Class 8 RD Sharma Solutions - Chapter 22 Mensuration III (Surface Area And Volume Of Right Circular Cylinder) - Exercise 22.1 | Set 2

Chapter 22 of RD Sharma's Class 8 Mathematics textbook delves into the fascinating world of mensuration, specifically focusing on the surface area and volume of right circular cylinders. This chapter builds upon students' previous knowledge of geometry and introduces them to more complex three-dimensional shapes.

The chapter covers various aspects of cylinders, including their lateral surface area, total surface area, and volume, providing students with a comprehensive understanding of how these measurements are interrelated and how they can be applied to real-world problems.

Question 11: Find the cost of plastering the inner surface of a well at Rs. 9.50 m². If it is 21 m deep and diameter of its top is 6 m.

Solution: 

The details given about well are - 

Height of well = 21 m

Diameter of well = 6 m

So radius of well = 6/2 = 3 m

Curved surface area of cylinder = 2 * (22 /7) * r * h

                                                   = 2 * (22/7) * 3 * 21

                                                   = 396 m²

Cost of plastering the inner surface of a well = 396 * 9.50

                                                                        = Rs 3762

Question 12: A cylindrical vessel open at the top has diameter 20 cm and height 14 cm. Find the cost of tin - plating it onthe inside at the rate of 50 paise per hundred square centimeters.

Solution: 

The details given about cylinder are -

Diameter of cylinder = 20 cm

So, radius = 20/2 = 10 cm

Height of cylinder = 14 cm

Total surface area of cylinder = 2 * (22/7) * r * h + (22/7) * r²

                                               = 2 * (22/7) * 10 * 14 + (22/7) * 14 * 14

                                               = 880 + 2200/7

                                               = ((880 * 7) + 2200)/7

                                               = (6160 + 2200)/7

                                               = 8360/7 cm²

It is given that code per 100 cm² = 50 paise

So, cost per 1 cm² = Rs 0.005

Cost of tin-plating the area inside the vessel = (8360/7) * 0.005

                                                                        = Rs 5.97

Question 13: The inner diameter of circular well is 3.5 m. It is 10 m deep. Find the cost of plastering its inner curvedsurface at Rs 4 square meter.

Solution:

The details given about well are -

Inner diameter of circular well = 3.5 m

So radius = 3.5/2 m

Height of well = 10 m

Curved surface area of well = 2 * (22/7) * r * h

                                            = 2 * (22/7) * (3.5/2) * 10

                                            = 110 m²

Cost of painting 1 m² = Rs. 4

Cost of painting 110 m² = 4 * 110 

                                       = Rs 440

Question 14: The diameter of roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions moving once overto level a playground. What is the area of the playground?

Solution: 

The details given about roller are -

Diameter of roller = 84 cm

So radius = 84/2 = 42 cm

Length of roller = 120 cm

Curved surface area of roller = 2 * (22/7) * r * h

                                              = 2 * (22/7) * 42 * 120

                                              = 31680 cm²

Area of playground = Number of revolution * Curved surface area of roller

                                = 500 * 31680

                                = 15840000 cm²

= 1584 m² (1 m = 100 cm)

Question 15: Twenty-one cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50m high and height is 4 m, what will be the cost of cleaning them at the rate of Rs 2. 50 per square meter?

Solution: 

The details given about pillar are -

Diameter of pillar = 0.50 m

So radius = 0.50/2 = 0.25 m

Height of roller = 4 m

Curved surface area of pillar = 2 * (22/7) * r * h

                                              = 2 * (22/7) * 0.25 * 4

                                              = 44/7 m²

Curved surface area of pillar = (21 * 44)/7

                                              = 132 m²

Cost of cleaning the pillars = 2.50 * 132 

                                            = Rs 330

Question 16: The total surface area of a hollow cylinder which is open from both sides if 4620 sq cm, area of base ring is 115.5 sq cm and height 7 cm. Find the thickness of the cylinder. 

