Volume and Surface Area of Cylinder Class 10 RS Aggarwal Exe-21A ICSE Maths Solutions Ch-21. In this article you would learn how to solve problems on Cylinder in a very easy way. Solved practice questions/problems has been given related cylinder. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

Volume and Surface Area of Cylinder Class 10 RS Aggarwal Exe-21A ICSE Maths Solutions Ch-21

Board	ICSE
Publications	Goyal Brothers Prakashan
Subject	Maths
Class	10th
Chapter-21	Volume and Surface Area of Solids (Cylinder , Cone and Sphere)
Writer	RS Aggarwal
Book Name	Foundation
Topics	Solved Questions on Cylinder

Practice / Questions on Cylinder with Answer / Solutions

 Class 10 RS Aggarwal Exe-21A ICSE Maths Solutions Ch-21 Volume and Surface Area of Solids (Cylinder , Cone and Sphere)

Que-1: Find the curved surface area and the total surface area of the cylinder for which:
(i) h = 16cm , r = 10.5 cm
(ii) h = 5 cm , r = 21 cm
(iii) h = 20cm , r = 14 cm
(iv) h =1 m , r = 1.4 cm

Sol: (i) Curved Surface Area (CSA) of cylinder  = 2πrh
= 2×22/7×(10.5)×16 = 1056 cm²

Total Surface Area (TSA) of a cylinder  = 2πr(h+r)
= 2×22/7×(10.5)×(16+10.5) = 2×22/7×(10.5)×(26.5) = 1749 cm²

(ii) Curved Surface Area (CSA) of cylinder  = 2πrh
= 2×22/7×(21)×5 = 660 cm²

Total Surface Area (TSA) of a cylinder  = 2πr(h+r)
= 2×22/7×(21)×(5+21) = 2×22/7×(21)×(26) = 3432 cm²

(iii) Curved Surface Area (CSA) of cylinder  = 2πrh
= 2×22/7×(14)×20 = 1760 cm²

Total Surface Area (TSA) of a cylinder  = 2πr(h+r)
= 2×22/7×(14)×(20+14) = 2×22/7×(14)×(34) = 2992 cm²

(iv) Curved Surface Area (CSA) of cylinder  = 2πrh
= 2×22/7×(1.4)×100 = 880 cm²

Total Surface Area (TSA) of a cylinder  = 2πr(h+r)
= 2×22/7×(1)×(100+1.4) = 2×22/7×(1.4)×(101.4) = 891.12 cm²

Que-2: Find the volume of the cylinder in which
(i) Height = 21 cm and Base Radius = 5 cm
(ii) Diameter 28 cm and Height 40 cm

Sol: (i) Volume of cylinder = πr²h
= 22/7 × (5cm)² × (21cm)
= 1650 cm³

(ii) Diameter = 2 × radius = 28 cm
radius = 14 cm
Volume of cylinder = πr²h
= 22/7 × (14cm)² × (40cm) =24640 cm³

Que-3: A cylindrical tank has a capacity of 6160m³. Find its depth, if ito radius is 14 m. Also, find the cost of painting its curved surface at 30 per m²

Sol: Let us assume that the depth of the tank is h. Given,
Radius of tank = 14 m Volume of tank = 6160m³ Now,
6160 m³ = 22/7 × (14m²) × h
Volume of tank = πr²h
616h = 6160 m
h = 6160/616
∴ Depth of the tank = 10 m Now,
Curved surface area of cylinder = 2πrh
= 880 m²
Cost of painting the curved outer surface of tank = 3 × 880 Rs. = Rs. 2640

Que-4: (i) The curved surface area of a cylinder is 4400cm² and the circumference of its base is 110 cm. Find the height and the volume of the cylinder.
(ii)The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the radius of the cylinder and also its volume.

Sol: (i) Given, curved surface area of cylinder = 4400 cm².
We know that curved surface area of cylinder = 2πrh.
∴ 2πrh = 4400 …..(i)
Given, circumference of base = 110 cm.
We know that circumference = 2πr.
∴ 2πr = 110 …..(ii)
Dividing equation (i) by (ii),
⇒ 2πrh / 2πr = 4400/110
⇒ h = 40 cm.
Hence, the height of the cylinder = 40 cm.

We know that circumference = 2πr.
Given, circumference = 110 cm.
∴ 2πr = 110
⇒2 × 22/7 × r = 110
r = 110 ×7 / 2×22 = 770/44 = 17.5 cm
Volume of cylinder = πr²h.
Putting values we get,
Volume of cylinder = 22/7 × (17.5)² × 40
= 22/7 × 306.25 × 40 / 7
= 269500/7
= 38500 cm³
Hence, the volume of the cylinder = 38500 cm³.

(ii) Circumference = 2πr.
Given, circumference = 132 cm.
∴ 2πr = 132
⇒ 2×22/7×r = 132
r = 132 × 7/2×22 = 924/44 = 21cm
Hence, the radius of the cylinder = 21 cm.

