Rd Sharma 2021 Solutions for Class 9 Maths Chapter 19 Surface Area And Volume Of A Right Circular Cylinder are provided here with simple step-by-step explanations. These solutions for Surface Area And Volume Of A Right Circular Cylinder are extremely popular among Class 9 students for Maths Surface Area And Volume Of A Right Circular Cylinder Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2021 Book of Class 9 Maths Chapter 19 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2021 Solutions. All Rd Sharma 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 19.20:

#### 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 diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

#### Answer:

Given data is as follows:

(i) For tin can with rectangular base:

Length = 5 cm

Width = 4 cm

Height = 10 cm

(ii) For plastic cylinder with circular base

Diameter = 7 cm

Height = 10 cm

We have to find out which container has greater capacity and also the difference in their capacities.

(i) Volume of tin can =

=

=200

(ii) Volume of cylinder =

Diameter is given as 7cm. Therefore, *r* =

Volume of cylinder =

= 385 cm^{3}

From the above it can be concluded that plastic cylinder has greater capacity.

Difference in their capacities = 385 – 200 = 185 cm^{3}

#### Page No 19.20:

#### 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?

#### Answer:

Given data is as follows:

Number of pillars = 14

We have to find the total amount of concrete present in all 14 pillars.

Radius of the pillar is given in centimeters, so let us convert it to meters.

Let us first find the amount of concrete present in one pillar, which is nothing but the volume of the pillar.

Therefore, total amount of concrete mixture in 14 pillars is 17.6 m^{3}

#### Page No 19.21:

#### 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.

#### Answer:

Given data is as follows:

Inner diameter = 24cm

Outer diameter = 28cm

*h* = 35cm

Mass of 1 cm^{3} of wood = 0.6gm

We have to find the mass of the pipe.

In this problem the inner and outer diameter of the pipe is given. Let us first find out the radius.

Inner radius (*r*) = 12cm

Outer radius (R) = 14cm

Volume of the hollow pipe =

$=\frac{22}{7}\times \left({14}^{2}-{12}^{2}\right)\times 35\phantom{\rule{0ex}{0ex}}=22\times 5\times 2\times 26$

=5720 cm^{3}

It is given that,

1 cm^{3}^{ }of wood weighs 0.6gm

Therefore, 5720 of wood will weigh = 3432gm

= 3.432kg

Therefore, weight of the wooden pipe = 3.432kg

#### Page No 19.21:

#### 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 $\mathrm{\pi}$ = 3.14]

#### Answer:

Given data is as follows:

Lateral Surface Area = 94.2 cm^{2}

*h* = 5cm

We have to find:

(i) Radius of the base

(ii) Volume of the cylinder

(i) We know that,

Lateral Surface Area =

That is,

2*πrh* = 94.2

=94.2

=94.2

(ii) Volume of a cylinder =

=

#### Page No 19.21:

#### 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?

#### Answer:

Given data is as follows:

Volume of the cylinder = 15.4 litres

*h* = 1m

We have to find the area of the sheet required to make this cylinder.

We know that 1 liter = 1000 cm^{3}

Therefore, 15.4 liters = 15400 cm^{3}

Also, *h* = 1m

=100cm

We know that,

Volume =

Therefore,

=15400

Now, using this radius we have to find the Total Surface Area.

Total Surface Area = +

#### Page No 19.21:

#### 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 server 250 patients?

#### Answer:

Given data is as follows:

Diameter = 7cm

*h* = 4cm

Number of patients = 250

We have to find the total volume of soup required to serve all 250 patients.

Given is the diameter, which is equal to 7cm. Therefore,

Volume of soup given to each patient =

=

= 154 cm^{3}

Volume of soup for all 250 patients = 154 × 250

=38500 cm^{3}

We know that, 1000 cm^{3}^{ }= 1 litre.

Therefore,

Volume of soup for all 250 patients = 38.5 litres

#### Page No 19.21:

#### 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.

#### Answer:

Given data is as follows:

*h* = 63 cm

Girth is nothing but the outer circumference of the roller, which is 440 cm.

Thickness of the roller = 4 cm

We have to find the volume of the roller.

We have been given the outer circumference of the roller. Let R be the external radius.

We have,

= 440

Also, thickness of the cylinder is given which is 4 cm. So we can easily find out the inner radius ‘*r*’.

Now, since we know both inner and outer radii, we can easily find out the volume of the hollow cylinder.

Volume =

#### Page No 19.21:

#### Question 8:

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.

#### Answer:

Data given is as follows:

Total cost of painting=Rs.198

Painting rate= Rs.0.50 per square decimeter

We have to find the volume of the cylinder.

We know that,

Total Surface Area of the cylinder = $2\mathrm{\pi rh}+2{\mathrm{\pi r}}^{2}$

Also, it is given that,

Total cost of painting = 198

That is,

= 198

$\left(2\pi rh+2\pi {r}^{2}\right)\times \mathrm{painting}\mathrm{rate}$=198

$\left(2\pi rh+2\pi {r}^{2}\right)\times 0.50$=198

$\left(2\pi rh+2\pi {r}^{2}\right)$= 396

In the above equation, let us replace with 6.

=396

=396

=396

=3 decimeters

==18 decimeters

Volume of the cylinder =

#### Page No 19.21:

#### 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.

#### Answer:

Data given is as follows:

Ratio of radii of two cylinders = 2:3

Ratio of heights of two cylinders = 5:3

We have to find out the following:

(i) Ratio of the volumes of the two cylinders

(ii) Ratio of the Curved Surface Area of the two cylinders

Let and be the radii of the two cylinders respectively.

Let and be the heights of the two cylinders respectively.

Therefore we have,

(i) Since we have to find the ratio of the volumes of the two cylinders, we have

(ii) Since we have to find the ratio of the curved surface areas of the two cylinders, we have,

=

#### Page No 19.21:

#### 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}.

#### Answer:

Data given is as follows:

We have to find the volume of the cylinder.

From the given data we have,

Also,

We have found out the Curved Surface Area of the cylinder which is.

Curved Surface Area =

=308

Now, let us replace with in the above equation since in the previous step we have found that .

=308

Since , is also equal to 7

Volume =

=

#### Page No 19.21:

#### 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.

#### Answer:

Given data is as follows:

Curved Surface Area = 1320 cm^{2}

Diameter=21cm

We have to find the height and volume of the cylinder.

First of all, we have been given the diameter so let us find out the radius.

=cm

We know that,

Curved Surface Area =

Therefore,

=1320

Now that we know both and, we can easily find out the volume.

Volume =

=

#### Page No 19.21:

#### 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, it its volume is 1617 cm^{3}.

#### Answer:

Data given is as follows:

Volume = 1617 cm^{3}

We have to find the Total surface area

From the given data, we have

Therefore,

Also, we know that

Volume = =1617

=1617

=1617

=7 cm

= cm

Therefore, total surface area is

#### Page No 19.21:

#### Question 13:

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

#### Answer:

Given data is as follows:

Dimensions of the rectangular sheet =

We have to find the volume of the cylinder that is made out of this rectangular strip.

Since the rectangular strip is rolled along its length, we have,

Circumference of the cylinder = length of the rectangular strip

We know that,

Circumference = 2*πr*

Therefore,

Also, the width of the rectangular strip will be the height of the cylinder. Therefore,

Now that we know the values of *r* and *h*, we can easily find the volume of the cylinder.

Volume =

=

#### Page No 19.21:

#### Question 14:

The curved surface area of a cylindrical pillar is 264 m^{2} and its volume is 924 m^{3}. Find the diameter and the height of the pillar.

#### Answer:

Given data is as follows:

Curved Surface Area = 267 m^{2}

Volume = 924 m^{3}

We have to find the height and diameter of this cylinder.

We know that,

Volume =

= 924

=924 ……(1)

Also, it is given that

Curved Surface Area = 267

That is,

=264

……(2)

Now let us replace the value of in equation (1). We get,

=924

Therefore, diameter =

= 14 cm

Substitute the value of in equation (2). We get,

Therefore, the answer to this question is,

Diameter of the cylinder = 14 m

Height of the cylinder = 6 m

#### Page No 19.21:

#### Question 15:

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

#### Answer:

Given data is as follows:

We have to find the ratio of their radii

Since the volumes of the two cylinders are equal,

But it is given that,

Therefore,

Therefore, the ratio of the radii of the two cylinders is

#### Page No 19.21:

#### 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.

#### Answer:

Given data is as follows:

We have to find the volume of the cylinder.

