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.
Correct Answer : Magnetic Ink Character Recognition
Explanation : Magnetic ink character recognition is used in banking.
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.
Correct Answer : All of the above
Explanation : They are of three types :CISC : Complex Instruction Set ComputerRISC : Reduced Instruction Set ComputerEPIC : Explicitly Parallel Instruction Computing
Correct Answer : -0
Correct Answer : Magnetic tape
Explanation : Magnetic tape can be accessed only sequentially.
Correct Answer : 13.625
Explanation : Binary .101 equals 0.625 in decimal.
Correct Answer : i = i + 1
Explanation : i is incremented by 1.
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.
Correct Answer : PMOS
Explanation : PMOS technology was used for designing the processor because this technology was slow but simple.
Correct Answer : Circuit A has more gates than circuit B
Correct Answer : True
JCOKE = 3
JCOKE = JCOKE + 1
GO TO (5, 8, 9, 11, 15, 16 18, 20) JCOKE.
Correct Answer : 11
Explaination : It is a computed GO TO statement. Since J COKE = 3 + 1 = 4, the control is transferred to statement 11.
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.
Correct Answer : thrice
Explanation : When RAL instruction is used once the number is doubled.
Correct Answer : IC = FC + EC
Explanation : Instruction cycle consists of fetch and execute cycles.
Correct Answer : 1 a
Correct Answer : .4546 E 5 is changed .004546 E 7 and .5433 E 7 is not changed
Correct Answer : a few kilobytes
Explanation : Size of Cache memory varies from about 16 K bytes to about 256 K bytes.
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.
Correct Answer : 246
Explanation : In an 8-bit microprocessor, maximum 28 = 256 opcodes are possible. But it consists of only 246 opcodes.
Correct Answer : The same as
Correct Answer : Z, CY, S, P and AC
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.
Correct Answer : Primary memory
Explanation : Primary memory holds these during processing of instructions.
Correct Answer : 350 mW and 5 mW respectively
Explanation : Power requirement in stand by mode is very low 1000.
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).
Correct Answer : 4
Explanation : C has four storage classes : 1. automatic storage class2. register storage class3. static storage class4. external storage class.
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.
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.
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.
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.
Correct Answer : PIC1x
Explanation : Z8000, Motorola 6809, and Zilog Z8 are microprocessors but PIC1x is an 8-bit microcontroller.
Correct Answer : 11 decimal places
Correct Answer : Forces the assembler to reserve a specified number of consecutive bytes in the memory
Correct Answer : BYTE
Correct Answer : Set of {AND,OR,NOT}
Correct Answer : Arithmetic, relational, assignment
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.
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.
Correct Answer : need 5 V supply
Explanation : Digital chips need 5 V supply.
Correct Answer : 8255
Explanation : 8255 is interface chip for 8086 and ADC.
Correct Answer : AB * 2
Explanation : Operator is not allowed.
Correct Answer : C
Explanation : Whenever keyword 'break' is encountered inside any C loop control passes to the first statement after the loop.
C
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.
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.
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).
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.
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.
Correct Answer : One million
Correct Answer : A pair of cross coupled NAND
Correct Answer : A is wrong R is correct
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.
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.
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.
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.
Explaination : Crystal has to be connected externally to pins 1 and 2 of 8085.
X = 2.57
X = (X + 0.06) * 10
I = TRUNC (X)
X = I
X = X / 10.0
Correct Answer : 2.6
Explaination :
Correct Answer : else
Correct Answer : 8 address pins and 4 data pins
Explanation : 28 = 256
Correct Answer : + + + - * =
Correct Answer : 2's complement
Explanation : 2's complement is universally used in computers.
Explanation : All these characteristics are important.
Correct Answer : read, byte erase, byte write and chip erase operations
Correct Answer : 6
Explanation : It gives remainder in division operation.
Explanation : All are valid.
IF (JCOKE - 3 * LPEPSI) 5, 6, 7
5 JCOKE = JCOKE + 5
6 JCOKE = JCOKE + 8
7 JCOKE = JCOKE + 11
Correct Answer : 103
Explaination : Since (J COKE - 3 x L PEPSI) is negative, statement 5 is executed and the result is 98 + 5 = 103.
Correct Answer : x = 7
Explanation : The operator + = means add and assign.
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.
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.
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.
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.
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.
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.
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.
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.
Correct Answer : 127
A2 + B2 - 3 AB
Correct Answer : A * A + B * B - 3.0 * A * B
Correct Answer : 210
Explanation : The result should not contain decimal.
