Chapter 8 Right Circular Cylinder Exercise 8.2
Question 1. Let us look at the picture of the solid below and answer the following
1. Solid has 
surfaced.
Answer. 3
2. Number of the curved surfaces is and number of plane surface is
Answer. 1, 2.
Read and Learn More WBBSE Solutions For Class 10 Maths
Question 2. Let us write the names of five solid objects of my house, the shapes of which are right circular cylinder.
Solution: Glass,
Pipe,
Tube light,
Pencil,
Gas cylinder.
Question 3. The length of diameter of a drum made of steel covered with a lid is 28 cm. If a 2816 sq cm steel sheet is required to make the drum, let us write by calculating the height of the drum.
Solution: Let the height = h cm
& radius of the drum = 28/2 = 14 cm.
According to the problem,
2π(14+ h) x 14 = 2816
or, 2 x 22/7 x 14 (14+ h) = 2816
or, 14+h=2816 x 7/2 x 22 x 14
= 32
∴ h=32-14
18 cm.
∴Height of the drum = 18 cm.
Question 4. Let us write by calculating how many cubic decimetres of concrete materials will be necessary to construct two cylindrical pillars, each of whose diameter is. 5.6 decimetres and height are 2.5 meters.
Solution: Height of each pillar = h = 2.5 m = 25 dcm.
Radius of each pillar = (r) = 5.6/2 = 2.8 dcm.
The volume of plaster materials required to cover the two pillars
= 2 × π x (2.8)2 × 25 cu dcm.
= 2x 22/7 x 28/10 x 28/10 25
= 56 x 22
= 1232 cu dcm.
The curved surface area of two pillars = 2 x π x 2.8 x 25 sq dcm.
= 2 x 2 x 22/7 x 28/10
= 880 Sq dcm.
= 8.80 sqm.
Total cost at the rate of Rs. 125 per sqm for the two pillars
= Rs. 8.8 x 125
= Rs. 1100.
Question 5. If a gas cylinder for fuel purposes having a length of 7.5 dcm and the length of an inner diameter of 2.8 dcm carries 15.015 kg of gas, let us write by calculating the weight of the gas of per cubic dcm.
Solution: Weight of gas in the cylinder 15.015 kg = 15015. gm.
The radius of the cylinder = 2.8/2
= 1.4 dcm.
∴The volume of the cylinder = (1.4)2 x 7.5 cu dcm.
=22/7 x 14/10 x 14/10 x 75/10
=462/10
= 46.2 cu dcm.
∴ Weight of 1 cu dcm gm = 15015/46.2gm
= 325 gm.
Question 6. Out of three jars of equal diameter and height, 2/3 part of the first, 5/6 part of the second, and 7/9 part of the third were filled with dilute sulphuric acid. The whole of the acid in the three jars was poured into a jar of 2.1 dcm diameter, as a result the height of acid in the jar becomes 4.1 dcm. If the length of the diameter of each of the three equal jars is 1.4 dcm, let us write by calculating the height of the three jars.
Solution: Let the height of each jar = h dcm.
According to problem,(2/3 + 5/6 + 7/9) x h = π(2.1/2)2 x 4.1
Or,12+15+14/18 x h
= 22/7 x 21/20 x 21/20 41/10
.. h = 21/7 x 21/20 x 21/20 x 41/10 c 18/41
4.05 dcm.
∴ The height of each jar is 4.05 dcm.
Question 7. The total surface area of a right circular pot open at one end is 2002 sq cm. If the length of diameter of base of the pot is 14 cm, let us write by calculating how many liters of water the drum will contain.
Solution: Let the height of the cylinder = h cm.
& the radius of the cylinder =14/2 = 7 cm.
According to the problem,
or, 2π x 7 x h + л(7)2 = 2002
or, 2 x 22/7 x 7 x h+ 22/7 x 7 x 7 = 2002
or, 44h+ 154 = 2002
∴ 44h = 2002 – 154 = 1848
∴ h= 1848/44
= 42
∴ Height of the cylinder = 42 cm. = 4.2 dcm.
The radius of the cylinder = 7 cm. = 0.7 dcm.
∴ The volume of water in the cylinder = 7(0.7)2 x 4.2 cu dcm.
= 22/7 x 7/10 x 7/10 x 42/10
=6468 / 1000
= 6.468 cu dcm.