From the given data, we have

But we know from the given data, that

Therefore,

Since we know and , we can easily find the volume of the cylinder.

Volume =

=

Volume = 1617 m^{3}

Therefore, the volume of the cylinder is 1617 m^{3}.

#### Page No 19.21:

#### 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 Rs 9.50 per m^{2}.

#### Answer:

Given data is as follows:

Diameter = 6 m

Plastering rate = Rs.9.50/m^{2}

We have to find the volume and the cost of plastering the inner surface of this well.

Given is the diameter, which is 6 m. Therefore,

We know that,

Volume =

=

Volume = 594 m^{3}

We know that,

Curved Surface Area =

=

Curved Surface Area =396 m^{2}

Therefore, the volume of this well is 594 m^{3}^{ }and cost of plastering its inner surface is Rs.3762.

#### Page No 19.21:

#### Question 18:

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

#### Answer:

Given data is as follows:

Circumference = 176 cm

We have to find the volume of the trunk.

We know that,

Circumference =

Therefore,

We know,

Volume =

=

Volume =0.7392 m^{3}

Therefore, the volume of timber that can be obtained from this trunk is 0.7392 m^{3}

#### Page No 19.21:

#### 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 × 22 cm × 14 cm. Find the rise in the level of the water when the solid is completely submerged.

#### Answer:

Given data is as follows:

Diameter of cylinder = 56 cm

Dimensions of rectangular block =

We have to find the raise in the level of water in the cylinder.

First let us find the radius of the cylinder. Diameter is given as 56 cm. Therefore,

= 28 cm

We know that the raise in the volume of water displaced in the cylinder will be equal to the volume of the rectangular block.

Let the raise in the level of water be. Then we have,

Volume of cylinder of height and radius 28 cm = Volume of the rectangular block

Therefore the raise in the level of water when the rectangular block is immersed in the cylinder is 4 cm.

#### Page No 19.21:

#### 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.

#### Answer:

Given data is as follows:

Internal diameter = 10.4 cm

Thickness of the metal = 8 mm

Length of the pipe = 25 cm

We have to find the volume of the metal used in the pipe.

We know that,

Volume of the hollow pipe =

Given is the internal diameter which is equal to 10.4cm. Therefore,

=

Also, thickness is given as 8 mm. Let us convert it to centimeters.

Thickness = 0.8 cm

Now that we know the internal radius and the thickness of the pipe, we can easily find external radius ‘’.

= 5.2 + 0.8

=6 cm

Therefore,

Volume of metal in the pipe =

=704

Therefore, the volume of metal present in the hollow pipe is 704

#### Page No 19.21:

#### 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.

#### Answer:

Given data is as follows:

= 0.75 cm

Water flow rate = 7 m/sec

Time = 1 hour

We have to find the volume of water the flows through the pipe for 1 hour.

Let us first convert water flow rate from m/sec to cm/sec, since radius of the pipe is in centimeters. We have,

Water flow rate = 7 m/sec

= 700 cm/sec

Volume of water delivered by the pipe is equal to the volume of a cylinder with =7 m and = 0.75 cm. Therefore,

Volume of water delivered in 1 second =

We have to find the volume of water delivered in 1 hour which is nothing but 3600 seconds. Therefore, we have

Volume of water delivered in 3600 seconds =

= 4455000

We know that 1000 = 1 liter

Therefore,

Volume of water delivered in 1 hour = 4455 liters

Therefore, volume of water delivered by the pipe in 1 hour is equal to 4455 liters.

#### Page No 19.21:

#### Question 22:

A rectangular sheet of paper 30 cm × 18 cm can be transformed into the curved surfaced 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.

#### Answer:

Given data is as follows:

Dimensions of the rectangular sheet of paper =

We have to find the ratio of the volumes of the cylinders formed by rolling the sheet along its length and along its breadth.

Let be the volume of the cylinder which is formed by rolling the sheet along its length.

When the sheet is rolled along its length, the length of the sheet forms the perimeter of the base of the cylinder. Therefore, we have,

The width of the sheet will be equal to the height of the cylinder. Therefore,

=18 cm

Therefore,

Let be the volume of the cylinder formed by rolling the sheet along its width.

When the sheet is rolled along its width, the width of the sheet forms the perimeter of the base of the cylinder. Therefore, we have,

=18

The length of the sheet will be equal to the height of the cylinder. Therefore,

= 30 cm

Now,

=

=

=

Now that we have the volumes of the two cylinders, we have,

=

=

Therefore, the ratio of the volumes of the two cylinders is 5:3

#### Page No 19.22:

#### Question 23:

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

#### Answer:

Given data is as follows:

Area of cross-section of the pipe = 5

Speed of water = 30 cm/sec

We have to find the volume of water that flows through the pipe in 1 minute.

Volume of water that flows through the pipe in one second =

Here, is nothing but the cross section of the pipe and is 30 cm.

Therefore, we have,

Volume of water that flows through the pipe in one second =

= 150

Volume of water that flows through the pipe in one minute =

= 9000

We know that 1000 = 1 liter. Therefore,

Volume of water that flows through the pipe in one minute = 9 liters

Hence, the volume of water that flows through the given pipe in 1 minute is 9 liters.

#### Page No 19.22:

#### Question 24:

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

#### Answer:

Given data is as follows:

Height of the tube well = 280 m

Diameter = 3 m

Rate of sinking the tube well = Rs.3.60/

Rate of cementing = Rs.2.50/

Given is the diameter of the tube well which is 3 meters. Therefore,

m

Volume of the tube well =

=

= 1980

Cost of sinking the tube well =

=

= Rs. 7128

Curved surface area =

=

=2640

Cost of cementing =

=

= Rs.6600

Therefore, the total cost of sinking the tube well is Rs.7128 and the total cost of cementing its inner surface is Rs.6600.

#### Page No 19.22:

#### 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.

#### Answer:

Given data is as follows:

Weight of copper wire = 13.2 kg

Diameter = 4 mm

Density = 8.4 gm/

We have to find the length of the copper wire.

Given is the diameter of the wire which is 4 mm. Therefore,

= 2 mm

Let us convert from millimeter to centimeter, since density is in terms of gm/. Therefore,

= cm

Also, weight of the copper wire is given in kilograms. Let us convert into grams since density is in terms of gm/. Therefore, we have,

Weight of copper wire = 13.2 1000 gm

= 13200 gm

We know that,

Volume Density = Weight

Therefore,

Hence, the length of the copper wire is 125 meters.

#### Page No 19.22:

#### 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.

#### Answer:

Given data is as follows:

Total Surface Area = 231 cm^{2}

Curved Surface Area =

We have to find the volume of the cylinder.

We have,

Total Surface Area = 231 cm^{2}

+ =231

Where, is nothing but the Curved Surface Area.

Curved Surface Area =

Curved Surface Area= =154

Let us replace in the above equation with the value of Curved Surface Area we have just obtained.

154+=231

=77

=77

=

Now, let us find the value of by using the Curved Surface Area.

Curved Surface Area=154 cm^{2}

=154

Since we know that,

=154

*h* = 7

Now that we know the value of both and , we can easily find the volume of the cylinder.

Volume of the cylinder =

=

#### Page No 19.22:

#### 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.

#### Answer:

Given data is as follows:

We have to find the height of the embankment.

Let the height of the embankment be H meter.

From the given data we have,

Volume of earth in embankment = Volume of earth dug out

Volume of embankment =

Volume of earth dug out =

Therefore, we have

Here,

But,

Therefore,

Substituting the values in the above equation, we have

Therefore the height of the embankment is equal to 53.3 cm

#### Page No 19.22:

#### 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.

#### Answer:

Given data is as follows:

Outer Curved Surface Area − Inner Curved Surface Area =

Volume = 176

We have to find the inner and outer radii of the tube.

