinder is 21 cm.

Sol: Let R and r be the outer and inner radii of hollow metallic cylinder.
Let h be height of the metallic cylinder.
It is given that
Outer curved surface area + Inner curved surface area = 1056
⇒ 2πRh + 2πrh = 1056
⇒ 2πh(R + r) = 1056
⇒2 × (22/7) × 21⁢(𝑅+𝑟) =1056
⇒ 𝑅 +𝑟 = (1056×7)/(2×22×21)
⇒ R + r = 8   …(i)
Volume of material in it = 1056 cm³
⇒ πR²h – πr²h = 1056
⇒ πh(R² – r²) = 1056
⇒ (22/7) × 21⁢(𝑅²−𝑟²) = 1056
⇒ 𝑅² − 𝑟² = (1056×7)/(22×21)
⇒ (R + r)(R – r) = 16
⇒ 8 × (R – r) = 16
⇒ R – r = 2   …(ii)
Adding (i) and (ii), we get
2R = 10 ⇒ R = 5 cm
⇒ 5 – r = 2
⇒ r = 3 cm
∴ Internal radius = 3 cm and External radius = 5 cm

Que-20: The difference between the outer curved surface area and the inner curved surface area of a hollow cylinder is 352 cm². If its height is 28 cm and the volume of material in it is 704 cm³;find its external curved surface area.

Sol: Let R and r be the outer and inner radii of hollow metallic cylinder.
Let h be the height of the metallic cylinder.
It is given that
Outer curved surface area – Inner curved surface area = 352
⇒ 2πRh – 2πrh = 352
⇒ 2πh(R – r) = 352
⇒2 × (22/7) × 28⁢(𝑅−𝑟) = 352
⇒ 𝑅 −𝑟 = (352×7)/(2×22×28)
⇒ R – r = 2   …(i)
Volume of material in it = 704 cm³
⇒ πR²h – πr²h = 704
⇒ πh(R² – r²) = 704
⇒ (22/7) × 28⁢(𝑅²−𝑟²) = 704
⇒ 𝑅² − 𝑟² = (704×7)/(22×28)
⇒ (R + r)(R – r) = 8
⇒ (R + r) × 2 = 8
⇒ R + r = 4    …(ii)
Adding (i) and (ii) we get
2R = 6
⇒ R = 3 cm
∴ External curved surface area = 2πRh
= 2 × (22/7) × 3 × 28
= 2 × 22 × 3 × 4
= 528 cm²

Que-21: The sum of the heights and the radius of a solid cylinder is 35 cm and its total surface area is 3080 cm², find the volume of the cylinder.

Sol: Let r and h be the radius and height of a solid cylinder
Then, r + h = 35 cm
Total surface area of a cylinder = 3080 cm²
⇒ 2πr(h + r) = 3080
⇒2 × (22/7) × 𝑟 × 35 = 3080
⇒ 2 × 22 × r × 5 = 3080
⇒𝑟 = 3080/(2×22×5) = 14 𝑐⁢𝑚
⇒ h = 35 – r
= 35 – 14
= 21 cm
∴ Volume of cylinder = πr²h
= (22/7) ×14 ×14 ×21
= 12936 cm³

Que-22: The total surface area of a solid cylinder is 616 cm². If the ratio between its curved surface area and total surface area is 1 : 2; find the volume of the cylinder.

Sol: Let r and h be the radius and height of a solid cylinder.
Total surface area of a cylinder = 616 cm²
⇒ 2πr(h + r) = 616
⇒ 𝜋⁢𝑟⁢(ℎ+𝑟) = 616/2
⇒ πr(h + r) = 308 …(i)
Curved surface area of a cylinder = 2πrh
Now, Curved surface area of a cylinderTotal surface area of a cylinder =12
⇒2⁢𝜋⁢𝑟⁢ℎ2⁢𝜋⁢𝑟⁢(ℎ+𝑟) =12
⇒ℎℎ+𝑟 =12
⇒ 2h = h + r
⇒ 2h – h = r
⇒ h = r
Substituting h = r in (i), we get
⇒ πr(r + r) = 308
⇒ πr.2r = 308
⇒ 2πr² = 308
⇒ πr² = 154
⇒227 ×𝑟2 =154
⇒𝑟2 =154×722
⇒ r² = 49
⇒𝑟 =√49
⇒ r = 7 cm
⇒ h = 7 cm.
∴ Volume of cylinder = πr²h
= 227 ×7 ×7 ×7
= 1078 cm³

Que-23: A cylindrical vessel of height 24 cm and diameter 40 cm is full of water. Find the exact number of small cylindrical bottles, each of height 10 cm and diameter 8 cm, which can be filled with this water.

Sol: For a large cylindrical vessel,
Height = H = 24 cm
Radius = R = 40/2 = 20 cm
∴ Volume of large cylindrical vessel = πR²H
= (π × 20 × 20 × 24) cm³
For each small cylindrical bottle,
Height = h = 10 cm
Radius = r = 8/2 = 4 cm
∴ Volume of each small cylindrical bottle = πr²h
= (π × 4 × 4 × 10) cm³
Now, number of small cylindrical bottles which can filled
= Volume of large cylindrical vessel / Volume of each small cylindrical bottle
= (𝜋×20×20×24)/(𝜋×4×4×10)
= 60

Que-24: Two solid cylinders, one with diameter 60 cm and height 30 cm and the other with radius 30 cm and height 60 cm, are metled and recasted into a third solid cylinder of height 10 cm. Find the diameter of the cylinder formed.

