Read RD Sharma Solutions Class 9 Chapter 19 Surface Area and Volume of A Right Circular Cylinder below, students should study RD Sharma class 9 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 9 Mathematics have been prepared by teacher of Grade 9. These RD Sharma class 9 Solutions have been designed as per the latest NCERT syllabus for class 9 and if practiced thoroughly can help you to score good marks in standard 9 Mathematics class tests and examinations
Exercise 19.1
Question 1: Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m. Find its height.
Solution:
It is given that,
Radius of the cylinder (r) = 0.7m
Curved surface area of cylinder = 4.4m2
Curved surface area of a cylinder = 2πrh
2πrh = 4.4
2 x 3.14 x 0.7 x h = 4.4
4.4 x h = 4.4
h = 4.4/4.4
h = 1
Hence, the height of the cylinder is 1 m.
Question 2: In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Solution:
Height of cylinder (h) = 28m × 100 = 2800cm
Diameter = 5 cm
r = diameter/2 = 5/2 cm
Curved surface area = 2πrh
Curved surface area = 2 × 3.14 × 5/2 × 2800
Curved surface area = 44000
Hence, the total radiating surface in the system is 44000cm2.
Question 3: A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs 12.50 per m2.
Solution:
Height of the pillar (h) = 3.5 m
Radius of end of pillar (r) = 50/2 cm = 25 cm = 0.25m
CSA of cylindrical pillar = 2πrh
CSA of cylindrical pillar = 2 x 3.14 x 0.25 x 3.5
CSA of cylindrical pillar = 5.5m
It is given that, cost of whitewashing is Rs. 12.50 for 1m2
Cost of whitewashing 5.5 m2 area = Rs. 12.50 × 5.5
Cost of whitewashing 5.5 m2 area = Rs. 68.75
Hence, the cost of whitewashing the pillar is Rs. 68.75.
Question 4: It is required to make a closed cylindrical tank of height 1 m and the base diameter of 140 cm from a metal sheet. How many square meters of the sheet are required for the same?
Solution:
Height of cylindrical tank (h) = 1 m
Radius of cylindrical tank (r) = diameter/2 = 140/2 cm = 70 cm = 0.7 m
TSA of tank = 2πr(h + r)
TSA of tank = 2 x 3.14 x 0.7(1 + 0.7)
TSA of tank = 7.48
Hence, metal sheet is required to make required closed cylindrical tank 7.48 m2.
Question 5: A solid cylinder has a total surface area of 462 cm2. Its curved surface area is one-third of its total surface area. Find the radius and height of the cylinder.
Solution:
Total surface area of a cylinder = 462 cm2
Curved or lateral surface area = 1/3 (TSA)
2πrh = 1/3 (462)
2πrh = 154
h = 154/2πr
h = (154 × 7)/(2 × 22 × r)
h = (7 × 7)/2r
h = 49/2r ______(1)
Put the value of h in below equation
TSA = 462cm2
2πr(h + r)=462
2× 22/7×r(49/2r+r)=462
44r/7×((49 + 2r2)/2r)=462
22/7×(49 + 2r2)=462
(49 + 2r2)=462×7/22
49 + 2r2 = 42 × 7
49 + 2r2 = 147
2r2 = 147 - 49
2r2 = 98
r2 = 98/2
r2 = 49
r2 = √49
r = 7
Put the value of r in equation (1)
h = 49/2r
h = 49/(2(7))
h = 49/14
h = 7/2
Hence, Height (h) of the cylinder is 7/2 cm and radius of the cylinder is 7cm.
Question 6: The total surface area of a hollow cylinder which is open on both the sides is 4620 sq.cm and the area of the base ring is 115.5 sq.cm and height is 7 cm. Find the thickness of the cylinder.
Solution:
TSA of hollow cylinder = 4620 cm2
Height of cylinder (h) = 7 cm
Area of base ring = 115.5 cm2
To find: Thickness of the cylinder
Let ‘r1’ and ‘r2’ are the inner and outer radii of the hollow cylinder respectively.