As given in the problem we have,

Also, from the given data we have,

We have already found out that

Therefore,

Now let us solve these two equations, by adding them

We get

Substituting for in, we get

= 1.5

Thus, inner radius of the pipe is equal to 1.5 cm and outer radius of the pipe is equal to 2.5 cm.

#### Page No 19.22:

#### 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?

#### Answer:

Given data is as follows:

Internal diameter of the pipe = 2 cm

Water flow rate through the pipe = 6 m/sec

Radius of the tank = 60 cm

Time = 30 minutes

The volume of water that flows for 1 sec through the pipe at the rate of 6 m/sec is nothing but the volume of the cylinder with.

Also, given is the diameter which is 2 cm. Therefore,

Since the speed with which water flows through the pipe is in meters/second, let us convert the radius of the pipe from centimeters to meters. Therefore,

Volume of water that flows for 1 sec =

Now, we have to find the volume of water that flows for 30 minutes.

Since speed of water is in meters/second, let us convert 30 minutes into seconds. It will be

Volume of water that flows for 30 minutes =

Now, considering the tank, we have been given the radius of tank in centimeters. Let us first convert it into meters. Let radius of tank be ‘’.

= 60 cm

=

Volume of water collected in the tank after 30 minutes=

We know that,

Volume of water collected in the tank after 30 minutes= Volume of water that flows through the pipe for 30 minutes

Therefore, the height of the tank is 3 meters.

#### Page No 19.22:

#### 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?

#### Answer:

Given data is as follows:

Diameter of the tank = 1.4 m

Height of the tank = 2.1 m

Diameter of the pipe = 3.5 cm

Water flow rate = 2 m/sec

We have to find the time required to fill the tank using this pipe.

The diameter of the tank is given which is 1.4 m. Let us find the radius.

=

=0.7 m

Volume of the tank =

=

Given is the diameter of the pipe which is 3.5 cm. Therefore, radius is cm. Let us convert it to meters. It then becomes, m.

Volume of water that flows through the pipe in 1 second =

Let the time taken to fill the tank be seconds. Then we have,

Volume of water that flows through the pipe in seconds =

We know that volume of the water that flows through the pipe in seconds will be equal to the volume of the tank. Therefore, we have

Volume of water that flows through the pipe in seconds= Volume of the tank

=

=1680 seconds

= minutes

=28 minutes

Hence, it takes 28 minutes to fill the tank using the given pipe.

#### Page No 19.22:

#### 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 cm^{2}. Find the volume of the cylinder.

#### Answer:

Given data is as follows:

Total surface area of the cylinder = 1628

We have to find the volume of the cylinder.

It is given that,

Total surface area = 1628

That is,

But it is already given in the problem that,

Therefore,

= 1628

= 1628

= 7 cm

Since

We have,

cm

Now that we know both height and radius of the cylinder, we can easily find the volume.

Volume =

Volume =

Volume = 4320

Hence, the volume of the given cylinder is 4620.

#### Page No 19.22:

#### 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.

#### Answer:

Given data is as follows:

Inner diameter of the well = 10 m

Height = 8.4 m

Width of embankment = 7.5 m

We have to find the height of the embankment.

Given is the diameter of the well which is 10 m. Therefore,

= 5 m

The outer radius of the embankment,

= Inner radius of the well + width of the embankment

= 5 + 7.5

= 12.5 m

Let be the height of the embankment.

The volume of earth dug out is equal to the volume of the embankment. Therefore,

Volume of embankment = Volume of earth dug out

Thus, height of the embankment is 1.6 m

#### Page No 19.22:

#### Question 33:

*C*_{1} and *C*_{2} are two cylinders having equal total surface areas. The radius of each cylinder is equal to the height of the other. The sum of the volumes of both the cylinders is 250π cm^{3}. Find the sum of their curved surface areas.

#### Answer:

Let *r*_{1} be the radius and *h*_{1}be the height of cylinder *C*_{1}* *& *r*_{2} be the radius and *h*_{2} be the height of cylinder *C*_{2}.

Given:

*r*_{1} = *h*_{2} .....(1)

*r*_{2} = *h*_{1} .....(2)

Also,

Total surface area of cylinder *C*_{1} = Total surface area of cylinder *C*_{2} (Given)

$\Rightarrow 2\mathrm{\pi}{r}_{1}\left({r}_{1}+{h}_{1}\right)=2\mathrm{\pi}{r}_{2}\left({r}_{2}+{h}_{2}\right)$

$\Rightarrow {r}_{1}\left({r}_{1}+{r}_{2}\right)={r}_{2}\left({r}_{2}+{r}_{1}\right)$ [Using (1) and (2)]

$\Rightarrow {r}_{1}={r}_{2}$ .....(3)

From (1), (2) and (3), we have

*r*_{1} = *r*_{2} = *h*_{1} = *h*_{2} .....(4)

It is given that, the sum of the volumes of both the cylinders is 250$\mathrm{\pi}$ cm^{3}.

$\therefore \mathrm{\pi}{r}_{1}^{2}{h}_{1}+\mathrm{\pi}{r}_{2}^{2}{h}_{2}=250\mathrm{\pi}{\mathrm{cm}}^{3}$

$\Rightarrow 2{r}_{1}^{3}=250$ [Using (4)]

$\Rightarrow {r}_{1}^{3}=125={\left(5\right)}^{3}$

$\Rightarrow {r}_{1}=5\mathrm{cm}$

So, *r*_{1} = *r*_{2} = *h*_{1} = *h*_{2} = 5 cm [Using (4)]

∴ Sum of the curved surface areas of *C*_{1} and *C*_{2}

$=2\mathrm{\pi}{r}_{1}{h}_{1}+2\mathrm{\pi}{r}_{2}{h}_{2}$

$=2\mathrm{\pi}\times 5\mathrm{cm}\times 5\mathrm{cm}+2\mathrm{\pi}\times 5\mathrm{cm}\times 5\mathrm{cm}$ (*r*_{1} = *r*_{2} = *h*_{1} = *h*_{2} = 5 cm)

$=100\mathrm{\pi}{\mathrm{cm}}^{2}$

Thus, the sum of the curved surface areas of the two cylinders is 100$\mathrm{\pi}$ cm^{3}.

#### Page No 19.28:

#### Question 1:

*Mark the correct alternative in each of the following*:

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

#### Answer:

Let S_{1} be the curved surface area of the cylinder with radius r_{1} and height h_{1}, then

……. (1)

Now, let S_{2} be the surface area after changing the dimension, then

So,

Hence, the correct option is (c)

#### Page No 19.28:

#### 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

#### Answer:

Let V_{1} and V_{2} be the volume of the two cylinders with radius r_{1} and height h_{1}, and radius r_{2} and height h_{2}, where

So,

……. (1)

Now,

…… (2)

From equation (1) and (2), we have

Hence, the correct option is (c)

#### Page No 19.28:

#### Question 3:

The number of surfaces in right cylinder is

(a) 1

(b) 2

(c) 3

(d) 4

#### Answer:

In a cylinder there are three surfaces, one is curved surface area and other two are circular discs.

Hence, the correct option is (c)

#### Page No 19.28:

#### Question 4:

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

(a) square

(b) rectangle

(c) rhombus

(d) trapezium

#### Answer:

From the following figure of cylinder we observe that the vertical cross section is the rectangle PQRS.

Hence, the correct option is (b)

#### Page No 19.28:

#### Question 5:

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

(a) $\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}$

(b) ${\mathrm{\pi r}}^{2}\mathrm{h}$

(c) 2$\mathrm{\pi r}(\mathrm{h}+\mathrm{r})$

(d) 2$\mathrm{\pi rh}$

#### Answer:

It is given that r is the radius and h is the height of the cylinder.

We know that the volume of cylinder is given by the formula

Hence, the correct option is (b)

#### Page No 19.28:

#### Question 6:

The number of surfaces of a hollow cylindrical object is

(a) 1

(b) 2

(c) 3

(d) 4

#### Answer:

In a hollow cylinder, there are two curved surface areas: inner and outer and one circular base with inner and outer surface area.