=6.468 litre [1 cu dcm = 1 litre]
Question 8. If a pump set with a pipe of 14 cm diameter can drain 2500 meters of water per minute, let us write by calculating how much kilometer of water that pump will drain per hour. [1 liter = 1 cubic dcm.]
Solution: Radius of the pipe = 14/2 = 7 cm = 0.7 dcm.
Height of the pipe = 2500 m = 25000 dcm.
The cross-sectional area of the mouth of the pipe
= 22/7 x 72 sq cm. = 154 Sq cm. 1.54 sq dm.
The volume of water irrigated by the pipe in one minute
= (1.54 x 25000) cu dcm
= 38500 cu dcm
∴ The volume of water irrigated by the pipe in one hour
= (38500 x 60) cu dcm
= 2310000 cu dcm.
Now, 1 cu dcm = 1 liter
∴ The pump set can irrigate 2310000 liters
= 2310 Kilolitre in 1 hour.
Question 9. There is some water in a long gas jar of 7 cm in diameter. If a solid right circular cylindrical pipe of iron having 5 cm length and 5.6 cm diameter be immersed completely in that water, let us write by calculating how much the level of water will rise.
Solution: Let the water level will rise by h cm.
According to the problem,
π(7/2)² x h = π(5.6/2)² x 5
49/4 h = 28/10 x 28/10 x 5
H = 28 x 28 x 5 / 10 x 10 x 49 x 4
= 32/10
= 3.2
∴ The water level will rise by 3.2 cm.
Question 10. If the surface area of a right circular cylindrical pillar is 264 sq meters and the volume is 92 cubic meters, let us write by calculating the height and length of the diameter of this pillar.
Solution: Let the radius of the pillar = rm & height = h m.
According to 1st condition, 2πth = 264 ——– (1)
According to 2nd condition, πr²h = 924———(2)
πr²h/2πrh = 924/264
Or, r = 7
2πrh 2 x 22/7 x 7 h = 264
H = 264/44
= 6
∴ Diameter of the pillar = 2 x 7 = 14 m
& height of the pillar = 6 m.
Question 11. A right circular cylindrical tank of 9 meters in height is filled with water. Water comes out from there through a pipe having a length of 6 cm diameter with a speed of 225 meters per minute and the tank becomes empty after 2 hr. 24 minutes, let us write by calculating the length of the diameter of the tank.
Solution: Let the radius of the tank = R m
& radius of the pipe = 6/2
= 3 cm.
= 0.03m.
According to the problem,
π R² x9= 36 (0.03)∴ x 225 (Where radius of pipe = R)
R² = 36 x 3/100 x 3/100 x 225 x 1/9
= 81/100
∴ R = 9/10
Diameter of the tank = 2 x 9/10
= 1.8 m.
Question 12. The curved surface area of the right circular cylindrical log of wood of uniform density is 440 sq dcm. If I cubic dcm of wood weighs 1.5 kg and the weight of the log is 9.24 quintals. Let us write by calculating the length of the diameter of the log and its height.
Solution: Weight of the wooden log = 9.24 quintal = 924 kg.
∴ The volume of the log = 924 / 1.5 cu dcm 616 cu dcm.
Let the radius of the log = r dcm
& height of the log = h dcm.
According to 1st condition, 2лth = 440———-(1)
According to 2nd condition, лr²h = 616———(2)
= лr²h/2лrh = 616/440
Or, r/2 = 616/440
R = 2 x 616 / 440
= 28/10
Or, 2 x 22/7 x 28/10
h = 100/4
= 25
∴ Diameter of the log = 2 x 28/10
= 56/10
= 5.6
Height of the log = 25. dcm.
Question 13. The lengths of the inner and outer diameter of a right circular cylindrical pipe open at two ends are 30 cm and 26 cm respectively and the length of the pipe is 14.7 metres. Let us write by calculating the cost of painting its all surfaces with coal tar at Rs. 2.25 per dcm.
Solution: Internal radius of the pipe = 26/20
= 1.3 dcm.
The external radius of the pipe = 30/20
= 1.5 dcm.
Height of the pipe 14.7 m = 147 dcm.
The total surface area of the pipe
= [27(1.5+ 1.3) 147+ 2π((1.5)2- (1.3)2)] sq dcm..
= [2π x 2.8 × 147 + 2 x 2.8 x 0.2] sq dcm.
= 2π x 2.8(147+ 0.2)
= 2x 22/7 x 28/10 x 147.2 sq dcm.
