Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions -Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

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

Solution:

Given data is as follows :

Diameter of cylinder = 56 cm

Dimensions of rectangular block = 32 cm × 22 cm × 14 cm

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

First let us find the raise in the level of water in the cylinder . Diameter is given as 56 cm . Therefore ,

r = 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 h . Then we have ,

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

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

h = 4 cm

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


Q. 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 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 = π(R2−r2)h

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

r = 10.4/2

r = 5.2 cm

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 ‘R’ .

R = 5.2 + 0.8

R = 6 cm

Therefore , Volume of metal in the pipe Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= 704 cm3

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


Q. 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 data is as follows :

r = 0.75 cm

Water flow rate = 7 m/sec

Time = 1 hour

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

Let us first convert water flow 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 h = 7 m and r = 0.75 cm . Therefore ,

Volume of water delivered in 1 second  Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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

Therefore , we have

Volume of water delivered in 3600 seconds Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

We know that 1000 cm= 1 litre

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.


Q. 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 metre per second . In how much time the tank will be filled ?

Solution:

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 the pipe .

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

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Volume of the tank = πr2h

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Given is the diameter of the pipe which is 3.5 cm . Therefore , radius is  Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics Let us convert it into metres . It then becomes  Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Volume of water that flows through the pipe in 1 second  Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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

Volume of water that flows through the pipe in x seconds  Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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

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

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 MathematicsEx-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

x = 1680 seconds

x = 1680/60 minutes

x = 28 minutes

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


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

Solution:

Given data is as follows :

Dimensions of the rectangular sheet of paper = 30 cm × 18 cm

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 V1 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 cylinder . Therefore , we have ,

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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

h= 18 cm

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Let V2 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 ,

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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

h= 30 cm

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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


Q.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 data is as follows :

Area of cross section of the pipe = 5 cm2

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 = πr2h

Here , πr2 is nothing but the cross section of the pipe and h is 30 cm .

Therefore , we have ,

Volume of water that flows through the pipe in one second = 5×30=150cm3

Volume of water that flows through the pipe in one minute = 150×60=9000cm3

We know that 1000 cm3 = 1 litre . Therefore ,

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

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


Q.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 data is as follows :

h + r = 37 cm

Total surface area of the cylinder = 1628 cm2

That is ,

2πrh+2πr2 = 1628

2πr(h+2r) = 1628

But it is already given in the problem that ,

h + r = 37 cm

Therefore , 2πr×37 = 1628

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

r = 7 cm

Since , h + r = 37 cm

We have , h + 7 = 37 cm

H = 30 cm

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

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Volume = 4620 cm 3

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


Q.30. Find the cost of sinking a tube well 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 .

Solution:

Given data is as follows :

Height of the tube well = 280 m

Diameter = 3 m

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

Rate of cementing = Rs. 2.50/m2

Given is the diameter of the tub well which is 3 metres . Therefore

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Volume of the tube well = πr2h

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= 1980 m2

Cost of sinking the tube well = Volume of the tube well × Rate of sinking the tube well  = 1980 × 3.60

= Rs . 7128

Curved surface area = 2πrh

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= 2640 m2

Cost of cementing = Curved Surface area × Rate of cementing

= 2640 × 2.50

= 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 .


Q.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 data is as follows :

Weight of copper  wire = 13.2 kg

Diameter = 4 mm

Density = 8.4 gm /cm3

We have to find the length of the copper wire .

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

r = 2 mm

Let us convert r from millimeter to centimeter , since density is in terms of gm/cm3 . Therefore ,

r = 2/10 cm

Also , weight of the copper wire is given in kilograms . Let us convert into grams since density is in terms of gm/cm3 . Therefore , we have ,

Weight of copper wire = 13.2 × 1000 gm

= 13200 gm

We know that

Volume × Density = Weight

Therefore , πr2h×8.4 = 13.2

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

h = 12500 cm

h = 125 m

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


Q.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  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 ,

r = 5 m

The outer radius of the embankment,

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

= 5 + 7.5

= 12.5 m

Let H 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

Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

H = 1.6 m

Thus , height of the embankment is 1.6 m.

The document Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on Ex-19.2, (Part -3), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions - RD Sharma Solutions for Class 9 Mathematics

1. What is the formula for finding the surface area of a right circular cylinder?
Ans. The formula for finding the surface area of a right circular cylinder is given by 2πr(r + h), where r is the radius of the base and h is the height of the cylinder.
2. How do you find the volume of a right circular cylinder?
Ans. The volume of a right circular cylinder can be found using the formula V = πr²h, where r is the radius of the base and h is the height of the cylinder.
3. Can you explain the concept of lateral surface area in a right circular cylinder?
Ans. The lateral surface area of a right circular cylinder refers to the sum of the areas of the curved surface (excluding the top and bottom circles). It can be calculated using the formula LSA = 2πrh, where r is the radius of the base and h is the height of the cylinder.
4. How can we calculate the total surface area of a right circular cylinder?
Ans. The total surface area of a right circular cylinder includes both the curved surface area and the areas of the top and bottom circles. It can be calculated using the formula TSA = 2πr(r + h), where r is the radius of the base and h is the height of the cylinder.
5. Is there any relationship between the volume and surface area of a right circular cylinder?
Ans. Yes, there is a relationship between the volume and surface area of a right circular cylinder. If the height of the cylinder is kept constant, increasing the radius will result in an increase in both the volume and surface area. Similarly, if the radius is kept constant, increasing the height will also lead to an increase in both the volume and surface area.
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