💻 Computer Fundamentals & Information Technology (BCA101T)

UNIT 1 — Fundamentals of Computers

1.1 Data vs Information

Types of Data (structured, semi-structured, unstructured)

Data is not always plain numbers in neat columns. Depending on how well it is organised, data is classified into three types — and modern systems handle all three together.

DIKW Model (Data → Information → Knowledge → Wisdom)

The DIKW model explains how raw readings climb up to good decisions. Each higher level requires one more step of processing and judgement.

Qualities of Good Information

Information ki value uski quality par depend karti hai. Exam me "qualities of good information" ka chhota question aa sakta hai — 6–8 points yaad rakho.

1.2 Information Processing Cycle

Block diagram of the Information Processing Cycle

   ┌──────────────┐     ┌──────────────┐     ┌──────────────┐     ┌──────────────┐
   │  1. INPUT    │     │ 2. PROCESS   │     │  3. OUTPUT   │     │  4. COMMU-   │
   │ keyboard     │     │ CPU (ALU+CU) │     │ monitor      │     │    NICATION  │
   │ scanner,mic  │     │ RAM me data  │     │ printer,spe- │     │ network card │
   │ camera       │     │ + program    │     │ aker         │     │ internet     │
   └──────┬───────┘     └──────┬───────┘     └──────┬───────┘     └──────┬───────┘
          │  data & instructions  │   result        │   report        │  messages
          ▼                       ▼                 ▼                 ▼
   ┌────────────────────────────────────────────────────────────────────────┐
   │             STORAGE  (HDD / SSD / pen drive / cloud / database)        │
   │   input data, intermediate results aur final output — sab save hota hai│
   └────────────────────────────────────────────────────────────────────────┘

   (Ek cycle ka OUTPUT aksar agle cycle ka INPUT ban jaata hai — isliye "cycle" 🔁)

1.3 Introduction to Computers

Characteristics of a Computer

1.4 Evolution and History of Computers

Generations of Computers

History ko samajhne ka sabse easy tarika generations hai — har generation ka main technology alag, aur machine poori badal jaati hai.

1.5 Types of Computers (by data & by size)

Computers ko do tarah se classify karte hain — (A) kaam karne ke tarike se (which type of data they handle: analog, digital, hybrid) aur (B) size aur power ke hisaab se (micro, mini, mainframe, super). Ye dono classification exam me alag-alag poochhe jaate hain — inhe mix na karo.

A) Working ke hisaab se (type of data)

B) Size aur power ke hisaab se

1.6 Basic Organization of a Digital Computer (von Neumann Architecture)

Five units — kaam kya hai

Buses — units ke beech ki "sadkein"

Stored Program Concept (von Neumann)

Program aur data dono ek hi memory me store hote hain; CPU memory se instruction laata hai, decode karta hai, execute karta hai. Isi idea ne modern computer banaya — pehle machines ko wire badal kar reprogram karna padta tha (jaise ENIAC ke plugboards). Ab sirf naya program memory me load karo, machine ka behaviour badal jaata hai. Aaj tak ke computers "von Neumann architecture" follow karte hain — bas performance features (cache, pipelining) add hue hain.

1.7 Number Systems High-yield syllabus topic

Yaad rakhne wala table (0–15) — isse har conversion fast ho jaata hai:

Conversion Methods (step-by-step, exam me aise likho)

1) Decimal → Binary (2 se baar-baar divide karo; remainders ulta likho — bottom to top):

(25)₁₀ → Binary
25 ÷ 2 = 12  remainder 1  ← LSB (last likhne wala)
12 ÷ 2 =  6  remainder 0
 6 ÷ 2 =  3  remainder 0
 3 ÷ 2 =  1  remainder 1
 1 ÷ 2 =  0  remainder 1  ← MSB (pehla likhne wala)
Read bottom → top: 1 1 0 0 1
=> (25)₁₀ = (11001)₂ ✔
Verify: 16 + 8 + 0 + 0 + 1 = 25 ✔

Second example: (100)₁₀ → Binary
100 ÷ 2 = 50 r0
 50 ÷ 2 = 25 r0
 25 ÷ 2 = 12 r1
 12 ÷ 2 =  6 r0
  6 ÷ 2 =  3 r0
  3 ÷ 2 =  1 r1
  1 ÷ 2 =  0 r1
=> (100)₁₀ = (1100100)₂ ✔ [64+32+4 = 100]

2) Binary → Decimal (positions ka weight = 2ⁿ, phir jodo):

(11001)₂ = 1×2⁴ + 1×2³ + 0×2² + 0×2¹ + 1×2⁰
         = 16 + 8 + 0 + 0 + 1 = 25  => (25)₁₀ ✔

(1001011)₂ = 64 + 0 + 0 + 8 + 0 + 2 + 1 = 75 => (75)₁₀ ✔
(position weights right se: 1, 2, 4, 8, 16, 32, 64)

3) Decimal → Octal / Hex — same tareeka, bas 8 ya 16 se divide karo:

(345)₁₀ → Octal                 (500)₁₀ → Hex
345 ÷ 8 = 43 r1                 500 ÷ 16 = 31 r4   (4)
 43 ÷ 8 =  5 r3                 31 ÷ 16 =  1 r15   (F)
  5 ÷ 8 =  0 r5                  1 ÷ 16 =  0 r1    (1)
=> (531)₈ ✔                     => (1F4)₁₆ ✔
Check: 5×64 + 3×8 + 1 = 345     Check: 1×256 + 15×16 + 4 = 500 ✔
(745)₁₀ → Hex = (2E9)₁₆ ✔

4) Binary → Octal (3-3 bits ka group, right se):

(11010110)₂ → groups from right: 011 | 010 | 110
                                 3  |  2  |  6   => (326)₈ ✔
Check: 192 + 16 + 6 = 214 = binary 11010110 ✔
Note: group banate waqt left me 0 add karke 3 bits poore karo.

5) Binary → Hex (4-4 bits ka group):

(11001011)₂ → groups from right: 1100 | 1011
                                  C   |  B    => (CB)₁₆ ✔
Check: 12×16 + 11 = 203 ✔

6) Octal / Hex → Binary — ulta karo: har digit ko 3 (octal) ya 4 (hex) bits me kholo:

(72)₈   = 111 010        = (111010)₂ ✔  [= 58 decimal]
(2F)₁₆  = 0010 1111      = (00101111)₂ ✔ [= 47 decimal]
(345)₈  = 011 100 101    = (011100101)₂ ✔ [= 229 decimal]
(A5)₁₆  = 1010 0101      = (10100101)₂ ✔ [= 165 decimal]

Binary Arithmetic

ADDITION: 1010 + 0111          SUBTRACTION: 1101 − 1011
   1 0 1 0  (10)                   1 1 0 1  (13)
 + 0 1 1 1  ( 7)                 − 1 0 1 1  (11)
 --------                         --------
 1 0 0 0 1  (17) ✔                 0 0 1 0  ( 2) ✔
 (carry ko jodna na bhoolo)      (haan, borrow lagega: 10−1 = 1)

MULTIPLICATION: 101 × 11        DIVISION: 1100 ÷ 10
     1 0 1   (5)                     1 1 0    (answer 110 = 6)
  ×    1 1   (3)                  ┌────────
  ---------                       1 0 ) 1 1 0 0
     1 0 1   (×1)                      − 1 0
   1 0 1     (×1, ek left shift)      ─────
  ---------                             1 0 0
   1 1 1 1   (15) ✔                      − 1 0 0
                                        ─────
                                            0 0  → remainder 0
                                     => quotient (110)₂ = 6 ✔

Signed Numbers & 2's Complement important

Negative numbers represent karne ke 3 tarike hain: Sign-Magnitude (pehla bit sign: 0=+, 1=−, baaki 7 bit magnitude), 1's Complement (saare bits ulto), 2's Complement (= 1's complement + 1). Computer 2's complement use karta hai kyunki isme subtraction ko addition se kar sakte hain — isliye CPU ko alag "subtract circuit" ki zaroorat nahi padti, aur -0 jaisa confusion bhi nahi hota.

