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ECE : Microprocessors - Quiz(MCQ)
A)
A multipurpose PLD that accepts an integer as input
B)
A multipurpose PLD that accepts binary data as input
C)
A multipurpose PLD that accepts whole numbers as input
D)
A multipurpose PLD that accepts prime numbers as input

Correct Answer :   A multipurpose PLD that accepts binary data as input


Explanation : A microprocessor is a multipurpose PLD that accepts binary data as input and processes data according to the instructions, and outputs the result. Binary instructions are read from memory.

A)
Magnetic Ink Chart Receipt
B)
Magnetic Ink Character Receipt
C)
Magnetic Ink Chart Recognition
D)
Magnetic Ink Character Recognition

Correct Answer :   Magnetic Ink Character Recognition


Explanation : Magnetic ink character recognition is used in banking.

A)
Intel's first x86 processor
B)
Motrola’s first x86 processor
C)
NanoXplore x86 processor
D)
STMICROELECTRONICS’s first x86 processor

Correct Answer :   Intel's first x86 processor


Explanation : The Intel 8086 is Intel’s first x86 processor. They launched the most powerful processor in terms of advanced architecture i.e. 8086 processor in 1978. It has larger memory addressing capability and a powerful instruction set.

A)
RISC
B)
EPIC
C)
CISC
D)
All of the above

Correct Answer :   All of the above


Explanation : They are of three types :

CISC : Complex Instruction Set Computer
RISC : Reduced Instruction Set Computer
EPIC : Explicitly Parallel Instruction Computing

A)
-0
B)
+0
C)
-1
D)
+1

Correct Answer :   -0

A)
ROM
B)
Hard disk
C)
Floppy disk
D)
Magnetic tape

Correct Answer :   Magnetic tape


Explanation : Magnetic tape can be accessed only sequentially.

A)
13.5
B)
13.75
C)
13.625
D)
13.875

Correct Answer :   13.625


Explanation : Binary .101 equals 0.625 in decimal.

A)
i = i - 1
B)
i = i + 1
C)
i = i + 2
D)
i = i + i + i

Correct Answer :   i = i + 1


Explanation : i is incremented by 1.

A)
Flag
B)
Main Memory
C)
Secondary Memory
D)
Program Counter

Correct Answer :   Main Memory


Explanation : If the information isn’t in the computer’s main store, the microprocessor can’t do anything with it.

The primary storage area in a computer, also known as main storage or memory, is where data is stored for easy access by the computer’s processor.

Random-access memory (RAM) and memory are frequently used interchangeably to refer to primary or main storage.

A)
TTL
B)
NMOS
C)
PMOS
D)
HMOS

Correct Answer :   PMOS


Explanation : PMOS technology was used for designing the processor because this technology was slow but simple.

11 .
Circuit A is a 1-bit adder; circuit B is a 1 bit multiplier.
[Hint: Construct the truth table for the adder and the multiplier]
A)
Circuit A has more gates than circuit B
B)
Circuit B has more gates than circuit A
C)
Circuit A has the same number of gates as circuit B
D)
None of the above

Correct Answer :   Circuit A has more gates than circuit B

A)
True
B)
False
C)
Can Not Say
D)
None of the above

Correct Answer :   True

13 .
JCOKE = 3
JCOKE = JCOKE + 1
GO TO (5, 8, 9, 11, 15, 16 18, 20) JCOKE.

After the execution of above statement, the control is transferred to statement number
A)
20
B)
16
C)
11
D)
8

Correct Answer :   11


Explaination : It is a computed GO TO statement. Since J COKE = 3 + 1 = 4, the control is transferred to statement 11.

A)
direct addressing mode
B)
register addressing mode
C)
register relative addressing mode
D)
register indirect addressing mode

Correct Answer :   register indirect addressing mode


Explanation : In register indirect addressing mode the address of the operand is stored in the register. Since the instruction specifies that the register used to refer to the address is accessed indirectly, it represents the register indirect addressing mode.

A)
once
B)
twice
C)
thrice
D)
four times

Correct Answer :   thrice


Explanation : When RAL instruction is used once the number is doubled.

A)
EC = IC + EC
B)
IC = FC + EC
C)
IC = FC - EC
D)
IC = FC + 2EC

Correct Answer :   IC = FC + EC


Explanation : Instruction cycle consists of fetch and execute cycles.

A)
1 a
B)
a 1 2
C)
a b 123
D)
a b c 123

Correct Answer :   1 a

A)
none of the numbers is changed to any other form
B)
.4546 E 5 is changed .004546 E 7 and .5433 E 7 is not changed
C)
.5433 E 7 is changed to 54.33 E 5 and .4546 E 5 is not changed
D)
both the numbers are changed and their exponents are made equal to 6

Correct Answer :   .4546 E 5 is changed .004546 E 7 and .5433 E 7 is not changed

A)
a few bytes
B)
a few kilobytes
C)
a few megabytes
D)
a few gigabytes

Correct Answer :   a few kilobytes


Explanation : Size of Cache memory varies from about 16 K bytes to about 256 K bytes.

A)
4-bits – 32 bits
B)
8-bits – 16 bits
C)
8-bits – 32 bits
D)
8-bits – 64 bits

Correct Answer :   8-bits – 64 bits


Explanation : At a time, an 8-bit CPU can process 8 bits of data. Depending on the type of microcomputer, the word length might range from 4 to 64 bits.

A)
246
B)
250
C)
256
D)
278

Correct Answer :   246


Explanation : In an 8-bit microprocessor, maximum 28 = 256 opcodes are possible. But it consists of only 246 opcodes.

A)
Less than
B)
Greater than
C)
The same as
D)
Cannot be determined

Correct Answer :   The same as

A)
Z, C, S, P, AC
B)
D, Z, S, P, AC
C)
Z, CY, S, D, AC
D)
Z, CY, S, P and AC

Correct Answer :   Z, CY, S, P and AC

A)
Primary
B)
Cache
C)
Mass storage
D)
Off line back up

Correct Answer :   Cache


Explanation : Cache memory has a speed of about four times the speed of main memory. It is a small high speed memory.

A)
ALU
B)
CPU
C)
Primary memory
D)
Secondary memory

Correct Answer :   Primary memory


Explanation : Primary memory holds these during processing of instructions.

A)
5 mW each
B)
350 mW each
C)
350 mW and 100 mW respectively
D)
350 mW and 5 mW respectively

Correct Answer :   350 mW and 5 mW respectively


Explanation : Power requirement in stand by mode is very low 1000.

A)
-2.3 E 2
B)
-2.3 e 2
C)
-2.3 e - 2
D)
1.2 e 2.0

Correct Answer :   -2.3 E 2


Explanation : In C real constant expressed in exponential form, mantissia and exponent should be separated by e (and not E).

A)
3
B)
4
C)
5
D)
6

Correct Answer :   4


Explanation : C has four storage classes :

1. automatic storage class
2. register storage class
3. static storage class
4. external storage class.

A)
It is a bidirectional bus
B)
It is 16 bits in length
C)
It consists of control PIN 21 to 28
D)
Lower address bus lines (AD0 – AD7) are called “Line number”

Correct Answer :   It is a bidirectional bus


Explanation : Data bus in the microprocessor is bidirectional but the address bus is unidirectional. AD0 – AD7 are the address lines that can be used for both address and data bus lines.

A)
It has interfacing circuits
B)
It has an internal memory
C)
It uses Harvard architecture
D)
It contains ALU, CU, and registers

Correct Answer :   It contains ALU, CU, and registers


Explanation : Microprocessors don’t have memory and interfacing circuits. They follow Princeton architecture and they contain ALU, CU, and registers inside them.

A)
Opcode fetch, memory write, memory read, I/O read, I/O write
B)
I/O read, opcode fetch, memory read, memory write, I/O write
C)
Opcode fetch, memory read, memory write, I/O read, I/O write
D)
I/O read, opcode fetch, memory write, memory read, I/O write

Correct Answer :   Opcode fetch, memory read, memory write, I/O read, I/O write


Explanation : Initially, the opcode is fetched from memory, then memory read and write operations are performed followed by I/O read and I/O write operations.

A)
ORG
B)
LABEL
C)
GROUP
D)
OFFSET

Correct Answer :   ORG


Explanation : When an ORG is written, the assembler starts the location counter to keep track of the module’s allotted address as specified in the directive.
The location counter is initialized to 0000H if the directive is not present.

A)
Z8000
B)
PIC1x
C)
Motorola 6809
D)
Zilog Z8

Correct Answer :   PIC1x


Explanation :  Z8000, Motorola 6809, and Zilog Z8 are microprocessors but PIC1x is an 8-bit microcontroller.

A)
6 decimal places
B)
8 decimal places
C)
10 decimal places
D)
11 decimal places

Correct Answer :   11 decimal places

A)
Forces the assembler to reserve one byte of memory
B)
Forces the assembler to reserve a specified number of bytes in the memory
C)
Forces the assembler to reserve a specified number of consecutive bytes in the memory
D)
None of the above

Correct Answer :   Forces the assembler to reserve a specified number of consecutive bytes in the memory

A)
BYTE
B)
NIBBLE
C)
WORD (16 bits)
D)
DOUBLEWORD (32 bits)

Correct Answer :   BYTE

37 .
We say that a set of gates is logically complete if we can build any circuit without using any other kind of gates. Which of the following sets are logically complete
A)
Set of {AND,OR}
B)
Set of {EXOR, NOT}
C)
Set of {AND,OR,NOT}
D)
None of the above

Correct Answer :   Set of {AND,OR,NOT}

38 .
In a C expression using assignment operators, relational operators and arithmetic operators, the hierarchy of operations (in the absence of parenthesis) is
A)
Arithmetic, assignment, relational
B)
Arithmetic, relational, assignment
C)
Assignment, relational, arithmetic
D)
Relational, assignment, arithmetic

Correct Answer :   Arithmetic, relational, assignment

A)
It is of highest priority
B)
It is a vectored interrupt
C)
It uses edge-triggered signal
D)
It is a non-maskable interrupt

Correct Answer :   It uses edge-triggered signal


Explanation : TRAP interrupt in 8085 microprocessor uses both level and edge-triggered clock because it is of highest priority among all the interrupts.

A)
It has 3rd highest priority
B)
It uses level-triggered signal
C)
It is a non-maskable interrupt
D)
Its vectored address is 003C H

Correct Answer :   Its vectored address is 003C H


Explanation : RST 7.5 is a maskable interrupt with 2nd highest priority after RST-4.5. It uses only the edge-triggered signal. It is a vectored interrupt and its vectored address is 003C H.

A)
need 2 V supply
B)
need 5 V supply
C)
need 12 V supply
D)
do not require any supply

Correct Answer :   need 5 V supply


Explanation : Digital chips need 5 V supply.

A)
8255
B)
8251
C)
8253
D)
8259

Correct Answer :   8255


Explanation : 8255 is interface chip for 8086 and ADC.

A)
A
B)
BETA
C)
ALPHA
D)
AB * 2

Correct Answer :   AB * 2


Explanation : Operator is not allowed.

A)
C
B)
Pascal
C)
Fortran 77
D)
None of the above

Correct Answer :   C


Explanation : Whenever keyword 'break' is encountered inside any C loop control passes to the first statement after the loop.

A)
zero flag
B)
sign flag
C)
interrupt flag
D)
auxiliary carry flag flag

Correct Answer :   interrupt flag


Explanation : If the interrupt flag, IF=1, is set, the microprocessor will serve any interrupt. The processor ignores the service if the interrupt flag, IF=0, is set to 0.

A)
Accumulator
B)
Program counter
C)
Temporary register
D)
Instruction register

Correct Answer :   Program counter


Explanation : Instruction register, accumulator, and temporary register are general-purpose registers but program counter is a special-purpose register because it holds the address of the next instruction.

