# Class 9 RD Sharma Solutions – Chapter 19 Surface Area And Volume of a Right Circular Cylinder – Exercise 19.2 | Set 2

### Question 17. The height of a right circular cylinder is 10.5 m. Three times the sum of the areas of its two circular faces is twice the area of the curved surface. Find the volume of the cylinder.

Solution:

Given that

Height of the cylinder = 10.5 m

According to the question

3(2Ï€r2) = 2(2Ï€rh)

3r = 2h

r = 2/3 x h

r = 2/3 x 10.5 = 7cm

So, the radius of the well is 7cm

Now we find the volume of the cylinder

V = Ï€r2 h

= 22/7 x 7 x 7 x 10.5

= 154 x 10.5 = 1617 cm3

Hence, the volume of the cylinder is 1617 cm3

### Question 18. How many cubic meters of earth must be dug out to sink a well 21m deep and 6 m diameter? Find the cost of plastering the inner surface as well at Rs.9.50 per m2.

Solution:

Given that

Height of the well = 21 m

Diameter of the well = 6 m

So, the radius of the well = 3 m

Now,

Volume of the cylinder = Ï€r2 h

= 22/7 x 3 x 3 x 21

= 66 x 9 = 594 cm3

Cost of plastering = 9.5 per m3

Cost of plastering inner surface = (594 x 9.50) = Rs. 5643

### Question 19. The trunk of a tree is cylindrical and its circumference is 176 cm. If the length of the tree is 3 m. Find the volume of the timber that can be obtained from the trunk.

Solution:

Given that,

The length of the tree = 3m = 300cm

The circumference of the trunk = 176 cm

We have to find the volume of the timber that can be obtained from the trunk

So, According to the circumference formula

C= 2Ï€r

176 = 2Ï€r

r = 176/2Ï€ = 28 cm

Hence, the radius of the trunk = 28 cm

Now we find the volume of timber

V= Ï€r2 h

= 22/7 x 28 x 28 x 300

= 44 x 8400 = 739200 cm3 or 0.7392 m3

Hence, the volume of timber is 739200 cm3 or 0.7392 m3

### Question 20. A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment.

Solution:

Given that

The diameter of well = 14 m

so the radius = 7 m

Height of the well = 8 m

So,

Volume of well = Ï€r2 h

= 22/7 x 7 x 7 x 8 = 22 x 56

= 1232 m3

Now, Let’s assume that r1 be the radius of embankment, and h1 be the height of embankment

So, Volume of well = Volume of embankment

1232 m3 = Ï€ x r1 x h1

1232 = 22/7 x (282 âˆ’ 72) x h1

h1 = 1232 x 7 / 22(784 – 49)

h1 = 0.533 m

Hence, the height of the embankment is 0.533 m or 53.3 cm

### Question 21. The difference between inside and outside surfaces of a cylindrical tube is 14 cm long is 88 sq.cm. If the volume of the tube is 176 cubic cm, Find the inner and outer radii of the tube.

Solution:

Let’s assume that R be the outer radius and r be the inner radius

Given that

The height of the cylindrical tube = 14 cm

The difference between inside and outside surfaces of a cylindrical tube = 88 cm2

2Ï€Rh – 2Ï€rh = 88

2Ï€h(R â€“ r) = 88

2 x 22/7 x 14(R â€“ r) = 88

(R â€“ r) = 1cm    ——————-(i)

Now,

Volume of tube = Ï€R2 h – Ï€r2 h

176 = Ï€h(R2 – r2)

176 = 22/7 x 14(R2 – r2)

(R2 – r2) = 4

(R + r)(R – r) = 4

Put the value of (R – r) from eq(1)

(R + r) (1) = 4

(R + r) = 4 cm

R = 4 – r                       ————(ii)

Here, R – r = 1

R = 1 + r

Now substitute value of R in eq(ii)

1 + r = 4 â€“ r

2r = 3

r = 3/2 = 1.5 cm

Substitute value of r in eq(i)

R â€“ 1.5 = 1

R = 1 + 1.5 = 2.5 cm

Hence, the value of inner radii is 1.5 cm and radius of outer radii is 2.5 cm.

### Question 22. Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 meters per second into a cylindrical tank. The water is collected in a cylindrical vessel radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes?

Solution:

Given that,

The internal diameter of the pipe = 2 cm,

So, the radius of the pipe = 1 cm = 1/100 m

Water flow rate through the pipe in the cylindrical tank = 6 m/sec,

Radius of the cylindrical tank = 60 cm = 60/100 m

Time = 30 minutes.