(-5) ko 8-bit me kaise likhein?
   5            = 0000 0101
   1's complement = 1111 1010        (saare bits ulto)
   2's complement = 1111 1011       ✔ yahi hai (-5)

Check subtraction 7 − 5 using 2's complement:
   7            = 0000 0111
  -5 (2's form) = 1111 1011
  -----------------------------
   sum          = 1 0000 0010
   → carry (left wala 1) ko chhod do
   → 0000 0010 = 2 ✔  (7 − 5 = 2)

Another example: 12 − 9
   12 = 0000 1100
    9 = 0000 1001 → 1's = 1111 0110 → 2's = 1111 0111 = (−9)
   0000 1100 + 1111 0111 = 1 0000 0011 → carry hatao → 0000 0011 = 3 ✔

Negative answer: 25 − 40 = −15
   25 = 0001 1001,  -40 = 1101 1000
   sum = 1111 0001 → sign bit 1 = answer negative hai
   magnitude nikaalne ke liye 2's complement wapas karo:
   1111 0001 → 1's = 0000 1110 → +1 = 0000 1111 = 15
   => answer = −15 ✔

Related terms — BCD, ASCII, EBCDIC, Unicode

Number system practice set (solutions ke saath) Exam practice

These 10 problems practise core conversion skills; the actual paper may differ — khud likh kar solve karo, phir solution se milao.

Q1. (156)₁₀ → Binary
156 ÷ 2 = 78 r0
 78 ÷ 2 = 39 r0
 39 ÷ 2 = 19 r1
 19 ÷ 2 =  9 r1
  9 ÷ 2 =  4 r1
  4 ÷ 2 =  2 r0
  2 ÷ 2 =  1 r0
  1 ÷ 2 =  0 r1
=> (10011100)₂   Check: 128+16+8+4 = 156 ✔

Q2. (10110110)₂ → Decimal
= 128 + 0 + 32 + 16 + 0 + 4 + 2 + 0 = 182
=> (182)₁₀ ✔

Q3. (999)₁₀ → Hex
999 ÷ 16 = 62 r7
 62 ÷ 16 =  3 r14 (E)
  3 ÷ 16 =  0 r3
=> (3E7)₁₆    Check: 3×256 + 14×16 + 7 = 768+224+7 = 999 ✔

Q4. (1110101)₂ → Octal
groups from right: 1 | 110 | 101  →  1 | 6 | 5
=> (165)₈ ✔  (decimal check: 117 = 64+48+5 ✔)

Q5. (A3F)₁₆ → Decimal and Binary
Decimal: 10×256 + 3×16 + 15 = 2560 + 48 + 15 = 2623
Binary : A=1010, 3=0011, F=1111 → (1010 0011 1111)₂
=> 2623₁₀ = (101000111111)₂ ✔

Q6. (77)₈ → Decimal / Binary
Decimal: 7×8 + 7 = 63
Binary : 111 111 → (111111)₂ ✔

Q7. (45)₁₀ → Octal
45 ÷ 8 = 5 r5
 5 ÷ 8 = 0 r5
=> (55)₈ ✔ (5×8+5 = 45)

Q8. (−20) ko 8-bit 2's complement me likho
  20 = 0001 0100
1's  = 1110 1011
2's  = 1110 1100  ← yahi hai (−20) ✔

Q9. 2's complement se subtract karo: 20 − 13
20  = 0001 0100
13  = 0000 1101 → (−13) = 1111 0010 + 1 = 1111 0011
Sum = 0001 0100 + 1111 0011 = 1 0000 0111
carry chhodo → 0000 0111 = 7 ✔

Q10. 2's complement se subtract karo: 15 − 25 (answer negative aayega)
15  = 0000 1111
25  = 0001 1001 → (−25) = 1110 0110 + 1 = 1110 0111
Sum = 0000 1111 + 1110 0111 = 1111 0110
Sign bit 1 = negative. Magnitude: 1's of 1111 0110 = 0000 1001, +1 = 0000 1010 = 10
=> answer = (−10) ✔

Number system MCQs (answers neeche)

1.8 Boolean Algebra & Logic Gates High-yield syllabus topic

Basic Operators with Truth Tables (0=Low/False, 1=High/True)

Aur ek useful derived truth table (NAND, NOR, XOR, XNOR) — ye 4 har exam me kisi na kisi form me aate hain:

Logic Gates & Symbols

Gate symbols (as drawn in exams):
           A ──▷o── Y            A ──┐
      NOT (inverter)             B ──┘D── Y     AND  (D-shape)
           A ──╲
           B ──╱── Y             A ──╲
      OR (shield shape)          B ──╱─o─ Y     NOR (OR + small circle)
           A ──┐
           B ──┘D──o── Y         NAND (AND + small circle)
           A ──╲
           B ──╱═)── Y           XOR (OR with extra curved line)

Universal Gates — NAND se sab kuch

NOT using NAND:   tie dono inputs:  A ──┬──NAND── Y = (A·A)' = A'
                                      A ──┘

AND using NAND:   A ──NAND──o────NAND── Y = ((A·B)')' = A·B
                  B ──┘     └──  (dono input same signal)
                  (do NAND: pehla NAND = (AB)', doosra NAND = inverter)

OR using NAND:    A ──NAND(o)──┐
                  A ──NAND(o)──┤ NAND ── Y = (A'·B')' = A + B  (De Morgan!)
                  B ──NAND(o)──┤
                  B ──NAND(o)──┘
                  (inputs ko pehle invert karo, phir NAND)

XOR using NAND:   A ──┬NAND──┐
                  B ──┘      ├── NAND ── (A·(AB)')' = A ⊕ B  (4 NAND gates chahiye)
                  A ──────────┤
                  B ──────────┘

Laws of Boolean Algebra

Worked example — De Morgan ko truth table se verify karo (exam me aise likho):

Boolean algebra practice set (solutions ke saath)

Q1. Simplify: F = A·B + A·B'
    = A·(B + B')          (distributive law)
    = A·1                 (complement law: B+B' = 1)
    = A ✔                 (identity law)

Q2. Simplify: F = (A + B)·(A + B')
    = A + (B·B')          (distributive law: (A+B)(A+C) = A + BC)
    = A + 0
    = A ✔

Q3. Simplify: F = A'B + AB + AB'
    = B(A' + A) + AB'     (pehle 2 terms me B common)
    = B·1 + AB'
    = B + AB'
    = B + A               (absorption: X + X'Y = X + Y → B + B'A = B + A) ✔
    => F = A + B  (yani OR gate hi kaafi hai!)