Given height = 25 cm.
Volume of cylinder = πr²h.
Putting values we get,
Volume of cylinder = 22/7×(21)²×25
= 22×441×25/7
= 242550/7
= 34650 cm³
Hence, the volume of the cylinder = 34650 cm³.

Que-6: The total surface area of a solid cylinder is 462cm² and its curved surface arisa i one-third of its total surface area. Find the volume of the cylinder.

Sol: We have
Curved surface area = 1/3 × total surface area
⇒ 2πrh = 1/3(2 πrh + 2πr²)
⇒ 6πrh = 2πrh +2πr²
⇒ 4πrh =2πr²
⇒ 2h = r
We know that,
Total surface area = 462
⇒ Curved surface Are = 1/3× 462
2πrh =154
⇒ 2× 3.14× 2h² = 154
⇒ h²=(154× 7)/(2× 22× 2)
⇒ 49/4
⇒ h= 7/2cm
⇒ r= 2h
⇒ r=2× 7/2cm`
⇒ r=7cm.

Que-7: The sum of the radius of the base and the height of a solid cylinder is overline 37m. If the total surface area of the cylinder be 1628m² find its volume

Sol:  Let r and h be the radius and height of the solid cylinder .
Given:
𝑟+ℎ=37⁢𝑚
Total surface area, 𝑆=2⁢𝜋⁢𝑟⁢(𝑟+ℎ)
1628=2⁢𝜋×𝑟×37
𝑟 = 1628/2⁢𝜋×37
= 1628/232.477
=7⁢𝑚
Circumference of its base ,𝑆1=2⁢𝜋⁢𝑟
=(2×22/7×7)⁢𝑚
=44⁢𝑚

Que-8: Find the height of a solid circular cylinder having total surface area of 660 cm² and radius 5 cm. 

Sol: Total Surface Area (A) = 660 cm²
A=2πr(h+r)
660=2π(5)(h+5)
660/10π = h+5
h = 660/10π – 5
h = 660/10×7/22 – 5 = 3×7 – 5
h = 21 – 5
h = 16 cm
∴ height of the cylinder is 16 cm.

Que-9: Find the total surface area of a hollow cylinder open at both ends, if its length is 12 cm, external diamter is 8 cm and the thickness is 2cm.[Hint: Total surface area = [2πRh+ 2πrh + 2(πR²-πr²) cm² , where h= 12 cm , R= 4 cm , r= 2cm]

Sol: Here, length is given which is basically height h of cylinder. External diameter d is given 8cm which when divided by 2 becomes radius R i.e. R = d/2 = 8/2 = 4 cm

Now, from this figure we can see that the external radius of the cylinder becomes the diameter of the inner cylinder. So, we need to find radius of inner cylinder i.e. r = R/2 = 4/2 = 2 cm
Now, we have all the parameters with us h=12 cm , R = 4 cm , r = 2 cm

We want the total area between dotted cylinder and external cylinder. So, we will take lateral surface area of both cylinder which will be 2πRh + 2πrh and subtracting the base which is circle from external cylinder which will get 2πR² – 2πr²
So, Total surface area of hollow cylinder = Lateral surface area + area of solid bases
= 2πRh + 2πrh + 2πR² – 2πr²
= 2πh(R+r) + 2π(R²-r²)
= 2 × 22/7 × 12(4+2) + 2 × 22/7 × (4²-2²)
taking 2 × 22/7 common , so
= 2 × 22/7 ×[12(4+2) + (4²-2²)]
= 2 × 22/7 × [72 + (16-4)]
= 2 × 22/7 × [72 + 12]
= 2 × 22/7 × 84
= 2 × 22 × 12
= 528 cm²
∴ Thus, the total surface area of the hollow cylinder is 528 cm²

Que-10: Water is flowing at the rate of 3km/hr through a circular pipe of 20cm internal diameter into a circular cistern of diameter 10 m and depth 2 m . In how much time will be cistern filled ?

Sol: Suppose the cistern is filled in x hours. Since water is flowing at the rate of 3km/hr. Therefore length of the water column in x hours = 3x km = 3000x metres
Clearly, the water column forms a cylinder of radius = r =20/2 cm = 10 cm = 1/10 m
and h = height = 3000x metres Volume of the water that flows in the cistern in x hours = πr²h = 22/7 × (1/10)² × 3000x = 22/7 × 1/10 × 1/10 × 3000x

Also, Volume of cistern = 22/7 × 5 × 5 × 2

Since the cistern is filled in x hours ,
So, Volume of the water that flows in the cistern in x hours = volume of the cistern
22/7 × 1/10 × 1/10 × 3000x = 22/7 × 5 × 5 × 2
x = (5 × 5 × 2 × 10 × 10)/3000
= 1 hour 40 minutes
∴ Cistern id filled in 1 hour 40 minutes

Que-11: A swimming pool 70 m long , 44 m wide and 3m deep , is filled by water issuing from a pipe of diameter 35 cm , at 6 m per second. How many hours does it take to fill the pool ?