Now we find the volume of water that flows in 1 sec

= 22/7 Ã— 1/100 Ã— 1/100 Ã— 6

Now we have to find the volume of water flows in 30 minute.

= 22/7 Ã— 1/100 Ã— 1/100 Ã— 6 Ã— 30 Ã— 60

So, the volume of water collected in the tank after 30 minutes = Volume of water that flows through the pipe in 30 minutes

22/7 Ã— 60/100 Ã— 60/100 Ã— h = 22/7 Ã— 1/100 Ã— 1/100 Ã— 6 Ã— 30 Ã— 60

h = 3 m

Hence, the height of the tank is 3 meter.

### Question 23. A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm Ã— 22 cm Ã— 14 cm. Find the rise in the level of the water when the solid is completely submerged.

Solution:

Given that,

The diameter of the cylindrical container = 56 cm

So, the radius of cylindrical container = 28 cm

The dimensions of rectangular block = 32 cm Ã— 22 cm Ã— 14 cm

Find: the raise in the level of water in the cylinder.

Let’s assume that the raise in the level of water be h, then

So, the volume of cylinder of height h and radius 28 cm = Volume of the rectangular block

22/7 Ã— 28 Ã— 28 Ã— h = 32 Ã— 22 Ã— 14

h = 4 cm

Hence, the rise in the level of the water when the solid is completely submerged is 4 cm.

### Question 24. A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal.

Solution:

Given that,

Internal diameter of the tube = 10.4 cm,

So, the radius of the tube = 10.4/2 = 5.2 cm

Thickness of the metal of the tube = 8 mm = 0.8 cm

Length of the pipe = 25 cm,

Find: the volume of the metal used in the pipe.

Let us assume that the external radius be ‘R’.

So, now we find the external radius

R = 5.2 + 0.8 = 6 cm

So, the volume of metal in the pipe

= Ï€(R2 âˆ’ r2)h

= 22/7 Ã— (62 âˆ’ 5.22) Ã— 25

= 704 cm3

Hence, the volume of metal present in the hollow pipe is 704 cm3.

### Question 25. From a tap of inner radius 0.75 cm, water flows at the rate of 7 m per second. Find the volume in litres of water delivered by the pipe in one hour.

Solution:

Given that,

Radius of the tap = 0.75 cm

Water flow rate = 7 m/sec = 700 cm/sec

So height of the cylinder = 7 m

Time = 1 hour = 60 min = 3600 sec

Find: the volume of water that flows through the pipe for 1 hour.

Now we find the volume of water delivered in 1 second

= 22/7 Ã— 0.75 Ã— 0.75 Ã— 700

Now, we have to find the volume of water delivered in 1 hour

= 22/7 Ã— 0.75 Ã— 0.75 Ã— 700 Ã— 3600 = 4455000 cm3 = 4455 liters

Hence, the volume of water delivered by the pipe is 4455 liters.

### Question 26. 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 meters per second. In how much time the tank will be filled?

Solution:

Given that,

Diameter of the tank = 1.4 m,

So, the radius = 1.4/2 = 0.7 m,

Height of the tank = 2.1 m,

Diameter of the pipe = 3.5 cm,

So, the radius of pipe = 3.5/2 cm = 3.5/200 m

Water flow rate = 2 m/sec,

Find: the time required to fill the tank using the pipe.

So, the volume of the tank

V = Ï€r2h

= 22/7 Ã— 0.7 Ã— 0.7 Ã— 2.1

Now we find the volume of water that flows through the pipe in 1 second

= 22/7 Ã— 3.5/200 Ã— 3.5/200 Ã— 2

Let’s assume that the time taken to fill the tank be x seconds,

So, the volume of water that flows through the pipe in x seconds

= 22/7 Ã— 3.5/200 Ã— 3.5/200 Ã— 2 Ã— x

As we know that the volume of water that flows through the pipe in x seconds = Volume of the tank

22/7 Ã— 3.5/200 Ã— 3.5/200 Ã— 2 Ã— x = 22/7 Ã— 0.7 Ã— 0.7 Ã— 2.1

So, x = 1680 seconds = 1680/60 minutes = 28 minutes

Hence, the tank is filled in 28 minutes.

### Question 27. A rectangular sheet of paper 30 cm Ã— 18 cm can be transformed into the curved surface of a right circular cylinder in two ways i.e. either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders thus formed.

Solution:

Given that,

The dimensions of the rectangular sheet of paper = 30 cm Ã— 18 cm

Let’s assume that V1 be the volume of the cylinder which is formed by rolling the sheet along its length.