Hence, the correct option is (d)

#### Page No 19.28:

#### 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

#### Answer:

Let V_{1} be the volume of the cylinder with radius r_{1} and height h_{1}, then

……. (1)

Now, let V_{2} be the volume after changing the dimension, then

So,

Hence, the correct option is (d)

#### Page No 19.28:

#### 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

#### Answer:

Let V_{1} be the volume of the cylinder with radius r_{1} and height h_{1}, then

……. (1)

Now, let V_{2} be the volume after changing the dimension, then

So,

Hence, the correct option is (a)

#### Page No 19.28:

#### 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

#### Answer:

Let V_{1} be the volume of the cylinder with radius r_{1} and height h_{1}, then

……. (1)

Now, let V_{2} be the volume after changing the dimension, then

So,

Hence, the correct option is (c)

#### Page No 19.28:

#### 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

(a) $2{\mathrm{\pi h}}^{2}$

(b) $\frac{3}{2}{\mathrm{\pi h}}^{2}$

(c) $\frac{4}{3}{\mathrm{\pi h}}^{2}$

(d) ${\mathrm{\pi h}}^{2}$

#### Answer:

Let r be the radius of the cylinder and h be its height.

It is given that:

Therefore, total surface area S is:

Hence, the correct option is (b)

#### Page No 19.28:

#### 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) 40$\mathrm{\pi}$

(b) 80$\mathrm{\pi}$

(c) 160$\mathrm{\pi}$

(d) 200$\mathrm{\pi}$

#### Answer:

Let r be the radius of the cylinder and h be its height.

It is given that:

And,

Therefore, total surface area S of the sheet required is:

Hence, the correct option is (b)

#### Page No 19.29:

#### Question 12:

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

(a) $1:\sqrt{2}$

(b) $\sqrt{2}:1$

(c) 1 : 2

(d) 1 : 4

(d)

#### Answer:

Let V_{1} and V_{2} be the volume of the two cylinders with h_{1} and h_{2} as their heights:

Let r_{1} and r_{2} be their base radius.

It is given that

Hence, the correct option is (b)

#### Page No 19.29:

#### 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

#### Answer:

Let V_{1} and V_{2} be the volume of the two cylinders with h_{1} and h_{2} as their heights:

Let r_{1} and r_{2} be their base radius.

It is given that

Hence, the length will become 9 times.

Hence, the correct option is (c)

#### Page No 19.29:

#### 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?

(a) 4

(b) $\frac{1}{\sqrt{2}}$

(c) 2

(d) $\frac{1}{2}$

#### Answer:

Let V_{1} be the volume of the cylinder with radius r_{1} and height h_{1}, then

……. (1)

Now, let V_{2} be the volume after changing the dimension, then

So,

It is given that V_{1} =V_{2}Therefpre,

Hence, the correct option is (b)

#### Page No 19.29:

#### 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*?

(a) $\frac{{x}^{2}}{2\mathrm{\pi}}$

(b) $\frac{x}{2\sqrt{\mathrm{\pi}}}$

(c) $\frac{\sqrt{2x}}{\mathrm{\pi}}$

(d) $\frac{\mathrm{\pi}}{2\sqrt{x}}$

#### Answer:

Let V_{1} be the volume of the cylinder with radius r and height h, then

……. (1)

Now, let V_{2} be the volume of the box, then

It is given that V_{1} =1/4 V_{2}Therefpre,

Hence, the correct option is (b)

#### Page No 19.29:

#### Question 16:

The height *h* of a cylinder equals the circumference of the cylinder. In terms of *h* what is the volume of the cylinder?

#### Answer:

Let h be the height of the cylinder with radius r.

It is given that:

Therefore, the volume of the cylinder is

Hence, the correct option is (a)

#### Page No 19.29:

#### Question 17:

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

(a) $2\mathrm{\pi r}(\mathrm{r}+\mathrm{h})$

(b) $\mathrm{\pi r}(\mathrm{r}+2\mathrm{h})$

(c) $\mathrm{\pi r}+(2\mathrm{r}+\mathrm{h})$

(d) $2{\mathrm{\pi r}}^{2}+\mathrm{h}$

#### Answer:

Let S be the total surface area of the closed cylinder with radius r and height h, then

Hence, the correct option is (a)

#### Page No 19.29:

#### 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?

(a) $\frac{24}{\mathrm{\pi}}$

(b) $\frac{48}{\mathrm{\pi}}$

(c) $\frac{32}{\mathrm{\pi}}$

(d) $\frac{48}{\mathrm{\pi}}$

#### Answer:

Let r be the radius of the cylinder. It is given that the height drops 3 inches, when 1 cubic foot of sand is poured out and 1 foot = 12 inch

So,

Hence, the correct option is (b)

#### Page No 19.29:

#### 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,

(a) *v*_{1} = *v*_{2}

(b) *a*_{1}*v*_{1} = *a*_{2}*v*_{2}

(c) *a*_{2}*v*_{1} = *a*_{1}*v*_{2}

(d) $\frac{{v}_{1}}{{a}_{1}}=\frac{{v}_{2}}{{a}_{2}}$

#### Answer:

Let the radius of the base of the cylinders be r and R.

Now, let sheet with length a_{1} be used to form a cylinder with volume v_{1}

So,

$2\mathrm{\pi r}={\mathrm{a}}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{r}=\frac{{\mathrm{a}}_{1}}{2\mathrm{\pi}}$

Volume ${v}_{1}={\mathrm{\pi r}}^{2}\mathrm{h}=\mathrm{\pi}{\left(\frac{{\mathrm{a}}_{1}}{2\mathrm{\pi}}\right)}^{2}{\mathrm{a}}_{2}=\frac{{{\mathrm{a}}_{1}}^{2}{\mathrm{a}}_{2}}{4\mathrm{\pi}}$

Similarly, let sheet with length a2 be used to form a cylinder with volume v2

So,

$2\pi R={\mathrm{a}}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{{\mathrm{a}}_{2}}{2\mathrm{\pi}}$

Volume ${v}_{2}=\pi {R}^{2}\mathrm{h}=\mathrm{\pi}{\left(\frac{{\mathrm{a}}_{2}}{2\mathrm{\pi}}\right)}^{2}{\mathrm{a}}_{1}=\frac{{{\mathrm{a}}_{2}}^{2}{\mathrm{a}}_{1}}{4\mathrm{\pi}}$

Now,

$\frac{{v}_{1}}{{v}_{2}}=\frac{{{a}_{1}}^{2}{a}_{2}}{{{a}_{2}}^{2}{a}_{1}}=\frac{{a}_{1}}{{a}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{1}{a}_{2}={v}_{2}{a}_{1}$

Hence, the correct option is (c).

