RD Sharma 2020 solution class 9 chapter 19 Surface Area and Volume of a Right Circular Cylinder Exercise 19.2

Exercise 19.2

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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 1:

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 cm³

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

Difference in their capacities = 385 – 200 = 185 cm³

Question 2:

The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius 20 cm and height 10 m. How much concrete mixture would be required to build 14 such pillars?

Answer 2:

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 m³

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Question 3:

The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³  of wood has a mass of 0.6 gm.

Answer 3:

Given data is as follows:

Inner diameter = 24cm

Outer diameter = 28cm

h = 35cm

Mass of 1 cm³ 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 = 

$$\begin{gathered} =\frac{22}{7}\times \left({14}^2-{12}^2\right)\times 35 \\ =22\times 5\times 2\times 26 \end{gathered}$$
=5720 cm³

It is given that,

1 cm³ of wood weighs 0.6gm

Therefore, 5720 of wood will weigh = 3432gm

= 3.432kg

Therefore, weight of the wooden pipe = 3.432kg

Question 4:

If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, find:

(i) radius of its base

(ii) volume of the cylinder
[Use $\mathrm{\pi }$ = 3.14]

Answer 4:

Given data is as follows:

Lateral Surface Area = 94.2 cm²

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,

2πrh = 94.2

=94.2

=94.2

(ii) Volume of a cylinder =

=

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 5:

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 cm³

Therefore, 15.4 liters = 15400 cm³

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 = +

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 6:

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 cm³

Volume of soup for all 250 patients = 154 × 250

=38500 cm³

We know that, 1000 cm³ = 1 litre.

Therefore,

Volume of soup for all 250 patients = 38.5 litres

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 7:

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 =

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 8:

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 =

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 9:

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,

=

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 cm².

Answer 10:

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 = 

=

Question 11:

The curved surface area of a cylinder is 1320 cm² and its base had diameter 21 cm. Find the height and the volume of the cylinder.

Answer 11:

Given data is as follows:

Curved Surface Area = 1320 cm²

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 =

=

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 cm³.

Answer 12:

Data given is as follows:

Volume = 1617 cm³

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

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 13:

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 = 2πr

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 =

=

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.

Answer 14:

Given data is as follows:

Curved Surface Area = 267 m²

Volume = 924 m³

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

Question 15:

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

Answer 15:

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

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 16:

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 m³

Therefore, the volume of the cylinder is 1617 m³.

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 m².

Answer 17:

Given data is as follows:

Diameter = 6 m

Plastering rate = Rs.9.50/m²

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 m³

We know that,

Curved Surface Area =

=

Curved Surface Area =396 m²

Therefore, the volume of this well is 594 m³ and cost of plastering its inner surface is Rs.3762.

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 18:

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 m³

Therefore, the volume of timber that can be obtained from this trunk is 0.7392 m³

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 19:

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.

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 20:

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

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 21:

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.

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 22:

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

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Question 23:

How many litres of water flow out of a pipe having an area of cross-section of 5cm² in one minute, if the speed of water in the pipe is 30 cm/ sec?

Answer 23:

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.

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 24:

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.

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 25:

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.

Question 26:

A solid cylinder has a total surface area of 231 cm². Its curved surface area is $\frac{2}{3}$ of the total surface area. Find the volume of the cylinder.

Answer 26:

Given data is as follows:

Total Surface Area = 231 cm²

Curved Surface Area =

We have to find the volume of the cylinder.

We have,

Total Surface Area = 231 cm²

+ =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 cm²

=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 =

=

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 27:

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

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 28:

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.

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 29:

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.

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 30:

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.

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 cm². Find the volume of the cylinder.

Answer 31:

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.

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 32:

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

Question 33:

C₁ and C₂ 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 cm³. Find the sum of their curved surface areas.

Answer 33:

Let r₁ be the radius and h₁be the height of cylinder C₁ & r₂ be the radius and h₂ be the height of cylinder C₂.

Given:

r₁ = h₂        .....(1)

r₂ = h₁        .....(2)

Also,

Total surface area of cylinder C₁ = Total surface area of cylinder C₂         (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

r₁ = r₂ = h₁ = h₂         .....(4)

It is given that, the sum of the volumes of both the cylinders is 250$\mathrm{\pi }$ cm³.

$\therefore \mathrm{\pi }{r}_1^2{h}_1+\mathrm{\pi }{r}_2^2{h}_2=250\mathrm{\pi } {\mathrm{cm}}^3$

$\Rightarrow 2r_1^3=250$             [Using (4)]

$\Rightarrow r_1^3=125={\left(5\right)}^3$

$\Rightarrow r_1=5 \mathrm{cm}$

So, r₁ = r₂ = h₁ = h₂ = 5 cm       [Using (4)]

∴ Sum of the curved surface areas of C₁ and C₂

$=2\mathrm{\pi }{r}_1{h}_1+2\mathrm{\pi }{r}_2{h}_2$

$=2\mathrm{\pi }\times 5 \mathrm{cm}\times 5 \mathrm{cm}+2\mathrm{\pi }\times 5 \mathrm{cm}\times 5 \mathrm{cm}$          (r₁ = r₂ = h₁ = h₂ = 5 cm)

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

Thus, the sum of the curved surface areas of the two cylinders is 100$\mathrm{\pi }$ cm³.