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Class 8 Maths NCERT Solutions for Ganita Prakash Part 1 Chapter 1 – A Square and a Cube provide clear and easy-to-follow solutions to the questions and activities given in the new NCERT Mathematics textbook for Class 8. This chapter introduces students to square numbers, perfect squares, square roots, cube numbers and cube roots through patterns, puzzles and mathematical exploration. Students also learn different methods for finding square roots and cube roots and explore useful patterns involving squares and cubes. The chapter is part of Ganita Prakash Part 1, the new Class 8 NCERT Mathematics textbook being used from the 2026–27 academic year. Watch Video:Class 8 Maths Ganit Praksh NCERT Solutions-1 | A Square & A Cube What You’ll Learn In – A Square and a Cube In this chapter, students will learn about: These concepts form the foundation for working with squares, cubes and roots in higher classes. Important Concepts from A Square and a Cube 1. Square Numbers A number obtained by multiplying a number by itself is called its square. For example: 52=5×5=25 Therefore, 25 is a square number. Similarly: 102=10×10=100 So, 100 is a perfect square. 2. Perfect Squares Numbers such as: 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100 are perfect squares because each can be expressed as the product of a whole number with itself. 3. Square Root The square root of a number is the number which, when multiplied by itself, gives the original number. For example: √(64) = 8​ because: 8×8=64 4. Cube Numbers A number obtained by multiplying a number by itself three times is called its cube. For example: 4=4×4×4=64 Therefore, 64 is a perfect cube. Some perfect cubes are: 1, 8, 27, 64, 125, 216, 343, 512 and 729. The textbook also explores patterns in cube numbers and asks students to identify relationships between numbers and their cubes Therefore, 64 is a perfect cube.Some perfect cubes are: 1, 8, 27, 64, 125, 216, 343, 512 and 729. The textbook also explores patterns in cube numbers and asks students to identify relationships between numbers and their cubes. 5. Cube Root The cube root of a number is the number which, when multiplied by itself three times, gives the original number. For example: Because: 5×5×5=125 Watch Video:Class 8 Maths Ganit Praksh NCERT Solutions-2 | A Square & A Cube Class 8 Ganita Prakash Chapter 1 – Figure It Out Questions The chapter contains Figure It Out questions that encourage students to apply the concepts of squares, square roots, cubes and cube roots rather than simply memorising formulas. For example, students are asked to identify numbers that are not perfect squares and investigate patterns in square and cube numbers. How to solve these questions How to Use These Class 8 Ganita Prakash Solutions Try to solve every question yourself before looking at the solution. If you cannot solve a question, read the explanation carefully and identify the mathematical concept or method being used. Then try solving the question again without looking at the answer. This approach will help you: Class 8 maths Ganita Prakash chapter 1 | A Square and A Cube | Exam Questions: Watch Full video Conclusion Class 8 Maths Ganita Prakash Part 1 Chapter 1 – A Square and a Cube introduces important concepts related to squares, cubes, square roots and cube roots. By practising the textbook questions and understanding the patterns explained in the chapter, students can develop a strong foundation in number concepts. Use the chapter-wise and question-wise solutions on Future Study Point to check your answers, understand the correct method and practise more questions.

