Table of Contents

**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. [NCERT]**

Solution: Curved surface area of a cylinder = 4.4 m^{2}

Radius (r) = 0.7 m

**Also Check: NCERT Solutions for Class 9**

**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. [NCERT]**

Solution: Diameter of the pipe = 5 cm

**Also Check: NCERT Solutions for Class 9 Maths**

**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 12.50 per m ^{2}. [NCERT]**

Solution: Diameter of cylindrical pillar = 50 cm

**Also Check: Extra Questions for Class 9 Maths with Solutions Chapter Wise**

**Question 4. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same? [NCERT]**

Solution: Height of cylinder (h) = 1 m = 100 cm

Diameter of box = 140 cm

**Also Check: NCERT Solutions Class 9 Maths Chapter 1**

**Question 5. The total surface area of a hollow cylinder which is open from both sides is 4620 sq. cm, area of base ring is 115.5 sq. cm and height 7 cm. Find the thickness of the cylinder.**

Solution: Total surface area of a hollow cylinder open from both sides = 4620 cm^{2}

Area of base of ring = 115.5 cm^{2}

Height (h) = 7 cm

Let outer radius (R) = R

and inner radius = r

**Also Check: Important Questions for CBSE Class 9 Mathematics Circles**

**Question 6. Find the ratio between the total surface area of a cylinder to its curved surface area, given that its height and radius are 7.5 cm and 3.5 cm.**

Solution: Radius of the cylinder (r) = 3.5 cm

and height (h) = 7.5 cm

Total surface area = 2πr (h + r)

and curved surface area = 2πrh

**Question 7. 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 ₹3.50 per 1000 cm ^{2}.**

Solution: Radius of the base of a cylindrical vessel (r) = 70 cm

and height (h) = 1.4 m = 140 cm

Total surface area (excluding upper lid) on both sides = 2πrh x 2 + πr

^{2}x 2

**Also Check: NCERT Solutions Class 9 Maths Chapter 2**

**Question 8. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find:**

**(i) inner curved surface area.**

**(ii) the cost of plastering this curved surface at the rate of ₹40 per m ^{2}. [NCERT]**

Solution: Inner diameter of a well = 3.5 m

**Question 9. The students of a Vidyalaya were asked to participate s a competition for making and decorating pen holders in the shape of a cylinder with a base, using cardboard. Each pen holder 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? [NCERT]**

Solution: Radius of cylinderical pen holder (r) = 3 cm

Height (h) = 10.5 cm

∴ Surface area of the pen holder

**Question 10. The diameter of roller 1.5 m long is 84 cm. If it takes 100 revolutions to level a play¬ground, find the cost of levelling this ground at the rate of 50 paise per square metre.**

Solution: Diameter of a roller = 1.5 m

∴ Radius = \(\frac { 1.5 }{ 2 }\) = 0.75 m = 75 cm

and length (h) = 84 cm

**Question 11. Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and height is 4 m. What will be the cost of cleaning them at the rate of ₹2.50 per square metre? [NCERT]**

Solution: Number of pillars = 20

Diameter of one pillar = 0.50 m

**Question 12. A solid cylinder has 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 of solid cylinder = 462 cm

^{2}

Curved surface area = \(\frac { 1 }{ 3 }\) of total surface area

**Question 13. The total surface area of a hollow metal cylinder, open at both ends of external radius 8 cm and height 10 cm is 338 π cm ^{2}. Taking r to be inner radius, obtain an equation in r and use it to obtain the thickness of the metal in the cylinder.**

Solution: Total surface area of a hollow metal cylinder = 338π cm

^{2}

Let R be the outer radius, r be inner radius and h be the height of the cylinder of the cylinder

∴ 2πRh + 2πrh + 2πR

^{2}– 2πr

^{2}= 338π

R = 8 cm, h = 10 cm

⇒ 2πh (R + r) + 2π(R

^{2}– r

^{2}) = 338π

⇒ Dividing by 2π , we get

⇒ h(R + r) + (R

^{2}– r

^{2}) = 169

⇒ 10(8 + r) + (8 + r) (8 – r) = 169

⇒ 80 + 10r + 64 – r

^{2}= 169

⇒ 10r – r

^{2}+ 144 – 169 = 0

⇒ r

^{2}– 10r + 25 = 0

⇒ (r-5)

^{2}= 0

⇒ r = 5

∴ Thickness of the metal = R – r = 8 – 5 = 3 cm

**Question 14. Find the lateral curved surface area of a cylinderical petrol storage tank that is 4.2m in diameter and 4.5 m high. How much steel was actually used, if \(\frac { 1 }{ 12 }\) of steel actually used was wasted in making the closed tank? [NCERT]**