Q4. De Morgan se simplify: F = (A·B + C)'
    = (A·B)' · C'         (De Morgan 2: (X+Y)' = X'·Y')
    = (A' + B') · C'      (De Morgan 1: (A·B)' = A'+B') ✔

Q5. Truth table se verify: A ⊕ B = A'B + AB'
(A=0,B=0): A'B=0, AB'=0 → 0   |  (A=0,B=1): 1 + 0 = 1
(A=1,B=0): 0 + 1 = 1         |  (A=1,B=1): 0 + 0 = 0
=> XOR ka output sirf alag inputs par 1 — verified ✔

Q6. NAND se NOT, AND, OR banao (3 gate-level circuits)
NOT  : dono inputs jodo → (A·A)' = A'
AND  : NAND + NOT (do NAND gates)
OR   : dono inputs ko pehle NOT karo, phir NAND → (A'·B')' = A + B ✔

1.9 Types of Software

Utility Software (systems ka helper)

Utility software are small programs that maintain, secure and optimise the system — ye "system software" ke parivaar ka hissa hain (OS ke saath aate hain ya alag se milte hain).

Software vs Hardware

1.10 Computer Languages & Translators

Extra points (exam me 1–2 line add karna achha lagta hai): low-level languages hardware ke bahut kareeb hoti hain isliye fast aur memory-efficient, par har machine ke liye alag code likhna padta hai. High-level languages portable, readable aur maintainable hoti hain — ek hi program Windows/Linux/Mac par (recompile karke) chalta hai. Isi liye aaj ka saara commercial software HLL me banta hai, aur sirf speed-critical hisse (drivers, kernel) assembly me.

Translators: Compiler vs Interpreter vs Assembler

1.11 Planning a Computer Program — Algorithm, Flowchart, Pseudocode

Coding se pehle programmer problem ko plan karta hai — is planning ke teen tools hain: algorithm (steps in words), flowchart (steps in diagram), aur pseudocode (steps in English + programming keywords). Ye teeno "program design" ke tools hain, aur BCA ke exams me inpar 5–10 marks ka question is an important practice topic.

Algorithm

Examples — likh kar yaad karo, exam me daigram ke saath aate hain:

Algo 1: Sum of two numbers           Algo 2: Largest of two numbers
Step 1: Start                        Step 1: Start
Step 2: Read A, B                    Step 2: Read A, B
Step 3: SUM = A + B                  Step 3: If A > B then print "A is largest"
Step 4: Print SUM                    Step 4: Else print "B is largest"
Step 5: Stop                         Step 5: Stop

Algo 3: Even / Odd                   Algo 4: Factorial of N
Step 1: Start                        Step 1: Start
Step 2: Read N                       Step 2: Read N
Step 3: If (N mod 2 = 0)             Step 3: F = 1, I = 1
        print "Even"                 Step 4: While I <= N
        else print "Odd"                    F = F * I
Step 4: Stop                                I = I + 1
                                     Step 5: Print F
                                     Step 6: Stop

Algo 5: Sum 1 to N (loop needed)     Algo 6: Swap two numbers (with temp)
Step 1: Start                        Step 1: Start
Step 2: Read N                       Step 2: Read A, B
Step 3: S = 0, I = 1                 Step 3: TEMP = A
Step 4: While I <= N                       A = B
        S = S + I                          B = TEMP
        I = I + 1                    Step 4: Print A, B
Step 5: Print S                      Step 5: Stop
Step 6: Stop

Algo 7: Largest of three numbers      Algo 8: Check prime
Step 1: Start                        Step 1: Start
Step 2: Read A, B, C                 Step 2: Read N
Step 3: If A > B and A > C           Step 3: I = 2, FLAG = 0
        print A                      Step 4: While I <= N/2
Step 4: Else if B > C                       If N mod I = 0 then FLAG = 1
        print B                                    break
Step 5: Else print C                        I = I + 1
Step 6: Stop                         Step 5: If FLAG = 0 print "Prime" else "Not Prime"
                                     Step 6: Stop

Flowchart

Example 1 — Largest of two numbers (branches):

        ┌──────────┐
        │  START   │   (oval)
        └────┬─────┘
             ▼
        ┌──────────┐
        │ Read A,B │   (parallelogram)
        └────┬─────┘
             ▼
          ╱╲
         ╱  ╲  A > B ?         (diamond)
         ╲  ╱
          ╲╱
      Yes  │  No
     ┌─────┴─────┐
     ▼           ▼
┌─────────┐  ┌─────────┐
│ Print   │  │ Print   │
│ "A big" │  │ "B big" │
└────┬────┘  └────┬────┘
     └─────┬──────┘
           ▼
      ┌──────────┐
      │   STOP   │
      └──────────┘

Example 2 — Sum 1 to N (loop = diamond se wapas jaana):

   START ─▶ Read N ─▶ S=0, I=1 ─▶ ◇ I ≤ N? ──No──▶ Print S ─▶ STOP
                                    │Yes
                                    ▼
                        S = S + I, I = I + 1  ──(wapas ◇ par)──┐
                                    └─────────────────────────────┘

Pseudocode

BEGIN                       BEGIN // factorial
  READ A, B                   READ N
  IF A > B THEN               F = 1
     PRINT "A is largest"     FOR I = 1 TO N DO
  ELSE                            F = F * I
     PRINT "B is largest"     END FOR
  END IF                      PRINT F
END                         END

BEGIN // even/odd           BEGIN // sum 1 to N
  READ N                      READ N
  IF N MOD 2 = 0 THEN         S = 0
     PRINT "Even"             FOR I = 1 TO N DO
  ELSE                            S = S + I
     PRINT "Odd"              END FOR
  END IF                      PRINT S
END                         END

Algorithm vs Flowchart vs Pseudocode

Algorithm & flowchart practice set (solutions)

Q1. Sum of even numbers from 1 to N
Step 1: Start
Step 2: Read N
Step 3: S = 0, I = 2
Step 4: While I <= N
            S = S + I
            I = I + 2
Step 5: Print S
Step 6: Stop
(Flowchart me: diamond "I <= N?", Yes branch me process box S=S+I, I=I+2,
 wapas diamond par; No branch me Print S → STOP)

Q2. Table of a number N (1 se 10 tak)
Step 1: Start
Step 2: Read N
Step 3: I = 1
Step 4: While I <= 10
            Print N × I
            I = I + 1
Step 5: Stop

Q3. Count digits of a number N (jaise 4567 me 4 digits)
Step 1: Start
Step 2: Read N, COUNT = 0
Step 3: While N > 0
            N = N / 10   (integer division — last digit hat gaya)
            COUNT = COUNT + 1
Step 4: Print COUNT
Step 5: Stop

Q4. Smallest of three numbers (nested decision)
Step 1: Start
Step 2: Read A, B, C
Step 3: If A < B and A < C then print A
Step 4: Else if B < C then print B
Step 5: Else print C
Step 6: Stop

Q5. Print first N natural numbers
Step 1: Start
Step 2: Read N, I = 1
Step 3: While I <= N
            Print I
            I = I + 1
Step 4: Stop

Unit 1 ke mixed MCQs (answers neeche)

Unit 1 short-note practice (model answers ki outline ke saath)

Unit 1 — key terms glossary (MCQ ke liye)

Model answers — Unit 1 ke 5 sabse important questions

Ye "ready-made" answers hain — exam me isi dhaanche me apne shabdon me likh do, Model answer outline—adapt it to the exact question and required depth.