#### Page No 19.29:

#### Question 20:

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

(a) $\frac{2}{3}$

(b) $\frac{1}{2}$

(c) $\frac{3}{2}$

(d) 2

#### Answer:

$\mathrm{Let}\mathrm{the}\mathrm{original}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{base}\mathrm{of}\mathrm{cylinder}=\mathrm{r}\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{the}\mathrm{original}\mathrm{height}\mathrm{of}\mathrm{cylinder}=\mathrm{h}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{original}\mathrm{base}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{cylinder},\mathrm{S}={\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{original}\mathrm{LSA}\mathrm{of}\mathrm{cylinder},\mathrm{A}=2\mathrm{\pi rh}............\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{When}\mathrm{the}\mathrm{altitude}\mathrm{is}\mathrm{increased}\mathrm{to}6\mathrm{times}\mathrm{of}\mathrm{its}\mathrm{initial}\mathrm{value}\mathrm{and}\mathrm{base}\mathrm{area}\mathrm{is}\mathrm{decreased}\mathrm{one}-\mathrm{ninth}\mathrm{of}\mathrm{its}\mathrm{initial}\mathrm{value}:\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{the}\mathrm{new}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{cylinder}=\mathrm{h}\text{'}\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{the}\mathrm{new}\mathrm{radius}\mathrm{of}\mathrm{base}\mathrm{of}\mathrm{cylinder}=\mathrm{r}\text{'}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{new}\mathrm{base}\mathrm{area}\mathrm{of}\mathrm{cylinder},\mathrm{S}\text{'}=\mathrm{\pi}{\left(\mathrm{r}\text{'}\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that},\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{new}\mathrm{height}\mathrm{of}\mathrm{cylinder}=6\times \mathrm{original}\mathrm{height}\mathrm{of}\mathrm{cylinder}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \overline{)\mathrm{h}\text{'}=6\mathrm{h}}\phantom{\rule{0ex}{0ex}}\mathrm{Also},\mathrm{new}\mathrm{base}\mathrm{area}\mathrm{of}\mathrm{cylinder}=\frac{1}{9}\left(\mathrm{original}\mathrm{base}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{cylinder}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{S}\text{'}=\frac{1}{9}\mathrm{S}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\pi}{\left(\mathrm{r}\text{'}\right)}^{2}=\frac{1}{9}\left({\mathrm{\pi r}}^{2}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\mathrm{r}\text{'}\right)}^{2}={\left(\frac{\mathrm{r}}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \overline{)\mathrm{r}\text{'}=\frac{\mathrm{r}}{3}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{new}\mathrm{LSA}\mathrm{of}\mathrm{cylinder},\mathrm{A}\text{'}=2\mathrm{\pi r}\text{'}\mathrm{h}\text{'}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{A}\text{'}=2\mathrm{\pi}\times \left(\frac{\mathrm{r}}{3}\right)\times \left(6\mathrm{h}\right)=2\times \left(2\mathrm{\pi rh}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{A}\text{'}=2\mathrm{A}\left[\mathrm{Using}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{lateral}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{cylinder}\mathrm{becomes}\mathrm{twice}\mathrm{of}\mathrm{the}\mathrm{original}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{option}\left(\mathrm{d}\right)\mathrm{is}\mathrm{correct}.$

#### Page No 19.30:

#### Question 1:

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

#### Answer:

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

∴ Curved surface of the original cylinder, *C*_{1} = 2$\mathrm{\pi}$*rh*

If the radius is doubled and height is halved, then

Radius of the new cylinder, *R* = 2*r*

Height of the new cylinder, *H* = $\frac{h}{2}$

∴ Curved surface of the new cylinder, *C*_{2} = $2\mathrm{\pi}RH=2\mathrm{\pi}\times 2r\times \frac{h}{2}$ = 2$\mathrm{\pi}$*rh *

Thus, the curved surface area of the new cylinder is same as the curved surface area of the original cylinder.

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

#### Page No 19.30:

#### Question 2:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio of 5 : 3. The ratio of their volumes is _________.

#### Answer:

Let* **r*_{1} be the radius and *h*_{1} be the height of first cylinder & *r*_{2} be the radius and *h*_{2} be the height of second cylinder.

It is given that,

$\frac{{r}_{1}}{{r}_{2}}=\frac{2}{3}$ and $\frac{{h}_{1}}{{h}_{2}}=\frac{5}{3}$ .....(1)

Now,

$\frac{\mathrm{Volume}\mathrm{of}\mathrm{first}\mathrm{cylinder}}{\mathrm{Volume}\mathrm{of}\mathrm{second}\mathrm{cylinder}}$

$=\frac{\mathrm{\pi}{r}_{1}^{2}{h}_{1}}{\mathrm{\pi}{r}_{2}^{2}{h}_{2}}$

$={\left(\frac{{r}_{1}}{{r}_{2}}\right)}^{2}\times \frac{{h}_{1}}{{h}_{2}}$

$={\left(\frac{2}{3}\right)}^{2}\times \frac{5}{3}$ [Using (1)]

$=\frac{4}{9}\times \frac{5}{3}$

$=\frac{20}{27}$

∴ Volume of first cylinder : Volume of second cylinder = 20 : 27

Thus, the ratio of volumes of two cylinders is 20 : 27.

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio of 5 : 3. The ratio of their volumes is _____20 : 27_____.

#### Page No 19.30:

#### Question 3:

If the radius of a right circular cylinder is doubled and its curved surface area is not changed, then its height must be _________.

#### Answer:

Let* **r* be the radius and *h* be the height of the cylinder.

∴ Curved surface of the cylinder, *C*_{1} = 2*$\mathrm{\pi}$rh*

Let the height of the new cylinder be *H*.

Radius of the new cylinder, *R* = 2*r* (Given)

∴ Curved surface of the new cylinder, *C*_{2} = 2*$\mathrm{\pi}$RH *= 2*$\mathrm{\pi}$*(2*r*)*H*

It is given that, the curved surface areas of the new cylinder is not changed.

∴ *C*_{2} = *C*_{1}

⇒ 2*$\mathrm{\pi}$*(2*r*)*H* = 2*$\mathrm{\pi}$rh*

⇒ $H=\frac{h}{2}$

Thus, the height of the new cylinder is half of the height of given cylinder.

If the radius of a right circular cylinder is doubled and its curved surface area is not changed, then its height must be ____halved____.

#### Page No 19.30:

#### Question 4:

The difference between the total surface area and curved surface area of a cylinder of base radius 10 cm is _________.

#### Answer:

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

*r* = 10 cm (Given)

Total surface area of the cylinder = 2*$\mathrm{\pi}$r*(*r* + *h*)

Curved surface area of the cylinder = 2*$\mathrm{\pi}$rh*

∴ Difference between the total surface area and curved surface area of the cylinder

= Total surface area of the cylinder − Curved surface area of the cylinder

= 2*$\mathrm{\pi}$r*(*r* + *h*) − 2*$\mathrm{\pi}$rh*

= 2*$\mathrm{\pi}$r*^{2} + 2*$\mathrm{\pi}$rh *− 2*$\mathrm{\pi}$rh*

= 2$\mathrm{\pi}$× (10 cm)^{2 }(*r* = 10 cm)

= 2$\mathrm{\pi}$× 100 cm^{2}

= 200$\mathrm{\pi}$ cm^{2}

Thus, the difference between the total surface area and curved surface area of a cylinder of base radius 10 cm is 200$\mathrm{\pi}$ cm^{2}.

The difference between the total surface area and curved surface area of a cylinder of base radius 10 cm is _____200$\mathrm{\pi}$ cm ^{2}_____.

#### Page No 19.30:

#### Question 5:

The cost of painting the total outside surface of a closed cylinder at ₹ 3 per cm^{2} is ₹ 2772. If the height of the cylinder is 2 times the radius, then its volume is __________.

#### Answer:

Let *r* cm and *h* cm be the radius and height of the closed cylinder, respectively.

*h* = 2*r* (Given)

Total surface area of the cylinder

$=2\mathrm{\pi}r\left(r+h\right)$

$=2\mathrm{\pi}r\left(r+2r\right)$ (*h* = 2*r*)

$=2\mathrm{\pi}r\times 3r$

$=6\mathrm{\pi}{r}^{2}{\mathrm{cm}}^{2}$

Rate of painting the surface of closed cylinder = ₹ 3 per cm^{2}

∴ Total cost of painting the total outside surface of closed cylinder = ₹ 3 per cm^{2 }× $6\mathrm{\pi}{r}^{2}{\mathrm{cm}}^{2}$ = ₹ $18\mathrm{\pi}{r}^{2}$

⇒ ₹ $18\mathrm{\pi}{r}^{2}$ = ₹ 2772

$\Rightarrow 18\times \frac{22}{7}\times {r}^{2}=2772$

$\Rightarrow {r}^{2}=\frac{2772\times 7}{22\times 18}=49$

$\Rightarrow r=\sqrt{49}=7\mathrm{cm}$

So, *h* = 2*r* = 2 × 7 cm = 14 cm

∴ Volume of the cylinder

$=\mathrm{\pi}{r}^{2}h$

$=\frac{22}{7}\times {\left(7\right)}^{2}\times 14$

$=2156{\mathrm{cm}}^{3}$

Thus, the volume of cylinder is 2156 cm^{3}.

The cost of painting the total outside surface of a closed cylinder at ₹ 3 per cm^{2} is ₹ 2772. If the height of the cylinder is 2 times the radius, then its volume is _____2156 cm ^{3}____.

#### Page No 19.30:

#### Question 6:

The ratio between the radius of base and the height of a cylinder is 2 : 3. If its volume is 1617 cm^{3}, the total surface area of the cylinder is _________.