Then, πr22 – πr12 = 115.5 …….(1)
And,
2πr1h +2πr2h+ 2(πr22 – πr12) = 4620
Or 2πh (r1 + r2 ) + 2 x 115.5 = 4620
(Using equation (1) and h = 7 cm)
or 2π7 (r1 + r2 ) = 4389
or π (r1 + r2 ) = 313.5 ….(2)
Again, from equation (1),
πr22 – πr12 = 115.5
or π(r2 + r1) (r2 – r1) = 115.5
[using identity: a2 – b2 = (a – b)(a + b)]
Using result of equation (2),
313.5 (r2 – r1) = 115.5
or r2 – r1 = 7/19 = 0.3684
Therefore, thickness of the cylinder is 7/19 cm or 0.3684 cm.
Question 7: Find the ratio between the total surface area of a cylinder to its curved surface area, given that height and radius of the tank are 7.5 m and 3.5 m.
Solution:
Height of cylinder (h) = 7.5 m
Radius of cylinder (r) = 3.5 m
Total Surface Area of cylinder (TSA) = 2πr(r + h)
Curved surface area of a cylinder (CSA) = 2πrh
The Ratio between TSA and CSA
TSA/CSA = (2πr(r+h))/2πrh
TSA/CSA = ((r + h))/h
TSA/CSA = ((3.5 + 7.5))/7.5
TSA/CSA = 11/7.5
TSA/CSA = 22/15
TSA : CSA= 22 : 15
Hence, the ratio of TSA and CSA is 22 : 15.
Question 8: The total surface of a hollow metal cylinder, open at both ends of an external radius of 8cm and height 10cm is 338cm2. Take r to be the inner radius, obtain an equation in r and use it to obtain the thickness of the metal in the cylinder.
Solution:
The external radius of the cylinder (R) = 8 cm
Height of the cylinder (h) = 10 cm
TSA of the hollow cylinder = 338π cm2
CSA of External cylinder + CSA of Internal cylinder + Circumference of Internal Circle - Circumference of External Circle = 338π cm2
2πrh + 2πRh + 2πR2 − 2πr2 = 338π cm2
2πh(R + r) + (R2 − r2) = 338π cm2
10(R + r) + (R2 − r2) = 338π/2πcm2
10(8 + r) + (82 − r2) = 169
80 + 10r + 64 − r2 = 169
10r − r2 = 169 – 80 - 64
10r − r2 = 25
r2 − 10r + 25 = 0
r2 − 5r − 5r + 25 = 0
r(r – 5) – 5(r – 5) = 0
(r – 5) (r – 5) = 0
r = 5
Thickness of the metal cylinder = External Radius of cylinder - Internal Radius of cylinder
Thickness of the metal cylinder = 8cm – 5cm
Thickness of the metal cylinder = 3cm
Hence, the thickness of the metal cylinder is 3cm.
Question 9: A cylindrical vessel, without lid, has to be tin-coated on its both sides. If the radius of the base is 70 cm and its height is 1.4 m, calculate the cost of tin-coating at the rate of Rs 3.50 per 1000 cm2.
Solution:
It is given that,
Radius of the vessel (r) = 70 cm
Height of the vessel (h) = 1.4 m = 140 cm
Area to be tin coated = 2(2πrh + πr2)
Area to be tin coated = 2πr (2h + r)
Area to be tin coated = 2 × 3.14 × 70[(2 × 140) + 70]
Area to be tin coated = 154000 cm2
Also, it is given that the rate of Rs 3.50 per 1000 cm2.
= 154000 × 3.5/1000
= Rs. 154 × 3.5
= Rs. 539
Hence, the cost of tin-coating is Rs. 539
Question 10: The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find:
(a) Inner curved surface area
(b) The cost of plastering this curved surface at the rate of Rs. 40 per m2.
