Three Dimensional Solids Class 10 OP Malhotra Exe-15B ICSE Maths Solutions Ch-15 questions as latest prescribe guideline for upcoming exam. In this article you would learn how to solve problems on Volume of Solid / Hollow Cylinder. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Three Dimensional Solids Class 10 OP Malhotra Exe-15B ICSE Maths Solutions Ch-15
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 10th |
| Chapter-15 | Three Dimensional Solids |
| Writer | OP Malhotra |
| Exe-15B | Volume of Cylinder |
| Edition | 2024-2025 |
Volume of Solid Cylinder
The formula for the volume of a cylinder is V = πr²h where “r” is the radius of the circular base and “h” is the height of the cylinder
Volume of Hollow Cylinder
Suppose, r1 and r2 are the two radii of the given hollow cylinder with ‘h’ as the height, then the volume of this cylinder V = πh(r12 – r22)
Exercise- 15B
(Three Dimensional Solids Class 10 OP Malhotra Exe-15B ICSE Maths Solutions Ch-15)
Que-1: Find the volume of the cylinders whose radii and height are given below : Take π to be 22/7. (i) r = 7cm, h = 8cm (ii) r = 7cm, h = 12cm (iii) r = 14cm, h = 16cm (iv) r = 21cm, h = 40cm
Sol: (i) Radius, r = 7 cm
Height, h = 8 cm
Volume of the cylinder = π × r2 × h
Volume of the cylinder = (22/7) × 72 × 8 = 1232 cm3
(ii) Radius, r = 7 cm
Height, h = 12 cm
Volume of the cylinder = π × r2 × h
Volume of the cylinder = (22/7) × 72 × 12 = 1848 cm3
(iii) Radius, r = 14 cm
Height, h = 16 cm
Volume of the cylinder = π × r2 × h
Volume of the cylinder = (22/7) × 142 × 16 = 9856 cm3
(iv) Radius, r = 7 cm
Height, h = 8 cm
Volume of the cylinder = π × r2 × h
Volume of the cylinder = (22/7) × 212 × 40 = 55440 cm3
Que-2: Find the diameter of the circular cylinders if : (a) Volume is 44 cm³, height 3.5 cm; (b) Volume 385 cm³, height 1 dm
Sol: (a) Volume = 44 cm³
Height, h = 3.5 cm
Volume of the cylinder = π × r2 × h
44 = (22/7) × r2 × 3.5
14/3.5 = r2
4 = r2
r = 2 cm.
D = 2 × r = 2 × 2 = 4 cm.
(b) Volume = 385 cm³
Height, h = 1 dm = 10 cm
Volume of the cylinder = π × r2 × h
385 = (22/7) × r2 × 10
245/20 = r2
12.25 = r2
r = 3.5 cm.
D = 2 × r = 2 × 3.5 = 7 cm.
Que-3: Find the height of the circular cylinder if : (a) Volume is 66 cm³, radius 2 cm; (b) Volume 4 litres, radius 5 cm
Sol: (a) Volume = 66 cm³
Radius r = 2 cm
Volume of the cylinder = π × r2 × h
66 = (22/7) × 22 × h
21/4 = h
4 = r2
r = 5.25 cm.
(b) Volume = 4 L = 4000 cm³
Radius r = 5 cm
Volume of the cylinder = π × r2 × h
4000 = (22/7) × 52 × h
(4000×7)/(22×25) = h
(80×7)/11 = h
r = 50.9 cm.
Que-4: A wooden pole is 7m high and 20cm in diameter. Find its weight if the wood weighs 225 kg per m³.
Sol: Given weight = 225 kg/m3
Given height = 7 m and diameter = 20 cm = 20/100 = 0.2 m
Radius = Diameter/2 = 0.2/2 = 0.1 m
Volume of cylinder = πr2h.
Putting values in the formula we get,
Volume of wooden pole = (22/7)×(0.1)²×7 = 22×0.1×0.1 = 0.22 m³.
Since weight of 1 m3 of wooden pole = 225 kg.
∴ Weight of 0.22 m3 of wooden pole = 225 × 0.22 = 49.5 kg.
Que-5: Find the volume of metal in the hollow pipe of internal radius 3cm, metal 1cm thick and length 6cm.
Sol: Outer Radius = 3cm
Inner Radius = 1cm
Length = 6cm
Volume = πh (R² – r²)
= 22/7 × 6 × (3^2 – 1^2)
= 22/7 × 6 × (9-1)
= 22/7 × 6 × 8
= 132 cm³
Que-6: The sum of the radius of the base and height of a solid cylinder is 37cm. If total surface area of the solid cylinder is 1628 cm², find the volume of the cylinder.