#### Answer:

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

It is given that, the ratio between the radius of base and the height of a cylinder is 2 : 3.

So, *r* = 2*x* and *h* = 3*x*, where *x* is constant

Volume of the cylinder = 1617 cm^{3}

$\Rightarrow \mathrm{\pi}{r}^{2}h=1617$

$\Rightarrow \frac{22}{7}\times {\left(2x\right)}^{2}\times 3x=1617$

$\Rightarrow \frac{22}{7}\times 4{x}^{2}\times 3x=1617$

$\Rightarrow {x}^{3}=\frac{1617\times 7}{12\times 22}={\left(\frac{7}{2}\right)}^{3}$

$\Rightarrow x=\frac{7}{2}\mathrm{cm}$

$\therefore r=2x=2\times \frac{7}{2}=7\mathrm{cm}$ and $h=3x=3\times \frac{7}{2}=\frac{21}{2}\mathrm{cm}$

Total surface area of the cylinder

$=2\mathrm{\pi}r\left(r+h\right)$

$=2\times \frac{22}{7}\times 7\times \left(7+\frac{21}{2}\right)$

$=44\times 17.5$

$=770{\mathrm{cm}}^{2}$

Thus, the total surface area of the cylinder is 770 cm^{2}.

The ratio between the radius of base and the height of a cylinder is 2 : 3. If its volume is 1617 cm^{3}, the total surface area of the cylinder is _____770 cm ^{2}_____.

#### Page No 19.30:

#### Question 7:

The ratio of the total surface area to the lateral surface area of a cylinder of base radius *r *and height *h *is __________.

#### Answer:

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

Total surface of the cylinder = 2$\mathrm{\pi}$*r*(*r* + *h*)

Lateral surface area of the cylinder = 2$\mathrm{\pi}$*r**h*

$\therefore \frac{\mathrm{Total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{cylinder}}{\mathrm{Lateral}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{cylinder}}$

$=\frac{2\mathrm{\pi}r\left(r+h\right)}{2\mathrm{\pi}rh}$

$=\frac{r+h}{h}$

So,

Total surface area of the cylinder : Lateral surface area of the cylinder = *r* + *h* : *h*

Thus, the ratio of the total surface area to the lateral surface area of a cylinder of base radius *r *and height *h *is *r* + *h* : *h*.

The ratio of the total surface area to the lateral surface area of a cylinder of base radius *r *and height *h *is _____ r + h : h_____.

#### Page No 19.30:

#### Question 8:

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

#### Answer:

Let *r* units and *l* units be the radius and length of the wire, respectively.

∴ Original volume of the wire = $\mathrm{\pi}$*r*^{2}*l *cu. units

When the radius of the wire is decreased to one-third, then the new radius of the wire, *R* = $\frac{r}{3}$ units.

Suppose the new length of the wire is *L* units.

∴ New volume of the wire = $\mathrm{\pi}{\left(\frac{r}{3}\right)}^{2}L$ cu. units

It is given that, the volume of the wire remains the same.

$\therefore \mathrm{\pi}{\left(\frac{r}{3}\right)}^{2}L=\mathrm{\pi}{r}^{2}l$

$\Rightarrow \frac{{r}^{2}}{9}\times L={r}^{2}\times l$

$\Rightarrow L=9l$

Thus, the length of the wire will be increased 9 times.

The radius of a wire is decreased to one-third. If volume remains the same, the length will be increased _____9 times_____.

#### Page No 19.30:

#### Question 9:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their curved surface areas is __________.

#### Answer:

Let* **r*_{1} be the radius and *h*_{1} be the height of first cylinder & *r*_{2} be the radius and *h*_{2} be the height of second cylinder.

$\therefore \frac{{r}_{1}}{{r}_{2}}=\frac{2}{3}$ and $\frac{{h}_{1}}{{h}_{2}}=\frac{5}{3}$ .....(1) (Given)

Now,

$\frac{\mathrm{Curved}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{first}\mathrm{cylinder}}{\mathrm{Curved}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{second}\mathrm{cylinder}}$

$=\frac{2\mathrm{\pi}{r}_{1}{h}_{1}}{2\mathrm{\pi}{r}_{2}{h}_{2}}$

$=\frac{{r}_{1}}{{r}_{2}}\times \frac{{h}_{1}}{{h}_{2}}$

$=\frac{2}{3}\times \frac{5}{3}$ [Using (1)]

$=\frac{10}{9}$

∴ Curved surface area of first cylinder : Curved surface area of second cylinder = 10 : 9

Thus, the ratio of the curved surface areas of the two cylinders is 10 : 9.

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their curved surface areas is _____10 : 9_____.

#### Page No 19.30:

#### Question 10:

A rectangular tin sheet is 12 cm long and 5 cm broad. It is rolled along its length to form a cylinder by making the opposite edges just to touch each other. The volume of the cylinder is __________.

#### Answer:

It is given that, a rectangular tin sheet is 12 cm long and 5 cm broad. When it is rolled along its length to form a cylinder by making the opposite edges just to touch each other, then the circumference of the base of the cylinder so formed is same as the length of the rectangular sheet and the height of the cylinder is same as the breadth of the rectangular sheet.

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

Height of the cylinder, *h* = 5 cm .....(1)

Also,

Circumference of base of cylinder = Length of the rectangular tin sheet = 12 cm

⇒ 2*$\mathrm{\pi}$r* = 12 cm

$\Rightarrow r=\frac{6}{\mathrm{\pi}}$ .....(2)

∴ Volume of the cylinder

$=\mathrm{\pi}{r}^{2}h$

$=\mathrm{\pi}\times {\left(\frac{6}{\mathrm{\pi}}\mathrm{cm}\right)}^{2}\times 5\mathrm{cm}$ [Using (1) and (2)]

$=\frac{180}{\mathrm{\pi}}{\mathrm{cm}}^{3}$

Thus, the volume of the cylinder so formed is $\frac{180}{\mathrm{\pi}}{\mathrm{cm}}^{3}$.

A rectangular tin sheet is 12 cm long and 5 cm broad. It is rolled along its length to form a cylinder by making the opposite edges just to touch each other. The volume of the cylinder is $\overline{)\frac{180}{\mathrm{\pi}}{\mathrm{cm}}^{3}}$.

#### Page No 19.30:

#### Question 11:

If the area of the curved surface and the circumference of the base of a right circular cylinder are 4400 cm^{2} and 110 cm respectively, then the volume of the cylinder is _________.

#### Answer:

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

Curved surface area of the cylinder = 4400 cm^{2} (Given)

∴ 2$\mathrm{\pi}$*rh* = 4400 cm^{2} .....(1)

Also, circumference of the base = 110 cm (Given)

∴ 2$\mathrm{\pi}$*r* = 110 cm .....(2)

$\Rightarrow 2\times \frac{22}{7}\times r=110\mathrm{cm}$

$\Rightarrow r=\frac{110\times 7}{44}=\frac{35}{2}\mathrm{cm}$

Now, dividing (1) by (2), we get

$\frac{2\mathrm{\pi}rh}{2\mathrm{\pi}r}=\frac{4400{\mathrm{cm}}^{2}}{110\mathrm{cm}}$

$\Rightarrow h=40\mathrm{cm}$

∴ Volume of the cylinder

$=\mathrm{\pi}{r}^{2}h$

$=\frac{22}{7}\times {\left(\frac{35}{2}\mathrm{cm}\right)}^{2}\times 40\mathrm{cm}$

$=\frac{22}{7}\times \frac{35}{2}\times \frac{35}{2}\times 40{\mathrm{cm}}^{3}$

$=38500{\mathrm{cm}}^{3}$

Thus, the volume of the cylinder is 38500 cm^{3}.

If the area of the curved surface and the circumference of the base of a right circular cylinder are 4400 cm^{2} and 110 cm respectively, then the volume of the cylinder is _____38500 cm ^{3}_____.