A)
Decoder
B)
Demultiplexer
C)
Address Latch Enable
D)
Priority Encoder

Correct Answer :   Address Latch Enable


Explanation : Address Latch Enable is a positive pulse and it is generated whenever the microprocessor starts an operation to latch the lower order address lines (AD7 to AD0).

A)
4
B)
5
C)
7
D)
12

Correct Answer :   5


Explanation : There are five flags or flip-flops in a flag register in 8085 microprocessor that shows the status after ALU operation. They are mainly affected by the content of the accumulator.

A)
Sign flag
B)
Zero flag
C)
Parity flag
D)
Auxiliary carry flag

Correct Answer :   Auxiliary carry flag


Explanation : Among all the five flag conditions in microprocessor, the auxiliary carry flag is used internally for BCD arithmetic operations. Based on the auxiliary flag, the instruction set doesn’t contain the conditional jump operation.

A)
1000
B)
10000
C)
100000
D)
One million

Correct Answer :   One million

A)
A pair of cross coupled NAND
B)
A pair of cross copled AND
C)
A pair of cross coupled OR
D)
A cross coupled NAND/OR

Correct Answer :   A pair of cross coupled NAND

52 .
Assertion (A): Each memory chip has its own address latch.
Reason (R): ALE signal comes out of microprocessor 8085 and goes to memory chip.
A)
A is wrong R is correct
B)
A is correct R is wrong
C)
Both A and R are correct and R is correct explanation of A
D)
Both A and R are correct but R is not correct explanation of A

Correct Answer :   A is wrong R is correct

A)
2C H
B)
3C H
C)
24 H
D)
36 H

Correct Answer :   24 H


Explanation : TRAP interrupt is a non-maskable interrupt with the highest priority. When it occurs, its vectored address 0024 H is placed on the program counter. 24 H is the lower address bus which also acts as the data bus.

A)
Go to memory
B)
Length of the instruction and number of operations
C)
Perform ALU operation
D)
Go to the output device

Correct Answer :   Length of the instruction and number of operations


Explanation : After decoding the opcode of the instruction, a microprocessor understands the length of the instruction and the total number of operations to be performed.

A)
16
B)
20
C)
32
D)
40

Correct Answer :   20


Explanation : 8086 microprocessor is a 16-bit microprocessor that uses 20 address lines and 16 data lines. AD0 to AD15 are 16 lower order address lines that can be operated in both address and data bus mode.

A)
Index flag
B)
Interrupt flag
C)
Direction flag
D)
Overflow flag

Correct Answer :   Index flag


Explanation : Overflow flag represents whether the result is out of scope or not. Direction flag is used in string operations and interrupt flag is used to enable the interrupts.

57 .
Assertion (A) : Microprocessor 8085 has on-chip oscillator with inbuilt crystal.

Reason (R) : For frequency stability crystal oscillator is preferred.
A)
A is correct R is wrong
B)
A is wrong R is correct
C)
Both A and R are correct and R is correct explanation of A
D)
Both A and R are correct but R is not correct explanation of A

Correct Answer :   A is wrong R is correct


Explaination : Crystal has to be connected externally to pins 1 and 2 of 8085.

58 .
Consider the following program in Pascal

X = 2.57
X = (X + 0.06) * 10
I = TRUNC (X)
X = I
X = X / 10.0

The final value of X would be
A)
2
B)
2.6
C)
2.63
D)
26.3

Correct Answer :   2.6


Explaination : Microprocessors

A)
coal
B)
ram
C)
else
D)
None of the above

Correct Answer :   else

A)
4 address pins and 4 data pins
B)
4 address pins and 8 data pins
C)
8 address pins and 8 data pins
D)
8 address pins and 4 data pins

Correct Answer :   8 address pins and 4 data pins


Explanation : 28 = 256

A)
* + + + - =
B)
+ + + - * =
C)
+ + - * + =
D)
- + + + * =

Correct Answer :   + + + - * =

A)
binary
B)
2's complement
C)
1's complement
D)
None of the above

Correct Answer :   2's complement


Explanation : 2's complement is universally used in computers.

A)
capacity
B)
ac characteristics
C)
dc characteristics
D)
All of the above

Correct Answer :   All of the above


Explanation : All these characteristics are important.

A)
read operation only
B)
read and byte erase operations
C)
read, byte erase and byte write operations
D)
read, byte erase, byte write and chip erase operations

Correct Answer :   read, byte erase, byte write and chip erase operations

A)
6
B)
3
C)
2
D)
-1

Correct Answer :   6


Explanation : It gives remainder in division operation.

A)
DO 161 = 1, 15
B)
DO 16 1 = 1, 10, 1
C)
DO 16 I = - 5, 5, 2
D)
All of the above

Correct Answer :   All of the above


Explanation : All are valid.

67 .
Let JCOKE = 98 and LPEPSI = 42 Consider the statement

IF (JCOKE - 3 * LPEPSI) 5, 6, 7
5 JCOKE = JCOKE + 5
6 JCOKE = JCOKE + 8
7 JCOKE = JCOKE + 11

The value of JCOKE after the execution of above statement will be
A)
103
B)
106
C)
109
D)
112

Correct Answer :   103


Explaination : Since (J COKE - 3 x L PEPSI) is negative, statement 5 is executed and the result is 98 + 5 = 103.

A)
x = 2
B)
x = 3
C)
x = 7
D)
None of the above

Correct Answer :   x = 7


Explanation : The operator + = means add and assign.

A)
Zero flag
B)
Trap flag
C)
Parity flag
D)
Auxiliary carry flag

Correct Answer :   Trap flag


Explanation : Trap, direction, and interrupt are the control flags. Carry, parity, auxiliary carry, zero, sign and overflow flags are the condition flags.

A)
Flag register
B)
Program counter
C)
Memory data register
D)
Memory address register

Correct Answer :   Flag register


Explanation : Flag register is not used in opcode fetch operations. It is used to represent the status of ALU operation which is performed after decoding of the opcode.

A)
8 KB
B)
2 MB
C)
16 MB
D)
64 MB

Correct Answer :   2 MB


Explanation : Total number of locations of the memory possible for 20 address lines is 220 and every location has 16/8 = 2 bytes. So, total memory capacity will be 220 × 2 = 2 MB.

A)
8-bit
B)
16-bit
C)
32-bit
D)
64-bit

Correct Answer :   64-bit


Explanation : The Pentium-II microprocessor is a 64-bit microprocessor. It was introduced in 1997 and designed with 7.5 million transistors. Its word length is 64-bit.

A)
It is a 40 pin DIP chip
B)
It has 16 address lines
C)
It is an 8-bit microprocessor
D)
It is manufactured using PMOS technology

Correct Answer :   It is manufactured using PMOS technology


Explanation : 8085 microprocessor is manufactured using NMOS technology. PMOS technology is used in a 4004 microprocessors. NMOS technology is faster than PMOS technology.

A)
TRAP
B)
RST-6.5
C)
INTR
D)
RST-7.5

Correct Answer :   INTR


Explanation : TRAP, RST-7.5, and RST-6.5 are vectored inputs but INTR is a non-vectored input. It has the least priority and its address is provided by the user using an external device.

A)
Immediate
B)
Direst
C)
Indexed
D)
Base-Register

Correct Answer :   Immediate


Explanation : Indexed addressing mode is used for array and list operations. It uses a constant value and index register to execute the instruction.

A)
Opcode
B)
Address of memory
C)
Temporary data
D)
Address of next instruction

Correct Answer :   Address of memory


Explanation : H and L are 8-bit general-purpose registers. They are used to store the address of memory. Together they can store a 16-bit address.

A)
127
B)
127.0
C)
125 + 3
D)
None of the above

Correct Answer :   127

78 .
The expression A2 + B2 - 3 AB when written is Pascal should be written as
A)
A * A + B * B - 3 A * B
B)
A * A + B * B - 3.0 A * B
C)
A * A + B * A - 3 * A * B
D)
A * A + B * B - 3.0 * A * B

Correct Answer :   A * A + B * B - 3.0 * A * B

A)
210
B)
210.5
C)
210.50
D)
210.5000

Correct Answer :   210


Explanation : The result should not contain decimal.

A)
RAM
B)
Diskette
C)
Hard disk
D)
Magnetic tape

Correct Answer :   RAM


Explanation : RAM has volatile memory and therefore the matter has to be saved frequently.

A)
4.0 / 2
B)
5 / 2
C)
4 / 2.0
D)
4.0 / 2.0

Correct Answer :   5 / 2


Explanation : Since both 5 and 2 are integers, the result will be 2.

A)
BRG
B)
MBR
C)
LOTS
D)
JAMMU

Correct Answer :   BRG


Explanation : The first letter is not I, J, K, L, M, N.

83 .
Consider the following tasks in a micro computer

1. Receiving data and instruction
2. Performing arithmetic computation
3. Performing logical computations
4. Storing data and instructions

Which of the above are performed by ALU?
A)
2 only
B)
2 and 3 only
C)
1, 2 and 3 only
D)
All of the above

Correct Answer :   2 and 3 only


Explaination : ALU performs arithmetic and logical computations.

84 .
Consider the following DO statement in FORTRAN 77

DO 25 J = 1, 7

The number of DO loop executions in the above statement is
A)
1
B)
6
C)
7
D)
8

Correct Answer :   7


Explaination : Do statement is executed when J = 1, 2, 3, 4, 5, 6, 7, i.e., a total of 7 times.

A)
32
B)
36
C)
37
D)
39

Correct Answer :   37


Explanation : 90 GB is greater than 64 GB and less than 128 GB. For 64 GB (236 Bytes), a minimum of 36 address lines are needed. So, for 90 GB, we need 36 + 1 = 37 address lines.

A)
INTR
B)
TRAP
C)
RST-6.5
D)
RST-5

Correct Answer :   RST-5


Explanation : TRAP, INTR, and RST-6.5 are the hardware interrupts but RST-5 is a software interrupt present. All software interrupts are vectored interrupts.

A)
0010 H
B)
0032 H
C)
0028 H
D)
0030 H

Correct Answer :   0028 H


Explanation : Vectored address of RST-n is calculated by the formula n × 8. So, vectored address of RST-5 is (40)10. (40)10 in hexadecimal is (28)16 = 0028 H.

A)
Stack pointer is an 8-bit register
B)
Stack pointer stores data permanently
C)
Stack pointer is initialized after stack operation
D)
Stack pointer contains the address of the top of the stack memory

Correct Answer :   Stack pointer contains the address of the top of the stack memory


Explanation : Stack pointer is initialized before stack operation, it is a 16-bit register that stores data temporarily. It follows a LIFO operation. So, it contains the address of the top of the stack memory.

A)
10
B)
12
C)
16
D)
20

Correct Answer :   12


Explanation : 4 KB RAM is equal to 4096 × 8 bits. Total number locations are (4096 × 8)/8 = 4096 and 4096 = 212. So, 12 address lines are required to connect a 4 KB RAM to a microprocessor.

A)
It is a 2-byte instruction
B)
It doesn’t affect the flag register
C)
It uses immediate addressing mode
D)
It means move the content of register A to register B

Correct Answer :   It doesn’t affect the flag register


Explanation : MOV A, B means moving the content of register B to register A. It is a 1-byte instruction and it uses register addressing mode. No flags are affected because operations of this instruction are not performed in the ALU.

A)
It has 13 T-states
B)
It is a 3-byte instruction
C)
It doesn’t affect any flags
D)
It uses indirect addressing mode

Correct Answer :   It uses indirect addressing mode


Explanation : LDA is a direct addressing mode instruction which stands for Load Accumulator Direct. It is a 3-byte instruction and it has 13 T-states. No flags are affected by this instruction.

