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

Last Updated : 20 Sep, 2024
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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 on the 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 curved surface 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 over to 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 solid cylinder 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 height and 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 and its 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.

FAQs On Surface Area And Volume Of Right Circular Cylinder

What is the difference between curved surface area and total surface area of a cylinder?

The curved surface area includes only the lateral surface of the cylinder, while the total surface area includes the lateral surface plus the areas of both circular bases.

How does changing the radius affect the volume of a cylinder compared to changing its height?

The volume of a cylinder is proportional to the square of its radius but only directly proportional to its height. This means that small changes in radius have a more significant impact on volume than equal changes in height.

Can the formulas for cylinder measurements be applied to containers that aren’t perfectly cylindrical?

While the formulas are designed for perfect cylinders, they can often be used as approximations for near-cylindrical objects. However, for precise measurements of irregular shapes, more advanced techniques may be needed.

How are cylinder calculations relevant in real life?

Cylinder calculations are used in various fields, including engineering (designing tanks and pipes), manufacturing (determining material needs for cylindrical products), and everyday life (estimating storage capacity of containers).

What’s the relationship between the surface area and volume formulas for cylinders?

Both formulas involve the radius and height of the cylinder, but the surface area formula is linear with respect to these dimensions, while the volume formula is cubic. This relationship is crucial for understanding how changes in dimensions affect different aspects of the cylinder.



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