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RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4 Volume and Surface Areas of Solids

Here, we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4. Students can view these RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4 before exams for better understanding.

Neha Tanna30 Jul, 2024

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4: The Entrancei academic team has produced a comprehensive answer for Chapter 19 Volume and Surface Area of Solids in the RS Aggarwal textbook for Class 10. Complete the NCERT exercise questions and utilise them as a guide. Solutions for Entrancei NCERT Class 10 Maths problems in the exercise require assistance to be completed. For maths in class 10, Entrancei published NCERT answers.

The RS Aggarwal class 10 solution for chapter 19 Volume and Surface Area of Solids Exercise-19D is uploaded for reference only. Before going through the solution of chapter-19 Volume and Surface Area of Solids Exercise-19D, one must have a clear understanding of the chapter-19 Volume and Surface Area of Solids. Read the theory of chapter-19 Volume and Surface Area of Solids and then try to solve all numerical of exercise-19D.

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4(D) Volume and Surface Areas of Solids Overview

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4 focuses on the volume and surface areas of solids. This chapter provides comprehensive problems that help students understand the calculations involved in determining the volume and surface area of various geometric shapes like spheres, cylinders, cones, and cuboids. The exercise emphasizes practical application, making it easier for students to grasp the concepts by solving real-world problems. The solutions provided in this exercise are step-by-step, ensuring that students can follow along and understand each calculation's rationale. This approach helps reinforce the learning process and builds a solid foundation for solving more complex problems in geometry. Overall, Exercise 19.4 is crucial for mastering the concepts of volume and surface areas of solids, which are fundamental in higher mathematics and practical applications.

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4 provides a comprehensive learning experience that equips students with the necessary knowledge and skills to tackle questions confidently. Here we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4 for the ease of students so that they can prepare better for their upcoming exams –

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4

Below we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4 for the ease of the students –

Question

The diameter of a copper sphere is $18 c m$ . The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is $108 m$ , find its diameter.

Solution

Let radius of wire be 'r',then diameter

= 2 r

. Now, volume of wire=volume of sphere

\Rightarrow π r^{2} h = \frac{4}{3} π R^{3} \Rightarrow r^{2} \times 10800 = \frac{4}{3} \times {(9)}^{3} \Rightarrow r^{2} = \frac{4 \times 9 \times 9 \times 9}{10800 \times 3} = 0.09

\Rightarrow r = \sqrt{0.09} = 0.3 ㎝

, therefore, diameter

= 2 \times 0.3 = 0.6 ㎝

Question

A spherical cannon ball, $28 cm$ in diameter is melted and cast into a right circular conical mould, the base of which is $35 cm$ in diameter. Find the height of the cone.

Solution

Radius of cannon ball

= 14 cm

Volume of cannon ball

= \frac{4 π r^{3}}{3}

Radius of conical mould

= \frac{35}{2} cm

Volume of conical mould

= \frac{1}{3} π r^{2} h

Since material of cannon ball is used to make right circular mould, Hence

⟹ \begin{matrix} V o l u m e of s p h e r e = v o l u m e o f c o n e \frac{4}{3} π r^{3} = \frac{1}{3} π r^{2} h 4 \times 14 \times 14 \times 14 = \frac{35}{2} \times \frac{35}{2} \times h h = \frac{4 \times 2 \times 2 \times 14 \times 14 \times 14}{35 \times 35} h = 35.84 cm \end{matrix}

Question

A hemispherical bowl of internal radius $9 c m$ is full of liquid. This liquid is to be filled into a cylindrical shaped small bottle each of diameter $3 c m$ and height $4 c m$ . How many bottles are necessary to empty the bowl.

Solution

Number of bottles = Volume of bowl / Volume of bottle = (23π) / (9.3π) = 1.5 / 0.24 = 54

Question

Find the volume and surface area of a sphere whose radius is:

$4.2 c m$

Solution

It is given that

Radius of the sphere

= 4.2 c m

We know that

Volume of the sphere

= \frac{4}{3} π r^{3}

By substituting the values

Volume of the sphere

= \frac{4}{3} \times \frac{22}{7} \times {4.2}^{3}

So we get

Volume of the sphere

= 310.464 c m^{3}

We know that

Surface area of the sphere

= 4 π r^{2}

By substituting the values

Surface area of the sphere

= 4 \times \frac{22}{7} \times {4.2}^{2}

So we get

Surface area of the sphere

= 221.76 c m^{2}

Question

Find the volume and surface area of a sphere whose radius is: $5 m$

Solution

It is given that

Radius of the sphere

= 5 c m

We know that

Volume of the sphere

= \frac{4}{3} π r^{3}

By substituting the values

Volume of the sphere

= \frac{4}{3} \times \frac{22}{7} \times 5^{3}

So we get

Volume of the sphere

= 523.81 m^{3}

We know that

Surface area of the sphere

= 4 π r^{2}

By substituting the values

Surface area of the sphere

= 4 \times \frac{22}{7} \times 5^{2}

So we get

Surface area of the sphere

= 314.28 m^{2}

Question

The volume of a sphere is $38808 c m^{3}$ . Find its radius and hence its surface area.