Sol: Let the radius of base and height of the solid right circular cylinder be r cm and h cm, respectively.
According to the question,
r+h = 37 …..(1)
Total surface area=1628 sq cm
2πr(r+h) = 1628 …..(2)
From (1) and (2), we get
2πr(37) = 1628
⇒ 2πr = 44
⇒ 2×(22/7)×r = 44
⇒r = (44×7)/(2×22)
⇒ r = 7 cm
Substituting the value of r in (1), we get
7+h = 37
⇒ h = 30 cm
Now,
Volume of the cylinder = πr2h
= (22/7)×7×7×30
= 4,620 cm³
Hence, the volume of the cylinder is 4,620 cm³.
Que-7: A cylindrical tank has capacity 6160 cu m. Find its depth if the diameter of its base is 28m. Also, find the area of the inside curved surface of the tank.
Sol: Capacity of tank = 6160 m3
Diameter of tank = 28 m
Radius of tank = 28/2 m
⇒ 14 m
Volume of tank = πr2h
⇒ 22/7 × (14)2 × h = 6160
⇒ 22/7 × 196 × h = 6160
⇒ 22 × 28 × h = 6160
⇒ 616 × h = 6160
⇒ h = 10 m
Inner Curved Surface area of tank = 2πrh
= 2×(22/7)×14×10
= 880 m²
Que-8: The area of the curved surface of a cylinder is 4400 cm², and the circumference of its base is 110 cm. Find (i) the height of the cylinder (ii) the volume of the cylinder
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.
(ii) 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
= (22×306.25×40)/7
= 269500/7
= 38500 cm³.
Que-9: (i) How many cubic metres of earth must be dug out to make a well 20 metres deep and 2 metres in diameter ? (ii) If the inner curved surface of the well in part (i) above is to be plastered at the rate of Rs 5 per m², find the cost of plastering.
Sol: (i) A well needs to be formed which is 20 meters deep and has a diameter of 2 meters.
Height of well = 20 m
Radius = Diameter/2 = 2/2 = 1 m.
Volume of soil to be dug out = Volume of well to be formed.
Volume of cylinder = πr2h.
Putting values we get,
Volume of well = (22/7)×(1)²×20
= (22×1×20)/7
= 440/7
= 62*(6/7) m³.
Hence, 62*(6/7) m³ of soil must be dug out for the formation of well.
(ii) Curved surface area of cylinder = 2πrh
= 2×(22/7)×1×20
= 880/7 m².
Rate of plastering = ₹ 50/m2.
So, plastering well will cost = (880/7)×50 = 44000/7 = ₹ 6285.70
Hence, the cost of plastering the inner curved surface area of the well = ₹ 6285.70.
Que-10: A cylinder has a diameter of 20cm. The area of curved surface is 1000 cm². 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 = 1000 cm2
⇒ 2×(22/7)×10×h=1000
⇒ h = (1000×7)/(22×10×2)
= 350/22
⇒ h = 15.9 cm
(ii) Volume of cylinder = πr2h.
Putting values we get,
Volume of cylinder = 3.14×(10)²×15.9
= 3.14×100×15.9
= 4992.6 cm³.
Que-11: An electric geyser is cylindrical in shape, having a diameter of 35cm and height of 1.2 m. Neglecting the thickness of its walls, calculate (i) the outer lateral surface area; (ii) its capacity in litres.
Sol: (i) Given diameter = 35 cm, height = 1.2 m = 1.2 × 100 = 120 cm.
We know,
radius = diameter/2 = 35/2 = 17.5 cm
Curved surface area = 2πrh = 2×(22/7)×120×17.5
= 92400/7 = 13200 cm².
Hence, the outer lateral surface area of electric geyser = 13200 cm².
(ii) Volume of cylinder = πr²h.
Putting values we get,
Volume of electric geyser = (22/7)×(17.5)²×120
= (22×306.25×120)/7
= 808500/7
= 115500 cm³.
Hence, the capacity of electric geyser = 115.5 litres.
Que-12: How many cylindrical glasses, diameter 8cm, height 15cm, can be filled from cylindrical vessel, diameter 30cm, height 80cm, full of milk.
Sol: Given
Diameter of cylindrical vessel = 30cm
Radius = 30/2 = 15cm
Height = 80cm
Volume of vessel = πr²h.