Solution:

Diameter of a cylinderical tank = 4.2 m

### RD Sharma Class 9 Solution Chapter 19 Surface Areas and Volume of a Circular Cylinder Ex 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? [NCERT]**

**Solution:**

In first case, in a rectangular container of soft drink, the length of base = 5 cm

and Width = 4 cm

Height = 15 cm

∴ Volume of soft drink = lbh = 5 x 4 x 15 = 300 cm^{3}

and in second case, in a cylindrical container, diameter of base = 7 cm

∴ The soft drink in second container is greater and how much greater = 385 cm – 380 cm^{2} = 85 cm^{2}

**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? [NCERT]**

Solution:

Radius of each pillar (r) = 20 cm

**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. [NCERT]**

Solution:

Inner diameter of a cylindrical wooden pipe = 24 cm

**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 [Use π = 3.14] [NCERT]**

Solution:

Lateral surface area of a cylinder = 94.2 cm2

**Question 5. The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it? [NCERT]**

Solution:

The capacity of a closed cylindrical vessel = 15.4 l

**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? [NCERT]**

Solution:

Diameter of the cylindrical bowl = 7 cm

∴ Radius (r) = \(\frac { 7 }{ 2 }\)cm

Level of soup in it = 4 cm

∴ Volume of soup in one bowl for one patient

**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:

Width of hollow cylinder (A) = 63 cm

Girth = 440 cm

**Question 8. The cost of painting the total outside surface of a closed cylindrical oil tank at 50 paise per square decimetre is ₹ 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:

Rate of painting = 50 paise per dm^{2}

Total cost = ₹198

**Question 9. 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:

Ratio in radii of two cylinders = 2:3

and ratio in their heights = 5:3

Let radius of the first cylinder (r_{1}) = 2x

and radius of second cylinder (r_{2}) = 3x

and height of first cylinders (h_{1}) = 5y

and height of second cylinder (h_{2}) = 3y

(i) Now volume of the first cylinder = πr^{2}h = π(2x)^{2} x 5y = 20πx^{2}2y

and volume of tlie second cylinder = π(3x)^{2} x 3y = π x 9×2 x 3y = 27πx^{2}y

Now ratio in their volume

= 20πx^{2}y : 21πx^{2}y = 20 : 27

(ii) Curved surface area of first cylinder = 2πrh = 2π x 2x x 5y =20πxy

and curved surface area of second cylinder = 2π x 3x x = 1 8πxy

∴ Ratio in their curved surface area

= 20πxy : 18πxy = 10 : 9

**Question 10. 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:

Ratio in curved surface area and total surface area of a cylinder =1:2

Total surface area = 616 cm

^{2}

**Question 11. The curved surface area of a cylinder is 1320 cm ^{2} and its base had diameter 21 cm. Find the height and the volume of the cylinder. [Use π = 22/7]**

Solution:

Curved surface area of a cylinder = 1320 cm

^{2}

Diameter of the base = 21 cm

**Question 12. 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 1617 cm ^{3}.**

Solution:

Ratio between radius and height of a cylinder = 2:3

Volume =1617 cm

^{3}

Let radius (r) = 2x

Then height (h) = 3x

∴ Volume = πr

^{2}h

**Question 13. A rectangular sheet of paper, 44 cm x 20 cm, is rolled along its length of form a cylinder. Find the volume of the cylinder so formed.**

Solution:

Length of sheet = 44 cm

Breadth = 20 cm

By rolling along length, the height of cylinder (h) = 20cm

and circumference of the base = 44cm

**Question 14. The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the diameter and the height of the pillar.**

Solution:

Curved surface area of a pillar = 264 m^{2}

and volume = 924 m^{3}

Let r be the radius and It be height, then 2πrh = 264

**Question 15. Two circular cylinders of equal volumes have their heights in the ratio 1 : 2. Find the ratio of their radii.**

Solution:

Volumes of two cylinders are equal Ratio in their height h_{1} :h_{2} = 1: 2

**Question 16. The height of a right circular cylinder is 10.5 m. Three times the sum of the areas of its two circular faces is twice the” area of the curved surface. Find the volume of the cylinder.**

Solution:

Height of a right circular cylinder = 10.5 m

3 x sum of areas of two circular faces

= 2 x area of curved surface

Let r be that radius,

**Question 17. How many cubic metres of earth must be dugout to sink a well 21 m deep and 6 m diameter? Find the cost of plastering the inner surface of the well at ₹9.50 per m ^{2}.**