A)
NAND-NAND
B)
OR-NAND
C)
NAND-NOR
D)
NOR-NAND

Correct Answer :   NAND-NAND

A)
A smaller floating point number
B)
A negative floating point number
C)
A larger floating point number
D)
Either A or C depending on the actual number

Correct Answer :   A larger floating point number

A)
Mode 0 and mode 1
B)
Mode 0 and mode 2
C)
Mode 0, mode 1 and mode 2
D)
Mode 0, mode 2 and mode 3

Correct Answer :   Mode 0, mode 1 and mode 2

A)
Class words
B)
Reserved words
C)
Special words
D)
Character words

Correct Answer :   Reserved words

A)
Logical IF and Block IF
B)
Logical IF and arithmetic IF
C)
Logic IF, block IF, arithmetic IF and negate IF
D)
Logical IF, block IF and arithmetic IF

Correct Answer :   Logical IF, block IF and arithmetic IF

A)
R is asserted, S is asserted
B)
R is negated, S is negated
C)
R is negated, S is asserted
D)
R is asserted, S is negated

Correct Answer :   R is asserted, S is asserted

A)
decimal 2
B)
decimals 8
C)
decimal 16
D)
decimal 32

Correct Answer :   decimal 32


Explanation : If instruction RAL is executed once the number gets doubled.

A)
3
B)
4
C)
3 / 4
D)
4 / 3

Correct Answer :   4


Explanation : Modulus operator gives the remainder of division as answer.

100 .
Assertion (A) : In 8085 SP and PC are 8 bit each.

Reason (R) : 8085 is an 8 bit microprocessor.
A)
A is wrong R is correct
B)
A is correct R is wrong
C)
Both A and R are correct and R is correct explanation of A
D)
Both A and R are correct but R is not correct explanation of A

Correct Answer :   A is wrong R is correct


Explaination : SP and PC are 16 bit each in 8085.

101 .
Let JCOKE = 3 and LPEPSI = 5. Consider the statement

IF (3 * JCOKE. EQ. LPESPSI)
JCOKE = JCOKE + 2
JCOKE = JCOKE + 5

What will be the final value of JCOKE after the above statement is executed?
A)
16
B)
10
C)
8
D)
4

Correct Answer :   8


Explaination : The logical IF statement means that IF (3 x J COKE) is equal to (L PEPSI), J COKE = 3 + 2 = 5 otherwise J COKE = 3+5 = 8 Since (3 x J COKE) is more than (L PEPSI), the result is 8.

A)
time required to access cell is same as that to access cell 65536
B)
time required to access cell is less than that required to access cell 65536
C)
time required to access cell is more than that required to access cell 65536
D)
Either (A) or (B)

Correct Answer :   time required to access cell is same as that to access cell 65536


Explanation : Time required to access different cells is the same.

103 .
Consider the following logical IF statement in FORTRAN 77

IF (SALT. GE. PEPPER) GOTO 11
GOTO 13

The above statement using arithmetic IF statement would be
A)
IF (SALT - PEPPER) 11, 11, 13
B)
IF (SALT - PEPPER) 13, 11, 13
C)
IF (SALT - PEPPER) 13, 11, 11
D)
IF (SALT - PEPPER) 11, 13, 13

Correct Answer :   IF (SALT - PEPPER) 13, 11, 11


Explaination : The given statement is logical IF statement. If SALT greater than or equal to PEPPER control goes to statement 11 otherwise control goes to statement 13 Same is true of arithmetic IF statement (c). If (SALT - PEPPER) is negative, control goes to statement 13 and otherwise control goes to statement 11.

A)
It is a 3-byte instruction
B)
It required three machine cycles
C)
It uses immediate addressing mode
D)
Accumulator is loaded with the content of memory

Correct Answer :   It is a 3-byte instruction


Explanation : STA is a direct addressing mode instruction which stands for Store Accumulator Direct. It is a 3-byte instruction and it requires four machine cycles. In STA instruction memory is loaded with the content of the accumulator.

A)
Octal addition
B)
Binary addition
C)
BCD addition
D)
Excess-3 addition

Correct Answer :   BCD addition


Explanation : DAA instruction is used to perform BCD addition. DAA instruction changes the binary values of the contents of the accumulator to BCD.

A)
Decodes an opcode
B)
Converts decimal to binary
C)
Converts hexadecimal code to binary
D)
Increments the content of the program counter by 1

Correct Answer :   Converts hexadecimal code to binary


Explanation : Microprocessor understands only 0s & 1s. So, a loader first converts hexadecimal code to binary form and then loads it to the memory.

A)
40H, 40H
B)
50H, 40H
C)
50H, 50H
D)
60H, 40H

Correct Answer :   40H, 40H


Explanation : Initially, register A contains 50H & B contains 40H. Instruction MOV A, B means move the content of B to A. So, after this operation, both registers A & B will contain 40H.

A)
One time
B)
Two times
C)
Depends on the length of the instruction
D)
Depends on the memory capacity of the processor

Correct Answer :   Depends on the length of the instruction


Explanation : After decoding an opcode the microprocessor understands the length of the instruction. So, the content of the program counter is placed on the address bus as many times as the length of the instruction.

A)
It is a machine control instruction
B)
It is used to start the execution of the program
C)
PC is disconnected from the address bus
D)
A reset interrupt is required to come out of halt state

Correct Answer :   It is used to start the execution of the program


Explanation : HLT is a machine control statement. Internally, the PC is disconnected from the address bus so, the next fetch is not possible. It is used mainly to stop the execution of the program.

A)
XY
B)
X + Y
C)
X - Y
D)
X / Y

Correct Answer :   XY


Explanation : Adding X to PROD, Y times evidently gives the product XY.

A)
If A < = B then A : = C * D
B)
If A > = B then B : = C + D + X
C)
If (N = 45) THEN X : = X + 2 ELSEX : = 0
D)
If (A > = B) X : Y else X : = Z

Correct Answer :   If (A > = B) X : Y else X : = Z


Explanation : 'then' is missing.

112 .
If J = 12 and X = 24.4 the result of the following FORTRAN 77 program will be

READ *, J, X
K = J ** 2
Z = 3 * X
Print *, J, K, Z
A)
12, 144, 73.2
B)
15. 144. 73
C)
15.0, 144.0, 73
D)
15.0, 144.0, 73.2

Correct Answer :   12, 144, 73.2


Explaination :

J = 12, K = 122 = 144, Z = 3 x 24.4 = 73.2 In the result J and K will be without decimal point.

A)
9
B)
18
C)
90
D)
100

Correct Answer :   90


Explanation : SUM = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 = 90.

A)
MAX
B)
DSTDEV
C)
DCOUNT
D)
DAVERAGE

Correct Answer :   MAX


Explanation : MAX is not data base function.

115 .
Consider the following statements about Infix and Post fix notations in Pascal

1. In Infix notation operators appear between the operands on which they operate.
2. In Post fix notation operators appear after the operand.
3. The Infix expression A + B - C + D will appear as AB + C - D + in post fix notation.

Which of the above are correct?
A)
1 and 2 only
B)
1 and 3 only
C)
2 and 3 only
D)
All of the above

Correct Answer :   All of the above

A)
8.0
B)
80
C)
5.0
D)
50

Correct Answer :   80


Explanation : The result should be sum of 50 and 30 and decimal must be added.

A)
COBOL
B)
MS WORD
C)
Page maker
D)
DB III plus

Correct Answer :   DB III plus


Explanation : dB III plus is DBMS and not for financial accounting.

A)
Direct
B)
Relative
C)
Register
D)
Indexed

Correct Answer :   Register


Explanation : Register addressing mode is used when data is present inside the register of the microprocessor. For example, MOV B, C. Here, B and C are the registers of the microprocessor.

A)
8155
B)
8279
C)
8253/54
D)
8257/37

Correct Answer :   8257/37


Explanation : 8155 is a multipurpose programmable I/O device. 8253/54 is a programmable counter. 8279 is a keyboard/display controller and 8257/37 is a DMA controller.

A)
IN 01H
B)
MOV A, B
C)
LDA 2500H
D)
JMP 2085H

Correct Answer :   IN 01H


Explanation : LDA 2500H and JMP 2085H are 3-word instructions. MOV A, B is a 1-word instruction. IN 01H is a 2-word instruction, the first byte specifies the opcode and the second byte specifies the operand.

A)
RAM
B)
EEPROM
C)
UVEPROM
D)
Both (B) and (C)

Correct Answer :   EEPROM

A)
Gate
B)
Mux
C)
Register
D)
Decoder

Correct Answer :   Register

A)
int bar
B)
s = s + 1
C)
king = horse + 1
D)
prin = prin * prin

Correct Answer :   int bar

124 .
If the number of address bits in a memory is reduced by 2 and the addressability is doubled, the size of the memory (i.e., the number of bits stored in the memory)
A)
Halves
B)
Doubles
C)
Remains unchanged
D)
Increases by 2^(address bits)/addressability

Correct Answer :   Halves

A)
2
B)
4
C)
6
D)
8

Correct Answer :   6

126 .
Consider the following rules for constructing variable names in C

1. It can have alphabets, digits or underscores but no other special character.
2. The maximum number of characters is 8.
3. The first character may be an alphabet or digit but not underscore.

Which of the above are correct?
A)
1 and 2 only
B)
1 and 3 only
C)
2 and 3 only
D)
None of the above

Correct Answer :   1 and 2 only


Explaination : First character must be alphabet only.

A)
1 μs
B)
3 μs
C)
0.666 μs
D)
0.333 μs

Correct Answer :   0.333 μs


Explanation : Microprocessors

128 .
Assertion (A) : In 8085 WR and RD signals are active high.

Reason (R) : LOW WR means write operation and low RD means read operation.
A)
A is correct R is wrong
B)
A is wrong R is correct
C)
Both A and R are correct and R is correct explanation of A
D)
Both A and R are correct but R is not correct explanation of A

Correct Answer :   A is wrong R is correct


Explaination :

WR and RD are active low.

A)
AFH
B)
5 AFH
C)
64 H
D)
OCFH

Correct Answer :   AFH

A)
2 MOS transistors
B)
4 MOS transistors
C)
4 MOS transistors and 1 capacitor
D)
4 MOS transistors and 2 capacitors

Correct Answer :   4 MOS transistors


Explanation : Each SRAM cell has 4 MOS transistors.

A)
Logic group
B)
Branch group
C)
Arithmetic group
D)
Data transfer group

Correct Answer :   Branch group

A)
1
B)
2
C)
4
D)
8

Correct Answer :   4

A)
format of data transfer
B)
electrical characteristics of the two devices
C)
timing according to which data transfer is to take place
D)
All of the above

Correct Answer :   All of the above

A)
6
B)
7
C)
9
D)
10

Correct Answer :   10

A)
Address Latch Enable
B)
Address Latch Enter
C)
Accumulator Latch Enter
D)
Accumulator Latch Enable

Correct Answer :   Address Latch Enable

A)
4.37 E - 4
B)
4.37 E - 5
C)
43.7 E - 4
D)
43.7 E - 5

Correct Answer :   4.37 E - 5

A)
28
B)
-28
C)
-28.0
D)
None of the above

Correct Answer :   -28.0

A)
6400 B
B)
9600 B
C)
240000 B
D)
480000 B

Correct Answer :   480000 B

A)
DAA
B)
LDAX B
C)
ADI 36H
D)
LDA 2700H

Correct Answer :   LDAX B


Explanation : LDA 2700H is a direct addressing mode instruction. ADI 36H is an immediate addressing mode instruction. DAA is an implicit addressing mode and LDAX B is a register-indirect addressing mode instruction.