Unique ID codes for 8.5 billion milk packets is the focus of this interesting question from Class 8 Maths Ganit Prakash Chapter 2 – Power Play. In this question, a dairy plans to produce 8.5 billion packets milk packets in a year and wants to assign a unique ID code to each packet. The challenge is to determine the minimum number of digits required so that every packet can have a different code. Question:A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID (identifier) code for each packet. If they choose to use the digits 0–9, how many digits should the code consist of? Watch the video Solution Solution The dairy needs to produce: 8.5 billion packets We know that: 1 billion = 1,000,000,000 Therefore, 8.5 billion = 8,500,000,000 So, the dairy needs to assign a unique ID code to 8,500,000,000 packets. If code is of 1 digit,then codes generated are from 0 to 9 that are 10 codes If code is of 2 digit,then codes generated are =10 x 10 =102 See the Complete Solution Class 8 Maths – Ganit Prakash Chapter 2: Power Play This question is based on large numbers and the use of digits to create unique identifiers. It helps students understand how the number of digits changes as numbers become larger and how to determine the minimum number of digits needed to represent a large quantity. If you are studying Class 8 Maths Ganit Prakash Chapter 2 – Power Play, practise more questions involving large numbers, place value, and number representation. Subscribe to Future Study Point for more Class 8 Maths Ganit Prakash solutions, NCERT-based questions, important questions, and exam preparation videos. Watch complete Solution of Class 8 Maths Chapter 2 Power Play Similar Questions Class 8 Maths Ganit Prakash Chapter 2 Power Play Q1.Roxie has 7 dresses, 2 hats, and 3 pairs of shoes. How many different ways can Roxie dress up? (Page 26) Q2.Write the number of lotuses (in exponential form) when the pond was(i) fully covered(ii) half covered (Page 25) Q3.Estu says, “Next time I will buy a lock that has 6 slots with the letters A to Z. I feel it is safer.”How many passwords are possible with such a lock? (Page 27) Q4.Estu and Roxie came across a safe containing old stamps and coins that their great-grandfather had collected. It was secured with a 5-digit password. Since nobody knew the password, they had no option except to try every password until it opened. They were unlucky, and the lock only opened with the last password, after they had tried all possible combinations. How many passwords did they end up checking? (Page 26) Q5:Say you can fold a sheet of paper as many times as you wish. What would its thickness be after 30 folds? Make a guess. Let us find out how thick a sheet of paper will be after 46 folds. Assume that the thickness of the sheet is 0.001 cm. Q6:Which expression describes the thickness of a sheet of paper after it is folded 10 times? The initial thickness is represented by the letter-number v. (i) 10v (ii) 10 + v (iii) 2 × 10 × v (iv) 2¹⁰ (v) 2¹⁰v (vi) 10²v Q7:Express the number 32400 as a product of its prime factors and represent the prime factors in their exponential form. 8:What is (−1) 5 ? Is it positive or negative? What about (−1) 56 ? Q9:Is (−2) 4 = 16? Verify. Q10. Express the following in exponential form: (i) 6×6×6×6 (ii) y× y (iii) b×b×b×b (iv) 5×5×7×7×7 (v) 2×2×a×a (vi) a×a×a×c×c×c×c×d Class 10 Maths Most Important Questions on Real Numbers Class 10 Maths Most Important Questions Chapter 1 Real Numbers Class 10 Maths Basic Sample Paper Solution: Practise with the Complete CBSE Basic Sample Paper Solution Class 10 Maths NCERT Solutions CBSE 2026-27 Update Continue your Class 10 Maths preparation with these useful resources from Future Study Point: Class 10 Maths Chapter 1 – Real Numbers: Complete NCERT Exercise-Wise Video Solutions Class 10 Maths Chapter 2 – Polynomials: Complete NCERT Solutions for Polynomials Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables: Complete NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations: – Quadratic Equations complete NCERT video solutions Class 10 Maths Chapter 5 Arithmetic Progression: Arithmetic Progression: complete video solution 👉 Explore these resources to strengthen your concepts, revise important topics, and prepare effectively for the CBSE Board Examination 2026–27.

Total Internal Reflection (TIR) is an important phenomenon related to the refraction of light. It occurs when light travelling through an optically denser medium strikes the boundary with an optically rarer medium at an angle greater than the critical angle. In this article, we will understand Total Internal Reflection Class 10, its conditions, critical angle, examples, applications, and important questions for CBSE Class 10 Science Chapter 9 – Light Reflection & Refraction. What is Total Internal Reflection? When a ray of light travels from an optically denser medium to an optically rarer medium, it normally bends away from the normal. However, when the angle of incidence is increased beyond a certain limiting angle, called the critical angle, the refracted ray disappears and the entire light is reflected back into the denser medium. This phenomenon is called Total Internal Reflection. Definition Total Internal Reflection is the phenomenon in which a ray of light travelling from an optically denser medium to an optically rarer medium is completely reflected back into the denser medium when the angle of incidence is greater than the critical angle. Conditions Necessary for Total Internal Reflection There are two essential conditions for Total Internal Reflection: 1. Light must travel from a denser medium to a rarer medium The light must move from a medium having a higher refractive index to a medium having a lower refractive index. For example: TIR does not occur when light travels from a rarer medium to a denser medium. 