Solution:

Diameter of a well = 6 m

∴ Radius (r) = \(\frac { 6 }{ 2 }\) = 3 m

Depth (h) = 21 m

∴ Volume of earth dugout = πr

^{2}h

**Question 18. The trunk of a tree is cylindrical and its circumference is 176 cm. If the length of the trunk is 3 m. Find the volume of the timber that can be obtained from the trunk.**

Solution:

Circumference of a cylindrical trunk of a tree = 176 cm

**Question 19. A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm x 22 cm x 14 cm. Find the rise in the level of the water when the solid is completely submerged.**

Solution:

Diameter of cylindrical container = 56 cm

∴ Radius (r) = \(\frac { 56 }{ 2 }\) = 28 cm

Dimensions of a rectangular solid are = 32 cm x 22 cm x 14 cm

∴ Volume of solid = lbh

= 32 x 22 x 14 = 9856 cm^{3}

∴ Volume of water in the container = 9856 cm^{3}

Let h be the level of water, then

πr2h = 9856

**Question 20. A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal.**

Solution:

Length of metallic tube = 25 cm

Inner diameter = 10.4 cm

∴ Radius (r) = \(\frac { 10.4 }{ 2 }\) = 5.2 cm

Thickness of metal = 8 mm

∴ Outer radius (R) = 5.2 + 0.8 = 6.0 cm

Volume of metal used = π(R^{2} – r^{2}) x h

**Question 21. From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second. Find the volume in litres of water delivered by the pipe in one hour.**

Solution:

Inner radius of a tap = 0.75 cm

Speed of flow of water in it = 7 m/s

Time = 1 hour

∴ Length of flow of water (h)

= 7 x 60 x 60 m = 25200 m

∴ Volume of water = πr^{2}h

**Question 22. A rectangular sheet of paper 30 cm x 18 cm can be transformed into the curved surface of a right circular cylinder in two ways i.e., either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed.**

Solution:

Size of rectangular sheet = 30 cm x 18 cm

∴ Length of sheet = 30 cm

and breadth = 18 cm

By folding length wise,

Height = 18 cm

and circumference = 30 cm

Question 23.

How many litres of water flow out of a pipe having an area of cross-section of 5 cm^{2} in one minute, if the speed of water in the pipe is 30 cm/sec?

Solution:

Area of the cross-section of the pipe = 5 cm^{2}

Speed of water flow = 30 cm/sec

Period = 1 minute

∴ Flow of water in 1 minute = 30 x 60 cm = 1800 cm

Area of mouth of pipe = 5 cm^{2}

∴ Volume = 1800 x 5 = 9000 cm^{3}

Volume of water in litres = 9000 ml

**Question 24. Find the cost of sinking a tubewell 280 m deep, having diameter 3 m at the rate of ₹3.60 per cubic metre. Find also the cost of cementing its inner curved surface at ₹2.50 per square metre.**

Solution:

Depth of well (h) = 280 m

Diameter = 3 m

**Question 25. Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm.**

Solution:

Weights of copper wire = 13.2 kg

Diamter = 4 mm

**Question 26. A solid cylinder has a total surface area of 231 cm ^{2}. Its curved surface area is \(\frac { 2 }{ 3 }\) of the total surface area. Find the volume of the cylinder.**

Solution:

Surface area of solid cylinder = 231 cm

^{2}

and curved surface area = \(\frac { 2 }{ 3 }\) of 231 cm

^{2}

**Question 27. A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment.**

Solution:

Diameter of a well = 14 m

∴ Radius (r) = y = 7 m

Depth (h) = 8 m

∴ Volume of the earth dugout = πr^{2}h

**Question 28. The difference between inside and outside surfaces of a cylindrical tube 14 cm long is 88 sq. cm. If the volume of the tube is 176 cubic cm, find the inner and outer radii of the tube.**

Solution:

Length of cylindrical tube = 14 cm

Difference betveen the outer surface and inner surface = 88 cm^{2}

and volume of the tube = 176 cm^{3}

Let R and r be the outer and inner radius of the tube

**Question 29. Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 metres per second into a cylindrical tank. The radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes?**

Solution:

Internal diameter of the pipe = 2 cm

∴ Radius (r) = \(\frac { 2 }{ 2 }\) = 1 cm

Speed of water flow = 6m per second Water in 30 minutes (h) = 6 x 60 x 30 m = 10800 m

Volume of water = πr^{2}h

Question 30.

A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 metre per second. In how much time the tank will be filled?