Correct Answer : RAM
Explanation : RAM has volatile memory and therefore the matter has to be saved frequently.
Correct Answer : 5 / 2
Explanation : Since both 5 and 2 are integers, the result will be 2.
Correct Answer : BRG
Explanation : The first letter is not I, J, K, L, M, N.
Correct Answer : 2 and 3 only
Explaination : ALU performs arithmetic and logical computations.
DO 25 J = 1, 7
Correct Answer : 7
Explaination : Do statement is executed when J = 1, 2, 3, 4, 5, 6, 7, i.e., a total of 7 times.
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.
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.
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.
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.
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.
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.
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.
Correct Answer : NAND-NAND
Correct Answer : A larger floating point number
Correct Answer : Mode 0, mode 1 and mode 2
Correct Answer : Reserved words
Correct Answer : Logical IF, block IF and arithmetic IF
Correct Answer : R is asserted, S is asserted
Correct Answer : decimal 32
Explanation : If instruction RAL is executed once the number gets doubled.
Explanation : Modulus operator gives the remainder of division as answer.
Explaination : SP and PC are 16 bit each in 8085.
IF (3 * JCOKE. EQ. LPESPSI)
JCOKE = JCOKE + 2
JCOKE = JCOKE + 5
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.
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.
IF (SALT. GE. PEPPER) GOTO 11
GOTO 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.
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.
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.
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.
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.
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.
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.
Correct Answer : XY
Explanation : Adding X to PROD, Y times evidently gives the product XY.
Correct Answer : If (A > = B) X : Y else X : = Z
Explanation : 'then' is missing.
READ *, J, X
K = J ** 2
Z = 3 * X
Print *, J, K, Z
Correct Answer : 12, 144, 73.2
J = 12, K = 122 = 144, Z = 3 x 24.4 = 73.2 In the result J and K will be without decimal point.
Correct Answer : 90
Explanation : SUM = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 = 90.
Correct Answer : MAX
Explanation : MAX is not data base function.
Correct Answer : 80
Explanation : The result should be sum of 50 and 30 and decimal must be added.
Correct Answer : DB III plus
Explanation : dB III plus is DBMS and not for financial accounting.
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.
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.
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.
Correct Answer : EEPROM
Correct Answer : int bar
Correct Answer : Halves
Correct Answer : 1 and 2 only
Explaination : First character must be alphabet only.
Correct Answer : 0.333 μs
Explanation :
WR and RD are active low.
Correct Answer : AFH
Correct Answer : 4 MOS transistors
Explanation : Each SRAM cell has 4 MOS transistors.
Correct Answer : Branch group
Correct Answer : 10
Correct Answer : 4.37 E - 5
Correct Answer : -28.0
Correct Answer : 480000 B
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.
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.
Correct Answer : data copy/transfer instruction
Explanation : These instructions are used to copy and transfer the instructions.
Correct Answer : DAS
Explanation : DAS (Decimal Adjust after Subtraction) is an arithmetic instruction.
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.
Correct Answer : breaks the normal sequence of execution of instructions
Explanation : An interrupt function is to break the sequence of operation.
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.
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.
Correct Answer : two dimensional
Explanation : The semiconductor memories are organised as two dimensions of an array which consists of rows and columns.
Correct Answer : data bus
Explanation : The bits in a selected location are accessible using data bus.
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
Correct Answer : 1024
Explanation : Since for n address lines, the number of memory locations able to address is 2^n.
Correct Answer : odd address memory bank
Explanation : In static memory, the upper 8-bit bank is called odd address memory bank.
Correct Answer : joystick
Explanation : Since joystick is an input device, it reads data from the external devices.
Correct Answer : read, write
Explanation : The input activity is similar to read operation and the output activity is similar to write operation.
Correct Answer : write operation on output data
Explanation : IOWR (active low) operation means writing data to an output device and not an input device.
Correct Answer : parallel input-output port
Explanation : The parallel input-output port chip 8255 is also known as programmable peripheral input-output port.
Correct Answer : either input or output ports
Explanation : Port C can function independently either as input or as output ports.
Correct Answer : control word register
Explanation : By programming the bits of control word register, the operations of the ports are specified.
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.
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.
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.
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.
Correct Answer : 100 microseconds
Explanation : The conversion delay is 100microseconds which is low as compared to other converters.