A)
It is a 3-byte instruction
B)
It requires three T-states
C)
It uses indexed addressing mode
D)
Contents of HL pair is moved to SP

Correct Answer :   Contents of HL pair is moved to SP


Explanation : SPHL is a 1-byte instruction. It uses register addressing mode. It required 6 T-states and it means moving the contents of the HL pair to SP.

A)
data copy/transfer instruction
B)
string instruction
C)
branch instruction
D)
arithmetic/logical instruction

Correct Answer :   data copy/transfer instruction


Explanation : These instructions are used to copy and transfer the instructions.

A)
DAS
B)
POP
C)
MOV
D)
PUSH

Correct Answer :   DAS


Explanation : DAS (Decimal Adjust after Subtraction) is an arithmetic instruction.

A)
incremented by 1
B)
decremented by 1
C)
decremented by 2
D)
incremented by 2

Correct Answer :   decremented by 2


Explanation : The actual current stack-top is always occupied by the previously pushed data. So, the push operation decrements SP by 2 and then stores the two bytes contents of the operand onto the stack.

A)
stops executing the program
B)
jumps to instruction in other registers
C)
follows the next instruction in the program
D)
breaks the normal sequence of execution of instructions

Correct Answer :   breaks the normal sequence of execution of instructions


Explanation : An interrupt function is to break the sequence of operation.

A)
control unit
B)
Interrupt service routine
C)
Execution unit
D)
Counter word register

Correct Answer :   Interrupt service routine


Explanation : An interrupt transfers the control to interrupt service routine (ISR). After executing ISR, the control is transferred back again to the main program.

A)
multi-interrupt
B)
nested interrupt
C)
interrupt within interrupt
D)
nested interrupt and interrupt within interrupt

Correct Answer :   nested interrupt and interrupt within interrupt


Explanation : If an interrupt occurs while executing a program, and the processor is executing the interrupt, if one more interrupt occurs again, then it is called a nested interrupt.

A)
one dimensional
B)
two dimensional
C)
three dimensional
D)
None of the above

Correct Answer :   two dimensional


Explanation : The semiconductor memories are organised as two dimensions of an array which consists of rows and columns.

A)
data bus
B)
control bus
C)
address bus
D)
either address bus or data bus

Correct Answer :   data bus


Explanation : The bits in a selected location are accessible using data bus.

A)
log N (to the base 2)
B)
log N (to the base e)
C)
log N (to the base 10)
D)
log (2N) (to the base e)

Correct Answer :   log N (to the base 2)


Explanation : For n memory locations, log n(to the base of 2) address lines are required. For addressing 4K bytes of memory, 12 address lines are required since log(4KB) =log(4*1024)=log(212)=12

A)
512
B)
1024
C)
2048
D)
None of the above

Correct Answer :   1024


Explanation : Since for n address lines, the number of memory locations able to address is 2^n.

A)
static upper memory
B)
even address memory bank
C)
odd address memory bank
D)
upper address memory bank

Correct Answer :   odd address memory bank


Explanation : In static memory, the upper 8-bit bank is called odd address memory bank.

A)
reader
B)
display
C)
printer
D)
joystick

Correct Answer :   joystick


Explanation : Since joystick is an input device, it reads data from the external devices.

A)
read, read
B)
write, write
C)
read, write
D)
write, read

Correct Answer :   read, write


Explanation : The input activity is similar to read operation and the output activity is similar to write operation.

A)
write operation on output data
B)
read operation on input data
C)
write operation on input data
D)
read operation on output data

Correct Answer :   write operation on output data


Explanation : IOWR (active low) operation means writing data to an output device and not an input device.

A)
serial input port
B)
parallel output port
C)
serial input-output port
D)
parallel input-output port

Correct Answer :   parallel input-output port


Explanation : The parallel input-output port chip 8255 is also known as programmable peripheral input-output port.

A)
input port
B)
output port
C)
either input or output ports
D)
both input and output ports

Correct Answer :   either input or output ports


Explanation : Port C can function independently either as input or as output ports.

A)
data bus control
B)
control word register
C)
read logic control
D)
None of the above

Correct Answer :   control word register


Explanation : By programming the bits of control word register, the operations of the ports are specified.

A)
A1
B)
RESET
C)
WR(ACTIVE LOW)
D)
All of the above

Correct Answer :   All of the above


Explanation : RD(ACTIVE LOW), WR(ACTIVE LOW), A1, A0, RESET are the inputs provided by the microprocessor to the read/write control logic of 8255.

A)
edge time
B)
conversion delay
C)
time delay
D)
conversion time

Correct Answer :   conversion delay


Explanation : Broadly speaking, the time taken by the converter to calculate the equivalent digital data output from the moment of the start of conversion is called conversion delay.

A)
dual slope integration
B)
successive approximation
C)
successive approximation and dual slope integration
D)
None of the above

Correct Answer :   successive approximation and dual slope integration


Explanation : Successive approximation and dual slope integration are the most popular techniques that are used in the integrated ADC chips.

A)
ensuring stability of analog input
B)
issuing start of conversion pulse to ADC
C)
reading digital data output of ADC as equivalent digital output
D)
All of the above

Correct Answer :   All of the above


Explanation : The general algorithm for interfacing ADC contains ensuring the stability of analog input, issuing start of conversion pulse to ADC, reading end of conversion signal to mark the end of a conversion process, reading digital data output of ADC as equivalent digital output.

A)
100 microseconds
B)
100 milliseconds
C)
50 milliseconds
D)
None of the above

Correct Answer :   100 microseconds


Explanation : The conversion delay is 100microseconds which is low as compared to other converters.

A)
is mostly supplied externally
B)
is always supplied externally
C)
is always generated internally
D)
may be generated internally or supplied externally

Correct Answer :   may be generated internally or supplied externally

164 .
If x = 0.1396 radians, the values of 1 - cos x and 2 sin2 (x / 2) using floating point arithmetic with a 4 digit mantissa are respectively
A)
.1000 E - 1 each
B)
.1000 E - 1 and .9727 E - 2 respectively
C)
.9727 E - 2 each
D)
.9727 E - 2 and .1000 E - 1 respectively

Correct Answer :   .1000 E - 1 and .9727 E - 2 respectively

A)
2
B)
4
C)
8
D)
16

Correct Answer :   4

A)
DMA
B)
Programmed data transfer
C)
Either (A) or (B)
D)
Neither (A) nor (B)

Correct Answer :   DMA

A)
the value 20 is fetched into accumulator
B)
contents of memory location 20 are fetched into the accumulator
C)
Either (A) or (B)
D)
Neither (A) nor (B)

Correct Answer :   contents of memory location 20 are fetched into the accumulator

A)
-0.0032
B)
-0.032 E - 1
C)
-0.00032 EI
D)
-3.2 E - 3

Correct Answer :   -3.2 E - 3

A)
A - 4.0
B)
A / B + D
C)
- A + B + M
D)
A div B * mod C

Correct Answer :   - A + B + M

A)
cost of memory is reduced
B)
a larger memory is obtained
C)
a non-volatile memory is obtained
D)
effective speed of memory is increased

Correct Answer :   effective speed of memory is increased

A)
minimum mode and maximum mode
B)
low mode and high mode
C)
decrement mode and increment mode
D)
None of the above

Correct Answer :   minimum mode and maximum mode

A)
master unit and slave unit
B)
bus interface unit and execution unit
C)
command unit and follow unit
D)
memory unit and operation unit

Correct Answer :   bus interface unit and execution unit

A)
D S
B)
E Q U
C)
D B
D)
D W

Correct Answer :   E Q U

A)
I/O device
B)
memory chip
C)
programmable interrupt controller
D)
programmable peripheral interface

Correct Answer :   programmable interrupt controller

A)
The subscripts must be enclosed within parenthesis
B)
The subscripts in array variable must be separated by dots
C)
The general form of an array variable must be an integer name followed by subscripts enclosed within square brackets. The subscripts are separated by commas
D)
The general form of an array variable is an integer or a real variable name followed by subscripts enclosed within square brackets. The subscripts are separated by commas

Correct Answer :   The general form of an array variable is an integer or a real variable name followed by subscripts enclosed within square brackets. The subscripts are separated by commas

A)
Double the contents of register pair H L
B)
Shift the contents of register pair H L by one bit to left
C)
Shift the contents of register pair H L by one bit to the right
D)
Load the program counter with contents of register pair H L

Correct Answer :   Load the program counter with contents of register pair H L

A)
must be the same as that of host
B)
may overlap
C)
must be same
D)
must be disjoint

Correct Answer :   must be the same as that of host

A)
-32767 to + 32768
B)
-32768 to + 32767
C)
-65536 to + 65535
D)
-65535 to + 65536

Correct Answer :   -32768 to + 32767

A)
Logical group
B)
Arithmetic group
C)
Data transfer group
D)
Both (A) and (B)

Correct Answer :   Data transfer group

A)
Implict addressing
B)
Direct addressing
C)
Register addressing
D)
Immediate addressing

Correct Answer :   Implict addressing

A)
about 1 ns
B)
about 1 μs
C)
about 30 μs
D)
about 100 μs

Correct Answer :   about 30 μs

A)
Decimal Machine Address
B)
Direct Memory Address
C)
Direct Memory Accumulator
D)
Distributed Memory Address

Correct Answer :   Direct Memory Address

A)
program counter etc. are being set
B)
program counter etc. are being reset
C)
Either (A) or (B)
D)
None of the above

Correct Answer :   program counter etc. are being reset

A)
X + 2.5 / Y + 3.0
B)
(X + 2.5) / Y + 3.5
C)
(X + 2.5) / Y + 4.5
D)
(X + 2.5) / (Y + 4.5)

Correct Answer :   (X + 2.5) / (Y + 4.5)

A)
Hand shaking
B)
Direct transfer
C)
Synchronous transfer
D)
Interrupt driver transfer

Correct Answer :   Direct transfer

A)
THETA
B)
1 ST
C)
NUM
D)
$ BETA

Correct Answer :   THETA

187 .
Assertion (A) : Microprocessor 8085 has 16 address bus lines and 8 data bus lines.

