VEDANT CLASSES - A Pathway to Success

 Got it — I’ve reframed all 30 questions into slightly more elaborated, competency-based, student-friendly format while keeping them numerical-focused and beginner level 👇 Vedant Classes – A Pathway to Success Class 11 Chemistry – Chapter 1: Some Basic Concepts of Chemistry 📄 Competency-Based Worksheet (30 Questions – Elaborated Version) 🧠 Section A: Basic Understanding (Q1–Q6) Q1. Calculate the molecular mass of carbon dioxide (CO₂) and sulphuric acid (H₂SO₄) using atomic masses (C = 12, O = 16, H = 1, S = 32). Q2. Determine the number of moles present in 44 g of CO₂. Also identify what this mass represents in terms of molar mass. Q3. Calculate the total number of molecules present in 18 g of water. Explain the steps involved. Q4. A sample contains 1 mole of methane (CH₄). Find the total number of atoms present in the sample. Q5. Calculate the total number of electrons present in one molecule of ammonia (NH₃). (Atomic numbers: N = 7, H = 1) Q6. Determine the number of ...

Class 12 Chemistry – Chapter: Haloalkanes and Haloarenes (Short Notes) These notes are designed for CBSE, GSEB, JEE and NEET students. It includes detailed theory, mechanisms, reactions, and exam-oriented explanations. 1. Introduction Haloalkanes and haloarenes are organic compounds in which one or more hydrogen atoms of hydrocarbons are replaced by halogen atoms (F, Cl, Br, I). Haloalkanes (Alkyl halides): Halogen attached to sp³ hybridized carbon (R–X) Haloarenes (Aryl halides): Halogen attached to aromatic ring (Ar–X) Example: CH₃Cl → Chloromethane (Haloalkane) C₆H₅Cl → Chlorobenzene (Haloarene) 2. Classification of Haloalkanes Based on number of halogen atoms Mono halogen: CH₃Cl Di halogen: CH₂Cl₂ Tri halogen: CHCl₃ Tetra halogen: CCl₄ Based on carbon type Primary (1°): R–CH₂–X Secondary (2°): R₂CH–X Tertiary (3°): R₃C–X 3. Nomenclature (IUPAC System) Rules for naming haloalkanes: Select longest carbon chain Number ...

Class 9 Science – Chapter: Fundamental Unit of Life (Cell) These notes are designed for CBSE, GSEB and foundation students. It includes detailed theory, diagrams explanation, functions and exam-oriented concepts. 1. Discovery of Cell The cell was first discovered by Robert Hooke (1665) while observing a thin slice of cork under a microscope. He saw small box-like compartments and named them "cells" . Later, Anton van Leeuwenhoek observed living cells. Cell Theory was given by Schleiden and Schwann. Cell Theory States: All living organisms are made up of cells. Cell is the basic structural and functional unit of life. All cells arise from pre-existing cells (Virchow). 2. What is a Cell? A cell is the smallest unit of life capable of performing all vital functions such as respiration, digestion, and reproduction. Types of Organisms: Unicellular: Made of single cell (Amoeba, Paramecium) Multicellular: Made of many cells (Humans, Pl...

Class 10 Mathematics – Chapter: Probability (Detailed Notes) These notes are designed for CBSE, GSEB and board exam preparation. Includes all concepts, formulas, and exam-oriented explanations. 1. What is Probability? Probability means the chance of occurrence of an event. Probability ranges between 0 and 1 0 → Impossible event 1 → Certain event Example: Getting head in a coin toss → probability = 1/2 Sun rising in east → probability = 1 Getting 7 on a dice → probability = 0 2. Experiment, Outcome and Event Experiment: An action with uncertain result (e.g., tossing a coin) Outcome: Possible result (Head or Tail) Event: A group of outcomes Example: Rolling a die Sample space: {1,2,3,4,5,6} Event (getting even number): {2,4,6} 3. Sample Space The set of all possible outcomes is called sample space. Notation: S Example: Coin → S = {H, T} Dice → S = {1,2,3,4,5,6} 4. Classical Probability Formula If all outcomes ar...

VEDANT IGNITE TEST SERIES Class: 12 Subject: Physics Chapter: Electric Potential and Capacitance Board: CBSE Time: 1.5 Hours | Marks: 30 Section A (7 Marks) (5 MCQs + 2 Assertion-Reason, 1 mark each) Q1.  A charge of 2 × 10⁻⁶ C is placed in vacuum. Find the electric potential at a point 10 cm away. Options: (A) 1.8 × 10⁵ V (B) 3.6 × 10⁵ V (C) 9 × 10⁵ V (D) 1.8 × 10⁶ V Q2. A student observes that when a dielectric slab is inserted between capacitor plates (battery disconnected), potential difference decreases. What is the correct reason? Options: (A) Charge increases (B) Capacitance decreases (C) Capacitance increases (D) Energy increases Q3. In a laboratory setup, two capacitors 3µF and 6µF are connected in series. The equivalent capacitance is: Options: (A) 9µF (B) 2µF (C) 3µF (D) 1µF Q4. A charged capacitor stores energy. If voltage is doubled, the energy becomes: Options: (A) same (B) double (C) four times (D) half Q5. A metal sphere is given charge. Where is potenti...

VEDANT CLASSES – A Pathway to Success… Date: 11/04/2026 Class: 12 Subject: Physics Chapter: Electric Potential and Capacitance Board: GSEB Marks: 30 VEDANT IGNITE TEST SERIES Section A – MCQs (1 mark each) [7 Marks] Q1. Two charges (+2\mu C) and (-2\mu C) are placed 10 cm apart. Potential at midpoint is: (a) 0 V (b) 1.8×10⁵ V (c) 3.6×10⁵ V (d) 9×10⁵ V Q2. Work done in moving a charge of (5\mu C) through a potential difference of 200 V is: (a) 1×10⁻³ J (b) 2×10⁻³ J (c) 1×10⁻² J (d) 5×10⁻³ J Q3. A capacitor of 5 μF is charged to 10 V. Energy stored is: (a) 2.5×10⁻⁴ J (b) 5×10⁻⁴ J (c) 1×10⁻³ J (d) 2×10⁻³ J Q4. Two capacitors 4 μF and 6 μF are connected in series. Equivalent capacitance is: (a) 2.4 μF (b) 10 μF (c) 5 μF (d) 1 μF Q5. A parallel plate capacitor has plate area 2 m² and separation 1 mm. Capacitance is (ε₀ = 8.85×10⁻¹²): (a) 1.77×10⁻⁸ F (b) 1.77×10⁻⁹ F (c) 8.85×10⁻⁹ F (d) 3.54×10⁻⁸ F Q6. Three capacitors 2 μF, 3 μF and 6 μF are in parallel. Equivalent capacitance is...