UNIT 2 — Classification & Anatomy of a Computer

2.1 Classification of Digital Computer Systems

Digital computers ko unke size, power, cost aur number of users ke hisaab se chaar classes me baant te hain: microcomputer, minicomputer (mid-range), mainframe aur supercomputer. Order yaad rakho — chhote se bade: Micro → Mini → Mainframe → Super. Jaise-jaise size badhta hai, price, speed aur users ki sankhya badhti hai.

2.2 Anatomy of a Computer (Computer ke andar kya hota hai)

Desktop computer ke cabinet (CPU box) ke andar ye parts hote hain — sab motherboard par lage ya usse jude hote hain:

2.3 Central Processing Unit (CPU) — "Computer ka Dimaag"

(a) ALU — Arithmetic Logic Unit

Do tarah ke operations karta hai: Arithmetic (+, −, ×, ÷, increment/decrement, comparison bhi) aur Logical (>, <, =, AND/OR/NOT, true/false decisions). Saara calculation — is notes file me jo bhi total/average nikala jaata hai — ALU hi karta hai.

(b) CU — Control Unit

Poori machine ka traffic police 🚦. Na khud calculate karta hai na store karta hai — bas instructions fetch karke decode karta hai aur doosre units ko command deta hai ki kya karna hai, kab karna hai. Clock ke saath synchronise ho kar poora instruction cycle chalata hai (fetch → decode → execute → store).

(c) Registers (CPU ke andar ki smallest & fastest memory)

Registers sabse chhoti aur sabse tez memory hain — CPU ke andar hi hote hain, aur instruction ke chalte waqt kaam aane wala data inme aata hai. Space bahut kam (8/16/32/64 bit per register) par speed nanoseconds me.

Microprocessor — details & evolution

2.4 Storage Units & Memory

Units of storage

Memory Hierarchy (speed vs cost vs size)

Rule: jaise-jaise aap upar jaate ho (registers ki taraf), speed badhti hai, size ghatta hai, cost badhta hai. CPU har baar slow memory par depend na kare — isliye frequently used data cache me "copy" rakha jaata hai. Yehi "locality of reference" ka idea hai.

Primary Memory: RAM vs ROM

Storage & memory practice set (solutions ke saath) Exam practice

Q1. 4 GB kitne MB ke barabar hai?
4 × 1024 = 4096 MB ✔
(Aur agar decimal maane to 4 × 1000 = 4000 MB — isi farak se storage kam dikhta hai.)

Q2. 32 GB pen drive me 5 MB ke kitne photos aa jayenge?
32 GB = 32 × 1024 = 32,768 MB
32,768 ÷ 5 ≈ 6,553 photos ✔ (thoda OS/folder overhead hata ke ~6,500)

Q3. 2 GB ki movie 100 Mbps connection par kitni der me download hogi?
2 GB = 2,048 MB ≈ 16,384 Mb (megabits, ×8)
Time = 16,384 ÷ 100 ≈ 164 second ≈ 2 minute 44 second ✔

Q4. Ek laptop me 8 GB RAM aur 512 GB SSD hai. Iska matlab?
RAM (8 GB) = chalte hue programs ka temporary area (volatile)
SSD (512 GB) = permanent storage — Windows, files, photos
Dono alag cheezein hain; "RAM badhaane" se apps fast hote hain,
"storage badhaane" se zyada files rakh sakte ho. ✔

Q5. Aapka 300 MB ka game RAM (4 GB, 3 GB used) me load nahi ho raha.
Kya hoga? OS virtual memory use karega (SSD/HDD ka hissa) —
game chalega par "stuttering/lag" ke saath, kyunki SSD RAM se bahut slow hai. ✔

Q6. Kaun data power off hone par bacha rahega: RAM ka, ROM ka, ya cache ka?
Sirf ROM (aur permanent storage). RAM aur cache volatile hain — power gayi to khali. ✔

Unit 2 MCQs — storage/memory

2.5 Secondary Storage Devices

2.6 Input Devices

Input devices ke categories (short note ready)

2.7 Output Devices

Monitor ke types (short note ready)

Device selection practice (real-life scenarios)

2.8 How CPU and Memory Work Together (program kaise chalta hai)

Instruction cycle (step-by-step):

1. FETCH   → PC batata hai agla instruction ka address
           → MAR me address jata hai → address bus par memory ko
           → memory se instruction aata hai → MDR → IR me load
2. DECODE  → CU instruction ko samajhta hai: "ye ADD karna hai"
           → kaunse operands chahiye, kahan se laane hain — sab decode
3. EXECUTE → zarurat ka data ALU me jaata hai → calculation/logic hota hai
           → result ACC/register me aata hai
4. STORE   → result ko register se RAM ke location me likh diya jaata hai
           → PC = PC + 1 (agla instruction) → cycle repeat 🔁

(Interrupt aaye to CPU saari state save karke interrupt handler chalata hai,
 phir wahi se resume — isliye music chalti rehti hai jab aap typing karte ho 🎵)

2.9 Program Execution — C program ke perspective se

Source file hello.c (aapne likha: C language)
   │
   ▼ 1. PREPROCESSOR ( #include, #define expand karta hai; comments hataata hai)
Preprocessed code
   │
   ▼ 2. COMPILER (C → assembly) → syntax error yahin milti hai
Assembly code (hello.s)
   │
   ▼ 3. ASSEMBLER (assembly → object code/machine code)
hello.o (object file)
   │
   ▼ 4. LINKER (object file + libraries, jaise stdio.h ka code, jodta hai)
hello.exe (executable)
   │
   ▼ 5. LOADER (exe ko RAM me load karta hai, PC set karta hai)
RAM me program chal raha hai → CPU instruction cycle → OUTPUT 🎉

Instruction cycle practice (register-level trace)

Maan lo program ka instruction hai: ADD R1, R2 (R1 = 5, R2 = 3). Trace likho — exam me yahi style complete-answer checklist dilata hai:

2.10 Introduction to Microcontrollers

Microcontroller ke parts

Microprocessor & CPU practice (short notes + comparison)

Unit 2 — key terms glossary

Model answers — Unit 2 ke 5 sabse important questions

UNIT 3 — Operating System & Networking Fundamentals

3.1 Operating System — Introduction & Functions

Functions of an Operating System (study the listed functions and explain relevant ones)

Process aur uska life cycle (short note)

CPU scheduling ka kaam: ye decide karna ki ready queue me se kaunsa process kab CPU paye — simple policies jaise FCFS (jo pehle aaya), SJF (sabse chhota kaam pehle), Round Robin (har process ko thoda-thoda time), Priority (zaroori pehle). Isi se multi-tasking "feel" hota hai — har app ko seconds ke chhote tukde milte rehte hain. Isi ko context switching kehte hain: CPU ek process ki state save karke doosri load karta hai.