Solution:
Inner radius = diameter/2 = 3.52/2 = 1.75m
Height of the well = 10 m
(a) Inner curved surface area
Inner curved surface area = 2πrh
Inner curved surface area = 2 × 3.14 × 1.75 × 10
Inner curved surface area = 110m2
(b) It is given that, the cost of painting 1m2 area of the well is Rs. 40
Cost of painting 110m2 area of the well
Cost of painting = Rs. 40 × 110
Cost of painting = Rs. 4400
Question 11: Find the lateral surface area of a petrol storage tank is 4.2 m in diameter and 4.5 m high. How much steel was actually used, if 1/12th of the steel actually used was wasted in making the closed tank?
Solution:
It is given that,
Diameter of cylinder = 4.2 m
Radius of cylinder = 4.22/2m = 2.1 m
Height of cylinder = 4.5 m
CSA of cylinder = 2πrh
CSA of cylinder = 2 × 3.14 × 2.1 × 4.5
CSA of cylinder = 59.4 m2
TSA of cylinder = 2πr(r + h)
TSA of cylinder = 2× 22/7×2.1(2.1+4.5)m2
TSA of cylinder = 87.12 m2
Let, xm2 steel actually used in making the tank = x-1x/12=(12x-1x)/12=11x/12
Area of iron present in cylinder = 11x/12m2
11x/12 = Total surface area of cylinder
x = Total surface area ×12/11
x = 87.12×12/11
x = 95.04m2
Hence, m2 steel was actually wasted while constructing a tank.
Question 12: The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Take π = 3.14).
Solution:
It is given that
Radius of the circular part of the penholder (r) = 3 cm
The height of the penholder (h) = 10.5 cm
Surface area of one penholder
Surface area of one penholder = Curved surface area of penholder + Area of the circular base of penholder
Surface area of one penholder = 2πrh + πr2
Surface area of one penholder = (2 × 3.14 × 3 × 10.5) + 3.14 × 32
Surface area of one penholder = (2 × 22/7 × 3 × 10.5) + 22/7 × 9
Surface area of one penholder = 1386/7 + 198/7
Surface area of one penholder = (1386 + 198)/7
Surface area of one penholder = 1584/7
Cardboard used by 35 competitors = 1584/7 × 35 cm2
Cardboard used by 35 competitors = 7920 cm2
Hence, 7920 cm2 of cardboard sheet for the competition.
Question 13: The diameter of roller 1.5 m long is 84 cm.If it takes 100 revolutions to level a playground, find the cost of levelling this ground at the rate of 50 paise per square meter.
Solution:
Diameter of the roller (d) = 85 cm = 0.85m
Radius of the roller(r) = d/2 = 0.842/2 = 0.42m
Length of the roller = 1.5 m
CSA of the roller = 2πrh
CSA of the roller = 2 × 3.14 × 0.42 × 1.5
CSA of the roller = 0.12 × 22 × 1.5 m2
CSA of the roller = 3.96m2
Area of the playground = 100 × Area covered by roller in one revolution
Area of the playground = 100 × 3.96m2
Area of the playground = 396 m2
It is given that, Cost of levelling 1m2 = 50p = Rs. 0.5
Cost of levelling 396m2 = Rs. 396 × 0.5 = Rs. 198
Hence, the cost of levelling of the playground is Rs. 198
Question 14: Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50m and height is 4m.What will be the cost of cleaning them at the rate of Rs 2.50 per square meter?
Solution:
Diameter of each pillar = 0.5m
Radius of each pillar(r) = d/2 = 0.5/2 = 0.25m
Height of each pillar = 4m
Lateral surface area of one pillar = 2πrh
Lateral surface area of one pillar = 2 × 22/7 × 0.25 × 4
Lateral surface area of one pillar = 44/7m2
Lateral surface area of 20 pillars = 880/7m2
It is given that, Cost of cleaning one pillar is Rs 2.50/m2
Cost of cleaning 20 pillars = Rs. 2.50 × 880/7m2
Cost of cleaning 20 pillars = Rs. 314.28
Hence, the cost of cleaning 20 pillars is Rs. 314.28.