Reason (R) : Address bus in 8085 is unidirectional.
A)
A is wrong R is correct
B)
A is correct R is wrong
C)
Both A and R are correct and R is correct explanation of A
D)
Both A and R are correct but R is not correct explanation of A

Correct Answer :   Both A and R are correct but R is not correct explanation of A

A)
zero
B)
positive
C)
negative
D)
zero or negative

Correct Answer :   zero or negative

A)
strlen
B)
strcpy
C)
strbig
D)
strdup

Correct Answer :   strbig

A)
2 bytes long
B)
3 bytes long
C)
2 or 3 bytes long
D)
1 or 2 or 3 bytes long

Correct Answer :   1 or 2 or 3 bytes long

A)
directs the assembler to reserve a sequence of 105 bits in the memory
B)
directs the assembler to reserve a sequence of 105 bytes in the memory
C)
directs the assembler to reserve a sequence of consecutive 105 bytes in the memory
D)
directs the assembler to reserve a sequence of consecutive 105 bits in the memory

Correct Answer :   directs the assembler to reserve a sequence of consecutive 105 bytes in the memory

A)
about 1 / 64
B)
about one-eighth
C)
about 1 / 500
D)
about the same

Correct Answer :   about one-eighth

A)
Bus loading
B)
Timing characteristic
C)
Power consumption
D)
All of the above

Correct Answer :   Timing characteristic

A)
digits 0 to 9
B)
capital letters A to Z
C)
lower case letters a to z
D)
special characters +, -, *, / etc

Correct Answer :   special characters +, -, *, / etc

A)
Implicit
B)
Direct
C)
Register
D)
Immediate

Correct Answer :   Implicit

A)
+
B)
-
C)
*
D)
/

Correct Answer :   -

A)
μP and memory
B)
μP and I/O devices
C)
memory and I/O devices
D)
All of the above

Correct Answer :   All of the above

A)
1 and 1 respectively
B)
1 and 2 respectively
C)
40 and 20 respectively
D)
20 and 40 respectively

Correct Answer :   40 and 20 respectively

199 .
The normalised form of 1.81 x 10-11 is
A)
18.1 x 10⁻¹³
B)
18.1 x 10⁻¹²
C)
0.181 x 10⁻¹â°
D)
0.0181 x 10⁻⁹

Correct Answer :   0.181 x 10⁻¹â°

A)
Hidden refresh
B)
Distributed refresh
C)
Both (A) and (B)
D)
Concentrated refresh

Correct Answer :   Concentrated refresh

201 .
If CS = A15 A14 A13 is used as chip select logic of a 4 K RAM in 8085 system, its memory range is
A)
3000 H - 3 FFF H
B)
7000 H - 7 FFFH
C)
6000 H - 6 FFF H and 7000 H - 7 FFF H
D)
5000 H - 5 FFF H and 6000 H - 6 FFF H

Correct Answer :   6000 H - 6 FFF H and 7000 H - 7 FFF H

202 .
Consider the following part of a C program float a, b

b = (a > 4 ? 2 : 9);

Assuming that a has values of 3 and 6, the values stored respectively will be
A)
2 and 9
B)
2.0 and 2.0
C)
2.0 and 9.0
D)
9.0 and 2.0

Correct Answer :   9.0 and 2.0

A)
main memory and I/O devices
B)
two I/O devices
C)
cache and main memory
D)
cache memory and I/O devices

Correct Answer :   main memory and I/O devices

A)
the carry in of first stage connected to carry in of second stage
B)
the carry out of first stage is connected to carry in of second stage
C)
the carry in of first stage is connected to carry out of second stage
D)
the carry out of first stage is connected to carry out of second stage

Correct Answer :   the carry out of first stage is connected to carry in of second stage


Explanation : In cascading carry out of first stage is connected to carry in of second stage.

205 .
Consider the following statements about expressing real constants in exponential form in C

1. The mantissa and exponent are separated either by e or E.
2. The mantissa and exponent may have positive or negative sign.
3. Default sign of mantissa is positive.
4. Default sign of exponent is negative.

Which of the above statements are wrong?
A)
1 only
B)
1 and 4 only
C)
1, 3 and 4 only
D)
4 only

Correct Answer :   1 and 4 only


Explaination : E is not allowed and default sign of exponent is positive.

206 .
If F = 95.0 what will be the result of following FORTRAN 77 program?

REAL F , C
READ * , F
C = (5. / 9.) * (F - 32.)
A)
35
B)
35.0
C)
25
D)
25.0

Correct Answer :   35.0


Explaination : Microprocessors

207 .
Consider the following program in Basic

10 READ X, Y, Z
15 DATA 20, 30, 10
20 S = X + Y + Z
30 A = S/3
35 PRINT S, A, X, Y, Z
40 END

The output will be
A)
60, 20, 20, 30, 10
B)
10, 30, 20, 20, 60
C)
60, 20, 30, 10, 20
D)
60, 30, 10, 20, 20

Correct Answer :   60, 20, 20, 30, 10


Explaination : Microprocessors

208 .
Assuming that k is an integer what would the result of following expression in C
k = 2* 3 / 5 + 4 / 4 + 8 - 2 - 4 / 8
A)
7
B)
7.5
C)
8
D)
zero

Correct Answer :   8


Explaination : k = 1 + 1 + 8 - 2 - 0 = 8 because k is an integer.

A)
KM
B)
1281
C)
1281.0
D)
None of the above

Correct Answer :   1281.0


Explanation : It is the only one with decimal point.

A)
9.75
B)
11.50
C)
11.75
D)
13.75

Correct Answer :   13.75


Explanation : Convert to decimal.

A)
25 MB
B)
50 MB
C)
500 MB
D)
100 MB or 250 MB

Correct Answer :   100 MB or 250 MB


Explanation : Zip floppy has high memory 100 MB or 250 MB.

212 .
Consider the following program in Basic

5 READ R
10 D = 2 * R
20 A = 3.1416 * D * D / 4
30 PRINT A

Assuming that R = 3.0, the output A =
A)
2.827
B)
28.27
C)
282.7
D)
2827

Correct Answer :   28.27


Explaination : Microprocessors

A)
single chip LSI
B)
single chip SSI
C)
single chip MSI
D)
All of the above

Correct Answer :   single chip LSI


Explanation : It is LSI chip.

214 .
A computer program is used to read N and print the sum 12 + 22 + 32 +.....+ N2. If N = 10, the print out will show the number
A)
10
B)
100
C)
285
D)
385

Correct Answer :   385


Explaination :

SUM = 12 + 22 + 33 + 44 + 52 + 62 + 77 + 82 + 92 + 102 = 385.

A)
B(2, 4)
B)
B(M, R)
C)
B(3, 6)
D)
B(M, 3)

Correct Answer :   B(M, R)


Explanation : Subscript not proper.

A)
a c d (x * z)
B)
a* c* d* (x z)
C)
a * c * d (x * z)
D)
None of the above

Correct Answer :   a * c * d (x * z)

A)
1 a b
B)
56 a
C)
a 56
D)
5 a 6

Correct Answer :   a 56


Explanation : In all others first letter is not lower case.

A)
divided by 2
B)
divided by 4
C)
multiplied by 2
D)
multiplied by 4

Correct Answer :   divided by 2


Explanation : When RRC is used once the number is divided by 2.

A)
1 to 8095
B)
0 to 8095
C)
1 to 8096
D)
0 to 8096

Correct Answer :   0 to 8095


Explanation : The starting word address is 0.

A)
x = x + 1
B)
x = y * z
C)
z = y / x
D)
x * y = z

Correct Answer :   x * y = z


Explanation : Left hand side must be a variable name.

221 .
Consider the following expressions in Java
int a, b, c,
a = 40;
b = a++ ;
c = ++a ;
Now the values of a, b, c are
A)
40, 42, 42 respectively
B)
40, 40, 42 respectively
C)
42, 40, 42 respectively
D)
42, 42, 42 respectively

Correct Answer :   42, 40, 42 respectively


Explaination : The steps are : a is given the value 40, b is given the value of a before it is incremented (i.e., 40) and a is incremented to 41, a is incremented to 42 and c is given this value. Thus the result is 42, 40, 42.

A)
A * * B
B)
A mod B
C)
A div B - C
D)
L div M + P

Correct Answer :   A * * B


Explanation : Two operators cannot be together.

A)
ST
B)
STD
C)
STUD
D)
NSTUD

Correct Answer :   NSTUD


Explanation : It is an integer. Hence N as first character.

A)
0.4
B)
0.45
C)
40
D)
45

Correct Answer :   40


Explanation : 45/10 = 4.5 → 4 x 10 = 40.

A)
32
B)
16
C)
8
D)
4

Correct Answer :   16

A)
-7
B)
-73.2
C)
73
D)
-73

Correct Answer :   -73

A)
the value 20 is brought into accumulator
B)
contents of memory location 20 are brought into the accumulator
C)
Either (A) or (B)
D)
Neither (A) nor (B)

Correct Answer :   the value 20 is brought into accumulator

A)
font and footer
B)
portrait and landscape
C)
header and footer
D)
auto text and auto correct

Correct Answer :   portrait and landscape


Explanation : Portrait and worksheet are two orientations of worksheet page.

229 .
The correct form of expression B3 -2 A C in FORTRAN 77 is
A)
B * 3 - 2 * A AC
B)
B * * 3 - 2 * A * C
C)
B *** 3 - 2 * A * C
D)
B * * 3 - 2 ** A * C

Correct Answer :   B * * 3 - 2 * A * C


Explaination : It is the only expression with correct symbols for exponential and product.

A)
A (2 I)
B)
VEL(I)
C)
MET(J + 2, K - 4)
D)
KET(2* J, 3* 1 - 4)

Correct Answer :   A (2 I)

231 .
Consider the following statement in FORTRAN 77

READ*, [X(I) Y(I) = 0, 3, 2]

Which of the following is correct?
A)
2 values could be read and stored asX (0), Y (0)
B)
4 values could be read and stored asX (0), Y (0), X (2), Y (2)
C)
6 values could be read and stored as X (0), Y (0), Y (2), X (3), Y (3)
D)
None of the above

Correct Answer :   4 values could be read and stored asX (0), Y (0), X (2), Y (2)

232 .
Assertion (A) : If contents of F register in 8085 are 01010001, it means S = 0, Z = 1, AC= 1, P = 0 and CY = 1.

Reason (R) : When an instruction is called from memory, the op. code of instruction is stored in instruction register.
A)
A is wrong R is correct
B)
A is correct R is wrong
C)
Both A and R are correct and R is correct explanation of A
D)
Both A and R are correct but R is not correct explanation of A

Correct Answer :   Both A and R are correct but R is not correct explanation of A

A)
logical
B)
arithmetic
C)
data transfer
D)
program control

Correct Answer :   arithmetic

A)
9
B)
8
C)
6
D)
1

Correct Answer :   6

A)
the two 8 bit values in registers AL and CH are added and result placed in register CH
B)
the two 16 bit values in registers AL and CH are added and result placed in register AL
C)
the two 8 bit values in registers AL and CH are added and result placed in register AL
D)
the two 16 bit values in registers AL and CH are added and result placed in register CH

Correct Answer :   the two 8 bit values in registers AL and CH are added and result placed in register AL

A)
STC
B)
RRC
C)
CMA
D)
CMC

Correct Answer :   CMC

A)
100000 hours
B)
8100000 hours
C)
1800000 hours
D)
None of the above

Correct Answer :   100000 hours

A)
-285.560
B)
-2.8 E + 02
C)
-2.85 E + 02
D)
-2.85560 E + 02

Correct Answer :   -2.8 E + 02

A)
AB - C + D
B)
AB + C - D +
C)
+ AB - C + D
D)
None of the above

Correct Answer :   AB + C - D +

A)
Synchronous transfer
B)
Interrupt driver transfer
C)
Asynchronous transfer
D)
None of the above

Correct Answer :   Asynchronous transfer

A)
fetch
B)
execute
C)
Both (A) and (B)
D)
None of the above

Correct Answer :   Both (A) and (B)

A)
800
B)
1200
C)
1200 x 1200
D)
None of the above

Correct Answer :   1200 x 1200

A)
Z8000
B)
80286
C)
MC68000
D)
All of the above

Correct Answer :   All of the above

A)
INR r
B)
ADC r
C)
SEE M
D)
ACI data

Correct Answer :   INR r

A)
registers
B)
mass storage
C)
cache memory
D)
primary memory

Correct Answer :   registers

A)
9 and 9 respectively
B)
9 and 5 respectively
C)
13 and 9 respectively
D)
13 and 5 respectively

Correct Answer :   13 and 9 respectively

A)
2 bits
B)
4 bits
C)
8 bits
D)
16 bits

Correct Answer :   4 bits

A)
-3
B)
J + 2
C)
JA - 4
D)
3* IA - 2

Correct Answer :   -3

249 .
The instruction MOV M, r in 8085
A)
causes the content of memory location whose address is in register H, L to move to register r and uses immediate addressing mode
B)
causes the content of register r to be moved to memory location whose address is in register H, L and uses immediate addressing mode
C)
causes the content of memory location whose address is in register H, L to move to register r and uses register indirect addressing mode
D)
causes content of register r to be moved to memory location whose address is in register H, L and uses register indirect addressing mode

