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 m**

^{2}. 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 = 4.4m

^{2}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 44000cm

^{2}.**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 m**

^{2}.**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 1m

^{2}Cost of whitewashing 5.5 m

^{2}area = Rs. 12.50 × 5.5Cost of whitewashing 5.5 m

^{2}area = Rs. 68.75Hence, 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 m

^{2}.**Question 5: A solid cylinder has a total surface area of 462 cm**

^{2}. 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 cm

^{2}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 = 462cm

^{2}2πr(h + r)=462

2× 22/7×r(49/2r+r)=462

44r/7×((49 + 2r

^{2})/2r)=46222/7×(49 + 2r

^{2})=462(49 + 2r

^{2})=462×7/2249 + 2r

^{2}= 42 × 749 + 2r

^{2}= 1472r

^{2}= 147 - 492r

^{2}= 98r

^{2}= 98/2r

^{2}= 49r

^{2}= √49r = 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 cm

^{2}Height of cylinder (h) = 7 cm

Area of base ring = 115.5 cm

^{2}To find: Thickness of the cylinder

Let ‘r

_{1}’ and ‘r_{2}’ are the inner and outer radii of the hollow cylinder respectively.Then, πr

_{2}^{2}– πr_{1}^{2}= 115.5 …….(1)And,

2πr

_{1}h +2πr_{2}h+ 2(πr_{2}^{2}– πr_{1}^{2}) = 4620Or 2πh (r

_{1}+ r_{2}) + 2 x 115.5 = 4620(Using equation (1) and h = 7 cm)

or 2π7 (r

_{1}+ r_{2}) = 4389or π (r

_{1}+ r_{2}) = 313.5 ….(2)Again, from equation (1),

πr

_{2}^{2}– πr_{1}^{2}= 115.5or π(r

_{2}+ r_{1}) (r_{2}– r_{1}) = 115.5[using identity: a

^{2}– b^{2}= (a – b)(a + b)]Using result of equation (2),

313.5 (r

_{2}– r_{1}) = 115.5or r

_{2}– r_{1}= 7/19 = 0.3684Therefore, 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 338cm**

^{2}. 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 cm

^{2}CSA of External cylinder + CSA of Internal cylinder + Circumference of Internal Circle - Circumference of External Circle = 338π cm

^{2}2πrh + 2πRh + 2πR

^{2}− 2πr^{2}= 338π cm^{2}2πh(R + r) + (R

^{2}− r^{2}) = 338π cm^{2}10(R + r) + (R

^{2}− r^{2}) = 338π/2πcm^{2}10(8 + r) + (8

^{2}− r^{2}) = 16980 + 10r + 64 − r

^{2}= 16910r − r

^{2}= 169 – 80 - 6410r − r

^{2}= 25r

^{2}− 10r + 25 = 0r

^{2}− 5r − 5r + 25 = 0r(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 cm**

^{2}.**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 + πr

^{2})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 cm

^{2}Also, it is given that the rate of Rs 3.50 per 1000 cm

^{2}.= 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 m**

^{2}.**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 = 110m

^{2}(b) It is given that, the cost of painting 1m

^{2}area of the well is Rs. 40Cost of painting 110m

^{2}area of the wellCost 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 m

^{2}TSA of cylinder = 2πr(r + h)

TSA of cylinder = 2× 22/7×2.1(2.1+4.5)m

^{2}TSA of cylinder = 87.12 m

^{2}Let, xm

^{2}steel actually used in making the tank = x-1x/12=(12x-1x)/12=11x/12Area of iron present in cylinder = 11x/12m

^{2}11x/12 = Total surface area of cylinder

x = Total surface area ×12/11

x = 87.12×12/11

x = 95.04m

^{2}Hence, m

^{2}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 + πr

^{2}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 cm

^{2}Cardboard used by 35 competitors = 7920 cm

^{2}Hence, 7920 cm

^{2}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 m

^{2}CSA of the roller = 3.96m

^{2}Area of the playground = 100 × Area covered by roller in one revolution

Area of the playground = 100 × 3.96m

^{2}Area of the playground = 396 m

^{2}It is given that, Cost of levelling 1m

^{2}= 50p = Rs. 0.5Cost of levelling 396m

^{2}= Rs. 396 × 0.5 = Rs. 198Hence, 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 = 44/7m

^{2}Lateral surface area of 20 pillars = 880/7m

^{2}It is given that, Cost of cleaning one pillar is Rs 2.50/m

^{2}Cost of cleaning 20 pillars = Rs. 2.50 × 880/7m

^{2 }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)cm

^{3 }Volume of Tin Can = 300cm

^{3}(ii) Radius (r) = diameter/2

Radius (r) = 7/2cm

Radius (r) = 3.5cm

Height of plastic cylinder (H) = 10 cm

Volume of cylindrical container = πR

^{2}HVolume of cylindrical container = 22/7 × (3.5)

^{2}× 10cm^{3 }Volume of cylindrical container = 385 cm

^{3}The plastic cylinder has greater capacity

Difference in capacity = (385 – 300) cm

^{3 }Difference in capacity = 85 cm

^{3}**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 = πR

^{2}HVolume of the cylindrical pillar = (22/7 × (20)

^{2}× 1000) cm^{3}Volume of the cylindrical pillar = 8800000/7 cm

^{3}Volume of the cylindrical pillar = 8.87 m

^{3}Hence, volume of 14 pillars = 14 × 8.87m

^{3}= 17.6 m^{3}**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 cm**

^{3}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(R

^{2}– r^{2})hMass of pipe = 22/7 (14

^{2}– 12^{2})35Mass of pipe = 5720cm

^{3}It is given that, the mass of 1 cm

^{3}wood = 0.6 gmMass of 5720 cm

^{3}wood = 5720 x 0.6 = 3432gmHence, the mass of 5720 cm

^{3}wood is 3.432 kg.**Question 4: If the lateral surface of a cylinder is 94.2 cm**

^{2}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 cm

^{2}Height of the cylinder = 5 cm

Let ‘r’ be the radius.