#### Page No 19.30:

#### Question 12:

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

#### Answer:

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

∴ Area of the base of the cylinder = *$\mathrm{\pi}$r*^{2}

Lateral surface area of the original cylinder = 2$\mathrm{\pi}$*rh * .....(1)

Now,

Height of the new cylinder,* **H* = 6*h* (Given) .....(2)

Let the radius of the new cylinder be *R**.*

Area of the base of the new cylinder = $\frac{\mathrm{\pi}{r}^{2}}{9}$ (Given)

$\therefore \mathrm{\pi}{R}^{2}=\frac{\mathrm{\pi}{r}^{2}}{9}$

$\Rightarrow {R}^{2}=\frac{{r}^{2}}{9}$

$\Rightarrow R=\sqrt{\frac{{r}^{2}}{9}}=\frac{r}{3}$ .....(3)

Now,

Lateral surface area of the new cylinder

$=2\mathrm{\pi}RH$

$=2\mathrm{\pi}\times \frac{r}{3}\times 6h$ [From (1) and (2)]

$=4\mathrm{\pi}rh$

$=2\times 2\mathrm{\pi}rh$

= 2 × Lateral surface area of the original cylinder [From (1)]

Thus, the lateral surface area of the new cylinder increases 2 times.

The altitude of a right circular cylinder is increased six times and the base area is decreased to one-ninth of its value. The factor by which the lateral surface increases is _____2_____.

#### Page No 19.30:

#### Question 13:

Two rectangular sheets of paper each 30 cm × 18 cm are made into two right circular cylinders, one by rolling the paper along its length and the other along the breadth. The ratio of the volumes of the two cylinders, thus formed, is _________.

#### Answer:

The length and breadth of each rectangular sheet of paper is 30 cm and 18 cm, respectively.

When the sheet of paper is rolled along its length to form a cylinder, then the circumference of the base of the cylinder so formed is same as the length of the rectangular sheet and the height of the cylinder is same as the breadth of the rectangular sheet.

Let *r* be the radius and *h* be the height of the cylinder formed in this case.

∴ *h* = 18 cm

Circumference of base = 30 cm

⇒ 2$\mathrm{\pi}$*r* = 30 cm

$\Rightarrow r=\frac{15}{\mathrm{\pi}}$ cm

∴ Volume of the cylinder formed in this case, *V*_{1} = $\mathrm{\pi}{r}^{2}h=\mathrm{\pi}\times {\left(\frac{15}{\mathrm{\pi}}\mathrm{cm}\right)}^{2}\times 18\mathrm{cm}$ .....(1)

When the sheet of paper is rolled along its breadth to form a cylinder, then the circumference of the base of the cylinder so formed is same as the breadth of the rectangular sheet and the height of the cylinder is same as the length of the rectangular sheet.

Let *R* be the radius and *H* be the height of the cylinder formed in this case.

∴ *H* = 30 cm

Circumference of base = 18 cm

⇒ 2$\mathrm{\pi}$*R* = 18 cm

$\Rightarrow R=\frac{9}{\mathrm{\pi}}$ cm

∴ Volume of the cylinder formed in this case, *V*_{2} = $\mathrm{\pi}{R}^{2}H=\mathrm{\pi}\times {\left(\frac{9}{\mathrm{\pi}}\mathrm{cm}\right)}^{2}\times 30\mathrm{cm}$ .....(2)

Now,

$\frac{{V}_{1}}{{V}_{2}}=\frac{\mathrm{\pi}\times {\left({\displaystyle \frac{15}{\mathrm{\pi}}}\mathrm{cm}\right)}^{2}\times 18\mathrm{cm}}{\mathrm{\pi}\times {\left({\displaystyle \frac{9}{\mathrm{\pi}}}\mathrm{cm}\right)}^{2}\times 30}=\frac{5}{3}$

⇒ *V*_{1} : *V*_{2} = 5 : 3

Thus, the ratio of the volumes of the two cylinders so formed is 5 : 3.

Two rectangular sheets of paper each 30 cm × 18 cm are made into two right circular cylinders, one by rolling the paper along its length and the other along the breadth. The ratio of the volumes of the two cylinders, thus formed, is ____5 : 3____.

#### Page No 19.30:

#### Question 14:

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

#### Answer:

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

Now,

*r* + *h* = 37 m (Given) .....(1)

Total surface of the cylinder = 1628 m^{2} (Given)

$\therefore 2\mathrm{\pi}r\left(r+h\right)=1628{\mathrm{cm}}^{2}$

$\Rightarrow 2\times \frac{22}{7}\times r\times 37\mathrm{cm}=1628{\mathrm{cm}}^{2}$ [Using (1)]

$\Rightarrow r=\frac{1628\times 7}{44\times 37}=7\mathrm{cm}$ .....(2)

Substituting the value of *r* in (1), we get

7 cm + *h* = 37 cm

⇒ *h* = 37 − 7 = 30 cm .....(3)

∴ Volume of the solid cylinder

$=\mathrm{\pi}{r}^{2}h$

$=\frac{22}{7}\times {\left(7\mathrm{cm}\right)}^{2}\times 30\mathrm{cm}$ [From (2) and (3)]

$=4620{\mathrm{cm}}^{3}$

Thus, the volume of the solid cylinder is 4620 cm^{3}.

If the sum of the radius of the base and the height of a solid cylinder is 37 m and the total surface area of the cylinder is 1628 m^{2}, then its volume is _____4620 cm ^{3}_____.

#### Page No 19.30:

#### Question 15:

A largest possible right-circular cylinder is cut from a wooden cube of edge 7 cm. The volume of the wood left over after cutting the cylinder is __________.

#### Answer:

The largest possible right circular cylinder that can be cut off from a wooden cube of given edge is such that the diameter of the cylinder is same as the edge of the cube and height of the cylinder is same as the edge of the cube.

Let *r* be the radius and *h* be the height of the largest right circular cylinder cut from a wooden cube of edge 7 cm.

∴ Diameter of the cylinder = Edge of the cube = 7 cm

⇒ 2*r* = 7 cm

⇒ *r* = $\frac{7}{2}$ cm

Height of the cylinder = Edge of the cube = 7 cm

∴ *h* = 7 cm

Now,

Volume of the wood left over after cutting the cylinder

= Volume of the cube − Volume of the cylinder

= (Edge)^{3} − *$\mathrm{\pi}$r*^{2}*h*

$={\left(7\mathrm{cm}\right)}^{3}-\frac{22}{7}\times {\left(\frac{7}{2}\mathrm{cm}\right)}^{2}\times 7\mathrm{cm}$

= 343 cm^{3} − 269.5 cm^{3}

= 73.5 cm^{3}

Thus, the volume of the wood left over after cutting the cylinder is 73.5 cm^{3}.

A largest possible right-circular cylinder is cut from a wooden cube of edge 7 cm. The volume of the wood left over after cutting the cylinder is _____73.5 cm ^{3}_____.

#### Page No 19.30:

#### Question 16:

Four times the sum of the areas of the two circular faces of a cylinder is equal to the twice its curved surface area. The height of the cylinder is ________ the radius.

#### Answer:

Let *r* be the radius and *h* be the height of the cylinder.

It is given that,

4 × Sum of areas of the two circular faces of the cylinder = 2 × Curved surface area of the cylinder

$\Rightarrow 4\times \left(\mathrm{\pi}{r}^{2}+\mathrm{\pi}{r}^{2}\right)=2\times 2\mathrm{\pi}rh$

$\Rightarrow 8\mathrm{\pi}{r}^{2}=4\mathrm{\pi}rh$

$\Rightarrow 2r=h$

Or *h* = 2*r*

Thus, the height of the cylinder is twice the radius of the cylinder.

Four times the sum of the areas of the two circular faces of a cylinder is equal to the twice its curved surface area. The height of the cylinder is ____2 times____ the radius.

#### Page No 19.31:

#### Question 1:

Write the number of surfaces of a right circular cylinder.

#### Answer:

Let us list out the surfaces of a right circular cylinder.

1. Lateral surface

2. Lower circular base

3. Upper circular covering

Therefore, a right circular cylinder has a total of 3 surfaces.