Exercise 19.2
Question 1: A soft drink is available in two packs- (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm, Which container has greater capacity and by how much?
Solution:
(i) Length of a cubical tin (L) = 5 cm
Breadth of a cubical tin (B) = 4 cm
Height of a cubical tin (H) = 15 cm
Volume of Tin Can = l × b × h
Volume of Tin Can = (5 × 4 × 15)cm3
Volume of Tin Can = 300cm3
(ii) Radius (r) = diameter/2
Radius (r) = 7/2cm
Radius (r) = 3.5cm
Height of plastic cylinder (H) = 10 cm
Volume of cylindrical container = πR2H
Volume of cylindrical container = 22/7 × (3.5)2 × 10cm3
Volume of cylindrical container = 385 cm3
The plastic cylinder has greater capacity
Difference in capacity = (385 – 300) cm3
Difference in capacity = 85 cm3
Question 2: The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius 20 cm and height 10 m. How much concrete mixture would be required to build 14 such pillars?
Solution:
In this case, we have to find the volume of the cylinders.
Radius of cylinder = 20cm
Height of cylinder = 10m = 1000cm
Volume of the cylindrical pillar = πR2H
Volume of the cylindrical pillar = (22/7 × (20)2× 1000) cm3
Volume of the cylindrical pillar = 8800000/7 cm3
Volume of the cylindrical pillar = 8.87 m3
Hence, volume of 14 pillars = 14 × 8.87m3 = 17.6 m3
Question 3: The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 gm.
Solution:
Inner radius of a cylindrical pipe (r) = 24/7
Inner radius of a cylindrical pipe (r) = 12cm
Outer radius of a cylindrical pipe (R) = 24/7
Outer radius of a cylindrical pipe (R) = 14cm
Height of pipe (h) = 35cm
Mass of pipe = π(R2 – r2)h
Mass of pipe = 22/7 (142 – 122)35
Mass of pipe = 5720cm3
It is given that, the mass of 1 cm3 wood = 0.6 gm
Mass of 5720 cm3 wood = 5720 x 0.6 = 3432gm
Hence, the mass of 5720 cm3 wood is 3.432 kg.
Question 4: If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, find: (i) radius of its base (ii) volume of the cylinder
Solution:
Lateral surface of the cylinder = 94.2 cm2
Height of the cylinder = 5 cm
Let ‘r’ be the radius.
(i) Lateral surface of the cylinder = 94.2 cm2
2 πrh = 94.2
2 × 3.14 × r × 5 = 94.2
r = 94.2/(2 × 3.14 × 5)
r = 94.2/31.4
r = 3 cm
(ii) Volume of the cylinder = πr2h
Volume of the cylinder = 3.14 × (3)2 × 5 cm3
Volume of the cylinder = 141.3 cm3
Question 5: The capacity of a closed cylindrical vessel of height 1 m is 15.4 liters. How many square meters of the metal sheet would be needed to make it?
Solution:
It if given that,
The capacity of a closed cylindrical vessel of height 1m is 15.4 liters.
Height of the cylindrical vessel = 15.4 litres × 1000 = 0.0154 m3
Volume of the cylinder = πr2h
πr2h = 0.0154 m3
3.14 x r2 x 1 = 0.0154 m3
r2 = 0.0154/3.14
r = 0.0049
r = 0.07 m
TSA of a vessel = 2πr(r + h)
TSA of a vessel = 2(3.14(0.07)(0.07+1))m2
TSA of a vessel = 0.470 m2
Question 6: A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Solution:
Radius of cylindrical bowl (R) = diamete/2
Radius of cylindrical bowl (R) = 7/2 cm
Radius of cylindrical bowl (R) = 3.5 cm
Height = 4 cm
Volume of soup in 1 bowl = πr2h
Volume of soup = 22/7 × (3.5)2 × 4 cm3
Volume of soup = 154cm3
Volume of soup in 250 bowls = (250 × 154) cm3
Volume of soup in 250 bowls = 38500 cm3
Volume of soup in 250 bowls = 38500/1000
Volume of soup in 250 bowls 38.5 liters
Hence, to serve 250 people, the hospital must prepare 38.5 litres of soup per day.
