Competitive Exams

IIT JEE, NEET, ESE, GATE, AE/JE, Olympiad

Only IAS

UPSC, State PSC

School Preparation

Foundation (Class 6-12), Commerce, Arts, CuriousJr (1st - 8th), Science, International Boards

Govt Exam

Judiciary, SSC, Defence, Teaching, JAIIB & CAIIB, BIHAR EXAMS WALLAH, UP Exams, Railway, Nursing Exams, Banking, WB Exams

UG & PG Entrance Exams

MBA, IPMAT, IIT JAM, LAW, CUET UG, UGC NET, GMAT, Design & Architecture, Pharma, CSIR NET, CUET PG, NEET PG

FINANCE

CA, CS, Finance Courses, ACCA

Others

Online Degrees

English Proficiency Test

IELTS, TOEFL

Agriculture

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.3 Volume and Surface Areas of Solids

Here, we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.3. Students can view these RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.3 before exams for better understanding.

Neha Tanna26 Jul, 2024

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.3: 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 utilize them as a guide. Solutions for Entrance NCERT Class 10 Maths problems in the exercise require assistance to be completed. For maths in class 10, Entrance published NCERT answers.

The RS Aggarwal class 10 solution for Chapter 19 Volume and Surface Area of Solids Exercise-19C is uploaded for reference only; do not copy the solutions. Before going through the solution of Chapter 19 Volume and Surface Area of Solids Exercise-19C, 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-19C.

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.3 Overview

Chapter 19, Exercise 19.3 of RS Aggarwal's Class 10 Maths delves into the fundamental concepts of volume and surface areas of solids. This chapter aims to equip students with the necessary skills to calculate these properties for different geometric shapes such as cubes, cuboids, cylinders, cones, and spheres.

The exercises in this chapter are structured to provide step-by-step guidance on applying specific formulas for each type of solid.

By practicing these exercises, students not only understand the theoretical aspects but also gain proficiency in solving practical problems related to these solids. The problems vary in complexity, allowing students to gradually build their understanding and problem-solving skills.

Overall, Chapter 19.3 serves as a foundational module in geometry, preparing students to tackle more advanced topics while also providing practical knowledge applicable to real-life scenarios involving spatial measurements and calculations.

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.3 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.3 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 19.3 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 19.3 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.3

Below we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.3 for the ease of the students –

Question

The radii of the circular ends of a solid frustum of a cone are 18 cm and 12 cm and its height is 8 cm. Find its total surface area. $[U s e π = 3.14.]$

Solution

V = \frac{1}{3} π \times h \times (R^{2} + r^{2} + R \times r)

where: R = larger radius r = smaller radius h = height V = volume V =

\frac{1 / 3}{π} \times 8 [18^{2} + 12^{2} + 18 (12)] V = 5730.265 c m^{3}

Total surface area =

π [(R + r) s + (R^{2} + r^{2})

where: S = slant height R - r = 18 - 12 = 6 cm Pythagorean theorem:

S^{2} = (R - r)^{2} + h^{2}

S = \sqrt{6^{2} + 8^{2}}

S = 10 cm Total surface area =

π \times [(18 + 12) (10) + 18^{2} + 12^{2}]

Total surface area = 2412.743

c m^{2}

Question

A metallic bucket, open at the top, of height 24 cm is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are 7 cm and 14 cm respectively. Find

(i) the volume of water which can completely fill the bucket;

(ii) the area of the metal sheet used to make the bucket.

