RD Sharma Class 9 Solutions Chapter 19 VSAQS (Updated for 2021-22)

RD Sharma Class 9 Solutions Chapter 19 VSAQS are available here. These RD Sharma Class 9 Solutions Chapter 19 VSAQS are very beneficial for your class 9 exams. 

• RD Sharma Class 9 Maths Solutions

Download RD Sharma Class 9 Solutions Chapter 19 VSAQS

Access RD Sharma Class 9 Solutions Chapter 19 VSAQS

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

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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