Question 7: A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4 cm thick iron. Find the volume of the iron.
Solution:
Outer circumference = 440 cm
Thickness of the roller = 4 cm
Height (h) = 63 cm
Let ‘R’ be the external radius and ‘r’ be the inner radius of the roller.
Circumference of roller = 2πR
2πR = 440
2 × 22/7 × R = 440
44/7 × R = 440
R = 440 × 7/44
R = 70
It is given that, the inner radius ‘r’ is
r = R – 4
r = 70 – 4
r = 66
Inner radius is 66 cm
Volume of the iron = π(R2−r2)h
Volume of the iron = 22/7 {(70)2 − (66)2}63
Volume of the iron = 107712
Hence, the volume of the iron 107712 cm3
Question 8: A solid cylinder has a total surface area of 231 cm2. Its curved surface area is 2/3 of the total surface area. Find the volume of the cylinder.
Solution:
Total surface area = 231 cm2
CSA = 2/3 TSA
Curved surface area = 2/3 × 231
Curved surface area = 154 cm2
Curved surface area of cylinder = 2πrh + 2πr2
2πrh + 2πr2 = 231
154 + 2πr2 = 231 (2πrh = 154cm2)
2πr2 = 231- 154
2 × 22/7 × r2 = 77
r2 = (77 × 7)/(2 × 22)
r2 = 49/4
r = 7/2
CSA = 154 cm2
2πrh = 154
2 x 22/7 x 7/2 x h = 154
h = 154/22
h = 7
Volume of the cylinder = πr2h
Volume of the cylinder = 22/7 x 7/2 x 7/2 x 7
Volume of the cylinder = 269.5cm3
Hence, the volume of the cylinder is 269.5cm3
Question 9: The cost of painting the total outside surface of a closed cylindrical oil tank at 50 paise per square decimetre is Rs 198. The height of the tank is 6 times the radius of the base of the tank. Find the volume corrected to 2 decimal places.
Solution:
Let us assumed that, radius of the tank is ‘r’.
Height (h) = 6(Radius)
Height (h) = 6r dm
It is given that, cost of painting for 50 paisa dm2 is Rs. 198
2πr(r + h) × 1/2 = 198
2 × 22/7 × r(r + 6r) × 1/2 = 198
r = 3 dm
h = (6 × 3) dm = 18 dm
Volume of the tank = πr2h
Volume of the tank = 22/7 × 9 × 18
Volume of the tank = 509.14 dm3
Hence, the volume of the tank is 509.14 dm3
Question 10: The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their volumes and the ratio of their curved surfaces.
Solution:
Let us assumed that,
The radius of the first cylinders (r) = 2x
The radius of the first cylinders (R) = 3x
Height of the first cylinders (h) = 5y
Height of the first cylinders (H) = 3y
Question 11: The ratio between the curved surface area and the total surface area of a right circular cylinder is 1:2. Find the volume of the cylinder, if its total surface area is 616 cm2.
Solution:
It is given that, TSA = 616 cm2
Let us assumed that,
Radius of cylinder is = r
Height of cylinder is = h
CSA/TSA = 1/2
CSA = 1/2 TSA
CSA = 1/2 × 616
CSA = 308
CSA = 308cm2
TSA = 2πrh + 2πr2
616 = CSA + 2πr2
616 = 308 + 2πr2
616 – 308 = 2πr2
2πr2 = 616 – 308
2πr2 = 308
r2 = 308/2π
r2 = 49
r = 7cm
CSA = 308cm2
2πrh = 308
2 × 22/7 × 7 × h = 308
h = 7 cm
Volume of cylinder = πr2h
Volume of cylinder = 22/7 × 7 × 7 × 7
Volume of cylinder = 1078
Hence, the volume of cylinder is 1078 cm3.