Solution

(i). volume of frustum of cone

= \frac{1}{3} \times π \times h \times (R^{2} + R r + r^{2}) = \frac{1}{3} \times \frac{22}{7} \times 24 \times (14^{2} + 14 \times 7 + 7^{2}) = 22 \times \frac{8}{7} \times (196 + 98 + 49) = 22 \times \frac{8}{7} \times 343 = 22 \times 8 \times 49 = 8624 c m^{3}

(ii) slant height of frustum,

l = \sqrt{(R - r)^{2} + h^{2}} = \sqrt{(14 - 7)^{2} + 24^{2}} = \sqrt{7^{2} + 24^{2}} = \sqrt{49 + 576} = \sqrt{625} = 25 c m

area of metallic sheet = CSA of frustum + area of base

= π \times l \times (R + r) + π \times r^{2} = \frac{22}{7} \times 25 \times (14 + 7) + \frac{22}{7} \times 7^{2} = \frac{550}{7} \times 21 + 22 \times 7 = 550 \times 3 + 154 = 1650 + 154 = 1804 c m^{2}

Question

A container, open at the top, is in the form of a frustum of a cone of height 24 cm with radii of its lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container at the rate of Rs 21 per litre.

Solution

Sol : h = 24 cm lower end = r = 8 cm upper end = R = 20 cm V =

\frac{π \times h}{3} \times (R^{2} + R \times r + r^{2})

V =

π \times \frac{24}{3} \times (20^{2} + 20 \times 8 + 8^{2})

V = 176 / 7 x 624 V = 176 x 89.142

c m^{3}

V = 15688.992

c m^{3}

V =

\frac{15688.992 c m^{3}}{1000}

V = 15.688992 litre Cost of milk per litre fill the container = 21 Rs ∴ Total cost of milk fill the container = 15.688992 x21 = 329.4688rs

Question

A container made of a metal sheet open at the top is of the form of frustum of cone, whose height is 16 cm and the radii of its lower and upper circular edges are 8 cm and 20 cm respectively. Find

(i) the cost of metal sheet used to make the container if it costs Rs 10 per 100 $c m^{2}$

(ii) the cost of milk at the rate of Rs 35 per litre which can fill it completely.

Solution

(i)

Radius (

r_{1}

) of upper end of container = 20 cm Radius (

r_{2}

) of lower end of container = 8 cm Height (h) of container = 16 cm Slant height (l) of frustum,

l = \sqrt{(r_{1} - r_{2})^{2} + h^{2}} = \sqrt{(20 - 8)^{2} + 16^{2}} = \sqrt{12^{2} + 16^{2}} = \sqrt{144 + 256} = \sqrt{400} = 20 c m

Capacity of container = Volume of frustum

= \frac{1}{3} π h (r_{1}^{2} + r_{2}^{2} + r_{1} r_{2}) = \frac{1}{3} \times 3.14 \times 16 \times (20^{2} + 8^{2} + 20 \times 8) = \frac{1}{3} \times 3.14 \times 16 \times (400 + 64 + 160) = \frac{1}{3} \times 3.14 \times 16 \times 624 = 10449.92 c m^{3} = 10.45 l i t r e s

(ii) Cost of 1 litre milk = Rs 35 Cost of 10.45 litre milk = 10.45

\times

35 = Rs 365.75 (i) Area of metal sheet used to make the container

= π (r_{1} + r_{2}) l + π r_{2}^{2} = π (20 + 8) l + π 8^{2} = 560 π + 64 π = 624 π c m^{2}

Cost of 100

c m^{2}

metal sheet = Rs 10 cost of

624 π c m^{2}

metal sheet =

\frac{624 \times 3.14 \times 10}{100} = 195.94

Therefore, the cost of the milk which can completely fill the container is Rs 365.75 and the cost of metal sheet used to make the container is Rs 195.94.

Question

The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm, and its slant height is 10 cm. Find its capacity and total surface area. $[T a k e π = 22 / 7.]$

Solution

Greater radius = R = 33 cm Smaller radius = r = 27 cm Slant height = l = 10 cm Using the formula for the height of a frustum: Height

= h = \sqrt{l^{2} - (R - r)^{2}} = \sqrt{10^{2} - (33 - 27)^{2}} = \sqrt{10^{2} - (6)^{2}} = \sqrt{100 - 36} = \sqrt{64} = 8 c m

Capacity of the frustum

= \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 8 (33^{2} + 27^{2} + 33 \times 27) = \frac{22 \times 8}{3 \times 7} \times 2709 = 22704 c m^{3}