(i) Lateral surface of the cylinder = 94.2 cm

^{2}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 = πr

^{2}hVolume of the cylinder = 3.14 × (3)

^{2}× 5 cm^{3}Volume of the cylinder = 141.3 cm

^{3}**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 m

^{3}Volume of the cylinder = πr

^{2}hπr

^{2}h = 0.0154 m^{3}3.14 x r

^{2}x 1 = 0.0154 m^{3}r

^{2}= 0.0154/3.14r = 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))m

^{2}TSA of a vessel = 0.470 m

^{2}**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 = πr

^{2}hVolume of soup = 22/7 × (3.5)

^{2}× 4 cm^{3}Volume of soup = 154cm

^{3}Volume of soup in 250 bowls = (250 × 154) cm

^{3}Volume of soup in 250 bowls = 38500 cm

^{3 }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(R

^{2}−r^{2})hVolume of the iron = 22/7 {(70)

^{2}− (66)^{2}}63Volume of the iron = 107712

Hence, the volume of the iron 107712 cm

^{3}**Question 8: A solid cylinder has a total surface area of 231 cm**

^{2}. Its curved surface area is 2/3 of the total surface area. Find the volume of the cylinder.**Solution:**

Total surface area = 231 cm

^{2}CSA = 2/3 TSA

Curved surface area = 2/3 × 231

Curved surface area = 154 cm

^{2}Curved surface area of cylinder = 2πrh + 2πr

^{2}2πrh + 2πr

^{2}= 231154 + 2πr

^{2}= 231 (2πrh = 154cm^{2})2πr

^{2}= 231- 1542 × 22/7 × r

^{2}= 77r

^{2}= (77 × 7)/(2 × 22)r

^{2}= 49/4r = 7/2

CSA = 154 cm

^{2}2πrh = 154

2 x 22/7 x 7/2 x h = 154

h = 154/22

h = 7

Volume of the cylinder = πr

^{2}hVolume of the cylinder = 22/7 x 7/2 x 7/2 x 7

Volume of the cylinder = 269.5cm

^{3}Hence, the volume of the cylinder is 269.5cm

^{3}**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 dm

^{2}is Rs. 1982πr(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 = πr

^{2}hVolume of the tank = 22/7 × 9 × 18

Volume of the tank = 509.14 dm

^{3}Hence, the volume of the tank is 509.14 dm

^{3}**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 cm**

^{2}.**Solution:**

It is given that, TSA = 616 cm

^{2}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 = 308cm

^{2}TSA = 2πrh + 2πr

^{2}616 = CSA + 2πr

^{2}616 = 308 + 2πr

^{2}616 – 308 = 2πr

^{2}2πr

^{2}= 616 – 3082πr

^{2}= 308r

^{2}= 308/2πr

^{2}= 49r = 7cm

CSA = 308cm

^{2}2πrh = 308

2 × 22/7 × 7 × h = 308

h = 7 cm

Volume of cylinder = πr

^{2}hVolume of cylinder = 22/7 × 7 × 7 × 7

Volume of cylinder = 1078

Hence, the volume of cylinder is 1078 cm

^{3}.**Question 12: The curved surface area of a cylinder is 1320 cm**

^{2}and its base had diameter 21 cm. Find the height and volume of the cylinder.**Solution:**

CSA of a cylinder = 1320 cm

^{2}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 = πr

^{2}hVolume 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 cm

^{3}.**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 cm

^{3}Radius = 2x cm

Height = 3x cm

Volume of cylinder = πr

^{2}h1617= 22/7 (2x)

^{2}× 3x1617= 22/7 × 4x

^{2}× 3x1617 = 22/7 × 12x

^{3}(1617 × 7)/(22 × 12) = x

^{3}(1617 × 7)/(22 × 12) = x

^{3}11319/264 = x

^{3}x

^{3}= 343/8x = 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 = 770cm

^{2}Hence, The Total surface area of cylinder is 770 cm

^{2}.**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 = πr

^{2}hVolume of cylinder = 22/7 × 7 × 7 × 20

Volume of cylinder = 22 × 7 × 20

Volume of cylinder = 3080

Hence, the volume of cylinder is 3080 cm

^{3}.**Question 15: The curved surface area of cylindrical pillar is 264m**

^{2}and its volume is 924m^{3}. Find the diameter and the height of the pillar.**Solution:**

CSA of cylindrical pillar = 264m

^{2}2πrh = 264

πrh = 264/2

πrh = 132

Volume of the cylinder = 924 m

^{3}πr

^{2}h= 924πrh(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))/(2πr

^{2})TSA/CSA=(h+r)/r