Sol: For cylinder 1,
Height = h₁ = 30 cm
Radius = r₁ = 602 = 30 cm
Volume = V₁
= 𝜋⁢r²1⁢ℎ1
= π × 30 × 30 × 30
= 27000π cm³
For cylinder 2,
Height = h₂ = 60 cm
Radius = r₂ = 30 cm
Volume = V₂
= 𝜋⁢r²2⁢h2
= π × 30 × 30 × 60
= 54000π cm³
Let r be the radius of the third cylinder.
Height = h = 10 cm
Volume = V
= πr²h
= πr² × 10
Now,
V = V₁ + V₂
⇒ πr² × 10 = 27000π + 54000π
⇒ πr² × 10 = 81000π
⇒ r² = 8100
⇒ r = 90
⇒ Diameter = 2r = 180 cm

Que-25: The total surface area of hollow cylinder, which is open from both sides, is 3575 cm²; area of the base ring is 357.5 cm² and height is 14 cm. Find the thickness of the cylinder.

Sol: Total surface area of a hollow cylinder = 3575 cm²
Area of the base ring = 357.5 cm²
Height = 14 cm
Let external radius = R and internal radius = r
Let thickness of the cylinder = d = (R – r)
Therefore, Total surface area = 2πRh + 2πrh + 2π(R² – r²)
= 2πh(R + r) + 2π(R + r)(R – r)
= 2π(R + r)[h + R – r]
= 2π(R + r)(h + d)
= 2π(R + r)(14 + d)
But
2π(R + r)(14 + d) = 3575  …(i)
And area of base = π(R² – r²) = 357.5
⇒ π(R + r)(R – r) = 357.5
⇒ π(R + r)d = 357.5   …(ii)
Dividing (i) by (ii)
{2⁢𝜋⁢(𝑅+𝑟)⁢(14+𝑑)}/{𝜋⁢(𝑅+𝑟)⁢𝑑} = 3575/357.5
{2⁢(14+𝑑)}/𝑑 = 10
28 + 2d = 10d
8d = 28
d = 28/8 = 3.5 cm
Hence, thickness of the cylinder = 3.5 cm

Que-26: The given figure shows a solid formed of a solid cube of side 40cm and a solid cylinder of radius 20 cm and height 50 cm attached to the cube as shown.
Find the volume and the total surface area of the whole solid (Take π = 3.14)

Sol: Edge of a cube = I = 40 cm
∴ Volume of a cube = I³ = (40)³ = 64000 cm³
Radius of a solid cylinder = r = 20 cm
Height of a solid cylinder = h = 50 cm
∴ Volume of cylinder = πr²h
= 3.14 × 20 × 20 × 50
= 62800 cm³
∴ Volume of whole solid = Volume of cube + Volume of cylinder
= (64000 + 62800) cm³
= 126800 cm³
Total surface area of the whole solid
= Total surface area of a cube + Curved surface area of a cylinder
= 6l² + 2πrh
= 6 × (40)² + 2 × 3.14 × 20 × 50
= 6 × 1600 + 6280
= 9600 + 6280
= 15880 cm²

Que-27: Two right circular solid cylinders have radii in the ratio 3 : 5 and heights in the ratio 2 : 3, Find the ratio between their : (i) curved surface areas. (ii) volumes.

Sol: (i) Let the radii and height of two right circular cylinders be r₁, r₂ and h₁, h₂ respectively.
It is given that,
𝑟1/𝑟2 = 3/5 and ℎ1/ℎ2 = 2/3
Curved surface area of cylinder 1 / Curved surface area of cylinder 2 = (2⁢𝜋⁢𝑟1⁢ℎ1)/(2⁢𝜋⁢𝑟2⁢ℎ2)
= (𝑟1/𝑟2) × (ℎ1/ℎ2)
= (3/5) × (2/3)
= 2/5
∴ Ratio between their curved surface areas is 2 : 5.

(ii) Let the radii and height of two right circular cylinders be r₁, r₂ and h₁, h₂ respectively.
It is given that,
𝑟1/𝑟2 = 3/5  and  ℎ1/ℎ2 = 2/3
Volume of cylinder 1 / Volume of cylinder 2 = {𝜋⁢(𝑟1)²⁢ℎ1}/{𝜋⁢(𝑟2)²ℎ2}
= (𝑟1/𝑟2)² × (ℎ1/ℎ2)
= (3/5)² × (2/3)
= (9/25) × (2/3)
= 18/75 = 6/25
∴ Ratio between their volumes is 6 : 25.

Que-28: A dosed cylindrical tank, made of thin iron sheet, has diameter = 8.4 m and height 5.4 m. How much metal sheet, to the nearest m², is used in making this tank, if 1/15 of the sheet actually used was wasted in making the tank?

Sol: Diameter of cylinder = 8.4 m → Radius
𝑟 = 8.4/2 = 4.2 m
Height = 5.4 m
Tank is closed, so surface area = CSA + Top + Bottom
Step 1: Calculate the Total Surface Area of the Closed Cylinder
2πrh = 2 × π × 4.2 × 5.4
= 2 × 3.14 × 4.2 × 5.4
= 142.1 m²
Area of two bases:
2πr² = 2 × 3.14 × (4.2)²
= 6.28 × 17.64
= 110.8 m²
Total Surface Area (without waste):
142.1 + 110.8
= 252.9 m²
Step 2: Let total metal sheet used = X
14⁢𝑋/15
= 252.9
X = 252.9 × (15/14)
X = 271.9 m²
= 272 m²

–: Surface Area and Volume of Cylinder Class 10 Concise Exe-20A ICSE Maths Selina Solutions :–

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