Surface area of the frustum

= π R^{2} + π r^{2} + π l (R + r) = π [R^{2} + r^{2} + l (R + r)] = \frac{22}{7} [33^{2} + 27^{2} + 10 (33 + 27)] = \frac{22}{7} [1089 + 729 + 10 (60)] = \frac{22 \times 2418}{7} = 7599.43 c m^{2}

Question

A buket is in the form of a frustum of a cone. Its depth is 15 cm and the diameters of the top and the bottom are 56 cm and 42 cm respectively. Find how many litres of waer the bucket can hold. $[T a k e π = 22 / 7.]$

Solution

Greater diameter of the frustum = 56 cm Greater radius of the frustum = R = 28 cm Smaller diameter of the frustum = 42 cm Radius of the smaller end of the frustum = r = 21 cm Height of the frustum = h = 15 cm Capacity of the frustum

= \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 15 (28^{2} + 21^{2} + 28 \times 21) = \frac{22 \times 5}{7} \times 1813 = 28490 c m^{3} = 28.49 l i t r e s

Question

A bucket made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of the bucket if the cost of metal sheet used is Rs 15 per 100 $c m^{2}$ . $[U s e π = 3.14.]$

Solution

Height of the bucket, h = 16 cm (Given) Let the radii of the upper and lower ends of the bucket be

r_{1}

and

r_{2}

respectively. Given,

r_{1}

= 20 cm and

r_{2}

= 8 cm. Slant height of bucket , l

= \sqrt{h^{2} + (r_{1} - r_{2})^{2}} = \sqrt{16^{2} + (20 - 8)^{2}} = \sqrt{16^{2} + 12^{2}} = \sqrt{256 + 144} = = \sqrt{400} = 20 c m

Surface are of bucket = Curved surface area of the bucket + Area of base of the bucket

= π (r_{1} + r_{2}) l + π r 2^{2} = π [(r_{1} + r_{2}) l + r 2^{2}] = \frac{22}{7} [(20 c m + 8 c m) 20 c m + (20 c m)^{2}] = \frac{22}{7} (28 \times 20 c m^{2} + 400 c m^{2}) = \frac{22}{7} (560 c m^{2} + 400 c m^{2}) = \frac{22}{7} \times 960 c m^{2} = 3017. 14 c m^{2}

(Approx) Given, cost of 100

c m^{2}

metal sheet = Rs 15 Cost of the bucket

= \frac{15}{100} \times 3017.14 = 452.60

(Approx) Thus, the cost of the bucket is approximately Rs 452.60.

Question

A bucket made up of a metal sheet is in the form of frustum of a cone. Its depth is 24 cm and the diameters of the top and bottom are 30 cm and 10 cm respectively. Find the cost of milk which can completely fill the bucket at the rate of Rs 20 per litre and the cost of metal sheet used if it costs Rs 10 per 100 $c m^{2}$ $[U s e π = 3.14.]$

Solution

Greater diameter of the bucket = 30 cm Radius of the bigger end of the bucket = R = 15 cm Diameter of the smaller end of the bucket = 10 cm Radius of the smaller end of the bucket = r = 5 cm Height of the bucket = 24 cm Slant height, l

= \sqrt{h^{2} + (R - r)^{2}} = \sqrt{24^{2} + (15 - 5)^{2}} = \sqrt{24^{2} + (10)^{2}} = \sqrt{576 + 100} = \sqrt{676} = 26

cm Capacity of the frustum

= \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times 3.14 \times 24 (15^{2} + 5^{2} + 15 \times 5) = 3.14 \times 8 \times 325 = 8164 c m^{3} = 8.164

litres A litre of milk cost Rs. 20. So, total cost of filling the bucket with milk

= 8.164 \times 20 = R s . 163.28

Surface area of the bucket

= π r^{2} + π l (R + r) = π [5^{2} + 26 (15 + 5)] = 3.14 [25 + 26 (20)] = 3.14 [25 + 520] = 1711.3 c m^{2}