= (22/7)×15²×80
= (22×225×80)/7
= 56571.42cm³
Diameter of glass = 8cm
Radius = 4cm
Height = 15cm
Volume =πr²h.
= (22/7)×4²×15
= (22×16×15)/7
= 754.29 cm³
No of glass can be fill
= 56571.42/754.29
= 75.
Que-13: A path 2 m wide surrounds a circular pond of diameter 40 m. How many cubic metre of gravel are required to gravel the path to a depth of 7 cm ?
Sol: Diameter of circular pond = 40 m
Radius of pond(r) = 20m.
Thickness = 2m
Depth = 7cm = 0.07m
Since it is viewed as a hollow cylinder
Thickness (t) = R – r
2 = R – r
2 = R – 20
R = 22m
∴Volume of hollow cylinder = π(R2 – r2)h
= π(222 – 202)h
= π(222 – 202) x 0.07
= (22/7)(84) x 0.07
∴ Volume of hollow cylinder = 18.48 m3
∴ 18.48 m3 of gravel is required to have path to a depth of 20cm.
Que-14: An iron block is in the form of a cylinder of diameter 0.5 m and length 3.5 m. This block is to be rolled into the form of a bar having square section of side 25 cm. Find the length of the bar.
Sol: We know that,
Volume of cylinder = Volume of cuboid bar
πr²h = lbh
(22/7)×(0.5/2)×(0.5/2)×3.5 = l×0.25×0.25
l = (22×0.5×0.5×3.5)/(7×2×2×0.25×0.25)
l = 11 m.
Que-15: A swimming pool 70 m long, 44 m wide, 3 m deep is filled by water issuing from a pipe of diameter 14cm, at 2 m per second. How many hours does it take to fill the bath ?
Sol: Dimensions of swimming pool are: length =70 m
breadth = 44 m
depth(height) =3 m
Volume of pool = l × b × h
= 70 m× 44 m× 3 m
diameter of pipe = 14 cm
radius of pipe = 7 cm
Let the height (depth) be 2 m
Then, volume of water flowed in 1 second = πr²h
= (22/7)× 0.07 m × 0. 07m × 2 m
Volume of water flowed in 1 hour = πr²h ×3600
(1hour = 3600 seconds)
= (22/7) × 0.07 m ×0.07 m × 2 × 3600
Total time taken to fill the pool = (volume of pool) / (volume of pipe)
=(70 × 44 × 3) / 3.14× 0.07 × 0.07 × 2 × 3600
= 83.3 hours.
Que-16: What length of solid cylinder 2 cm in diameter must be taken to be cast into a hollow cylinder of external diameter 12cm, 0.25cm thick, 15 cm long,
Sol: External radius R = 12/2 = 6cm
inner radius = 6 – 0.25 = 5.75 cm [thickness = 0.25 cm]
length h = 15 cm
Volume of hollow cylinder = πR²h – πr²h
= πh(R²-r²)
= (22/7)×15(6²-5.75²)
= (330/7) (36-33.0625)
= (330/7) × 2.9375
= 969.375/7
Radius of solid cylinder = 2/2 = 1cm
Volume = πr²h
969.375/7 = (22/7)×1²×h
h = (969.375×7)/(22×7)
h = 969.375/22
h = 44.0625 cm.
Que-17: A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 11.2 cm, and its length is 21 cm. The metal every where is 0.4cm thick. Calculate the volume of the metal to 1 place of decimal.
Sol: Internal radius = 11.2/2 = 5.6 cm
length h = 21 cm
thickness = 0.4 cm
Outer radius = 0.4+5.6 = 6 cm
Volume of hollow cylinder = πR²h – πr²h
= πh(R²-r²)
= (22/7)×21(6²-5.6²)
= 66 (36-31.36)
= 66 × 4.64
= 306.2 cm³
Que-18: Water flows along a pipe of radii 0.6 cm at 8cm per second. This pipe is draining the water from a tank which holds 1000 litres of water when full. How long would it take to completely empty the tank ?
Sol: 1 litre = 1000 cm³
∴ volume of tank = 1000000 cm³
volume of water flowing out of the pipe per second = πr²l
where r is the radius of the pipe and l is the flow per second
r = 0.6cm and l = 8 cm/s
substituting the values we have volume of water flowing out of the pipe per second = 9.0477 cm³/s
∴ time taken to empty tank = volume of tank/volume of flowing out through the pipe per second
=1000000/9.0477 = 110525.32 seconds
i.e 30.7 hours.