2. Angle of incidence must be greater than the critical angle The angle of incidence must be greater than the critical angle (C). Therefore: Angle of incidence > Critical angle i > C Only then does complete internal reflection take place. What is the Critical Angle? The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium becomes 90°. In other words, when light travels from a denser medium to a rarer medium and the refracted ray just grazes the surface, the corresponding angle of incidence is called the critical angle. It is represented by C. At the critical angle: Angle of refraction = 90° For Total Internal Reflection: Angle of incidence > Critical angle What Happens When the Angle of Incidence Changes? Suppose light travels from glass to air. There are three important cases: Case 1: Angle of incidence is less than the critical angle When: i < C The ray is refracted into the rarer medium and bends away from the normal. Case 2: Angle of incidence is equal to the critical angle When: i = C The refracted ray travels along the boundary between the two media. Therefore: r = 90° Case 3: Angle of incidence is greater than the critical angle When: i > C No refracted ray emerges into the rarer medium. The complete ray is reflected back into the denser medium. This is Total Internal Reflection. Basis of Comparison Ordinary Reflection Total Internal Reflection (TIR) Meaning Light is reflected back from a surface when it strikes the surface. Light is completely reflected back into the denser medium at the boundary of two transparent media. Direction of Light It can occur for light travelling in any direction towards a reflecting surface. Light must travel from an optically denser medium to an optically rarer medium. Angle of Incidence No specific condition regarding the critical angle is required. The angle of incidence must be greater than the critical angle. Critical Angle Critical angle is not required. Critical angle is an essential concept for TIR. Refraction In ordinary reflection from a transparent surface, some light may be transmitted into the second medium. No refracted ray emerges into the rarer medium when TIR occurs. Amount of Light Reflected Generally, only a part of the incident light may be reflected. Ideally, 100% of the incident light is reflected back into the denser medium. Where It Occurs At a reflecting surface or boundary. At the boundary between an optically denser and an optically rarer transparent medium. Main Examples Reflection from a plane mirror, polished metal surface, etc. Optical fibres, sparkling of diamonds and totally reflecting prisms. Important Condition Follows the law of reflection: angle of incidence = angle of reflection. Requires denser → rarer medium and angle of incidence > critical angle. Refractive Index and Critical Angle For light travelling from a medium of refractive index n₁ to a rarer medium of refractive index n₂, where n₁ > n₂, the critical angle is related by: sin C = n₂ / n₁ For the special case of light travelling from glass to air, the refractive index of air is approximately 1. Therefore: sin C = 1 / n where n is the refractive index of glass with respect to air. Why Does Total Internal Reflection Occur? When light travels from an optically denser medium to an optically rarer medium, the refracted ray bends away from the normal. As the angle of incidence increases, the angle of refraction also increases. At a particular angle of incidence, the refracted ray makes an angle of 90° with the normal. This angle is called the critical angle. If the angle of incidence is increased further, the light can no longer emerge into the rarer medium. Instead, it is completely reflected back into the denser medium. This is Total Internal Reflection. Examples of Total Internal Reflection in Daily Life Total Internal Reflection has several important applications. 1. Optical Fibre One of the most important applications of Total Internal Reflection is in optical fibres. An optical fibre consists mainly of a core surrounded by cladding. The core has a higher refractive index than the cladding. Light entering the fibre undergoes repeated Total Internal Reflection and travels through the fibre over long distances. Uses of optical fibres Optical fibres are used in: 2. Sparkling of Diamonds Diamonds sparkle brightly because of Total Internal Reflection. Diamond has a very high refractive index and therefore a small critical angle. Light entering a diamond