Correct Answer :   causes content of register r to be moved to memory location whose address is in register H, L and uses register indirect addressing mode

A)
dot
B)
blank
C)
comma
D)
underscore

Correct Answer :   underscore

A)
2.3 E - 7
B)
-2.3 E - 7
C)
2.3 E - 8
D)
-2.3 E - 8

Correct Answer :   -2.3 E - 8

252 .
Consider the following program for 8085

MOV A, D
RAL
MOV D, A

If initial contents of register D is decimal number 20, the final content of register D is
A)
decimal 10
B)
decimal 20
C)
decimal 40
D)
None of the above

Correct Answer :   decimal 40

A)
a string of capital alphabets A to Z, included in question mark
B)
a string of capital alphabets A to Z, digits 0 to 9 and special characters
C)
a string of capital alphabets A to Z, digits 0 to 9 and special characters included in question mark
D)
a string of capital alphabets A to Z and digits 0 to 9 included in question mark

Correct Answer :   a string of capital alphabets A to Z, digits 0 to 9 and special characters included in question mark

A)
16 data bits are multiplexed with 16 bits of address bus
B)
8 data bits are multiplexed with 8 least significant bits of address bus
C)
8 data bits are multiplexed with 8 most significant bits of address bus
D)
None of the above

Correct Answer :   8 data bits are multiplexed with 8 least significant bits of address bus

A)
ADC M
B)
MOV M, r
C)
MVIM, data
D)
L X I rp, data 16

Correct Answer :   ADC M

A)
Data bus
B)
Control bus
C)
Address bus
D)
Exchange bus

Correct Answer :   Exchange bus

A)
only the sign of exponent
B)
only the sign of mantissa
C)
the sign of both mantissa and exponent
D)
Either (A) or (B)

Correct Answer :   only the sign of mantissa

A)
Crystal
B)
RC circuit
C)
LC resonant circuit
D)
Crystal and RC circuit

Correct Answer :   RC circuit

A)
00000000 01101100 10000100 00000000
B)
00000000 01101100 00000000 10000100
C)
01101100 00000000 00000000 10000100
D)
01101100 00000000 10000100 00000000

Correct Answer :   01101100 00000000 00000000 10000100

A)
8085
B)
8086
C)
partly 8085 and partly 8086
D)
None of the above

Correct Answer :   8086

A)
360 KB each
B)
1.2 MB each
C)
1.2 MB and 360 KB respectively
D)
360 KB and 1.2 MB respectively

Correct Answer :   360 KB and 1.2 MB respectively

A)
EEPROM
B)
UVEPROM
C)
Shadow RAM
D)
None of the above

Correct Answer :   Shadow RAM

A)
A ** B - C
B)
5.2 ** 4.3
C)
5.2 ** (- 4.3)
D)
(- 5.2) ** (4.3)

Correct Answer :   (- 5.2) ** (4.3)

A)
UV light intensity x exposure time in seconds
B)
UV light intensity / exposure time in seconds
C)
exposure time in seconds / UV light intensity
D)
None of the above

Correct Answer :   UV light intensity x exposure time in seconds

A)
OFH is stored at location 03 and 0 A 5 at location 02 and address of word in 02
B)
OFH is stored at location 02 and 0 A 5 at location 03 and address of word is 02
C)
OFH is stored at location 03 and 0 A 5 at location 02 and address of word is 03
D)
OFH is stored at location 02 and 0 A 5 at location 03 and address of word is 03

Correct Answer :   OFH is stored at location 02 and 0 A 5 at location 03 and address of word is 02

A)
bits
B)
bytes
C)
bubbles
D)
bits or bytes depending on type of memory

Correct Answer :   bits

A)
2 / 3.2
B)
4 / 8
C)
5*2
D)
5 * 2 / 3.2

Correct Answer :   5*2

A)
only when the logical expression controlling the loop becomes true for both repeat and while loops
B)
only when the logical expression controlling the loop becomes false for both repeat and while loops
C)
only when the logical expression controlling the loop becomes false for repeat loop and true for while loop
D)
only when the logical expression controlling the loop becomes true for repeat loop and false for while loop

Correct Answer :   only when the logical expression controlling the loop becomes true for repeat loop and false for while loop

A)
DO
B)
END
C)
ASSIGN
D)
DIMENSION

Correct Answer :   DIMENSION

A)
-2.3 e 37 to + 2.3 e 37
B)
-3.4 e 37 to + 3.4 e 37
C)
-2.3 e 38 to + 2.3 e 38
D)
-3.4 e 38 to + 3.4 e 38

Correct Answer :   -3.4 e 37 to + 3.4 e 37

271 .
The statement IMPLICIT INTEGER (A - F) in FORTRAN identifies
A)
all variable names which begin with A, B, C, D, E or F as being of integer type
B)
all variable names which begin with A and have 6 characters
C)
all character names which begin with A, B, C, D, E, or F as being of integer type
D)
None of the above

Correct Answer :   all variable names which begin with A, B, C, D, E or F as being of integer type

A)
Four 4 bit shift registers
B)
Four 8 bit shift registers
C)
Eight 4 bit shift registers
D)
Eight 8 bit shift registers

Correct Answer :   Eight 4 bit shift registers

A)
incremented by 2
B)
decremented by 2
C)
incremented by 1
D)
decremented by 1

Correct Answer :   incremented by 2


Explanation : The actual current stack top is poped into the specific operand as the contents of stack top memory is stored in AL&SP and further contents of the memory location pointed to by SP are copied to AH & SP.

A)
IN, MOV
B)
IN, OUT
C)
MOV, IN
D)
MOV, XCHG

Correct Answer :   IN, OUT


Explanation : The address of the input/output port may be specified directly or indirectly.
Example for input port: IN AX, DX; This instruction reads data from a 16-bit port whose address is in DX and stores it in AX
Example for output port: OUT 03H, AL; This sends data available in AL to a port whose address is 03H.

A)
interrupt handling ability
B)
interrupt processing ability
C)
multiple interrupt executing ability
D)
multiple interrupt processing ability

Correct Answer :   multiple interrupt processing ability


Explanation : The processor if handles more devices as interrupts then it has multiple interrupt processing ability.

A)
nonmaskable interrupt
B)
nonmultiple interrupt
C)
nonmovable interrupt
D)
None of the above

Correct Answer :   nonmaskable interrupt


Explanation : NMI is the acronym for nonmaskable interrupt.

A)
handle no interrupt request
B)
handle one interrupt request
C)
handle one or more interrupt requests at a time
D)
handle one or more interrupt requests with a delay

Correct Answer :   handle one or more interrupt requests at a time


Explanation : If more than one interrupt request (INTR) occurs at a time, then an external chip called programmable interrupt controller is required to handle them.

A)
static lower memory bank
B)
odd address memory bank
C)
even address memory bank
D)
lower address memory bank

Correct Answer :   even address memory bank


Explanation : In static memory, the lower 8-bit bank is called even address memory bank.

A)
address is odd and memory is in RAM
B)
address is even and memory is in RAM
C)
address is odd and memory is in ROM
D)
address is even and memory is in ROM

Correct Answer :   address is even and memory is in RAM


Explanation : If Ao=0 then address is even and is in RAM. If Ao=1 then, address is odd and is in RAM.

A)
RAM
B)
ROM
C)
ONLY RAM
D)
RAM and ROM

Correct Answer :   RAM and ROM


Explanation : If at a time Ao and BHE(active low) both are zero, then both RAM and ROM are selected.

A)
memory locations
B)
only input devices
C)
only output devices
D)
distinct I/O devices

Correct Answer :   memory locations


Explanation : In memory-mapped scheme, the devices are viewed as memory locations and are addressed likewise.

A)
sink to source
B)
source to sink
C)
inputs to outputs
D)
outputs to inputs

Correct Answer :   inputs to outputs


Explanation : If DIR is 1, then the direction is from A(inputs) to B(outputs).

A)
sinked from data lines
B)
sourced from data lines
C)
sinked from address lines
D)
sourced or sinked from data lines

Correct Answer :   sourced or sinked from data lines


Explanation : More current should not be sourced or sinked from data lines while reading to avoid loading.

A)
port A to data bus
B)
port B to data bus
C)
port C to data bus
D)
CWR to data bus

Correct Answer :   port B to data bus


Explanation : If A1=0, A0=1 then the input read cycle is performed from port B to data bus.

A)
CS(active low) = 0
B)
CS(active low) = 1
C)
CS(active low) = 0, RD(active low) = 1, WR(active low) = 1
D)
CS(active low) = 1 OR CS(active low) = 0, RD(active low) = 1, WR(active low) = 1

Correct Answer :   CS(active low) = 1 OR CS(active low) = 0, RD(active low) = 1, WR(active low) = 1


Explanation : The data bus is tristated when chip select pin=1 or chip select pin=0 and read and write signals are high i.e 1.

A)
SET
B)
CLK
C)
RESET
D)
CLEAR

Correct Answer :   RESET


Explanation : If reset pin is enabled then the control word register is cleared.

A)
low complexity
B)
low cost option
C)
slow practical applications
D)
All of the above

Correct Answer :   All of the above


Explanation : Compared to other 12-bit ADCs, it is of very low cost and useful for slow practical applications.

288 .
In the signal integrate phase, the differential input voltage between IN LO(input low) and IN HI(input high) pins is integrated by the internal integrator for a fixed period of
A)
256 clock cycles
B)
1024 clock cycles
C)
2048 clock cycles
D)
4096 clock cycles

Correct Answer :   2048 clock cycles


Explaination : The internal integrator needs 2048 clock cycles to integrate voltage difference between input low and input high.

A)
Direct addressing
B)
Register addressing
C)
Immediate addressing
D)
Register indirect addressing

Correct Answer :   Direct addressing

290 .
Consider the following statement in Fortran 77

READ *, (X.(I), I = 0, 4)

Which of the following is correct?
A)
3 values would be read and stored in X (0), X (1), X (2)
B)
4 values would be read and stored in X (0), X (1), X (2), X (3)
C)
5 values would be read and stored in X (0), X (1), X (2), (3)(4)
D)
5 values would be read and stored in X (I), X (2), X (3), X (4), X (5)

Correct Answer :   5 values would be read and stored in X (0), X (1), X (2), (3)(4)

A)
Using program counter
B)
Machine language programming
C)
Implicit source and destination address
D)
Common source and destination address

Correct Answer :   Machine language programming

A)
a negative number
B)
a positive number and closer to zero
C)
a negative number and closer to zero
D)
a positive number and away from zero

Correct Answer :   a positive number and closer to zero

A)
STOP
B)
GO TO
C)
PARAMETER
D)
None of the above

Correct Answer :   PARAMETER

294 .
The initial contents of ACC and CY in 8085 are

    CY = 1ACC = 10000001

After instruction RAL is executed once the contents of CY and ACC respectively will be
A)
1 and 00000011
B)
0 and 00000011
C)
1 and 00000001
D)
0 and 00000001

Correct Answer :   1 and 00000011

295 .
The instruction MOV r1, r2 in 8085
A)
causes contents of registers r1 to be moved to r2 and uses register addressing
B)
causes contents of registers r1 to be moved to r2 and uses register immediate addressing
C)
causes contents of registers r2 to be moved to r1 and uses register immediate addressing
D)
causes contents of registers r2 to be moved to r1, and uses register addressing

Correct Answer :   causes contents of registers r2 to be moved to r1, and uses register addressing

A)
A / D * CF AD - F * +
B)
AB / C * FAD - E * + *
C)
AB / CF * AD - E * + *
D)
None of the above