Solution: 

The details given about cylinder are -

Total surface area of hollow cylinder = 4620 cm²

Area of base ring = 115.5 cm²

Height of cylinder = 7 cm    

Let inner radius = r and outer radius = R

Area of hollow cylinder = 2 * (22/7) * (R² - r²) + 2 * (22/7) * R * h + 2 * (22/7) * r * h

                                      = 2 * (22/7) * (R + r) (R - r) + 2 * (22/7) * r * h (R + r)

                                      = 2 * (22/7) * (R + r) * (h + R - r)

Area of base = (22/7) * R * R - (22/7) * r * r 

                     = (22/7) (R + r) * (R - r)

Total Surface area/Area of base = 4620/115.5

(2 * (22/7) (R + r) * (h + R - r))/((22/7) (R + r) (R - r)) = 4620/115.5

2 * (h + R - r)/(R - r) = 4620/115.5

R - r = thickness of cylinder

Let thickness of cylinder = t

2 * (h + t)/t = 4620/115.5

2 * h + 2 * t = 40 * t

2h = 40t - 2t

2 * 7 = 38t

14/38 = t

t = 7/19 cm

Question 17: The sum of the radius of the base and height of a solid cylinder is 37 cm. If the total surface area of the solidcylinder is 1628 cm², find the circumference of its base.

Solution: 

The details about the cylinder are -

Sum of radius of base and height of cylinder = 37 cm

Total surface area pf cylinder = 1628 cm²

Total surface area of cylinder = 2 * (22/7) * r * (h + r)

                                      1628 = 2 * (22/7) * r * 37   

                                     (1628 * 7)/(2 * 22 * 37) = r

                                       11396/1628 = r

                                        r = 7 m

Circumference of base = 2 * (22/7) * r

                                     = 44 m

Question 18: Find the ratio between the total surface area of a cylinder to its curved surface area, given that its heightand radius are 7.5 cm and 3.5 cm.

Solution: 

The details given about the cylinder are -

Height of cylinder = 7.5 cm

Radius of cylinder = 3.5 cm

Total surface area of cylinder/Curved surface area of cylinder = (2 * (22/7) * r * (h + r))/(2 * (22/7) * r * h)

                                                                                                  = (h + r)/h

                                                                                                  = (7.5 + 3.5)/7.5

                                                                                                  = 11/7.5

                                                                                                  = 22/15

Question 19: A cylindrical vessel, without lid, has to be tin - coated on its both sides. If the radius of the base is 70 cm andits height is 1.4 m, calculate the cost of tin coating at the rate of Rs 3.50 per 1000 cm².

Solution: 

The details given about cylinder are -

Radius of base = 70 cm

Height of cylinder = 1.4 m = 140 cm (1 m = 100 cm)

Total surface area of vessel = Area of outer side of base + Area of inner and outer curved surface

                                            = 2 ((22/7) * r * r + 2 * (22/7) * r * h)

                                            = 2 * (22/7) * r * (r + 2 * h)

                                            = 2 * (22/7) * 70 * (70 + 280)

                                            = 2 * (22/7) * 70 * 350

                                            = 154000 cm²

Cost of tin coating at the rate of Rs 3.50 per 1000 cm² = (3.50 * 154000)/1000

                                                                                       = Rs 539.

Summary

This chapter equips students with essential skills for calculating various measurements of right circular cylinders, a fundamental shape in geometry and real-world applications. Through a series of carefully crafted problems and exercises, students learn to apply formulas for curved surface area, total surface area, and volume of cylinders. The chapter emphasizes the relationships between these measurements and how changing one dimension affects the others. By working through diverse problem types, including those involving hollow cylinders and practical scenarios, students develop their analytical and problem-solving abilities. This knowledge forms a crucial foundation for more advanced topics in mathematics and science, preparing students for future studies in fields such as engineering, architecture, and physics.