Solution

We know that

Volume of the sphere

= \frac{4}{3} π r^{3}

By substituting the values

38808 = \frac{4}{3} \times \frac{22}{7} \times r^{3}

On further calculation

r^{3} = \frac{(38808 \times 3 \times 7)}{88}

so we get

r^{3} = 9261

By taking cube root

r = 21 c m

We know that

Surface area of the sphere

= 4 π r^{2}

By substituting the values

Surface area of the sphere

= 4 \times \frac{22}{7} \times 21^{2}

So we get

Surface area of the sphere

= 5544 c m^{2}

Therefore, the radius of the sphere is

21 c m

and the surface area is

5544 c m^{2}

Question

From a solid cylinder whose height is 2.4 cm and diameter is 1.4 cm, a conical cavity of the

same height and same diameter is hollowed out. Find the total surface area of the

remaining solid to the nearest cm ² .

Solution

From the question, we know the following: The diameter of the cylinder = diameter of conical cavity = 1.4 cm So, the radius of the cylinder = radius of the conical cavity = 1.4/2 = 0.7 Also, the height of the cylinder = height of the conical cavity = 2.4 cm Ncert solutions class 10 chapter 13-11

Now, the TSA of the remaining solid = surface area of conical cavity + TSA of the cylinder = πrl+(2πrh+πr ² ) = πr(l+2h+r) = (22/7)× 0.7(2.5+4.8+0.7) = 2.2×8 = 17.6 cm ² So, the total surface area of the remaining solid is 17.6 cm ²

Question

Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Solution

For Sphere 1:

Radius (r ₁ ) = 6 cm ∴ Volume (V ₁ ) = (4/3)×π×r ₁ ³

For Sphere 2:

Radius (r ₂ ) = 8 cm ∴ Volume (V ₂ ) = (4/3)×π×r ₂ ³

For Sphere 3:

Radius (r ₃ ) = 10 cm ∴ Volume (V ₃ ) = (4/3)× π× r ₃ ³ Also, let the radius of the resulting sphere be “r” Now, The volume of the resulting sphere = V ₁ +V ₂ +V ₃ (4/3)×π×r ³ = (4/3)×π×r ₁ ³ +(4/3)×π×r ₂ ³ +(4/3)×π×r ₃ ³ r ³ = 6 ³ +8 ³ +10 ³ r ³ = 1728 r = 12 cm

Question

How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Solution

It is known that the coins are cylindrical in shape. So, height (h ₁ ) of the cylinder = 2 mm = 0.2 cm Radius (r) of circular end of coins = 1.75/2 = 0.875 cm Now, the number of coins to be melted to form the required cuboids be “n” So, Volume of n coins = Volume of cuboids n × π × r ² × h ₁ = l × b × h n×π×(0.875) ² ×0.2 = 5.5×10×3.5 Or, n = 400

Question

How many spheres $12 c m$ in diameter can be made from a metallic cylinder of diameter $8 c m$ and height $90 c m$ ?

Solution

It is given that

Diameter of the sphere

= 12 c m

Radius of the sphere

= \frac{12}{2} = 6 c m

We know that

Volume of the sphere

= \frac{4}{3} π r^{3}

By substituting the values

Volume of the sphere

= \frac{4}{3} \times \frac{22}{7} \times 6^{3}

So we get

Volume of the sphere

= 905.142 c m^{3}

It is given that

Diameter of the cylinder

= 8 c m

Radius of the cylinder

= \frac{8}{2} = 4 c m

Height of the cylinder

= 90 c m

We know that

Volume of the cylinder

= π r^{2} h

By substituting the values

Volume of the cylinder

= \frac{22}{7} \times 4^{2} \times 90

So we get

Volume of the cylinder

= 4525.714 c m^{3}

We know that

Number of spheres = Volume of cylinder / Volume of sphere

By substituting the values

Number of spheres

= \frac{4525.714}{905.142} = 5

Therefore,

5

spheres can be made from a metallic cylinder.

Question

The outer diameter of a spherical shell is $12 c m$ and its inner diameter is $8 c m$ . Find the volume of a metal contained in the shell. Also, find its outer surface area.