Correct Answer :   AB / CF * AD - E * + *

A)
5 to 10 KB of main memory
B)
10 to 100 KB of main memory
C)
100 to 256 KB of main memory
D)
256 KB to 1 MB of main memory

Correct Answer :   256 KB to 1 MB of main memory

A)
Organisation
B)
Physical dimensions
C)
Timing characteristics
D)
Power consumption and bus loading

Correct Answer :   Power consumption and bus loading

A)
F DISK
B)
A DISK
C)
D DISK
D)
H DISK

Correct Answer :   F DISK

300 .
In 8085 the instruction MOV A, M is an example of
A)
Direct addressing
B)
Register addressing
C)
Immediate addressing
D)
Register indirect addressing

Correct Answer :   Register indirect addressing

301 .
Consider the following statements

1. A 33 MHz 486 has higher MIPS than 33 MHz 386.
2. A 33 MHz 486 has higher MFLOPS than 33 MHz 386 of these statements.
A)
Both are true
B)
Both are false
C)
1 is true and 2 is false
D)
1 is false and 2 is true

Correct Answer :   Both are true

A)
one character set having 48 characters
B)
one character set having 60 characters
C)
one character set having 60 characters and another character set having 48 character
D)
None of the above

Correct Answer :   one character set having 60 characters

A)
-1
B)
-6
C)
6
D)
7

Correct Answer :   -1

A)
1 μs
B)
10 ns
C)
60 ns
D)
300 ns

Correct Answer :   60 ns

A)
I/O port
B)
Programmable interval timer
C)
Programmable CRT controller
D)
Programmable interrupt controller

Correct Answer :   Programmable CRT controller

A)
.5452 E - 3 is changed to 5.452 E - 4 but .9432 E - 4 is not changed
B)
.9432 E - 4 is changed to .09432 E - 3 and .5452 E - 3 is not changed
C)
both Ihe numbers are changed and their exponents are, made equal to -5
D)
None of the above

Correct Answer :   .9432 E - 4 is changed to .09432 E - 3 and .5452 E - 3 is not changed

307 .
Consider the following program for 8085

MVIA, 00H
LXIH, OEAOFG
V LDAZ ADI 0005H

What does this program do?
A)
Adds decimal number 5 to contents of memory
B)
Adds decimal number 15 to contents of memory
C)
Subtracts decimal number 5 from contents of memory
D)
Subtracts decimal number 15 from contents of memory

Correct Answer :   Adds decimal number 5 to contents of memory

A)
10
B)
12
C)
13
D)
15

Correct Answer :   13

A)
1 μs
B)
1 ns
C)
10 ns
D)
100 ns

Correct Answer :   100 ns

A)
ALE
B)
DEN
C)
INTR
D)
MN / MX

Correct Answer :   MN / MX

A)
IA * B
B)
IA + IB
C)
- K * I + J
D)
IB * IC / ID

Correct Answer :   IA * B

A)
8
B)
16
C)
20
D)
32

Correct Answer :   20

A)
sign magnitude
B)
1's complement
C)
2's complement
D)
Both (A) and (B)

Correct Answer :   2's complement

A)
MOVE
B)
MEM IM
C)
Both (A) and (B)
D)
MEM MAKER

Correct Answer :   MEM MAKER

A)
read operation
B)
write operation
C)
verifying DMA operation
D)
All of the above

Correct Answer :   All of the above


Explanation : The 8257 can accomplish three types of operations and they are

i) verify DMA operation
ii) write operation
iii) read operation.

A)
HRQ
B)
DACK
C)
HLDA
D)
All of the above

Correct Answer :   HLDA


Explanation : If the HLDA signal is received by the DMA controller, it indicates that the bus is available.

A)
HRQ signal
B)
DACK (active low)
C)
HLDA signal
D)
DACK (active high)

Correct Answer :   DACK (active low)


Explanation : The DACK (active low) line of the used channel is pulled down by the DMA controller to indicate the I/O device that its request for the DMA transfer has been honored by the CPU.

A)
burst transfer
B)
multi transfer
C)
simultaneous transfer
D)
None of the above

Correct Answer :   burst transfer


Explanation : If more than one channel requests service simultaneously, then the transfer occurs as a burst or continuous transfer.

A)
fixed priority scheme
B)
rotating priority scheme
C)
rotating priority and fixed priority scheme
D)
None of the above

Correct Answer :   fixed priority scheme


Explanation : In this scheme, the DRQ3 has the lowest priority followed by DRQ2 and DRQ1. The DRQ0 has the highest priority.

A)
Status register
B)
DMA address register
C)
Mode set register
D)
Terminal count register

Correct Answer :   Mode set register


Explanation : The selected register may be read or written depending on the instruction executed by the CPU. But only write operation can be performed on the mode set register.

A)
fixed priority scheme
B)
rotating priority scheme
C)
rotating priority and fixed priority scheme
D)
None of the above

Correct Answer :   rotating priority scheme


Explanation : In this scheme, the priorities assigned to the channels are not fixed.

A)
Return buffers
B)
Display address registers
C)
I/O control and data buffers
D)
Control and timing registers

Correct Answer :   Control and timing registers


Explanation : The control and timing register to store the keyboard and display modes and other operations programmed by CPU.

A)
display RAM
B)
control and timing register
C)
display address registers
D)
control and timing register and timing control

Correct Answer :   display address registers


Explanation : The display address registers holds the address of the word currently being written or read by the CPU to or from the display RAM.

A)
scanned keyboard special error mode
B)
sensor matrix mode
C)
scanned keyboard with N-key rollover
D)
scanned keyboard mode with 2 key lockout

Correct Answer :   scanned keyboard special error mode


Explanation : The scanned keyboard special error mode is programmed using end interrupt/error mode set command. This mode is valid only under the N-key rollover mode.

326 .
When a key is pressed, the debounce circuit waits for 2 keyboard scans and then checks whether the key is still depressed in
A)
sensor matrix mode
B)
scanned keyboard with N-key rollover
C)
scanned keyboard special error mode
D)
scanned keyboard mode with 2 key lockout

Correct Answer :   scanned keyboard with N-key rollover


Explaination : In this mode, When a key is pressed, the debounce circuit waits for 2 keyboard scans and then checks whether the key is still depressed. If it is still depressed, the code is entered in FIFO RAM.

A)
keyboard mode
B)
strobed input mode
C)
scanned sensor matrix mode
D)
keyboard and strobed input mode

Correct Answer :   keyboard and strobed input mode


Explanation : Overrun error occurs when an already full FIFO has attempted an entry. Underrun error occurs when an empty FIFO read is attempted.

A)
8 MB
B)
16 MB
C)
24 MB
D)
64 MB

Correct Answer :   16 MB


Explanation : The 80286 with its 24-bit address bus is able to address 16 Mbytes of physical memory.

A)
memory
B)
control unit
C)
bus interface unit
D)
memory management unit

Correct Answer :   memory management unit


Explanation : The memory management which is an important task of the operating system is now supported by a hardware unit called a memory management unit.

A)
mapping
B)
pipelining
C)
swapping out
D)
swapping in

Correct Answer :   swapping out


Explanation : In swapping out, a portion of the program or important partial results required for further execution, may be saved back on secondary storage to make the physical memory free, for further execution of another required portion of the program.

A)
virtual memory
B)
logical memory
C)
auxiliary memory
D)
imaginary memory

Correct Answer :   virtual memory


Explanation : To the user, there exists a very large logical memory space, which is actually not available called virtual memory. This does not exist physically in a system. It is however, possible to map a large virtual memory space onto the real physical memory.

A)
normal mode
B)
real address mode
C)
virtual address mode
D)
real and virtual address mode

Correct Answer :   real and virtual address mode


Explanation : The 80286 is operated in two modes, namely real address mode and virtual address mode. In both the modes, the 80286 is compatible with 8086/8088.

A)
bus control logic
B)
floating point unit
C)
data interface and control unit
D)
None of the above

Correct Answer :   None of the above


Explanation : The three blocks of an internal architecture of 80287 are:

1. bus control logic
2. data interface and control unit
3. floating point unit.

A)
instruction decoders
B)
status and control words
C)
tag words and error pointers
D)
All of the above

Correct Answer :   All of the above


Explanation : The data interface and control unit contains status and controls words, TAG words and error pointers.

A)
masking bits
B)
condition code bits
C)
error summary bits
D)
rounding control bits

Correct Answer :   condition code bits


Explanation : The condition code bits are similar to the flags of a CPU. These are modified depending upon the result of the execution of arithmetic instructions.

A)
masking bits
B)
infinity control bits
C)
rounding control bits
D)
precision control bits

Correct Answer :   infinity control bits


Explanation : The infinity control bit is initialized to zero after reset.

A)
barrel shifter
B)
data buffer
C)
error pointer
D)
None of the above

Correct Answer :   barrel shifter


Explanation : The barrel shifter arranges and presents the data to be shifted successively, whenever required for the execution.

A)
tightly coupled configuration
B)
loosely coupled configuration
C)
tightly and loosely coupled configurations
D)
None of the above

Correct Answer :   tightly coupled configuration


Explanation : In a tightly coupled configuration, the 8089 shares the system bus and memory with the host CPU using its RQ (active low) or GT (active low) pins.

A)
tightly coupled configuration
B)
loosely coupled configuration
C)
tightly and loosely coupled configurations
D)
None of the above

Correct Answer :   loosely coupled configuration


Explanation : In a loosely coupled configuration, the 8089 has its own local bus and communicates with the host CPU using bus arbiter and controller.

A)
8-bit IO
B)
16-bit IO
C)
64-bit IO
D)
16-bit and 8-bit IO

Correct Answer :   16-bit and 8-bit IO


Explanation : The 8089 handled IO devices need not have the same data bus width as that of 8089. This enables even 8-bit IO devices to be interfaced easily with 8089.

A)
adding 16 to the contents of CCP
B)
adding memory table address of channel-1
C)
adding 8 to the contents of CCP
D)
None of the above

Correct Answer :   adding 8 to the contents of CCP


Explanation : The address of the memory table for channel-2 is calculated by adding 8 to the contents of CCP or by adding memory table address for channel-1 to the contents of CCP.

A)
GB, GC
B)
GC, BC
C)
GA, GB
D)
GC, GA

Correct Answer :   GA, GB


Explanation : GA register is used as source and GB as destination pointers during DMA operations.

A)
EXT
B)
SINTR
C)
DRQ and EXT
D)
RQ (active low) or GT (active low)

Correct Answer :   DRQ and EXT


Explanation : The DRQ and EXT are used for data transfer control and operation termination signals during DMA operations.

A)
SINTR
B)
DRQ
C)
RQ (active low)
D)
GT (active low)

Correct Answer :   SINTR


Explanation : The SINTR pins are used by the channels either to inform the CPU that the previous operation is over or to ask for its attention or interference if required, before the completion of the task.

A)
8-bit data operand
B)
16-bit data operand
C)
32-bit data operand
D)
All of the above

Correct Answer :   All of the above


Explanation : The 80386DX is a 32-bit processor that supports, 8-bit/16-bit/32-bit data operands.

A)
8 address lines
B)
16 address lines
C)
32 address lines
D)
64 address lines

Correct Answer :   32 address lines


Explanation : The 80386, with its 32-bit address bus, can address up to 4 GB of physical memory.

A)
paging
B)
segmentation
C)
memory division
D)
None of the above

Correct Answer :   paging


Explanation : The concept of paging which is introduced in 80386, enables it to organize the available physical memory into pages of size 4 KB each, under the segmented memory.

A)
8086 and 80287
B)
80286 and 8086
C)
80286 and 80287
D)
80287 and 80387

Correct Answer :   80286 and 8086


Explanation : The 80386 in protected mode, supports all software written for 8086 and 80286 (to be executed under the control of memory management and protection abilities of 80386).