Question 12: The curved surface area of a cylinder is 1320 cm2 and its base had diameter 21 cm. Find the height and volume of the cylinder.
Solution:
CSA of a cylinder = 1320 cm2
r = diamter/2 = 21/2 cm = 10.5 cm
CSA = 2πrh
2πrh = 1320
2 × 22/7 × 10.5 × h = 1320
h = (1320 × 7)/(10.5 × 44)
h = 1320/(1.5 × 44)
h = 1320/66
h = 20 cm
Volume of cylinder = πr2h
Volume of cylinder = 22/7 × 10.5 × 10.5 × 20
Volume of cylinder = 22 × 1.5 × 10.5 × 20
Volume of cylinder = 6930
Hence, Volume of cylinder is 6930 cm3.
Question 13: The ratio between the radius of the base and the height of a cylinder is 2:3. Find the total surface area of the cylinder, if its volume is 1617cm3.
Solution:
It is given that, the ratio of radius of the cylinder and height of the cylinder is 2 : 3.
r : h = 2:3
And Volume of cylinder= 1617 cm3
Radius = 2x cm
Height = 3x cm
Volume of cylinder = πr2h
1617= 22/7 (2x)2 × 3x
1617= 22/7 × 4x2 × 3x
1617 = 22/7 × 12x3
(1617 × 7)/(22 × 12) = x3
(1617 × 7)/(22 × 12) = x3
11319/264 = x3
x3 = 343/8
x = 7/2
x = 3.5 cm
Radius of cylinder = 2x
Radius of cylinder = 2 × 3.5
Radius of cylinder = 7cm
Height of cylinder = 3x
Height of cylinder = 3 × 3.5
Height of cylinder = 10.5cm
TSA of cylinder = 2πr(h + r)
TSA of cylinder = 2 × 22/7 × 7(10.5+7)
TSA of cylinder = 2 × 22/7 × 7(17.5)
TSA of cylinder = 2 × 22 × (17.5)
TSA of cylinder = 44 × (17.5)
TSA of cylinder = 770cm2
Hence, The Total surface area of cylinder is 770 cm2.
Question 14: A rectangular sheet of paper, 44 cm x 20 cm, is rolled along its length of form cylinder. Find the volume of the cylinder so formed.
Solution:
Length of a rectangular sheet = 44 cm
Height of a rectangular sheet = 20 cm
The base of the cylinder in shape of circle
2πr = 44
r = 44/2π
r = 44 × 1/2 × 7/22
r = 7 cm
Volume of cylinder = πr2h
Volume of cylinder = 22/7 × 7 × 7 × 20
Volume of cylinder = 22 × 7 × 20
Volume of cylinder = 3080
Hence, the volume of cylinder is 3080 cm3.
Question 15: The curved surface area of cylindrical pillar is 264m2 and its volume is 924m3. Find the diameter and the height of the pillar.
Solution:
CSA of cylindrical pillar = 264m2
2πrh = 264
πrh = 264/2
πrh = 132
Volume of the cylinder = 924 m3
πr2h= 924
πrh(r) = 924
132r = 924
r = 924/132
r = 7m
Put the value of r in equation
22/7 × 7 × h = 132
h = (132 ×7)/(7 ×22)
h = 132/22
h = 6m
Diameter = 2r = 2(7) = 14 m
Height = 6m
Hence, the diameter and height is 14m and 6m.
Exercise VSAQs :-->
Question 1: Write the number of surfaces of a right circular cylinder.
Solution:
3 surfaces of a right circular cylinder.
Question 2: Write the ratio of total surface area to the curved surface area of a cylinder of radius r and height h.
Solution:
The Ratio of TSA to the CSA of a cylinder of radius (r) and height (h)
TSA/CSA=(2πr(h+r))/(2πr2 )
TSA/CSA=(h+r)/r