CPU scheduling — worked example (numeral-based question)

Teen processes (time 0 par aaye, CPU ek saath ek hi process chala sakta hai):
P1: 5 ms burst | P2: 3 ms burst | P3: 2 ms burst

(a) FCFS (jo pehle aaya, pehle chalega): P1 → P2 → P3
P1: 0 se 5    (wait 0)
P2: 5 se 8    (wait 5)
P3: 8 se 10   (wait 8)
Average waiting time = (0 + 5 + 8) ÷ 3 = 13 ÷ 3 ≈ 4.33 ms

(b) SJF (sabse chhota kaam pehle): P3 → P2 → P1
P3: 0 se 2    (wait 0)
P2: 2 se 5    (wait 2)
P1: 5 se 10   (wait 5)
Average waiting time = (0 + 2 + 5) ÷ 3 = 7 ÷ 3 ≈ 2.33 ms
→ SJF ne average waiting time kam kar diya (yehi iska fayda hai).

(c) Round Robin (time slice = 2 ms): P1 → P2 → P3 → P1 → P2 → P1
P1: 0–2, 6–8, 10–11  (total 5 ms mila)
P2: 2–4, 8–10        (total 3 ms)
P3: 4–6              (total 2 ms)
Sabko fair time mila, aur koi process zyada der "starve" nahi hua —
isliye time-sharing systems (Windows/Linux) me RR jaisa logic chalta hai.

3.2 Classification of Operating Systems

3.3 System Programs, Application Programs and Utilities

3.4 The Unix Operating System & Basic Commands

Unix File System (structure yaad rakho)

Root: /  (sab kuch iske neeche)
├── bin   → basic commands (ls, cp, mv)
├── etc   → configuration files
├── home  → users ki personal files (/home/ansh)
├── dev   → devices (keyboard, disk)
├── var   → logs, changing data
├── tmp   → temporary files
├── usr   → user programs aur libraries
├── opt   → extra software
└── root  → superuser (admin) ka home

Path: absolute path root se shuru hota hai (/home/ansh/notes/c.txt), relative path current directory se (notes/c.txt). . = current folder, .. = parent folder, ~ = aapka home. Windows me drive letters (C:\) hote hain, Unix me sab kuch ek hi tree me hota hai — pen drive bhi /media/usb jaisi jagah "mount" hoti hai.

Unix practice set — permissions aur commands (solutions ke saath) Exam practice

Q1. Permission likho octal me: owner = rwx, group = r-x, others = ---
owner: r+w+x = 4+2+1 = 7
group: r+x    = 4+0+1 = 5
others: none  = 0
=> chmod 750 ✔  (script ya private folder ke liye common)

Q2. Permission: owner = rw-, group = r--, others = r--  (normal file)
6, 4, 4 → chmod 644 ✔ (aaj kal har website file ka default)

Q3. Sabko saari permissions (exam me "security risk" likhna)
chmod 777 ✔

Q4. Sirf owner padh sake (private file — jaise password file)
owner = r-- = 4, group = 0, others = 0 → chmod 400 ✔

Q5. Ek script jo sab chala sake par sirf owner badal sake
owner = rwx = 7, group = r-x = 5, others = r-x = 5 → chmod 755 ✔

Q6. Ulta sawaal: chmod 640 ka kya matlab?
owner = 6 = rw-, group = 4 = r--, others = 0 = ---
"Owner read/write kar sakta hai, group ka sirf read, baaki koi kuch nahi." ✔

Q7. Commands likho (tasks → answers)
(a) Aap kahan ho, pata karo                     → pwd
(b) Saari files details ke saath dekho          → ls -l
(c) Naya folder "bca" banao                     → mkdir bca
(d) usme jao                                     → cd bca
(e) File a.txt banao aur usme "Hello" likho      → echo "Hello" > a.txt
(f) Uski copy b.txt banao                        → cp a.txt b.txt
(g) File ka content dekho                        → cat a.txt
(h) File me "BCA" text dhundho                   → grep "BCA" a.txt
(i) File ka naam badal kar note.txt karo          → mv a.txt note.txt
(j) File delete karo                             → rm note.txt
(k) Running processes dekho                      → ps -ef
(l) Disk space check karo                        → df -h
(m) File ka permission 755 karo                  → chmod 755 script.sh

3.5 Microkernel-Based Operating System

3.6 Booting

Booting ke steps (exam me yahi likhna):

1. Power ON → SMPS sab parts ko current deta hai (CPU reset se start)
2. POST (Power On Self Test) → BIOS/UEFI hardware check karta hai
   (RAM kitni hai, keyboard hai? galat ho to beep 🔔)
3. BIOS/UEFI boot device dhundhta hai (boot order: HDD/SSD/USB/DVD/CD-ROM)
4. Bootloader (GRUB/Windows Boot Manager) load hota hai (MBR/EFI se)
5. OS kernel RAM me load hota hai → hardware drivers initialise
6. OS services start karta hai → login screen aa jaati hai ✅
   (Fir background me: user login → desktop → apps ready)

3.7 Introduction to Networks & Types of Networks

3.8 Peer-to-Peer vs Client-Server Network

3.9 Network Topologies (layout ke hisaab se)

Bus:      PC ──┬── PC ──┬── PC ────── main cable (backbone) ── terminator

Star:          PC        PC
                ╲       ╱
                 SWITCH ──── PC         (aaj college lab/ghar WiFi isi shape me)
                ╱       ╲
              PC        PC

Ring:      PC ──▶ PC          (data ek direction me circle ghoomta hai)
           ▲       │
           └── PC ◀┘

Mesh:      PC ──── PC          (partial mesh: har PC kam se kam 2 links se juda)
           │ ╲  ╱ │
           │  ╳   │
           │ ╱  ╲ │
           PC ──── PC

Tree:      Backbone switch ─── dept switch (1) ─── lab switch ── PCs
                           └── dept switch (2) ─── office PCs

3.10 Networking Terms & Devices (glossary — MCQ ka khazana)

Network calculation practice (solutions ke saath)

Q1. 4 Mbps connection par 10 MB ka file?
Speed = 4 ÷ 8 = 0.5 MB/s
Time  = 10 ÷ 0.5 = 20 second ✔

Q2. 50 Mbps par 1 GB (1024 MB) file?
Speed = 50 ÷ 8 = 6.25 MB/s
Time  = 1024 ÷ 6.25 = 163.84 s ≈ 2 minute 44 second ✔

Q3. 100 Mbps par 2.5 GB (2560 MB) movie?
Speed = 12.5 MB/s → Time = 2560 ÷ 12.5 = 204.8 s ≈ 3 min 25 s ✔

Q4. 5 Mbps ka hotspot 4 log share kar rahe hain, sab download kar rahe hain.
Har ek ko approx 5 ÷ 4 = 1.25 Mbps ≈ 156 KB/s milta hai ✔
(Isliye "WiFi slow hai" ka ek karan share karna bhi hai.)