= 17.6 x 147.2 2590.72 sq dcm.
Total cost for painting with coal tar
= Rs. (2.25 x 2590.72)= Rs. 5829.12.
Question 14. The height of a hollow right circular cylinder, open at both ends, is 2.8 metres. If the length of the inner diameter of the cylinder is 4.6 dcm and the cylinder is made up of 84.48 cubic dcm of iron, let us calculate the length of the outer diameter of the cylinder.
Solution: Let the external radius = r dcm. of the cylinder & the height of the cylinder = 2.8 m = 28 dcm.
The internal radius of the cylinder = 4.6/2
= 2.3 dcm.
According to the problem,
л(r² (2.3)²) x 28 = 84.48
or,22/2(r2-5.29) × 28 = 84.48
r² – 5.29 = 84.48/88 = 0.96
R² = 5.29
= 5.29 + 0.96
= 6.25
External diameter = 2 x 2.5 5 dcm.
Question 15. Height of a right circular cylinder is twice its radius. If the height would be 6 times its radius, then the volume of the cylinder would be greater by 539 cubic dcm, let us write by calculating the height of the cylinder.
Solution: Let the height of the cylinder = h dcm & radius = r dcm.
∴ h = 2r
Volume of the cylinder = r2 x 2r= 2πr3
If height h=6r, volume = r2 x 6г = 6πr3
According to the problem, 6r²r³ = 539
4πr³ = 536
Or, 4 x 22/4 x r³ =539
∴ r³ = 539 x 7/88
=(7/3)³
r =7/2
2r = 7
∴ Height 2r= 7dcm.
Question 16. A group of fire brigade personnel carried a right circular cylindrical tank filled with water and pumped out water at a speed of 420 metres per minute to put out the fire in 40 minutes by three pipes of 2 cm in diameter each. If the diameter of the tank is 2.8 meters and its length is 6 metres, then let us calculate
1. what volume of water has been spent in putting out the fire and
2. the volume of water that still remains in the tank.
Solution: Radius of the tanker = 2.8/2
= 1.4
= 14 dcm.
Length of the tanker = 6 m = 60 dcm.
The volume of the tanker = π(14)² x 60 cu dcm.
Radius of each pipe = 2/2
= 1 cm
= 1/10 dcm.
Length of the pipe = 420 m = 4200 dcm.
Quantity of water ejected in 40 minutes by 3 pipes
= 40 x 3 x π(1/10)² x 4200 cu dcm.
= 40 x 3 x 22/7 x 1/10 x 1/10 x 4200 cu dcm.
= 15840 cu dcm.
∴ Quantity of water left in the tanker
= (3696015840) cu dcm.
= 21120 cu dcm.
Question 17. It is required to make a plastering of sand and cement with 3.5 cm thick, sur- rounding four cylindrical pillars, each of whose diameter is 17.5 cm.
1. If each pillar is of 3-meter height, let us write by calculating how many cubic dcm of plaster materials will be needed.
2. If the ratio of sand and cement in the plaster material be 4: 1, let us write how many cubic dcm of cement will be needed.
Solution: Length of each pillar = 3 m = 30 dcm.
Internal radius of each pillar = 17.5/2
175/20 cm.
= 7/8 dcm.
The external radius of each pillar = (7/8 + 75/10) dcm.
= (7/8 + 35/100)
= (7/8 + 7/20)dcm.
= 35+14 / 40
= 49/40 dcm.
1. Plaster material is required for 4 pillars
= 4 π {(49/40)² – (7/8 )²} x 30 cu dcm.
= 4x 22/7 {(49/40 + 7/8)(49/40 – 7/8)} x 30 cu dcm.
= 4 x 22/7 x 84/40 x 14/40 x 30 cu dcm.
= 2772/10
= 277.2 cu dcm.
2. Volume of cement required = 1/5x 277.2 cu dcm.
= 55.44 cu dcm.
Question 18. The length of the outer and inner diameter of a hollow right circular cylinder are 16 cm and 12 cm respectively. Height of the cylinder is 36 cm. Let us calculate how many solid cylinders of 2 cm radius and 6 cm length may be obtained by melting this cylinder.
Solution: External radius of the cylinder = 16/2
= 8 cm.
Internal radius of the cylinder = 12/2
= 6 cm.
Height of the cylinder = 36 cm.
