RD‌ ‌Sharma‌ ‌Chapter‌ ‌21 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ ‌21.2 ‌Solutions‌

RD‌ ‌Sharma‌ ‌Chapter‌ ‌21 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ ‌21.2 ‌Solutions‌ has been provided here. These questions have been prepared with thorough answers by the experts, thus assuring complete accuracy. Basically, the exercises cover the topics of the volume of a sphere. It also covers the topics around the hemisphere and also is based on spherical shells. These solutions are useful for preparing the examinations and enriching their fundamentals. Specifically, the stepwise and detailed analysis of the questions can help understand the underneath concepts.

The surface area and volume of a Sphere Exercise 21.2 do revolve around the following topics like- Volume of a sphere, Volume of a hemisphere, and Volume of a spherical shell.

Learn about RD Sharma Class 9 Chapter 21 (Surface Area And Volume of Sphere)

Download RD‌ ‌Sharma‌ ‌Chapter‌ ‌21 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ ‌21.2 ‌Solutions‌ PDF

Solutions for Class 9 Maths Chapter 21 Surface Area and Volume of A Sphere Exercise 21.2

Important Definitions RD‌ ‌Sharma‌ ‌Chapter‌ ‌21 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ ‌21.2 ‌Solutions‌

This exercise of Chapter 21- Surface Area and Volume of Sphere is based on the following topics. Also, know the definitions and formulas with examples below.

1. Volume of a sphere
2. Volume of a hemisphere and
3. Volume of a spherical shell

Volume of a sphere

The volume of a sphere is calculated through the formula V=4/3 πr3, where ‘r’ is the radius of the sphere.

Volume of a hemisphere

The volume of a hemisphere is calculated through formula (2/3) πr3 cubic units. Here π is a constant which is equal to 3.14 approximately, and “r” is the radius of the hemisphere.

Volume of a spherical shell

The volume of a Sphere is V = 4/3πr³. When it comes to a spherical shell, the formula becomes V = 4/3 • π • (r³ – (r-t)³), where ‘t’ is the difference between the radius of both.

Examples of RD‌ ‌Sharma‌ ‌Chapter‌ ‌21 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ ‌21.2 ‌Solutions‌

Ques: Find the volume of a sphere whose radius is:

(a) 2 cm (b) 3.5 cm (c) 10.5 cm.

Solution:

The volume of the sphere = 4/ 3πr3 Cubic Units

here, r = radius of a sphere

Volume = 4/ 3 × 22/ 7 × (2) x 3

= 33.52

Volume = 33.52 cm3

hence volume = 4/ 3 × 22/ 7 × (3.5) x 3

= 179.666

Volume = 179.666 cm3

Volume = 4 / 3 × 22/ 7 × (10.5) x 3

= 4851

Volume = 4851 cm3

Ques: Find the volume of a sphere whose diameter is:

(a) 14 cm (b) 3.5 dm (c) 2.1 m

Solution:

Volume of the sphere = 4/ 3πr3 Cubic Units

here, r = radius of a sphere

(a) diameter = 14 cm

So, radius = diameter / 2 = 14/ 2 = 7cm

Volume = 4/ 3× 22/ 7 × (7) x 3

= 1437.33

Volume = 1437.33 cm3

(b) diameter = 3.5 dm

As, radius = diameter / 2 = 3.5/ 2 = 1.75 dm

Volume = 4/ 3 × 22/ 7 × (1.75) x 3

= 22.46

Volume = 22.46 dm3

(c) diameter = 2.1 m

As, radius = diameter/ 2 = 2.1/ 2 = 1.05 m

Volume = 4/ 3× 22/ 7 × (1.05) x 3

= 4.851

Volume = 4.851 m3

Ques: A hemispherical tank has an inner radius of 2.8 m. Find its capacity in liters.

Solution:

The radius of the hemispherical tank = 2.8 m

Capacity of the hemispherical tank = 2/ 3 πr3

=2/ 3 × 22/ 7 × (2.8) x 3 m3

= 45.997 m3

[As 1m3 = 1000 liters]

Hence, capacity in liters = 45997 liters

Ques: How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?

Solution:

Edge of the cube = 22 cm

Diameter of the bullet = 2 cm

radius of the bullet (r) = 1 cm

Volume of cube = (side)3 = (22)3 cm3 = 10648 cm3

And,

Volume of each bullet which will be spherical in shape = 4/ 3 πr3

= 4/ 3 ×22/ 7 ×(1) x 3 cm3

= 4/ 3 ×22/ 7 cm3

= 88/ 21 cm3

Number of bullets = (Volume of the cube)/ (Volume of the bullet)

= 10648/ 88/ 21

= 2541

Therefore, 2541 bullets can be made.