Que-19: A cylindrical bucket 28 cm in diameter; 72 cm high and full of water, is emptied into a rectangular tank 66 cm long, 28 cm wide. Find the height of the water-level in the tank.
Sol: Let h be the height of rectangular tank = volume of cylindrical bucket
66×28×h = π(28/2)²×72
66×28×h = (22/7)×14×14×72
h = (22×2×14×72)/(66×28)
h = 24cm
Hence, the height of rectangular tank is 24 cm.
Que-20: There is some water in a cylindrical vessel of base diameter 14 cm. When an iron-cube is entirely immersed in it, the height of the water rises by 8*(9/14)cm. Find the length of the edge of the cube.
Sol: Diameter of cylinder = 14 cm
Radius of cylinder = 14/2 = 7 cm
Height of water Raised = 8*(9/14) = 121/14 cm
Volume of Water Displaced by iron cube
= π R² h
= (22/7) (7)² (121/14)
= 11 * 121 cm³
= 11 * 11 * 11 cm³
= 11³ cm³
Volume of iron cube = (length of edge of the cube)³
(length of edge of the cube)³ = 11³
length of edge of the cube = 11 cm.
Que-21: If the radius of the base of a right circular cylinder is halved, keeping the height same, what is the ratio of the volume of the reduced cylinder to that of the original one ?
Sol: If the radius of the base is halved, the new volume V’ of the cylinder with the halved radius becomes:
V′ = π(r/2)²h = π(r²/4)h
Now, let’s find the ratio of the volume of the reduced cylinder to the volume of the original cylinder:
Ratio = V′/V = {π(r²/4)h}/(πr²h) = 1/4
Therefore, if the radius of the base of a right circular cylinder is halved while keeping the height constant, the ratio of the volume of the reduced cylinder to the volume of the original cylinder is 1/4.
Hence the ratio of the volume of the reduced cylinder to the volume of the original cylinder is 1:4.
Que-22: A rectangular sheet of paper 22cm long and 12cm broad can be formed into the curved surface of a cylinder in two ways. Find the difference between the volumes of the two cylinders which can be formed.
Sol: Difference between the volumes of the two cylinders
= [πr²1h1−πr²2h2]
= π[(7/2)²×12 − (21/11)²×22]
= (22/7)×7×7×3[1−(3×2)/11]
= 210 cm.
Que-23: A 20 m deep well with diameter 7 m is dug up and the earth from digging is spread evenly to form a platform 22m × 14m. Determine the height of the platform.
Sol: Depth of the cylindrical well, h = 20 m
Radius of the cylindrical well, r = 7/2 m
Length of the cuboidal platform, l = 22 m
Breadth of the cuboidal platform, b = 14 m
Let the height of the cuboidal platform = H
Volume of the cylindrical well = Volume of the cuboidal platform
πr2h = lbH
H = πr2h / lb
= (22/7 × 7/2 m × 7/2 m × 20 m) / (22 m × 14 m)
= 5/2 m
= 2.5 m
Therefore, the height of the platform will be 2.5 m.
Que-24: The barrel of a fountain pen, cylindrical in shape, is 7cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre ?
Sol: Given, height of barrel h = 7 cm
Diameter of barrel = 5 mm
Radius = 2.5 mm
1 cm = 10 mm
So, radius r = 2.5/10 = 0.25 cm
Volume of cylindrical barrel = (22/7)(0.25)²(7)
= 22(0.25)(0.25)
= 1.375 cm³
With 1.375 cm³ of ink, 310 words can be written.
Volume of ink in bottle = 1/5 litre
= 0.2 litre
= 200 cm³
With 200 cm³ of ink, the number of words that can be written = 310(200)/1.375
= 62000/1.375
= 45100
Therefore, 45100 words can be written using 1/5 litre of ink in a bottle.
Que-25: A closed rectangular box 40cm long, 30cm wide and 25cm deep, has the same volume as that of a cylindrical tin of radius 17.5 cm. Calculate the height of the cylindrical tin correct to 1 decimal place.
Sol: Length = 40cm
breadth = 30cm
height = 25cm
Volume = l×b×h
= 40×30×25
= 30000 cm³
Radius of cylinder tin = 17.5 cm
Volume = 30000 cm³
Volume = πr²h
30000 = 3.14×17.5×17.5×h
30000 = 961.625 × h
h = 30000/961.325
h = 31.2 cm.