The radius of the solid cylinder = 2/2
= 1 cm.
Height of the cylinder = 6 cm.
Volume of hollow cylinders = π{(8)²- (6)²} x 36 cu cm.
The volume of x no. of solid cylinders = x. π(1)². 6 cu cm.
According to the problem,
2. π(1)². 6=π{(8)² – (6)²) x 36
or, x. π. 1.6 = πx (64-36) x 36
or, x = 28 x 36 / 6
= 168
∴ No. of solid cylinders = 168.
Multiple Choice Question
Question 1. If the lengths of radii of two solid right circular cylinders are in the ratio 2:3 and their heights are in the ratio 5: 3, then ratio of their lateral surface areas is
1. 2:5
2. 7:7
3. 10: 9
4. 16:9
Solution: Ratio of area of curved surfaces
=2π ×2×5:2π × 3 × 3
=10:9
Answer. 3. 10: 9
Question 2. If the lengths of radii of two solid right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5: 3, then the ratio of their volumes is
1. 27: 20
2. 20:27
3. 40:9
4. 9:4
Solution: Ratio of volumes of the cylinder
=π(2)² × 5: π(3)² x 3 = 20:27
Answer. 2. 20:27
Question 3. If volumes of two solid right circular cylinders are the same and their heights are in the ratio 1 2, then the ratio of lengths of radii is
1. 1:√2
2. √2:1
3. 1:2
4. 2:1
Solution: Ratio of the radius of the cylinder = √2:1
Answer. 2. √2:1
4. In a right circular cylinder, if the length of radius is halved and height is doubled, volume of the cylinder will be
1. Equal
2. Double
3. Half
4. 4 times
Question 5. If the length of the radius of a right circular cylinder is doubled and the height is halved, the lateral surface area will be
1. Equal
2. Half
3. Double
4. 4 times
True Or False
1. The length of a right circular drum is r cm and the height is h cm. If half part of the drum is filled with water then the volume of water will be r2h cubic cm.
False
2. If the length of the radius of a right circular cylinder is z unit, the numerical value of volume and surface area of the cylinder will be equal for any height.
True
Fill In The Blanks
1. The length of a rectangular paper is units and the breadth is b units. The rectangular paper is rolled and a cylinder is formed whose perimeter is equal to the length of the paper. The lateral surface area of the cylinder is lb sq unit.
2. The longest rod that can be kept in a right circular cylinder has a diameter of 3 cm and height of 4 cm, then the length of the rod is 5 cm.
3. If the numerical values of volume and lateral surface area of a right circular cylinder are equal then the length of the diameter of the cylinder is 4 units.
Short Answers
Question 1. If the lateral surface area of a right circular cylindrical pillar is 264 sq meters and the volume is 924 cubic meters, let us write the length of the radius of the base of the cylinder.
Solution: Let the radius & height of the pillar = r m & h m respectively.
According to 1st condition, 2rh = 264——–(1)
According to 2nd condition, r2h = 924 ——-(2)
(2) ÷ (1),
= πr²h/2πrh
= 924/264
Or, r/2 = 7/2
R = 7
∴ Radius = 7 cm.
Question 2. If the lateral surface area of a right circular cylinder is c square units, the length of the radius of the base is r units and the volume is cr/v cubic units, let us write the value of
Solution. Let height = h unit.
Cr/v= 2πrhxr / πr²h
= 2
Question 3. If the height of a right circular cylinder is 14 cm and the lateral surface area is 264 sq cm, let us write the volume of the cylinder.
Solution. Let the height of the cylinder = r cm.
∴ 2πгh = 264
or, 2 x 22/7 x r x 14= 264
r=3
∴ Volume = π(3)² x 14 cu cm
= 22/7 x 9 x 14 cu cm = 396 cu cm.
Question 4. If the heights of two right circular cylinders are in the ratio of 1: 2 and prim- meters are in the ratio of 3: 4, let us write the ratio of their volumes.
Solution. The Radius & height of the two cylinders are r unit & R unit & heights are h unit & 2h unit respectively.
∴ 2πr : 2πR 3:4
∴ r: R 3:4
h =3R / r
The ratio of volume = (r)² h: (R)²
2h = (3R/4)²: 2R²
Or, 9R/16 : 2R2 = 9:32
Question 5. The length of the radius of a right circular cylinder is decreased by 50% and height is increased by 50%, let us write by how much percent of the volume will be changed.
