RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS

In this chapter, we provide RD Sharma Solutions for Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS Maths pdf, free RD Sharma Solutions for Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 9
SubjectMaths
ChapterChapter 19
Chapter NameSurface Areas and Volume
of a Circular Cylinder
ExerciseVSAQS

RD Sharma Solutions for Class 9 Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS Download PDF

Question 1.
Write the number of surface of a right circular cylinder.
Solution:
Three, two circular and one curved.

Question 2.
Write the ratio of total surface area to the curved surface area of a cylinder of radius r and height h.
Solution:
∵ Radius = r
and height = h
∴ Curved surface area = 2πrh
and total surface area = 2πr(h + r)
∴ Ratio = 2πr(h + r) : 2πrh
= h + r : h

Question 3.
The ratio between the radius of the base and height of a cylinder is 2 : 3. If its volume is 1617 cm3, find the total surface area of the cylinder.
Solution:
Ratio in radius and height of the cylinder = 2 : 3
Let radius (r) = 2x
Then height (h) = 3x
∴ Volume = πr2h
Class 9 RD Sharma Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS
Class 9 Maths Chapter 19 Surface Areas and Volume of a Circular Cylinder RD Sharma Solutions VSAQS

Question 4.
If the radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3, then find the ratio of their volumes.
Solution:
Ratio of radii of two cylinder = 2:3
Let radius of first cylinder (r1) = 2x
and second cylinder (r2) = 3x
and ratio in their heights = 5:3
Let height of first cylinder (h1) = 5y
and height of second (h2) = 3y
∴ Volume of the first cylinder =πr2h
= π x (2x)2 x 5y = 20πx2y
and volume of second cylinder = π(3x)2 x 3y = 27πx2y
Now ratio between then,
= 20πx2y: 21πx2y
= 20 : 27

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