RD‌ ‌Sharma‌ ‌Chapter‌ ‌19 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ 19.2 ‌Solutions‌

RD‌ ‌Sharma‌ ‌Chapter‌ ‌19 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ 19.2 ‌Solutions‌ focuses on the problems and exercises concerned with surface area and volume of the right circular cylinder. In this context, the solutions provided here are the best ways for those willing to prepare for the tests. Here the questions have been solved in a way that can help the students deal with the questions easily. This exercise is based on the Volume of a Cylinder and Volume of a Hollow Cylinder, which is concerned under the Surface Area And Volume of A Right Circular Cylinder.

Moreover, we have attached the PDF in the article, which helps understand the problems by solving various questions. RD‌ ‌Sharma‌ ‌Chapter‌ ‌19 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ 19.2 ‌Solutions‌ PDF is prepared by our experts by taking the help of RD Sharma, Previous Year’s Question Paper, and Text Book of Class 9.

Learn about RD Sharma Class 9 Chapter 19 (Surface Area & Volume of A Right Circular Cylinder)

Download RD‌ ‌Sharma‌ ‌Chapter‌ ‌19 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ 19.2 ‌Solutions‌ PDF

Solutions for Class 9 Maths Chapter 19 Surface Area and Volume of A Right Circular Cylinder Exercise 19.2

Important Definitions RD‌ ‌Sharma‌ ‌Chapter‌ ‌19 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ 19.2 ‌Solutions‌

Chapter 19 (Surface Area & Volume of A Right Circular Cylinder) is based on the following topics. Also, we have given the definitions below with the stepwise solved examples.

1. Volume of a Cylinder
2. Volume of a Hollow Cylinder

Volume of a Cylinder

The cylinder volume can be calculated through the formula πr2h, in which ‘r’ is the radius of the circular base and ‘h’ is the height of the cylinder.

Volume of a Hollow Cylinder

When it comes to Hollow Cylinder, both interior and exterior circle radius is taken into account. If r and r1 are the radii of external and internal radius respectively, and he is the height, then the volume can be calculated through formula V = πh(r2 – r12)

Examples of RD‌ ‌Sharma‌ ‌Chapter‌ ‌19 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ 19.2 ‌Solutions‌

Ques- A soft drink is available in two packs-

(a) A tin can with a rectangular base of length 5cm and width 4cm, having a height of 15cm

(b) A plastic cylinder with a circular base of diameter 7cm and height 10cm.

Which container has greater capacity, and by how much?

Solution-

(a) Dimensions of a cubical tin can:

Length (L) = 5 cm

Height (H) = 15 cm

Capacity of tin can = Volume of Tin Can = L x B x H cubic units = (5 x 4 x 15) cm3 = 300 cm3

(b) Radius of the circular end of the plastic cylinder (r) = diameter/2 = 7/2 cm = 3.5cm

Height of plastic cylinder (h) = 10cm

Capacity of plastic cylinder = Volume of cylindrical container = πr2h = 3.14 × (3.5)2 × 10 cm3 = 385 cm3

From (a) and (b) results, the plastic cylinder has greater capacity.

Difference in capacity = (385 – 300) cm3 = 85cm3

Ques: The pillars of a temple are cylindrically shaped. Suppose each pillar has a circular base of radius 20 cm and height of 10 m. How much would concrete mixture be required to build 14 such pillars?

Solution:

We have to find the volume of the cylinders.

Radius of the cylinder, r = 20 cm

Height of the cylinder, H = 10 m = 1000cm

Volume of the cylindrical pillar = πr2H

= (3.14× 202× 1000) cm3

= 8800000/7 cm3 or 8.87m3

Therefore, volume of 14 pillars = 14 x 8.87 m3 = 17.6m3.

Ques: The inner diameter of a cylindrical wooden pipe is 24cm, and its outer diameter is 28cm. The length of the pipe is 35cm. Find the mass of the pipe if 1cm3 of wood has a mass of 0.6gm.

Solution-

Let ‘r’ and ‘R’ be the inner and outer radii of the pipe.

Inner radius of the pipe (r) = 24/2 = 12cm

Outer radius of the pipe (R) = 24/2 = 14cm

Height of the pipe (h) = length of the pipe = 35cm

Mass of pipe = volume x density = π x (R2 – r2) x h

= 22 / 7 (142 – 122) x 35

= 5720

Mass of the pipe = 5720cm3

Mass of 1cm3 wood is 0.6gm

Therefore, mass of 5720cm3 wood = 5720 x 0.6 = 3432gm = 3.432kg.