Cost of 100

c m^{2}

of the metal sheet is Rs. 10. So, cost of metal used for making the bucket

= 1711.3100 \times 10 = R s . 171.13

Question

The volume of a wall, 5 times as high as it is broad and 8 times as long as it is high, is 12.8 $m^{3}$ . The breadth of the wall is

(a) 30 cm (b) 40 cm (c) 22.5 cm (d) 25 cm

Solution

Let the breadth of the wall be x metres. Then, Height = 5x metres and Length = 40x metres. x * 5x * 40x = 12.8

\to x^{3} = \frac{12.8}{200}

$ x= 0.4m = 40cm $ (b) 40 cm

Question

The area of the base of a rectangular tank is 6500 $c m^{2}$ and the volume of water contained in it is 2.6 $m^{3}$ . The depth of water in the tank is

(a) 3.5 m (b) 4 m (c) 5 m (d) 8 m

Solution

Given l x b = 6500 $c m^{2} = 0.65 c m^{2}$

l x b x h = 2.6 $m^{3}$

0.65 x h = 2.6 h = $fraction numerator 2.6 over denominator 0.65 end fraction$

h = 4m

Question

How many bricks each measuring $(25 c m \times 11.25 c m \times 6 c m)$ will be required to construct a wall $(8 m \times 6 m \times 22.5 c m) ?$

(a) 8000 (b) 6400 (c) 4800 (d) 7200

Solution

Number of bricks = $fraction numerator v o l u m e space o f space w a l l over denominator v o l u m e space o f space 1 space b r i c k end fraction$ = (800 x 600 x 22.5)× $fraction numerator 1 over denominator 25 cross times 11.25 cross times 6 end fraction$ = 6400

Question.

A cylindrical bucket, 32 cm high and with a radius of a base of 18 cm, is filled with sand. This bucket is emptied on the ground, and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Solution:

The diagram will be as- Ncert solutions class 10 chapter 13-20

Given, Height (h ₁ ) of cylindrical part of the bucket = 32 cm Radius (r ₁ ) of circular end of the bucket = 18 cm Height of the conical heap ((h ₂ ) = 24 cm Now, let “r ₂ ” be the radius of the circular end of the conical heap. We know that volume of the sand in the cylindrical bucket will be equal to the volume of sand in the conical heap. ∴ The volume of sand in the cylindrical bucket = Volume of sand in the conical heap π×r ₁ ² ×h ₁ = (⅓)×π×r ₂ ² ×h ₂ π×18 ² ×32 = (⅓)×π ×r ₂ ² ×24 Or, r ₂ = 36 cm And, Slant height (l) = √(36 ² +24 ² ) = 12√13 cm.

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.

A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is drawn into a wire of diameter 1/16 cm, find the length of the wire.

Solution:

The diagram will be as follows NCERT Solutions Class 10 chapter 13-27

Consider AEG Ncert solutions class 10 chapter 13-29

Radius (r ₁ ) of upper end of frustum = (10√3)/3 cm Radius (r ₂ ) of lower end of container = (20√3)/3 cm Height (r ₃ ) of container = 10 cm Now, Volume of the frustum = (⅓)×π×h(r ₁ ² +r ₂ ² +r ₁ r ₂ ) Ncert solutions class 10 chapter 13-30

Solving this, we get Volume of the frustum = 22000/9 cm ³ The radius (r) of wire = (1/16)×(½) = 1/32 cm Now, Let the length of the wire be “l”. The volume of wire = Area of cross-section x Length = (πr ² )xl = π(1/32) ² x l Now, Volume of frustum = Volume of wire 22000/9 = (22/7)x(1/32) ² x l Solving this, we get, l = 7964.44 m

Question

A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it. $[U s e π = 22 / 7.]$

Radius

(r_{2})

at upper circular end = 4 cm Radius

(r_{1})

at lower circular end = 10 cm Slant height (l) of frustum = 15 cm Area of material used for making the fez = CSA of frustum + Area of upper circular end