Solution:

Diameter of cylindrical tank = 1.4 m

∴ Radius (r) = \(\frac { 1.4 }{ 2 }\) = 0.7 m

and height (h) = 2.1 m

∴ Volume of water in the tank = πr^{2}h

Question 31.

The sum of the radius of the base and height of a solid cylinder is 37 m. If the total surface area of the solid cylinder is 1628 m^{2}. Find the volume of the cylinder.

Solution:

Sum of radius and height of a cylinder = 37 m

Let r be the radius and h be the height, then r + h = 37m …(i)

Total surface area of a solid cylinder = 1628m^{3}

Question 32.

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 the embankment.

Solution:

Diameter of the well = 10 m 10

∴ Radius (r) = \(\frac { 10 }{ 2 }\) = 5 m

Depth (h) = 8.4 m

∴ Volume of earth dugout = πr^{2}h

### RD Sharma Class 9 Book Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS

Question 1.

Write the number of surface of a right circular cylinder.

Solution:

Three, two circular and one curved.

Question 2.

Write the ratio of total surface area to the curved surface area of a cylinder of radius r and height h.

Solution:

∵ Radius = r

and height = h

∴ Curved surface area = 2πrh

and total surface area = 2πr(h + r)

∴ Ratio = 2πr(h + r) : 2πrh

= h + r : h

Question 3.

The ratio between the radius of the base and height of a cylinder is 2 : 3. If its volume is 1617 cm3, find the total surface area of the cylinder.

Solution:

Ratio in radius and height of the cylinder = 2 : 3

Let radius (r) = 2x

Then height (h) = 3x

∴ Volume = πr^{2}h

Question 4.

If the radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3, then find the ratio of their volumes.

Solution:

Ratio of radii of two cylinder = 2:3

Let radius of first cylinder (r* _{1}*) = 2x

and second cylinder (r

_{2}) = 3x

and ratio in their heights = 5:3

Let height of first cylinder (h

_{1}) = 5y

and height of second (h

_{2}) = 3y

∴ Volume of the first cylinder =πr

^{2}h

= π x (2x)

^{2}x 5y = 20πx

^{2}y

and volume of second cylinder = π(3x)2 x 3y = 27πx

^{2}y

Now ratio between then,

= 20πx

^{2}y: 21πx

^{2}y

= 20 : 27

### RD Sharma Solutions Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS

Mark correct alternative in each of the following:

Question 1.

In a cylinder, if radius is doubled and height is halved, curved surface area will be

(a) halved

(b) doubled

(c) same

(d) four times

Solution:

Let radius of the first cylinder (r_{1}) = r

and height (h_{1}) = h

Surface area = 2πrh

If radius is doubled and height is halved

∴ Their surface area remain same (c)

Question 2.

Two cylindrical jars have their diameters in the ratio 3:1, but height 1:3. Then the ratio of their volumes is

(a) 1 : 4

(b) 1 : 3

(c) 3 : 1

(d) 2 : 5

Solution:

Sol. Ratio in the diameters of two cylinder = 3:1

and ratio in their heights = 1:3

Let radius of the first cylinder (r_{1}) = 3x

and radius of second (r_{2}) = x

and height of the first (h_{1}) = y

and height of the second (h_{2}) = 3y

Now volume of the first cylinder = πr^{2}h

= π(3x)^{2} x y = 9πx^{2}y

and of second cylinder = π(x^{2}) (3y)

∴ Ratio between then = 9πx^{2}y : 3πx^{2}y

= 3 : 1 (c)

Question 3.

The number of surfaces in right cylinder is

(a) 1

(b) 2

(c) 3

(d) 4

Solution:

The number of surfaces of a right cylinder is three. (c)

Question 4.

Vertical cross-section of a right circular cylinder is always a

(a) square

(b) rectangle

(c) rhombus

(d) trapezium

Solution:

The vertical cross-section of a right circular cylinder is always a rectangle. (b)

Question 5.

If r is the radius and h is height of the cylinder the volume will be

Solution:

Volume of a cylinder = πr^{2}h (b)

Question 6.

The number of surfaces of a hollow cylindrical object is

(a) 1

(b) 2

(c) 3

(d) 4

Solution:

The number of surfaces of a hollow cylindrical object is 4. (d)

Question 7.

If the radius of a cylinder is doubled and the height remains same, the volume will be

(a) doubled

(b) halved

(c) same

(d) four times

Solution:

If r be the radius and h be the height, then volume = πr2h

If radius is doubled and height remain same,

the volume will be

= π(2r)^{2}h = π x 4r^{2}h

= 4πr^{2}h = 4 x Volume

The volume is four times (d)

Question 8.