Que-26: Earth taken out on digging a circular tank of diameter 17.5 m is spread all around the tank uniformly to a width of 4m, to form an embankment of height 2m. Calculate the depth of the circular tank, correct two decimal places.
Sol: Diameter of circular tank = 17.5m
∴ Radius = 17.5/2 m = 8.75 m
Width of earth embankment = 4m
Height = 2m
Let it be the depth of the well then volume of earth taken out of πr²h = (22/7)×(8.75)²×h
= (22/7)×(875/100)×(875/100)h
= (22/7)×(35/4)×(35/4)h ….(1)
Volume of platform formed by earth taken out
= π{(51/4)²−(35/4)²}×2
= 2×(22/7)×{(51−35)(51+35)}/16
= 2×(22/7)×{16×86}/16 = 3784/7 sq.m ………(2)
From (1) and (2) we have
(22/7)×(35/4)×(35/4)h = 3784/7
⇒ h = 2.246m
∴Depth of the tank = 2.25m up to 2 decimal places
Que-27: Water is flowing at the rate of 7 metres 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 (1/2) hour.
Sol: Internal diameter of pipe = 2 cm
Internal radius of pipe = 1 cm
radius of base of tank = 40 cm
Speed of water flowing out from pipe = 7 m / sec
= 7×{100/(1/60)} = 7×100×60
= 42000 cm/min
Quantity of water flows out from pipe in 1 min = π×1²×42000 cm³
Quantity of water flows out from pipe in 30 min = π×1²××42000×30
= 3960000 cm³
Quantity of water filled in tank in 30 min = 3960000 cm³
= π×40²×H = 3960000
H = 3960000 × {7/(22×1600)}
H = 787.5 cm.
Que-28: Water flows through a cylindrical pipe of internal diameter 7cm at 36 km/hr. Calculate time in minutes it would take to fill a cylindrical tank, the radius of whose base is 35 cm and height is 1 m.
Sol: Radius of pipe = 7/2 cm
Rate of water flow = 36 km/h
= 36 × 5/18 m/s
= 10 m/s
= 10 x 100 cm/s
= 1000 cm/s
∴ Volume of water flowing in 1 second = πr2h
= π×(7/2)×(7/2)×1000 cm3
= π x 7 x 7 x 250 cm3 ….(1)
Radius of tank (R) = 35 cm
Height of tank (H) = 1 m = 100 cm
∴ Volume of tank = πr2H
= π x 35 x 35 x 100 cm3 …(2)
∴ Time taken to fill the tank = Volume of tank/Volume of water flowing in 1 second
= {π×35×35×100}/{π×7×7×250} seconds
= 10 seconds
= 10/60 minute
= 1/6 = 0.16 minute.
Que-29: Find the number of coins, 1.5cm in diameter and 0.2cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Sol: Radius = 1.5/2 = 3/4 cm
Its thickness = 0.2 cm
Volume of one coin = πr²h
= (22/7)×(3/4)×(3/4)×0.2
= 39.6/112
= 0.35 cm³
Radius of cylinder 4.5/2 = 9/4 cm
Height = 10cm
Volume of cylinder = πr²h
= (22/7)×(9/4)×(9/4)×10
= 17820/112
= 159.11 cm³
Volume of cylinder = volume of one coin × number of coins
159.11 = 0.35 × Number of coins
Number of coins = 159.11/0.35
Number of coins = 454.6
Que-30: If 1 cubic cm of cast iron weighs 21g, then find the weight of a cast iron pipe of length 1m with a bore of 3cm in which the thickness of the metal is 1 cm.
Sol: We know that,
Internal radius = 3/2 = 1.5 cm
External radius = 1.5+1 = 2.5cm
We know that
Volume of cast iron = (π×(2.5)²×100−π×(1.5)²×100)
Taking the common terms out
Volume of cast iron = π×100×(2.52−1.52)
On further calculation
Volume of cast iron = (22/7)×100×(6.25−2.25)
So we get
Volume of cast iron = (22/7)×100×4 = 1257.142 cm³
It is given that 1 cm³ of cast iron weighs 21g
We know that 1kg = 1000g
So the weight of cast iron pipe = 1257.142×21/100 = 26.4kg
Therefore, the weight of cast iron pipe is 26.4kg.
–: End of Three Dimensional Solids Class 10 OP Malhotra Exe-15B ICSE Maths Ch-15 :–
Return to : OP Malhotra S Chand Solutions for ICSE Class-10 Maths
Thanks
Please Share with Your Friends
