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Q1. Prove that √5 is an irrational number.
This question has appeared in 7 out of the last 10 CBSE Board Examinations. Students often lose marks for missing the co-prime assumption or Fundamental Theorem of Arithmetic citation.
1. Assume on the contrary that √5 is rational. Then √5 = a/b, where a and b are co-prime integers (HCF = 1) and b ≠ 0.
2. Squaring both sides: 5 = a²/b² ⇒ a² = 5b².
3. Since 5 divides a², by Theorem 1.3, 5 must divide a. Let a = 5c for some integer c.
4. Substitute a in equation: (5c)² = 5b² ⇒ 25c² = 5b² ⇒ b² = 5c².
5. This means 5 divides b², therefore 5 divides b.
6. Thus, both a and b have at least 5 as a common factor. But this contradicts our hypothesis that a and b are co-prime.
⇒ Hence, √5 is irrational. (Hence Proved - 3/3 Marks)
Q2. Derive the Lens Maker's Formula 1/f = (μ - 1)(1/R₁ - 1/R₂).
Derive refraction at spherical surfaces with radius R₁ and R₂ for a thin convex lens of refractive index μ₂ placed in a medium of refractive index μ₁.
1. Refraction at first surface (radius R₁): μ₂/v₁ - μ₁/u = (μ₂ - μ₁)/R₁ ... (Eq 1)
2. Refraction at second surface (radius R₂): μ₁/v - μ₂/v₁ = (μ₁ - μ₂)/R₂ = -(μ₂ - μ₁)/R₂ ... (Eq 2)
3. Adding Eq 1 and Eq 2: μ₁(1/v - 1/u) = (μ₂ - μ₁) [1/R₁ - 1/R₂].
4. When u = ∞, v = f. Therefore 1/v - 1/u = 1/f.
⇒ 1/f = [(μ₂/μ₁) - 1] (1/R₁ - 1/R₂). For air (μ₁ = 1): 1/f = (μ - 1)(1/R₁ - 1/R₂). (Q.E.D - 5/5 Marks)
Q3. Circuit calculation with resistors in series and parallel.
Three resistors of 5Ω, 10Ω, and 30Ω are connected in parallel across a 12V battery. Calculate (a) effective resistance, and (b) total current drawn.
1. Formula for parallel: 1/Rₛ = 1/R₁ + 1/R₂ + 1/R₃ = 1/5 + 1/10 + 1/30.
2. LCM of (5, 10, 30) = 30 ⇒ 1/Rₛ = (6 + 3 + 1) / 30 = 10 / 30 = 1 / 3.
3. Therefore, Equivalent Resistance Rₛ = 3 Ω.
4. Total Current I = V / Rₛ = 12V / 3Ω = 4.0 Amperes.
⇒ Rₛ = 3 Ω, Total Current = 4 A. (Full Marks)
Q4. Calculate EMF of Daniell Cell at 298 K using Nernst Equation.
Calculate the cell potential for: Zn(s) | Zn²⁺ (0.1 M) || Cu²⁺ (0.01 M) | Cu(s). Given: E° cell = +1.10 V.
1. Cell reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Number of electrons n = 2.
2. Nernst Equation: E_cell = E°_cell - (0.0591 / n) log([Zn²⁺] / [Cu²⁺]).
3. E_cell = 1.10 - (0.0591 / 2) log(0.1 / 0.01) = 1.10 - 0.02955 log(10).
4. Since log(10) = 1 ⇒ E_cell = 1.10 - 0.02955 = 1.07045 V.
⇒ Cell EMF = 1.07 V. (Full Marks)
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📋 Official 2025 Board Examination Blueprint Structure (80 Marks)
| Section | Question Type | No. of Questions | Marks Each | Total Marks | Competency Weight |
|---|---|---|---|---|---|
| Section A | Multiple Choice & Assertion-Reasoning | 20 Questions | 1 Mark | 20 Marks | 50% Application |
| Section B | Very Short Answer (Conceptual Proofs) | 5 Questions | 2 Marks | 10 Marks | Knowledge & Core |
| Section C | Short Answer (Step Derivations & Diagrams) | 6 Questions | 3 Marks | 18 Marks | Calculations |
| Section D | Long Answer (Complex Proofs & Theorems) | 4 Questions | 5 Marks | 20 Marks | Analytical Rigor |
| Section E | Case-Based Data Interpretation / Passage | 3 Case Studies | 4 Marks | 12 Marks | 100% Real-Life |
Case Study: Mathematical Modeling in Aerodynamics (Class 10 & 12 Quadratic Equations)
"A rocket is propelled upwards such that its height h (in meters) after t seconds is given by h(t) = -5t² + 40t + 100. (i) Find the time taken to reach maximum height. (ii) What is the maximum height attained? (iii) At what time does it strike the ground?"
Sub-part (i): Time for max height: Vertex formula t = -b/(2a) = -40 / (2 × -5) = 4 seconds. [1 Mark]
Sub-part (ii): Max height: h(4) = -5(16) + 40(4) + 100 = -80 + 160 + 100 = 180 meters. [1.5 Marks]
Sub-part (iii): Ground strike h(t) = 0: -5t² + 40t + 100 = 0 ⇒ t² - 8t - 20 = 0 ⇒ (t - 10)(t + 2) = 0. Since t > 0, t = 10 seconds. [1.5 Marks]
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