Practice question

Duration: 2 min

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Module outline

  1. About Course: About the Course
  2. Paper-2 | Programming & Data Structures: C Fundamentals, Control Flow, Functions, Arrays & Pointers, Storage Classes, Structures & Enums, DMA, Macros, Scoping & File Handling, Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing
  3. Paper-2 | Databases / DBMS & SQL: Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Decomposition Properties & 4NF, File Organization & Indexing, Relational Algebra, SQL, Relational Calculus, Transaction Management, Concurrency Control
  4. Paper-2 | Algorithms: Algorithm Analysis, Time Complexity Analysis, Sorting Algorithms, Greedy Algorithms, Dynamic Programming, Minimum Spanning Trees, Shortest Path Algos
  5. Paper-2 | Computer Networks & Security: Introduction to CN, Data Communication, DLL: Access Control, DLL: Flow Control, DLL: Error Control, DLL: Framing, Data Link Layer - Ethernet, Net Layer: IPv4 & Proto, Net Layer: IP Addressing, Net Layer:Routing Protocol, Transport Layer Services, TL: Congestion & UDP, Application Layer, Hardware Basics, Network Access Technologies, Network Security
  6. Paper-2 | Digital Logic: Digital Systems & Boolean Basics, Logic Gates & Hardware, Boolean Expression, Boolean Minimization, Combinational Circuit, Sequential Circuits, Number System, Number Representation, Floating Point Rep
  7. Paper-2 | Operating System: Introduction to OS, Process Management, CPU Scheduling, Process Synchronization, Threads & Process Creation, Deadlock, Memory Management, Virtual Memory, Disc Scheduling, File Management
  8. Paper-2 | Theory of Computation: Introduction to TOC, Deterministic FA (DFA), Non-Deterministic FA, Regular Expressions, Grammar, Regular Language Properties, Moore & Mealy Machines, Pushdown Automata & CFG, Turing Machines
  9. Paper-2 | Computer Organization & Architecture: Basics of COA, Cache Memory Organization, Input Output Organisation, Pipelining, Instr Formats & Modes, Control Unit Design, Programming the Basic Computer
  10. Paper-2 | Compiler Design: Intro to Compilers, Lexical Analysis, Grammar & CFG, Syntax Analysis: Top-Down, Syntax Analysis: Bottom-Up, Semantic Analysis & SDT, Intermediate Code Gen, Code Optimization, Run Time Environment
  11. Paper-1 | General Awareness: Polity, Physics, Biology, Chemistry, Static General Knowledge, GK- Current Affairs, History, Geography, Indian Economy, Environment
  12. Paper-1 | Logical Reasoning: Series (Number and Letter Series) (Numerical Relations and Reasoning), Coding Decoding, Non Verbal Reasoning (Spatial Aptitude) (Spatial Reasoning) (Visual Reasoning), Data Interpretation, Blood Relations, Syllogisms, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Selection Decision Table (Decision Making), Clock, Calendar, Cubes & Dices, Directions (Direction Test), Analogy, Seating Arrangements, Analytical Reasoning (Counting Figures Reasoning)
  13. Paper-1 | Quantitative Aptitude: Ratio and Proportion (Ratios), Mixture and Alligation, Permutation and Combination, Number System, HCF LCM, Speed Time and Distance, Average, Age Problems, Percentage, Profit and Loss, Time and Work, Pipe and Cistern, Simple Interest and Compound Interest, Powers and Exponents (Surds and Indices), Mensuration and Geometry, Probability, Statistics, Arithmetic, Algebra
  14. Paper-1 | English: Vocabulary, Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Sentence Correction (Error Correction), Sentence Completion (Fill in the blanks), Sentence Construction, Cloze Test (Cloze Passage)
  15. Coal India Previous Year Paper:
  16. Engineering Mathematics: Set Theory, Relations, Functions, Graph Theory, Group Theory, Propositional and Predicate Logic, Linear Algebra, Calculus
  17. Coal India Mock Test:
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This short practice-question video demonstrates how to compute the memory address of a specific element in a one-dimensional array. The on-screen question states that the base address of the first element is 250, each element occupies 3 bytes, and asks for the address of the fifth element in array a[10]. The instructor first identifies the given values, writing B = 250 and w = 3B in red handwriting. A diagram of the array is drawn as a rectangle with indices 0 through 9 written above it, emphasizing that the fifth element corresponds to index 4 because indexing starts at zero. The solution is then computed manually as 250 + (4 × 3) = 262. A formal array addressing formula, a[k] = B + w * (k - Lb), is also shown and applied with k = 4, B = 250, w = 3, and Lb = 0 to verify the same result. The final answer, 262 bytes, is underlined in red.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The slide displays the orange question: "Let the base address of the first element of the array is 250 and each element of the array occupies 3 bytes in the memory, then address of the fifth element of a one-dimensional array a[10]?" Red handwriting appears showing "B = 250" and "w = 3B", with a red underline under "first element". An empty red rectangle is drawn, then index numbers "0 1 2 3 4 5 6 7 8 9" are written above it to show the fifth element is at index 4. The working expands to "=250 + 4*3B", then "250 + 12", and finally "= 262B" underlined in red. A formal formula, a[k] = B + w * (k - Lb), is shown and applied with k=4, B=250, w=3 to yield 262.

The central concept is array element address calculation using the formula a[k] = B + w * (k - Lb), where B is the base address, w is the word size in bytes, k is the target index, and Lb is the lower bound (starting index). The key teaching point is that "fifth element" means index 4 in a zero-based array, not index 5. The instructor demonstrates both a manual offset calculation (base + index × word size) and the formal formula, arriving at 262 bytes. This reinforces that address computation depends on the index offset from the lower bound multiplied by element size, added to the base address.

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