Solution

It is given that

Outer diameter of spherical shell

= 12 c m

Radius of spherical shell

= \frac{12}{2} = 6 c m

Inner diameter of spherical shell

= 8 c m

Radius of spherical shell

= \frac{8}{4} = 2 c m

We know that

Volume of outer shell

= \frac{4}{3} π r^{3}

By substituting the values

Volume of outer shell

= \frac{4}{3} \times \frac{22}{7} \times 6^{3}

So we get

Volume of outer shell

= 905.15 c m^{3}

Volume of inner shell

= \frac{4}{3} π r^{3}

By substituting the values

Volume of outer shell

= \frac{4}{3} \times \frac{22}{7} \times 4^{3}

So we get

Volume of outer shell

= 268.20 c m^{3}

So the volume of metal contained in the shell = Volume of outer shell - Volume of inner shell

By substituting the values

Volume of metal contained in the shell

= 905.15 - 268.20 = 636.95 c m^{3}

We know that

Outer surface area

= 4 π r^{2}

By substituting the values

Outer surface area

= 4 \times \frac{22}{7} \times 6^{2}

On further calculation

Outer surface area

= 452.57 c m^{2}

Therefore, the volume of metal contained in the shell is

636.95 c m^{3}

and the outer surface area is

452.57 c m^{2}

Question

Water in a canal, 6 m wide and 1.5 m deep, flows at a speed of 10 km/h. How much area will it irrigate in 30 minutes if 8 cm of standing water is needed?

Solution

It is given that the canal is the shape of a cuboid with dimensions as: Breadth (b) = 6 m and Height (h) = 1.5 m It is also given that The speed of canal = 10 km/hr Length of canal covered in 1 hour = 10 km Length of canal covered in 60 minutes = 10 km Length of canal covered in 1 min = (1/60)x10 km Length of canal covered in 30 min (l) = (30/60)x10 = 5km = 5000 m We know that the canal is cuboidal in shape. So, The volume of the canal = lxbxh = 5000x6x1.5 m ³ = 45000 m ³ Now, The volume of water in the canal = Volume of area irrigated = Area irrigated x Height So, Area irrigated = 56.25 hectares ∴ The volume of the canal = lxbxh 45000 = Area irrigatedx8 cm 45000 = Area irrigated x (8/100)m Or, Area irrigated = 562500 m ² = 56.25 hectares.

Question

The slant height of a frustum of a cone is 4 cm, and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the surface area of the frustum.

Solution

Given, Slant height (l) = 4 cm Circumference of upper circular end of the frustum = 18 cm ∴ 2πr ₁ = 18 Or, r ₁ = 9/π Similarly, the circumference of the lower end of the frustum = 6 cm ∴ 2πr ₂ = 6 Or, r ₂ = 3/π Now, the surface area of the frustum = π(r ₁ +r ₂ ) × l = π(9/π+3/π) × 4 = 12×4 = 48 cm ²

Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4

The RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4, focusing on Volume and Surface Areas of Solids, offer several benefits to students:

Comprehensive Understanding : The solutions provide detailed step-by-step explanations, helping students understand the principles behind calculating volume and surface areas of various solids such as spheres, cylinders, cones, and cuboids.

Problem-Solving Skills : By working through these problems, students enhance their problem-solving abilities, learning how to approach and solve complex geometry problems systematically.

Concept Clarity : The solutions clarify fundamental concepts by breaking down complicated problems into manageable steps, ensuring that students grasp the underlying mathematical principles.

Exam Preparation : Practicing these exercises prepares students for exams by familiarizing them with the types of questions that may appear, improving their speed and accuracy.

Confidence Building : Successfully solving these problems boosts students' confidence in their mathematical abilities, encouraging them to tackle more challenging problems.

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4 FAQs

What topics are covered in Exercise 19.4 of RS Aggarwal Class 10 Maths?

Exercise 19.4 covers problems related to the volume and surface areas of various geometric solids, including spheres, cylinders, cones, and cuboids.

How can RS Aggarwal Solutions help in understanding volume and surface areas of solids?

The solutions provide step-by-step explanations and detailed calculations for each problem, helping students understand the concepts and methodologies involved in finding the volume and surface area of different solids.

Are the RS Aggarwal Solutions useful for board exam preparation?

Yes, these solutions are highly useful for board exam preparation as they cover a variety of problems similar to those that appear in exams, helping students practice and master the necessary skills.

Is RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4 Volume and Surface Areas of Solids helpful?

Yes, RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.4 Volume and Surface Areas of Solids is very helpful.

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