= π (r_{1} + r_{2}) l + π r_{2}^{2} = π (10 + 4) \times 15 + π (4)^{2} = π (14) \times 15 + 16 π = 210 π + 16 π = 226 π = 226 \times \frac{22}{7} = 710 \frac{2}{7} c m^{2}

Therefore, the area of material used for making it is

710 \frac{2}{7} c m^{2}

Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.3

The RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.3 on Volume and Surface Areas of Solids offer several benefits to students:

Structured Learning : The solutions provide a structured approach to learning about volumes and surface areas of different solids. Each type of solid (cube, cuboid, cylinder, cone, sphere) is covered with clear explanations and step-by-step solutions.

Clarity of Concepts : The solutions help clarify fundamental concepts such as how to calculate volume and surface area using specific formulas for each solid. This clarity aids in understanding the underlying principles of geometry.

Comprehensive Coverage : The exercises cover a range of problems from basic to advanced levels, ensuring that students develop a thorough understanding of the topic. This variety helps in reinforcing concepts and building problem-solving skills.

Practice Opportunities : Through numerous examples and exercises, students get ample practice applying formulas and solving problems related to volumes and surface areas of solids. This practice is crucial for improving proficiency and confidence in the subject.

Preparation for Exams : The exercises are designed to align with typical exam formats, preparing students effectively for their Class 10 Mathematics exams. They familiarize students with the types of questions they might encounter and equip them with the necessary skills to tackle them.

RS Aggarwal Solutions for Class 10 Maths Chapter 19 Exercise 19.3 FAQs

What are the three types of surface areas?

Surface Area of any figure is defined as the area of the faces of the figure. It is the total area of all the faces of the figure. Surface Area can be calculated for both 2-D figures and 3-D figures. For 3-D figures, we can have two types of Surface Areas, i.e. Lateral/Curved Surface Area, and Total Surface Area.

What is total surface area used for?

Total surface area refers to the area including the base(s) and the curved part. It is the total area covered by the surface of the object. If the shape has a curved surface and base, then the total area will be the sum of the two areas.

How to determine the surface area?

To calculate surface area, we add the areas of all the faces of the three-dimensional figure.

Is surface area TSA or CSA?

Curved Surface Area (CSA) – It includes the area of all the curved surfaces. Lateral Surface Area (LSA) – It includes the area of all the surface excluding the top and bottom areas. Total Surface Area (TSA) – It includes the area of all the surfaces of the object including the bases.

🔥 Trending Blogs

CBSE Class 10 Important Questions, Check Subject Wise

Smart Tips for Class 9th and 10th Students

CBSE Class 10 Social Science Weightage 2026, Chapter Wise Marks Distribution

CBSE Class 10 English Weightage 2026, Chapter Wise Marks Distribution

NCERT Solutions for Class 10 Social Science Civics

Talk to a counsellorHave doubts? Our support team will be happy to assist you!

08448982616

008448982616

Check out these Related Articles

CLASS 10 EXAM

NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.4 PDF

Read Full Story

28 Sept 2025, 08:47 pm

CLASS 10 EXAM

CBSE Class 10 Maths 60 Day Study Plan and Resources

Read Full Story

28 Sept 2025, 08:46 pm

CLASS 10 EXAM

NCERT Solutions for Class 10 Social Science Geography Chapter 3 Water Resources

Read Full Story

28 Sept 2025, 08:42 pm

CLASS 10 EXAM

RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.9 Pair of Linear Equations in Two Variables

Read Full Story

28 Sept 2025, 08:41 pm

CLASS 10 EXAM

RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.7 PDF Download

Read Full Story

28 Sept 2025, 08:40 pm

CLASS 10 EXAM

RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.8

Read Full Story

28 Sept 2025, 08:39 pm

CLASS 10 EXAM

RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.6

Read Full Story

28 Sept 2025, 08:38 pm

CLASS 10 EXAM

CBSE Class 10 English Sample Paper 2024-25 PDF Download

Read Full Story

28 Sept 2025, 08:34 pm

CLASS 10 EXAM

NCERT Solutions for Class 10 Social Science History

Read Full Story

28 Sept 2025, 08:24 pm

Join 15 Million students on the app today!

