Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions -Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q12. The curved surface area of a cylinder is 1320cm2 and its base had diameter 21 cm. Find the height and volume of the cylinder.

Solution:

Let, r be the radius of the cylinder

h be the height of the cylinder

⇒ 2r = 21cm

⇒ r  = 21/2

= 10.5cm

Given, Curved surface area(CSA) = 1320cm2

⇒ 2Πrh = 1320

⇒  Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

⇒ h = 1320/66

⇒ h = 20 cm

Volume of cylinder =  πr2*h

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= 22 * 1.5 * 10.5 * 20

= 6930cm2


Q13. The ratio between the radius of the base and the height of a cylinder is 2:3.Find the total surface area of the cylinder, if its volume is 1617cm2.

Solution:

Let, r be the radius of the cylinder

h be the height of the cylinder

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Volume of cylinder =  πr2*h

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

h = 10.5 cm

from, eq 1

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= 7 cm

Total surface area of cylinder = 2πr(h+r)

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics


Q14. A rectangular sheet of paper, 44 cm*20 cm, is rolled along its length of form cylinder. Find the volume of the cylinder so formed.

Solution:

Given, the dimensions of the sheet are 44cm*20cm

Here, length = 44 cm

Height = 20 cm

2πr = 44

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Volume of cylinder = r2*h

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= 154*20 = 3080cm3


Q15. The curved surface area of cylindrical pillar is 264m2 and its volume is 924m3. Find the diameter and the height of the pillar.

Solution:

Let, r be the radius of the cylindrical pillar

h be the height of the cylindrical pillar

CSA = 264m2

2πrh = 264m2 ——– 1

⇒ Volume of the cylinder = 924m2

Π*r2*h = 924

Πrh(r) = 924

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Substitute πrh in eq 1

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

substitute r value in eq 1

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

so, the diameter = 2r = 2(7) = 14 m and height = 6 m


Q16. Two circular cylinders of equal volumes have their heights in the ratio 1:2. Find the ratio of two radii.

Solution:

Let, r1,r2 be the radii of the cylinder

h1,h2 be the height of the cylinder

v1,v2 be the volume of the cylinder

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

⇒  Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

⇒  Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

⇒  Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Hence, the ratio of the radii are  Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics


Q17. 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:

Let, r be the radius of the right circular cylinder

h be the height of the right circular cylinder

h = 10.5 cm

⇒ 3(2πr2) = 2(2πrh)

⇒ 3r = 2h

⇒ Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

⇒  Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

⇒ r = 7cm

Volume of the cylinder = r2*h

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= 154*10.5

= 1617cm3


Q18. 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 perm2.

Solution:

Let, r be the radius

h be the height

here, h = 21m

2r = 6

⇒ r = 6/2

= 3 m

Volume of the cylinder = r2*h

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= 66*9

= 594cm3

Cost of plastering = 9.5 per m3

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


Q19. 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:

We know that, circumference = 2πr

⇒176 = 2πr

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics
⇒ r = 28 cm

Here, height(h) = 3m = 300cm

Volume of timber = r2*h

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= 44*8400 = 739200cm3 (or) 0.7392m3


Q20. 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:

Let, r be the radius of well

h be the height of well

here, h = 8m

2r = 14

⇒ r = 14/2

= 7m

Volume of well = r2*h

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= 22*56

= 1232m3

Let, re be the radius of embankment

hbe the height of embankment

Volume of well  = Volume of embankment

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics


Q21. 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, R be the outer radius

R be the inner radius

Here, h = 14cm

2πRh – 2πrh = 88

⇒ 2πh(R – r) = 88

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

⇒ (R – r) = 1cm    ——— 1

Volume of tube = π*R2*h – π*r2*h

176 = πh(R2 – r2)

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

⇒ (R2 – r2) = 4

⇒ (R + r)(R – r)  = 4

Here, (R –  r) = 1

⇒ (R + r)(1) = 4

⇒ (R + r) = 4 cm

⇒ R = 4 – r         ———— 2

Here , R – r = 1

⇒ R = 1 + r

Substitute R value in eq 2

⇒ 1 +  r = 4 – r

⇒  2r = 3

⇒ r = 32

= 1.5 cm

Substitute ‘r’ value in eq 1

⇒ R – 1.5 = 1

⇒ R = 1 + 1.5

⇒ R = 2.5 cm

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


Q.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 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 n = 6

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

R = 1 cm

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 ,

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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

Since , speed of water is in metres/second , let us convert 30 minutes into seconds . It will be 30×60

Volume of water that flows for 30 minutes  Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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

R = 60 cm

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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

Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

h = 3 m

Therefore , the height of the tank is 3 metres

The document Ex-19.2, (Part -2), 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 -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 RD Sharma Solutions - RD Sharma Solutions for Class 9 Mathematics

1. How do you find the surface area of a right circular cylinder?
Ans. To find the surface area of a right circular cylinder, you need to calculate the sum of the areas of its curved surface and its two circular bases. The formula for the surface area is given by: Surface Area = 2πrh + 2πr^2, 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: Volume = πr^2h, where r is the radius of the base and h is the height of the cylinder. This formula calculates the space occupied by the cylinder.
3. Can you explain the concept of a right circular cylinder?
Ans. A right circular cylinder is a three-dimensional geometric shape that consists of two parallel circular bases and a curved surface connecting the bases. The axis passing through the center of the bases is perpendicular to the bases, making the cylinder "right." The curved surface is a rectangle when unrolled.
4. What is the difference between the curved surface and the total surface area of a cylinder?
Ans. The curved surface area of a cylinder refers to the area of the lateral (curved) surface only. It does not include the areas of the circular bases. On the other hand, the total surface area of a cylinder includes the areas of both the curved surface and the two circular bases.
5. How can the surface area and volume of a cylinder be used in real-life situations?
Ans. The surface area and volume of a cylinder are used in various real-life situations. For example, in engineering, they are used in the design of pipes, tanks, and containers. In architecture, they are used to calculate materials needed for constructing cylindrical structures like pillars. Additionally, understanding these calculations can be useful in everyday situations such as estimating the capacity of cylindrical containers or calculating the amount of paint needed to cover a cylindrical surface.
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