#### Page No 19.31:

#### Question 2:

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

#### Answer:

Total Surface Area of a cylinder =

Curved Surface Area of a cylinder =

Ratio of Total Surface Area (TSA) to Curved Surface Area (CSA) is given by,

The ratio of Total Surface Area to Curved Surface Area is (h+r): h

#### Page No 19.31:

#### Question 3:

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

#### Answer:

Given data is as follows

We have to find the total surface area of this cylinder

It is given that

Therefore

Therefore;

(As)

So;

Therefore;

#### Page No 19.31:

#### Question 4:

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

#### Answer:

Let *r*_{1}_{ }and *r*_{2}_{ }be the radii of the two cylinders respectively and *h*_{1}_{ }and *h*_{2} are the heights of the two cylinders respectively. It is given that and

We are asked to find the ratio of the volumes of the two cylinders

Now;

Therefore the ratio of the volumes of the two cylinders is

#### Page No 19.8:

#### 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.

#### Answer:

In the problem it is given that the Curved Surface Area of the cylinder is.

We know that,

Curved Surface Area of a cylinder

Where, radius of the cylinder and height of the cylinder

In the given problem, and h is to be found out.

Let us substitute all the given values in the formula for Curved Surface Area of the cylinder.

We have,

For simplifying the, this can be written as,

Now, clearly m

Therefore, the height of the cylinder is 1 meter.

#### Page No 19.8:

#### 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.

#### Answer:

The following data is given in the problem:

*h* = 28 m

Diameter = 5 cm

We are asked to find the Total radiating surface, which is nothing but the Total Surface Area of the cylinder.

The height of the cylinder is in meters, so let us first convert it into centimeters.

*h* = 2800 cm

Also, the diameter of the cylinder is given, but we want the radius.

The formula for finding out the Total Surface Area is:

Total Surface Area

Substituting the above values in this equation, we have

Total Surface Area

Total Surface Area

Total Surface Area

Therefore, the answer to this problem is

#### Page No 19.8:

#### 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}.

#### Answer:

The data given in the problem is as follows:

Diameter of the cylinder =50cm

Height = 3.5m

Painting charges = Rs.12.50 per m^{2}

We have to find the total cost of painting the pillar

To find the cost of painting the pillar, we should first find the Curved Surface Area of the pillar using the given data.

Curved Surface Area

r =

Since the painting charges are given in terms of, we shall convert the radius from centimeters to meters.

r = 0.25m

Substituting the values in the formula for Curved Surface Area, we have

Curved Surface Area

Curved Surface Area = 5.5 m^{2}

It is given that, for 1 m^{2}^{ }the cost of painting is Rs.12.50

Therefore,

Total cost of painting the pillar =

= 68.75

Therefore, the answer to this questions is, Rs.68.75

#### Page No 19.8:

#### 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?

#### Answer:

The data given in the problem is as follows:

*h* = 1 m

Diameter = 140 cm

We are asked to find the area of the sheet in square meters required to make this cylinder.

Since it is a closed cylinder, the area of the sheet required to make this will be equal to the Total surface area of the cylinder.

Total surface area

= =70cm

Since area is asked in square meters, let us convert the radius from centimeters to meters.

r = 0.7m

Substituting the values in the formula for the Total Surface Area of a cylinder, we have

Total Surface Area

Total Surface Area = 7.48

Therefore, the area of sheet required to make this cylinder is 7.48 square meters.

#### Page No 19.8:

#### 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.

#### Answer:

Data given is the problem is as follows:

The cylinder is a hollow cylinder and is open on both sides

Total surface area of the cylinder is 4620 square centimeters

Area of the base ring = 115.5 square centimeters

Height = 7 cm

We are supposed to find the thickness of this cylinder.

We know that,

Total surface area of a hollow cylinder

Where, *r* is the inner radius and R is the outer radius of the cylinder.

Now we have,

= 4620

Also, *h* = 7cm

= 4620

Also, it is given that

Area of base ring = 115.5

That is, = 115.5 …..(1)

Substituting for in the above equation, we have

2(115.5) = 4620

=4389

Also, *h* = 7

Therefore,

.….(2)

Now let us again take up equation (1)

= 115.5

From equation (2) we have . Substitute this in the above equation.

(*R* – *r*) is nothing but the thickness of the cylinder.

Therefore, the thickness of the cylinder is cm

#### Page No 19.8:

#### 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.

#### Answer:

Data given in the problem is as follows:

*h* =7.5 cm

*r* = 3.5 cm

We are supposed to find the ratio between the Total Surface Area and the Curved Surface Area.

We know that,

Total Surface Area (TSA) =

Curved Surface Area (CSA) =

Therefore,

=

=

Substituting the values of h and r in the above expression, we have

=

= ==

Hence the ratio between Total Surface area and Curved Surface Area is 22 : 15.

#### Page No 19.8:

#### 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 Rs 3.50 per 1000 cm^{2}.

#### Answer:

It is given that,

*r* = 70 cm

*h* = 1.4 m

Tin coating rate = cm^{2}

We have to find the total cost of coating the cylinder with tin.

Let us first convert *h* from meters to centimeters.

*h* = 1.4 m

= 140 cm

Since the cylindrical vessel without lid has to be coated both on the inner side as well the outer side,

Area to be coated =

=

= 154000 cm^{2}

Now let us find the total cost of coating this area.

For 1000 cm^{2} the cost of coating is Rs.3.50

For 154000 cm^{2}^{ }the cost of coating is given by =539

Therefore the total cost of coating the vessel on both inner and outer sides is Rs.539

#### Page No 19.8:

#### 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 Rs 40 per m^{2}.

#### Answer:

Given data is as follows:

Inner diameter of the well = 3.5 m

*h* = 10 m

Rate of plastering = Rs.40 per square meter

We have to find two things,

1. Inner curved surface area

2. Total cost of plastering the inner curved surface

(i)

Inner curved surface area = 110

(ii) Now, let us find the total cost of plastering this area.

It is given that for 1the cost of plastering is Rs.40

Therefore, for 110 the cost of plastering =

= 4400

Cost of plastering = Rs 4400

#### Page No 19.8:

#### Question 9:

The students of a Vidyalaya were asked to participate in 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 carboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

#### Answer:

#### Page No 19.8:

#### Question 10:

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 metre.

#### Answer:

Given that height *h* = 1.5 m

Diameter = 84 cm = 0.84 m

Radius = $\frac{0.84}{2}=0.42\mathrm{m}$

Now, we have to find the area of the ground.

= 396 m^{2}

Cost of leveling for 1 m^{2}^{ }= 0.50

Cost of leveling for 396 m^{2}^{ }=

Cost of leveling for 396 m^{2 }= Rs.198

#### Page No 19.8:

#### 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 Rs 2.50 per square metre?

#### Answer:

Total cost of cleaning = Rs.314.28

#### Page No 19.8:

#### 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.

#### Answer:

The data given in this problem is as follows:

Total Surface Area of the cylinder = 462 cm^{2}

Curved Surface Area (CSA) = Total Surface Area (TSA)

We have to find the radius and the height of the cylinder.

Using the given data, we have

Substituting the formula for Curved Surface Area and Total Surface Area in the above equation, we have

=

=

We have,

Total Surface Area = 462 cm^{2}

Substituting the formula for Total Surface Area in the above equation, we get

= 462

= 462

Now, we know that

Substituting this in the above equation, we have

Since

Therefore, the final answer to this question is,

Radius of the cylinder = 7cm

Height of the cylinder = 3.5cm

#### Page No 19.8:

#### 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 p 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.

#### Answer:

Data given in the problem is as follows:

Given cylinder is a hollow cylinder

External radius (R) = 8cm

Height (h) = 10cm

Total Surface Area = 338*π*

We have to obtain an equation in r, where r is the inner radius of the cylinder and using this equation we have to find the thickness of the cylinder.

We know that,

Total Surface Area of a hollow cylinder =

Therefore,

=

= 338

Thickness = R – *r* = 8 – 5 = 3 cm

Thickness of the cylinder is 3 cm

#### Page No 19.9:

#### Question 14:

Find the lateral curved surface area of a cylinderical petrol storage tank that is 4.2 m 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?

#### Answer:

Actual area of steel used = 95.04 m^{2}

View NCERT Solutions for all chapters of Class 9