So,

2Ï€r1 = 30

r1 = 15/Ï€

h1 = 18 cm

So, the volume is

V1 = Ï€ Ã— 15/Ï€ Ã— 15/Ï€ Ã— 18

V1 = 225/Ï€ Ã— 18 cm3

Let’s assume that V2 be the volume of the cylinder formed by rolling the sheet along its width.

So, 2Ï€r2 = 18

r2 = 9/Ï€

h2 = 30 cm

So, the volume is

V2 = Ï€ r22 h2

= Ï€ x (9/Ï€)2 x 30

V2 = 81 x 30 / Ï€

Now we find the ratio of volumes of two cylinders:

V1 / V2 = 225 x 18 / 81 x 30

V1 / V2 = 5/3

Hence, the ratio of the volumes of the two cylinders is 5 : 3.

### Question 28. How many litres of water flow out of a pipe having an area of cross-section of 5 cm2 in one minute, if the speed of water in the pipe is 30 cm/sec?

Solution:

Given that,

Area of cross-section of the pipe = 5 cm2

Speed of water = 30 cm/sec

So, the volume of water that flows through the pipe in one second

= 5 Ã— 30 = 150 cm3

Now we find the volume of water that flows through the pipe in one minute

= 150 Ã— 60 = 9000 cm3 = 9 liter

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

### Question 29. 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.

Solution:

Given that,

The total surface area of the cylinder = 1628 cm2

The sum of the radius of the base and height of a solid cylinder = 37 m

h + r = 37 cm …(1)

Find: the volume of the cylinder.

Now we find the radius and height of the cylinder

So, total surface area of the cylinder = 2Ï€r(h + r) = 1628

Therefore, 2Ï€r Ã— 37 = 1628

2 Ã— 22/7 Ã— r Ã— 37 = 1628

r = 7 cm

Put the value of r in eq(1)

h = 30 cm

Now lets calculate volume of the cylinder

Volume = Ï€r2h

Volume = 22/7 Ã— 7 Ã— 7 Ã— 30 = 4620 cm3

Hence, the volume of the given cylinder is 4620 cm3.

### Question 30. Find the cost of sinking a tube well 280 m deep, having diameter of 3 m at the rate of Rs 3.60 per cubic meter. Find also the cost of cementing its inner curved surface at Rs 2.50 per square meter.

Solution:

Given that,

Height of the tube well = 280 m,

Diameter the tube well = 3 m,

So, the radius of tube well = 3/2 m

Rate of sinking of the tube well = Rs. 3.60/m3,

Rate of cementing = Rs. 2.50/m2k

Now we find the volume of the tube well

V = Ï€r2h

= 22/7 Ã— 3/2 Ã— 3/2 Ã— 280 = 1980 m2

Cost of Sinking the tube well = Volume of the tube well Ã— Rate of sinking the tube well

= 1980 Ã— 3.60 = Rs 7128

Now we find the curved surface area

CSA = 2Ï€rh

= 2 Ã— 22/7 Ã— 3/2 Ã— 280 = 2640 m2

Cost of cementing = Curved Surface area Ã— Rate of cementing

= 2640 Ã— 2.50 = Rs 6600

Hence, the total cost of sinking the tube well is Rs.7128 and

the total cost of cementing its inner surface is Rs. 6600.

### Question 31. Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm.

Solution:

Given that,

Weight of copper wire = 13.2 kg = 13.2 Ã— 1000 gm = 13200 gm

Diameter of copper wire = 4 mm,

So, the radius of the wire = 2 mm = 210 cm

Density = 8.4 gm / cm3,

Find: the length of the copper wire.

So, Volume Ã— Density = Weight

Therefore, Ï€r2h Ã— 8.4 = 13.2

22/7 Ã— 2/10 Ã— 2/10 Ã— h Ã— 8 .4

h = 12500 cm = 125 m

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

### Question 32. A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.

Solution:

Given that,

Inner diameter of the well = 10 m,

So, the radius of well = 5 m

The height of the well = 8.4 m,

Width of embankment = 7.5 m,

Find: the height of the embankment.

So, the outer radius of the embankment,

R = Inner radius of the well + width of the embankment

= 5 + 7.5 = 12.5 m

Let’s assume that H be the height of the embankment,

Volume of embankment = Volume of earth dug out

Ï€(R2 âˆ’ r2)H = Ï€r2h

22/7 Ã— (12.52 âˆ’ 52)H = 22/7 Ã— 5 Ã— 5 Ã— 8.4

H = 1.6 m

Hence, height of the embankment is 1.6 m.

Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape, GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out - check it out now!

Previous
Next