If the height of a cylinder is doubled and radius remains the same, then volume will be

(a) doubled

(b) halved

(c) same

(d) four times

Solution:

If r be the radius and h be the height, then volume of a cylinder = πr^{2}h

If height is doubled and radius remain same, then volume = πr^{2}(2h) = 2πr^{2}h

∴ Its doubled (a)

Question 9.

In a cylinder, if radius is halved and height is doubled, the volume will be

(a) same

(b) doubled

(c) halved

(d) four times

Solution:

Let r be radius and h be height, then Volume = πr^{2}h

If radius is halved and height is doubled

Question 10.

If the diameter of the base of a closed right circular cylinder be equal to its height h, then its whole surface area is

Solution:

Let diameter of the base of a cylinder (r) = h

Then its height (h) = h

Question 11.

A right circular cylindrical tunnel of diameter 2 m and length 40 m is to be constructed from a sheet of iron. The area of the iron sheet required in m^{2}, is

(a) 40π

(b) 80π

(c) 160π

(d) 200π

Solution:

Diameter of a cylindrical tunnel = 2 m

∴ Radius (r) = \(\frac { 2 }{ 2 }\) = 1m

and length (h) = 40 m

Curved surfae area = 2πrh = 2 x π x 1 x 40 = 80π (b)

Question 12.

Two circular cylinders of equal volume have their heights in the ratio 1 : 2. Ratio of their radii is

Solution:

Let r_{1} and h_{1} be the radius and height of the

first cylinder, then

Volume = πr_{1}^{2}h_{1}

Similarly r_{1} and h_{2} are the radius and height of the second cylinder

∴ Volume = πr^{2}h_{2}

But their volumes are equal,

Question 13.

The radius of a wire is decreased to one- third. If volume remains the same, the length will become

(a) 3 times

(b) 6 times

(c) 9 times

(d) 27 times

Solution:

In the first case, r and h1, be the radius and height of the cylindrical wire

∴ Volume = πr^{2}h_{1} …(i)

In second case, radius is decreased to one third

∴ In second case height is 9 times (c)

Question 14.

If the height of a cylinder is doubled, by what number must the radius of the base be multiplied so that the resulting cylinder has the same volume as the original cylinder?

Solution:

Let r be the radius and h be the height then volume = πr^{2}h

If height is doubled and volume is same and let x be radius then πr^{2}h = π(x)^{2} x 2h

Question 15.

The volume of a cylinder of radius r is 1/4 of the volume of a rectangular box with a square base of side length x. If the cylinder and the box have equal heights, what is r in terms of x?

Solution:

Let r be the radius and h be the height, then volume = πr^{2}h

This volume is \(\frac { 1 }{ 4 }\) of the volume of a rectangular box

∴ Volume of box = 4πr^{2}h

Let side of base of box = x and height h,

then volume = x^{2}h

∴ 4πr^{2}h = x^{2}h

Question 16.

The height ft of a cylinder equals the circumference of the cylinder. In terms of ft, what is the volume of the cylinder?

Solution:

In a cylinder,

h = circumference of the cylinder

Question 17.

A cylinder with radius r and height ft is closed on the top and bottom. Which of the following expressions represents the total surface area of this cylinder?

(a) 2πr(r + h)

(b) πr(r + 2h)

(c) πr(2r + h)

(d) 2πr^{2} + h

Solution:

r is the radius of the base and ft is the height of a closed cylinder

Then total surface area = 2πr(r + h ) (a)

Question 18.

The height of sand in a cylindrical-shaped can drops 3 inches when 1 cubic foot of sand is poured out. What is the diameter, in inches, of the cylinder?

Solution:

Let h be the height and d be the diameter of a cylinder, then

Question 19.

Two steel sheets each of length a_{1} and breadth a_{2} are used to prepare the surfaces of two right circular cylinders – one having volume v_{1 }and height a_{2} and other having volume v_{2} and height a_{1}. Then,

Solution:

Length of each sheet = a_{1}

and breadth = a_{2}

Volume of cylinder = πr^{2}h

In first case,

v_{1} is volume and a_{2} is the height

Question 20.

The altitude of a circualr cylinder is increased six times and the base area is decreased to one-ninth of its value. The factor by which the lateral surface of the cylinder increases, is

Solution:

In first case,

Let r be the radius and h be the height of the cylinder. Then,

∴ Lateral surface area = 2πrh

In second case,

### Surface Area and Volume of A Right Circular Cylinder Ex 19.1