Q5. Online game me ping 20 ms hai aur ek doosre server par 200 ms —
farak? 200 ms me lag/teleport feel hota hai; 20 ms me smooth ✔
Bandwidth aur latency alag hain: file download bandwidth se fast hoti hai,
par gaming latency se smooth hoti hai.

Topology practice (diagram-based questions)

OS type matching practice + booting questions

Booting sequence ko sahi order me lagao (jumbled steps ka question):
(a) OS kernel RAM me load hota hai
(b) POST hardware check karta hai
(c) Bootloader (GRUB) chalta hai
(d) Login screen aati hai
(e) Power ON + CPU reset
Sahi order: (e) → (b) → (c) → (a) → (d) ✔

Troubleshooting (short answers):
Q1. "Boot device not found" ka matlab?
   → BIOS ko bootable disk nahi mili — HDD/SSD kharab/cable loose/boot order galat.
Q2. Windows slow boot hota hai — do reasons?
   → HDD (SSD ki jagah), startup apps zyada, ya disk full (free space kam).
Q3. Ctrl+Alt+Del kaunsa boot hai?
   → Warm boot (restart — power band nahi hoti).

Unit 3 — key terms glossary

Model answers — Unit 3 ke 5 sabse important questions

UNIT 4 — Introduction to Database Management Systems

4.1 Database, DBMS & Why Database

File System vs DBMS (classic 10-mark question)

DBMS ke advantages & limitations

DBMS components (short-note practice)

4.2 Database Applications & Users

Applications (kahan-kahan chalta hai)

Database Users (kaun-kaun use karta hai)

4.3 Introduction to SQL

SQL ke 5 sub-languages (yaad karne ka golden trick — "DDL DML DCL TCL"): Define (structure) → Data (rows) → Domain/Login (control) → Transaction → Query (fetch).

4.4 Data Types in SQL

4.5 DDL with Constraints (structure + rules)

-- Table banana
CREATE TABLE student (
  roll_no   INT PRIMARY KEY,              -- unique + not null
  name      VARCHAR(50) NOT NULL,
  email     VARCHAR(60) UNIQUE,           -- duplicate email nahi
  age       INT CHECK (age >= 17),        -- rule: 17 se kam age allowed nahi
  city      VARCHAR(30) DEFAULT 'Kichha',
  marks     INT,
  course_id INT,
  FOREIGN KEY (course_id) REFERENCES course(course_id)
);
-- Structure change
ALTER TABLE student ADD phone VARCHAR(15);        -- naya column
ALTER TABLE student MODIFY name VARCHAR(80);      -- size badla
ALTER TABLE student DROP COLUMN phone;            -- column hatana
-- Hatana
TRUNCATE TABLE student;   -- saari rows (structure rahega)
DROP TABLE student;       -- table hi hat gaya (structure + data)
-- Constraints baad me bhi add ho sakte hain
ALTER TABLE student ADD CONSTRAINT chk_marks CHECK (marks BETWEEN 0 AND 100);

Keys in SQL (short-note practice)

4.6 DML — Practical SQL examples

Sab examples ke liye ye do tables maan lo (schema §4.5 me banaya tha):

-- 1. INSERT (nayi rows)
INSERT INTO student (roll_no, name, email, age, city, marks, course_id)
VALUES (1, 'Ansh', 'ansh@mail.com', 19, 'Kichha', 78, 101);
INSERT INTO student VALUES (2, 'Priya', 'priya@mail.com', 20, 'Haldwani', 85, 102);
-- (jab saari columns ka value de rahe ho to column list optional hai)

-- 2. SELECT (padhna) — kuch columns
SELECT name, city FROM student;

-- 3. WHERE (condition) — AND, OR, NOT
SELECT * FROM student WHERE age >= 19 AND city = 'Kichha';
SELECT * FROM student WHERE city = 'Kichha' OR city = 'Haldwani';

-- 4. BETWEEN, IN, LIKE, IS NULL
SELECT * FROM student WHERE marks BETWEEN 70 AND 90;      -- 70 se 90 ke beech
SELECT * FROM student WHERE city IN ('Kichha','Haldwani'); -- list me se koi
SELECT * FROM student WHERE name LIKE 'A%';               -- A se shuru (Ansh, Aman)
SELECT * FROM student WHERE email IS NULL;                -- NULL wale

-- 5. Sorting + limit
SELECT * FROM student ORDER BY marks DESC;                -- zyada marks pehle
SELECT * FROM student ORDER BY city ASC, marks DESC;      -- city se, phir marks
SELECT * FROM student LIMIT 3;                            -- sirf top 3 rows

-- 6. Aggregate (ganit wale functions) + GROUP BY
SELECT COUNT(*), AVG(marks), MAX(marks), MIN(marks), SUM(marks) FROM student;
SELECT city, COUNT(*) AS students FROM student GROUP BY city;
SELECT city, AVG(marks) FROM student GROUP BY city HAVING AVG(marks) > 60;

-- 7. UPDATE (rows badalna — WHERE ke bina sab rows badal jaayengi! ⚠️)
UPDATE student SET city = 'Rudrapur' WHERE roll_no = 2;
UPDATE student SET marks = marks + 5 WHERE city = 'Kichha';

-- 8. DELETE (rows hatana)
DELETE FROM student WHERE roll_no = 5;
-- DELETE FROM student;   -- ⚠️ saari rows gayi

-- 9. Do tables jodna (JOIN)
SELECT s.name, c.course_name, c.fees
FROM student s JOIN course c ON s.course_id = c.course_id;

Outputs (exam me queries ke saath result bhi likhna — yahi useful answer-writing step hai)

SQL practice set — queries aur expected output Exam practice

Neeche wale sawaal §4.6 ki sample tables (student, course) par hain. Pehle khud likho, phir output milao.

Q1. 70 se zyada marks wale students
SELECT * FROM student WHERE marks > 70;
Output: Ansh (78), Priya (85), Neha (91)               ✔ 3 rows

Q2. Kichha ke wo students jinki age 18 se zyada hai
SELECT name FROM student WHERE city = 'Kichha' AND age > 18;
Output: Ansh, Aman                                    ✔ (Rahul 18 hai — ">" me nahi aayega)

Q3. Marks ke hisaab se descending order me saare students
SELECT name, marks FROM student ORDER BY marks DESC;
Output: Neha 91, Priya 85, Ansh 78, Rahul 65, Aman 55  ✔

Q4. Kichha me kitne students hain?
SELECT COUNT(*) FROM student WHERE city = 'Kichha';
Output: 3                                             ✔

Q5. Har city ka average marks
SELECT city, AVG(marks) FROM student GROUP BY city;
Output: Kichha 66.0 (198/3), Haldwani 85.0, Rudrapur 91.0  ✔

Q6. Sirf wo cities jahan 2 ya zyada students hain
SELECT city, COUNT(*) FROM student GROUP BY city HAVING COUNT(*) >= 2;
Output: Kichha 3                                      ✔  (WHERE se nahi — HAVING se!)