Are you looking for Class 9 Maths Chapter 2 Full Video Solutions – Introduction to Linear Polynomials? You are at the right place. In this post, you’ll find complete video solutions to all the important questions and End of Exercises of Chapter 2. Every solution is explained in a simple, step-by-step manner so that CBSE Class 9 students can understand the concepts easily and score better in exams. Whether you’re revising for your school test or preparing for the annual examination, these video solutions will help you master every question from Introduction to Linear Polynomials. What You Will Learn in Chapter 2 In this chapter, you will understand: Topics Covered About This Chapter Welcome to Class 9 Maths Chapter 2 – Introduction to Linear Polynomials. This chapter is divided into 6 exercises (Exercise 2.1 to Exercise 2.6), followed by a comprehensive End of Exercises section. Each exercise introduces new concepts and gradually builds your understanding of linear polynomials, making it easier to solve more advanced problems. On this page, you will find 7 complete video solutions: We recommend watching the videos in sequence, starting from Exercise 2.1 and ending with the End of Exercises. This approach will help you understand every concept clearly, strengthen your problem-solving skills, and prepare effectively for your CBSE Class 9 Maths examinations. Exercise 2.1 – Introduction to Linear Polynomials Exercise 2.1 introduces the basic concept of linear polynomials. In this exercise, students learn how to identify a linear polynomial, determine its degree, and understand its general form. The questions build a strong foundation that is essential for the remaining exercises of the chapter. Exercise 2.2 – Graphs of Linear Polynomials In Exercise 2.2, students learn how to represent a linear polynomial graphically. This exercise explains how to prepare a table of values, plot points on the Cartesian plane, and draw the graph of a linear polynomial. It also helps students understand the relationship between algebraic expressions and their graphical representation. Exercise 2.3 – Zero of a Linear Polynomial Exercise 2.3 focuses on finding the zero (root) of a linear polynomial. Students learn different methods to determine where the graph intersects the x-axis and understand the significance of the zero of a polynomial. This exercise strengthens both algebraic and graphical concepts. Exercise 2.4 – Solving Application-Based Questions Exercise 2.4 contains interesting application-based questions where students apply the concepts of linear polynomials to solve real-life mathematical problems. These questions improve logical thinking and help students understand the practical use of algebra. Exercise 2.5 – Higher-Level Practice Questions Exercise 2.5 provides a variety of practice questions that test students’ understanding of the chapter. It includes graph-based questions, polynomial identification, and concept-based problems that prepare students for school examinations. Exercise 2.6 – Advanced Concept Practice Exercise 2.6 is designed to strengthen your command over the chapter with more challenging and concept-based questions. By solving these problems, students develop confidence and improve their analytical and problem-solving skills. End of Exercises – Complete Chapter Revision & Exam Questions The End of Exercises is the most important part of the chapter because it combines concepts from Exercises 2.1 to 2.6 into one comprehensive practice set. It includes important exam-oriented questions, mixed-concept problems, graph-based questions, application-based questions, and chapter revision. Solving these questions helps students revise the complete chapter and prepares them thoroughly for school examinations. Why Watch These Video Solutions? ✔ Easy-to-understand explanations ✔ Step-by-step solutions ✔ CBSE exam-oriented approach ✔ Covers every End of Exercises question ✔ Perfect for revision before exams ✔ Helpful for self-study Benefits for Students These video solutions will help you: Frequently Asked Questions (FAQs) 1. Is this video useful for CBSE Class 9 students? Yes. These video solutions are designed according to the latest CBSE syllabus. 2. Does this playlist cover all End of Exercises questions? Yes. Every End of Exercises question is explained in detail with complete solutions. 3. Are the solutions explained step by step? Yes. Each question is solved using an easy and student-friendly method. 4. Can beginners understand these videos? Absolutely. The explanations are simple and suitable for all Class 9 students. 5. Is this chapter important for exams? Yes. Introduction to Linear Polynomials is an important chapter and frequently appears in school examinations. Conclusion If you want to score high in Class 9 Maths, don’t miss these Class 9 Maths Chapter 2 Full Video Solutions – Introduction to Linear Polynomials. Watch every video carefully, practice the questions yourself, and revise regularly. With consistent practice, you’ll gain confidence and perform well in your CBSE examinations. Watch the complete playlist above and start mastering Chapter 2 today! Class 10 Maths Most Important Questions on Real Numbers Class 10 Maths Most Important Questions Chapter 1 Real Numbers Class 10 Maths Basic Sample Paper Solution: Practise with the Complete CBSE Basic Sample Paper Solution Class 10 Maths NCERT Solutions CBSE 2026-27 Update Continue your Class 10 Maths preparation with these useful resources from Future Study Point: Class 10 Maths Chapter 1 – Real Numbers: Complete NCERT Exercise-Wise Video Solutions Class 10 Maths Chapter 2 – Polynomials: Complete NCERT Solutions for Polynomials Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables: Complete NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations: – Quadratic Equations complete NCERT video solutions Class 10 Maths Chapter 5 Arithmetic Progression: Arithmetic Progression: complete video solution 👉 Explore these resources to strengthen your concepts, revise important topics, and prepare effectively for the CBSE Class 10 Maths Board Examination 2026–27.