Live & recorded classes available at ease

Dashboard for progress tracking

Millions of practice questions at your fingertips

Free Learning Resources

PW Books

Notes (Class 10-12)

Class 10 Math's Notes

Class 10 Chemistry Notes

Class 10 Physics Notes

Class 10 Biology Notes

Aptitude & Reasoning

Class 10 Geography

Physics Class 11 Notes

Class 11 Chemistry Notes

Maths Notes Class 11

Zoology Class 11

Class 11 Botany Notes

Physics Class 12 notes

Chemistry Class 12

Maths Notes Class 12

Zoology class 12

Class 12th Botany Notes

PW Study Materials

Arjuna JEE Study Material

JEE 7 PYQs

JEE Mind Maps

Yakeen Study Material

NCERT Punch

Arjuna Neet Study Material

NEET PYQs

CBSE Sample Paper

CBSE Question Bank

Udaan For Class 10th Study Material

Notes (Class 6-9)

Class-6 Theory & Notes

Math's Notes for class 7

Science Notes for class 7

Class 8 Math Notes

Class 8 Chemistry Notes

Class 8 Physics Notes

Class 8 Biology Notes

Class 8 SST Notes

Class 9 Math's Notes

Class 9 Physics Notes

Class 9 Chemistry Notes

Class 9 Biology Notes

Ncert Solutions

NCERT Solutions For Class 6

NCERT Solutions For class 7

NCERT Solutions for class 8

NCERT Solutions for class 9

NCERT Solutions for class 10

NCERT Solutions for class 11

NCERT Solutions for Class 12

Govt Exams

Entrance Exams

Our Other Websites

Class 6th to 12th Online Courses

Class 12 Online Course

Class 11 Online Course

Class 10 Online Course

Class 9 Online Course

Class 8 Online Course

Govt Job Exams Courses

SSC Online Coaching

Bank Exam Online Coaching

TET Online Coaching

JAIIB & CAIIB Online Coaching

Bihar Exam Online Coaching

UPSC Coaching

UPSC Online Coaching

UPSC Offline & Hybrid Coaching

State PSC Online Coaching

UPPSC Online Coaching

BPSC Online Coaching

Defence Exam Coaching

NDA Online Coaching

CDS Online Coaching

AFCAT Online Coaching

Agniveer Online Coaching

Gate Exam Coaching

Civil Engineering Courses

Mechanical Engineering Courses

Formulas

Other Exams

Company Secretary Exam

Know about Physics Wallah

Physics Wallah is an Indian edtech platform that provides accessible & comprehensive learning experiences to students from Class 6th to postgraduate level. We also provide extensive NCERT solutions, sample paper, NEET, JEE Mains, BITSAT previous year papers & more such resources to students. Physics Wallah also caters to over 3.5 million registered students and over 78 lakh+ Youtube subscribers with 4.8 rating on its app.

We Stand Out because

We provide students with intensive courses with India’s qualified & experienced faculties & mentors. PW strives to make the learning experience comprehensive and accessible for students of all sections of society. We believe in empowering every single student who couldn't dream of a good career in engineering and medical field earlier.

Our Key Focus Areas

Physics Wallah's main focus is to make the learning experience as economical as possible for all students. With our affordable courses like Lakshya, Udaan and Arjuna and many others, we have been able to provide a platform for lakhs of aspirants. From providing Chemistry, Maths, Physics formula to giving e-books of eminent authors like RD Sharma, RS Aggarwal and Lakhmir Singh, PW focuses on every single student's need for preparation.

What Makes Us Different

Physics Wallah strives to develop a comprehensive pedagogical structure for students, where they get a state-of-the-art learning experience with study material and resources. Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others