Q7. Kichha ke sab students ke marks 5 badha do
UPDATE student SET marks = marks + 5 WHERE city = 'Kichha';
Output: 3 rows updated → Ansh 83, Rahul 70, Aman 60      ✔

Q8. Wo students jo 23,000 se kam fees wale course me hain (JOIN)
SELECT s.name, c.course_name FROM student s
JOIN course c ON s.course_id = c.course_id WHERE c.fees < 23000;
Output: Priya BBA, Aman BBA, Neha BCom                 ✔

Q9. 'A' se shuru hone wale naam
SELECT name FROM student WHERE name LIKE 'A%';
Output: Ansh, Aman                                    ✔

Q10. Maximum, minimum aur total marks
SELECT MAX(marks), MIN(marks), SUM(marks) FROM student;
Output: 91, 55, 374                                   ✔

Q11. 60 se 90 ke beech marks
SELECT * FROM student WHERE marks BETWEEN 60 AND 90;
Output: Ansh 78, Priya 85, Rahul 65                    ✔

Q12. Haldwani ya Rudrapur ke students
SELECT name FROM student WHERE city IN ('Haldwani','Rudrapur');
Output: Priya, Neha                                   ✔

4.7 DCL & TCL

-- DCL (permission dena/lena)
GRANT SELECT, INSERT ON student TO user_ansh;      -- right do
GRANT ALL PRIVILEGES ON student TO admin_user;     -- sab kuch
REVOKE INSERT ON student FROM user_ansh;           -- right wapas lo

-- TCL (transaction control)
START TRANSACTION;                                  -- ya BEGIN;
  UPDATE account SET balance = balance - 500 WHERE acc_no = 'A101';
  UPDATE account SET balance = balance + 500 WHERE acc_no = 'A102';
SAVEPOINT after_transfer;                           -- beech ka point
  UPDATE account SET balance = balance - 100 WHERE acc_no = 'A101';
ROLLBACK TO after_transfer;                         -- sirf aakhri update cancel
COMMIT;   -- poora transaction pakka ho gaya (permanent)
-- agar beech me error aa jaye to:
ROLLBACK; -- sab cancel (jaise kuch hua hi nahi)

ACID properties — explain each property with a transaction example

Constraint violations — "ye query kyun fail hogi?" practice

Unit 4 — key terms glossary

Model answers — Unit 4 ke 4 sabse important questions

UNIT 5 — Internet Basics & Web Basics

5.1 Introduction to the Internet

Features of Internet

• Global reach: Duniya me kahin bhi — real-time connect.
• 24×7 availability: Din ho ya raat, chalta rehta hai (data centre + undersea cables).
• Fast & cheap communication: Email kisi bhi desh me, basically free; video call bhi.
• Huge information (WWW): Wikipedia se lekar YouTube tak sab — search karne par seconds me.
• Any device: PC, mobile, smart TV, watch, car — sab jud sakta hai (IoT).
• Entertainment & business: Video, gaming, online shopping, digital payments.
• Anonymity / open publishing: Koi bhi content publish kar sakta hai (blog, YouTube) — iske fayde-nuksaan dono (fake news bhi isi wajah se).
• Interactive: Sirf padhna nahi — comment, video call, online banking, gaming bhi.
• Scalable & resilient: Naye devices/connections lagte rehte hain; ek rasta fail ho to packet doosra rasta le leta hai.

Internet Applications & Services

5.2 Logical and Physical Addresses

5.3 Internet Service Providers (ISP)

5.4 Domain Name System (DNS)

DNS resolution process (exam answer):

1. Aap browser me likhte ho: www.surajmaluniversity.ac.in
2. Computer pehle apna local cache / hosts file dekhta hai
3. Nahi mila → ISP ke "Recursive DNS Resolver" ko poochhta hai
4. Resolver chain me poochhta hai:
      Root Server (.) → TLD Server (.in) → Authoritative Name Server (surajmaluniversity.ac.in)
5. IP address mila (jaise 103.21.xx.xx) → browser us server se connect
6. Page load 🎉 aur IP thodi der cache me save (TTL ke hisaab se — isliye next time fast)
   + OS, browser aur resolver — teeno level par caching hoti hai

Domain name ke parts (example: www.bca.surajmaluniversity.ac.in)

Common TLDs: .com (commercial), .org (organisation), .net (network), .edu (education, US), .gov (government), .in (India), .uk, .io, .ai. DNS records (short note): A = naam → IPv4, AAAA = naam → IPv6, CNAME = alias (www → main site), MX = mail server, NS = nameserver, TXT = verification/SPF records. URL vs Domain: Domain = naam ke hissa (google.com), URL = poora address (https://google.com/search?q=bca).

5.5 Web Basics — Introduction to Web

Web ka client-server model

How a web page loads (request-response cycle):

1. Aap URL type karte ho → browser URL ko todta hai
   (protocol + domain + path)
2. DNS se IP address milta hai
3. Browser web server ko HTTP/HTTPS request bhejta hai (GET /index.html)
   + browser headers bhejta hai (cookies, language, device info)
4. Server page (HTML/CSS/JS/images) response me bhejta hai (status code ke saath)
5. Browser HTML parse karta hai → DOM banata hai
   → CSS se style → images/JS load → page render (screen pe dikhta hai)
6. Tumhare clicks pe JS DOM update karta hai — bina page reload ke bhi kuch badal sakta hai

5.6 Web Browser

Examples: Google Chrome, Mozilla Firefox, Microsoft Edge, Safari, Opera, Brave, mobile me inbuilt browsers.

5.7 HTTP / HTTPS

HTTPS me connection kaise banta hai (TLS handshake, simple version):

1. Browser server se "hello" — kaun ho, kaunse encryption methods support karte ho?
2. Server apna SSL/TLS certificate bhejta hai (CA ne verify kiya hua)
3. Browser certificate check karta hai (domain match? valid? CA trusted?)
4. Dono ek secret session key banate hain (asymmetric → symmetric)
5. Ab saara data us session key se encrypt hokar aata-jaata hai (padlock 🔒)
Isliye https site par form bharne ka data public WiFi par bhi padha nahi ja sakta.