Sol: To solve this, we first find the volume of the pool, then find the volume flow rate of water from the pipe, and finally divide the volume of the pool by the flow rate to get time.

Volume of swimming pool = length × breadth × height
= 70 m × 44 m × 3 m = 9240 m³

Now , diameter of pipe = 35 cm = 0.35 m
radius of pipe = 0.35/2 = 0.175 m
Cross-sectional area of pipe = πr² = 22/7 × (0.175)² = 0.09625 m²

speed of water flow v = 2m/s
volume flow rate Q = A × v = 0.09625 × 6 = 0.5775 m³/s

Time taken to fill the pool ,
Time = Volume of pool / Volume of flow rate = 9240 / 0.5775 = 16000 s

In hours,
Time = 16000/60 × 60 = 4.44 hours (approx)

It takes approximately 4.44 hours to fill the swimming pool with water flowing through the pipe at the given rate.

Que-12: Water is flowing at the rate of 8 m per second through a circular pipe whose internal diameter is 2cm into a cylindrical tank the radius of whose base is 40 cm, Determine the increase in the water level in 30 minutes.

Sol: The volume of water flowing through the pipe is equal to the volume of a cylinder with a base area equal to the pipe’s cross-sectional area and a length equal to the distance the water travels in the given time.
radius of pipe = r = 2cm/2 = 1 cm = 0.01 m
time in seconds = 30 min × 60 = 1800 s

The volume (V) is given by the cross-sectional area multiplied by the distance traveled (v times t):
(where v×t = height of pipe (h))
V = πr²vt

The total volume (V) of water collected in the cylindrical tank must equal the volume of a cylinder with the tank’s base radius and the increase in water level (h).
The radius of the tank is (R}=40 cm=0.4 m
The increase in volume in the tank is = πR²h

Equating the two expressions:
πR²h = πr²vt

h= r²vt/R²
h = (0.01m)²×(8m/s)×1800/(0.4m)²
h =  0.0001 × 8 × 1800 / 0.16
h = 1.44/0.16
h = 9 m

∴ The increase in the water level in 30 minutes is 9 m.

Que-13: A 20m deep well with diameter 7 m is dug up and the earth from digging is spread evenly to form a platform 22 m × 14 m . Determine the height of the platform.

Sol: Radius of well – diameter/2 = 7/2 = 3.5 m
Volume of earth dug out = πr²h = 22/7 × 3.5 × 3.5 × 20 = 770 m³

let the height of platform be h m,
then volume of platform = length × breadth × height = 22 × 14 × h

According to the question volume of well = volume of platform
770 = 22 × 14 × h
h = 770 / 22 × 14
h = 2.5 m

∴ Height of the platform is 2.5 m.

Que-14: Find the mass of a metallic hollow cylindrical pipe 24 cm long with internal diameter 10 cm and made of 5mm thick metal , if 1 cm³ of the weighs 7.5 grams.

Sol: The internal radius of hollow cylindrical pipe (r_int) = 10cm/2 = 5 cm
The thickness t = 5mm = 0.5 cm
The external radius (r_ext) = (r_int) + t = 5 cm + 0.5 cm = 5.5 cm
The volume (V) of the metal in the pipe is calculated using the formula , V = π(r²_ext – r²_int)L , where L is the length:
V = 22/7 × ((5.5)²- 5²) × 24
V = 22/7 ×(30.25-25) × 24
V = 22/7 × 5.25 × 24 = 396 cm³

The mass M is found by multiplying the volume by the density h:(Ρ=7.5g/cm³)
M = V × Ρ
M = 396 × 7.5 = 2970 g

The mass of the metallic hollow cylindrical pipe is 2970 g.

Que-15: A well with 10 m inside diameter is dug 8.4 m deep . Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment . Find the height of trhe embankment .

Sol: The well is a cylinder with inside diameter D = 10 m , so radius r = D/2 = 10/2 = 5 m .
The depth is h = 8.4 m
The volume of the earth taken out is :
V_earth = πr²h = 22/7 × 5² × 8.4 = 22/7 × 5 × 5 × 8.4
V_earth = 660 m³

The earth is spread around the well to form an embankment with a width W = 7.5 m
The inner radius of the embankment is the radius of the well: R_inner = r = 5m
The outer radius of the embankment is the inner radius plus the width: R_outer  = R_inner + W = 5m + 7.5m = 12.5m

Now , according to the question volume of well = volume of embankment
let the height of the embankment be H,
V_embankment = π(R²_outer – R²_inner)H = 22/7 × ((12.5)² – (5)²) × H
660 = 22/7 × (156.25 – 25) × H
660 = 22/7 × 131.25 × H
H = 660×7/22×131.25
H = 1.6 m

∴ Height of the embankment is 1.6 m.

–: End of Volume and Surface Area of Cylinder Class 10 RS Aggarwal Exe-21A ICSE Maths Solutions  :–

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