A)
12 MHz and 20 MHz
B)
20 MHz and 33 MHz
C)
32 MHz and 12 MHz
D)
All of the above

Correct Answer :   20 MHz and 33 MHz


Explanation : The operating frequency of 80386DX is 20MHz and 33 MHz.

A)
on-chip address translation cache
B)
virtual memory space of 64TB
C)
instruction set of predecessors with upward compatibility
D)
All of the above

Correct Answer :   All of the above


Explanation : The 80386 has on-chip address translation cache, and the instruction set is upward compatible with all its predecessors.

A)
64 pin
B)
128 pin
C)
132 pin
D)
142 pin

Correct Answer :   132 pin


Explanation : The 80386DX is available in a 132-pin grid array package.

A)
E
B)
X
C)
XX
D)
32

Correct Answer :   E


Explanation : A 32 bit register, known as an extended register, is represented by the register name with a prefix of E.

A)
ES
B)
FS
C)
GS
D)
All of the above

Correct Answer :   All of the above


Explanation : The six segment registers available in 80386 are CS, SS, DS, ES, FS and GS, out of which DS, ES, FS and GS are the four data segment registers.

A)
virtual mode
B)
protected mode
C)
either virtual or protected mode
D)
All of the above

Correct Answer :   protected mode


Explanation : If VM flag is set, the 80386 enters the virtual 8086 mode within the protected mode. This is to be set only when the 80386 is in protected mode.

A)
IRET instruction
B)
Task switch operation
C)
IRET instruction or task switch operation
D)
None of the above

Correct Answer :   IRET instruction or task switch operation


Explanation : The VM flag can be set using the IRET instruction or any task switch operation, only in the protected mode.

A)
RF is set
B)
RF is cleared
C)
VM flag is set
D)
VM flag is cleared

Correct Answer :   RF is set


Explanation : If RF (resume flag) is set, any debug fault is ignored during the instruction cycle.

A)
reset
B)
read
C)
write
D)
terminate

Correct Answer :   reset


Explanation : If the 80386 enters the protected mode from the real address mode, then it cannot return back to the real mode without a reset operation.

A)
64
B)
128
C)
256
D)
512

Correct Answer :   256


Explanation : In virtual mode, the paging unit allows only 256 pages, each of 4Kbytes size. Each of the pages may be located anywhere within the maximum 4Gbytes physical memory.

A)
level 3
B)
level 2
C)
level 1
D)
level 0

Correct Answer :   level 0


Explanation : The real mode programs are executed at the highest privilege level i.e. level 0.

A)
protected mode
B)
synchronous mode
C)
real address mode
D)
asynchronous mode

Correct Answer :   synchronous mode


Explanation : If the CKM pin of 80387 is high, then 80387 is operated in synchronous mode. If it is low, then 80387 is operated in asynchronous mode.

A)
A30
B)
A31
C)
D31
D)
M/IO

Correct Answer :   M/IO


Explanation : The NPS1 and NPS2 lines are directly connected with M/IO and A31 respectively, to inform 80387 that the CPU wants to communicate with it (NPS1), and it is using one of the reserved I/O addresses for 80387 (NPS2).

A)
32-bit, 32-bit
B)
32-bit, 48-bit
C)
48-bit, 32-bit
D)
48-bit, 48-bit

Correct Answer :   48-bit, 48-bit


Explanation : 80387 consists of two 48-bit registers, known as instruction and data pointer registers.

A)
full sharing
B)
threshold sharing
C)
partitioned resources
D)
All of the above

Correct Answer :   All of the above


Explanation : Several resource sharing strategies have been investigated by the developers. Some of these are

1. Partitioned resources
2. Threshold sharing
3. Full sharing.

A)
HALT
B)
HLDA
C)
HOLD
D)
None of the above

Correct Answer :   HALT


Explanation : An optimization may require the use of HALT instruction, when either of the two logical processors is idle.

A)
control unit
B)
memory unit
C)
operating system
D)
execution unit

Correct Answer :   operating system


Explanation : The HALT instruction is a privileged instruction that can be only used by operating system.

A)
one logical processor per physical processor
B)
two logical processor per physical processor
C)
three logical processor per physical processor
D)
zero logical processor per physical processor

Correct Answer :   two logical processor per physical processor


Explanation : The Xeon TM processor on which hyperthreading technology was first implemented consists of two logical processor per physical processor.

A)
Branch History Table (BHT)
B)
Branch Target Buffer (BTB)
C)
Branch History Table and Branch Target Buffer
D)
None of the above

Correct Answer :   Branch History Table and Branch Target Buffer


Explanation : The dynamic branch prediction algorithms use two types of tables, namely Branch History Table (BHT) and Branch Target Buffer (BTB).

A)
one 32-byte streaming buffer
B)
one 64-byte streaming buffer
C)
two 32-byte streaming buffers
D)
two 64-byte streaming buffers

Correct Answer :   two 64-byte streaming buffers


Explanation : Each logic processor has its own set of two 64-byte streaming buffers, which store the instruction bytes and subsequently they are dispatched to the instruction decode stage.

A)
logical processors
B)
trace cache
C)
instruction decoder
D)
All of the above

Correct Answer :   logical processors


Explanation : Since there are two logical processors, there are two ITLBs. Thus each logical processor has its own ITLB and its own instruction pointer to track the progress of instruction fetch for each of them.

A)
Accumulator
B)
B register
C)
Data register
D)
Accumulator and B register

Correct Answer :   Accumulator and B register


Explanation : In some instructions, the Accumulator and B register are used to store the operands.

A)
8051
B)
8086
C)
80286
D)
80386

Correct Answer :   8051


Explanation : The 8051 stack is not a top-down data structure, like other Intel processors.

A)
serial-in serial-out register
B)
parallel-in serial-out register
C)
serial-in parallel-out register
D)
parallel-in parallel-out register

Correct Answer :   parallel-in serial-out register


Explanation : The transmit buffer of serial data buffer is a parallel-in serial-out register.

A)
IP and IE
B)
TMP1 and TMP2
C)
Instruction registers
D)
Accumulator and B register

Correct Answer :   TMP1 and TMP2


Explanation : The arithmetic operations are performed over the operands held by the temporary registers, TMP1 and TMP2. Users cannot access these temporary registers.

A)
IE
B)
IP
C)
TSCON
D)
PCON and SCON

Correct Answer :   PCON and SCON


Explanation :  The registers, PCON and SCON contain control and status information about serial port.

A)
ET0 and ET1
B)
ET1 and ET2
C)
EX0 and EX1
D)
EX1 and EX2

Correct Answer :   EX0 and EX1


Explanation : The bits, EX0 and EX1 individually control the external interrupts, INT0(active low) and INT1(active low). If INT0(active low) and INT1(active low) interrupts are to be enabled then the bits EX0 and EX1 must be set respectively.

A)
equal to one machine cycle
B)
equal to 2 machine cycles
C)
greater than one machine cycle
D)
greater than 2 machine cycles

Correct Answer :   equal to one machine cycle


Explanation : The minimum duration of the active low interrupt pulse should be equal to the duration of one machine cycle for being sensed, else it will be lost.

A)
one machine cycle
B)
two machine cycles
C)
three machine cycles
D)
four machine cycles

Correct Answer :   two machine cycles


Explanation : For an interrupt to be guaranteed served it should have duration of two machine cycles.

A)
read operation
B)
write operation
C)
read and write operations
D)
None of the above

Correct Answer :   write operation


Explanation : The control word register can only be written and cannot be read.

A)
1 clockcycle
B)
2 clockcycles
C)
3 clockcycles
D)
4 clockcycles

Correct Answer :   1 clockcycle


Explanation : After (N-1) cycles, the output becomes low for only 1 clockcycle. If the count N is reloaded and again the output becomes high and remains so for (N-1) clock pulses.

A)
read/load least significant byte only
B)
read/load most significant byte only
C)
read/load LSB first and then MSB
D)
read/load MSB first and then LSB

Correct Answer :   read/load LSB first and then MSB


Explanation : To access 16 bit, first LSB is loaded first, and then MSB.

A)
selection of counters
B)
choosing binary/BCD counters
C)
initializing the operating modes
D)
All of the above

Correct Answer :   All of the above


Explanation : The control word register contents are used for :

i) initializing the operating modes (mode 0-mode 4)
ii) selection of counters (counter0-counter2)
iii) choosing binary or BCD counters
iv) loading of the counter registers.

A)
1.4 E - 38 to 3.4 E + 45
B)
3.4 E - 38 to 1.4 E + 45
C)
3.4 E - 45 to 1.4 E + 38
D)
1.4 E - 45 to 3.4 E + 38

Correct Answer :   1.4 E - 45 to 3.4 E + 38

A)
A = B
B)
10 = N
C)
A = B = 1
D)
X + 2.1 = 5.6

Correct Answer :   A = B

A)
Logic operations can not be performed
B)
I/O devices have 16 bit addresses
C)
I/O devices are accessed during IN and OUT instructions
D)
There can be a maximum of 256 input and 256 output devices

Correct Answer :   I/O devices have 16 bit addresses

A)
Z 80
B)
6800
C)
6809
D)
8085

Correct Answer :   6809

A)
integer constants
B)
integer constants or real constants
C)
Integer constants or integer variable names
D)
integer constants or integer variable names or integer expressions

Correct Answer :   integer constants or integer variable names or integer expressions

A)
operation code, low order byte of data/address and high order byte of data / address
B)
low order byte of data / address, high order byte of data / address and operation code
C)
high order byte of data / address, low order byte of data / address and operation code
D)
high order byte of data / address, operation code and low order byte of data / address

Correct Answer :   operation code, low order byte of data/address and high order byte of data / address

A)
16 bit registers and 16 bit data bus
B)
32 bit register and 16 bit data bus
C)
16 bit registers and 32 bit data bus
D)
32 bit registers and 32 bit data bus

Correct Answer :   32 bit register and 16 bit data bus

A)
S = 1, Z = 1, AC = 1, P = 0 and CY=1
B)
S = 1, Z = 0, AC = 0, P = 0 and CY=1
C)
S = 0, Z = 1, AC = 1, P = 0 and CY=1
D)
S = 1, Z = 0, AC = 1, P = 0 and CY=1

Correct Answer :   S = 0, Z = 1, AC = 1, P = 0 and CY=1

A)
bit, byte, short and long
B)
byte, short, int and long
C)
nibble, byte, int and long
D)
nibble, byte, short and long

Correct Answer :   byte, short, int and long

A)
TRAP
B)
RST 5.5
C)
RST 7.5
D)
RST 9.5

Correct Answer :   RST 9.5

A)
111
B)
000
C)
001
D)
1111

Correct Answer :   111

A)
TRAP and RST lines are high
B)
RST line is high but not TRAP line
C)
TRAP and RST lines are not high
D)
TRAP line is high but not RST line

Correct Answer :   TRAP and RST lines are not high

A)
identify the variable as an array name
B)
reserve memory locations to store all the components of array
C)
check the values computed for subscripts expressions during program execution with the specified lower and upper limits of subscripts
D)
All of the above

Correct Answer :   All of the above

A)
Add
B)
Shift
C)
Multiply
D)
Subtract

Correct Answer :   Shift

A)
logic group
B)
branch group
C)
arithmetic group
D)
data transfer group

Correct Answer :   logic group

A)
30
B)
40
C)
50
D)
60

Correct Answer :   40

A)
Single stage device
B)
Digital state device
C)
Solid state device
D)
Surface stage device

Correct Answer :   Solid state device

A)
2500
B)
100000
C)
125000
D)
250000

Correct Answer :   125000

A)
5
B)
6
C)
7
D)
8

Correct Answer :   8