5.8 URL (Uniform Resource Locator)

https://www.surajmaluniversity.ac.in/bca/admission.html?year=2026#fees
└─┬─┘   └───────────┬───────────┘ └──────┬──────┘ └──┬──┘ └─┬─┘
protocol        domain name              path       query  fragment
  │                  │                     │           │       │
  1                  2                     3           4       5

1. Protocol     → https (kis tarah baat karni hai) — http/https/ftp
2. Domain/Host  → www.surajmaluniversity.ac.in (kis server se)
3. Port (optional) → :443 (default chhupa hota hai — http me 80)
4. Path         → /bca/admission.html (server me kaunsi file/page)
5. Query string → ?year=2026 (extra data bhejna; & se kai parameters)
6. Fragment     → #fees (page ke usi section par jump)

5.9 HTML5

<!DOCTYPE html>                        <!-- browser ko batao: ye HTML5 page hai -->
<html lang="en">                       <!-- page ki shuruaat, language -->
<head>                                 <!-- meta info — screen par nahi dikhta -->
  <meta charset="UTF-8">               <!-- Hindi/emoji ke liye encoding -->
  <meta name="viewport" content="width=device-width, initial-scale=1">
  <title>My First Page</title>         <!-- tab me dikhta hai -->
</head>
<body>                                 <!-- jo dikhta hai woh yahan -->
  <h1>Hello BCA!</h1>                   <!-- heading (h1 sabse bada) -->
  <p>Ye <b>mere</b> pehla page hai.</p>   <!-- paragraph -->
  <a href="https://google.com">Google</a> <!-- hyperlink -->
  <img src="photo.jpg" alt="mera photo" width="200">
  <ul><li>HTML</li><li>CSS</li></ul>     <!-- list -->
  <audio controls src="song.mp3"></audio>  <!-- HTML5 features -->
</body>
</html>

HTML5 practice set (code likh kar output socho)

Q1. Ek page banao jisme: heading "My College", ek paragraph, ek image,
    ek list (3 subjects) aur ek link ho.

<!DOCTYPE html>
<html lang="en">
<head>
  <meta charset="UTF-8">
  <title>My College</title>
</head>
<body>
  <h1>My College</h1>
  <p>Main <b>BCA</b> first year ka student hoon.</p>
  <img src="college.jpg" alt="college building" width="300">
  <ul>
    <li>Computer Fundamentals</li>
    <li>C Programming</li>
    <li>Mathematics</li>
  </ul>
  <a href="https://surajmaluniversity.ac.in">College website</a>
</body>
</html>
Output: browser me heading, bold "BCA" wala paragraph, photo, bullet list,
aur click karne par college site khulne wala link dikhega. ✔

Q2. Marksheet table banao (3 students, 3 columns: Roll, Name, Marks)
<table border="1">
  <tr><th>Roll</th><th>Name</th><th>Marks</th></tr>
  <tr><td>1</td><td>Ansh</td><td>78</td></tr>
  <tr><td>2</td><td>Priya</td><td>85</td></tr>
  <tr><td>3</td><td>Rahul</td><td>65</td></tr>
</table>
Note: <th> = heading cell (bold + center), <td> = data cell. ✔

Q3. Login form banao (email + password + submit button)
<form action="/login" method="post">
  <label>Email:</label>
  <input type="email" name="email" required><br>
  <label>Password:</label>
  <input type="password" name="pass" required><br>
  <button type="submit">Login</button>
</form>
Exam point: method="post" — password GET me URL me dikh jata (security issue). ✔

Q4. HTML5 ke semantic tags se page ka structure banao
<body>
  <header><h1>BCA Notes</h1></header>
  <nav><a href="#u1">Unit 1</a> | <a href="#u2">Unit 2</a></nav>
  <main>
    <section id="u1"><h2>Unit 1</h2><p>Fundamentals...</p></section>
    <section id="u2"><h2>Unit 2</h2><p>CPU...</p></section>
  </main>
  <footer><p>© 2026 Ansh</p></footer>
</body>
Fayda: search engine aur screen reader ko page ka matlab samajh aata hai. ✔

Q5. Audio/video embed karo (HTML5 ka bada feature)
<audio controls><source src="song.mp3" type="audio/mpeg"></audio>
<video width="320" controls><source src="clip.mp4" type="video/mp4"></video>

5.10 CSS (Cascading Style Sheets)

CSS ka syntax:   selector { property: value; property: value; }

<style>   /* internal CSS — head me likha */
  h1        { color: #6d28d9; text-align: center; }   /* tag selector */
  p         { font-family: Arial; font-size: 16px; }  /* tag selector */
  .card     { background: #fff; border-radius: 12px; padding: 16px; }  /* class */
  #main-title { font-size: 32px; font-weight: bold; }  /* id */
  h1, h2    { margin-bottom: 10px; }                   /* group */
  a:hover   { color: red; }                            /* pseudo-class */
</style>
<h1 id="main-title">BCA Notes</h1>
<p class="card">CSS se sundar ban gaya.</p>

Web scenario practice (HTTP/DNS/URL ke chhote sawaal)

CSS practice set (solutions ke saath)

Q1. Heading ko purple + center karo
h1 { color: #6d28d9; text-align: center; }

Q2. Ek "card" class banao: white background, rounded corners, shadow, padding
.card { background:#fff; border-radius:12px; box-shadow:0 2px 8px rgba(0,0,0,.1);
        padding:16px; margin:12px 0; }

Q3. Link par hover effect (mouse le jaane par colour badle)
a:hover { color: red; text-decoration: underline; }

Q4. Do classes: "pass" green, "fail" red (result page ke liye)
.pass { color:#059669; font-weight:bold; }
.fail { color:#dc2626; font-weight:bold; }
Use: <span class="pass">Pass</span> / <span class="fail">Fail</span>

Q5. Box model ka hisaab: width:200px; padding:10px; border:2px solid; margin:8px;
Total visible width = 200 + 10 + 10 + 2 + 2 = 224 px
Layout me jagah     = 224 + 8 + 8 = 240 px ✔
(Aur box-sizing: border-box; lagane par width 200 hi rehti hai — padding/border andar count.)

Q6. Mobile-friendly (responsive) banane ke liye
@media (max-width: 600px) {
  .card { padding: 10px; font-size: 15px; }
  nav a { display: block; }   /* nav links ek ke neeche ek */
}

Unit 5 — key terms glossary

Model answers — Unit 5 ke 5 sabse important questions

Rapid-fire 2-mark Q&A (poore syllabus ka nichod)

Diagram practice list (in sab ko ek baar haath se banao)

1. Basic organization of computer (5 units + buses) + von Neumann diagram
2. Memory hierarchy (registers → cache → RAM → secondary → cloud)
3. Generations timeline (V-TIM-A wala flow)
4. Instruction cycle (fetch→decode→execute→store ka loop)
5. C program ka build flow (preprocess→compile→assemble→link→load→run)
6. Booting steps ka sequence
7. Network topologies (bus, star, ring, mesh — chhote diagrams)
8. OSI-free version: data ka safar (client → DNS → server → response)
9. URL ke parts ka labelled diagram
10. CSS box model (content→padding→border→margin)

Sample question paper (70 marks) — practice set

Attempt karne ka order (exam hall strategy): pehle wahi questions karo jinke answers "pakke" hain (conversions, tables wale short notes) — confidence banta hai aur time bachta hai. Diagram wale question last me rakho agar time kam ho, par unhe chhodna nahi — diagram ke marks aasan hote hain. Aur har answer me heading + table/bullets use karo — examiner ko checking me aasani, aapko marks. 💪