Correct Answer : may be generated internally or supplied externally
Correct Answer : .1000 E - 1 and .9727 E - 2 respectively
Correct Answer : DMA
Correct Answer : contents of memory location 20 are fetched into the accumulator
Correct Answer : -3.2 E - 3
Correct Answer : - A + B + M
Correct Answer : effective speed of memory is increased
Correct Answer : minimum mode and maximum mode
Correct Answer : bus interface unit and execution unit
Correct Answer : E Q U
Correct Answer : programmable interrupt controller
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
Correct Answer : Load the program counter with contents of register pair H L
Correct Answer : must be the same as that of host
Correct Answer : -32768 to + 32767
Correct Answer : Data transfer group
Correct Answer : Implict addressing
Correct Answer : about 30 μs
Correct Answer : Direct Memory Address
Correct Answer : program counter etc. are being reset
Correct Answer : (X + 2.5) / (Y + 4.5)
Correct Answer : Direct transfer
Correct Answer : THETA
Correct Answer : Both A and R are correct but R is not correct explanation of A
Correct Answer : zero or negative
Correct Answer : strbig
Correct Answer : 1 or 2 or 3 bytes long
Correct Answer : directs the assembler to reserve a sequence of consecutive 105 bytes in the memory
Correct Answer : about one-eighth
Correct Answer : Timing characteristic
Correct Answer : special characters +, -, *, / etc
Correct Answer : Implicit
Correct Answer : -
Correct Answer : 40 and 20 respectively
Correct Answer : 0.181 x 10⁻¹⁰
Correct Answer : Concentrated refresh
Correct Answer : 6000 H - 6 FFF H and 7000 H - 7 FFF H
b = (a > 4 ? 2 : 9);
Correct Answer : 9.0 and 2.0
Correct Answer : main memory and I/O devices
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.
Correct Answer : 1 and 4 only
Explaination : E is not allowed and default sign of exponent is positive.
REAL F , C
READ * , F
C = (5. / 9.) * (F - 32.)
Correct Answer : 35.0
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
Correct Answer : 60, 20, 20, 30, 10
k = 2* 3 / 5 + 4 / 4 + 8 - 2 - 4 / 8
Explaination : k = 1 + 1 + 8 - 2 - 0 = 8 because k is an integer.
k = 1 + 1 + 8 - 2 - 0 = 8
Correct Answer : 1281.0
Explanation : It is the only one with decimal point.
Correct Answer : 13.75
Explanation : Convert to decimal.
Correct Answer : 100 MB or 250 MB
Explanation : Zip floppy has high memory 100 MB or 250 MB.
5 READ R
10 D = 2 * R
20 A = 3.1416 * D * D / 4
30 PRINT A
Correct Answer : 28.27
Correct Answer : single chip LSI
Explanation : It is LSI chip.
Correct Answer : 385
SUM = 12 + 22 + 33 + 44 + 52 + 62 + 77 + 82 + 92 + 102 = 385.
Correct Answer : B(M, R)
Explanation : Subscript not proper.
Correct Answer : a * c * d (x * z)
Correct Answer : a 56
Explanation : In all others first letter is not lower case.
Correct Answer : divided by 2
Explanation : When RRC is used once the number is divided by 2.
Correct Answer : 0 to 8095
Explanation : The starting word address is 0.
Correct Answer : x * y = z
Explanation : Left hand side must be a variable name.
int a, b, c,
a = 40;
b = a++ ;
c = ++a ;
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.
Correct Answer : A * * B
Explanation : Two operators cannot be together.
Correct Answer : NSTUD
Explanation : It is an integer. Hence N as first character.
Correct Answer : 40
Explanation : 45/10 = 4.5 → 4 x 10 = 40.
Correct Answer : 16
Correct Answer : -73
Correct Answer : the value 20 is brought into accumulator
Correct Answer : portrait and landscape
Explanation : Portrait and worksheet are two orientations of worksheet page.
Correct Answer : B * * 3 - 2 * A * C
Explaination : It is the only expression with correct symbols for exponential and product.
Correct Answer : A (2 I)
READ*, [X(I) Y(I) = 0, 3, 2]
Correct Answer : 4 values could be read and stored asX (0), Y (0), X (2), Y (2)
Correct Answer : arithmetic
Correct Answer : the two 8 bit values in registers AL and CH are added and result placed in register AL
Correct Answer : CMC
Correct Answer : 100000 hours
Correct Answer : -2.8 E + 02
Correct Answer : AB + C - D +
Correct Answer : Asynchronous transfer
Correct Answer : Both (A) and (B)
Correct Answer : 1200 x 1200
Correct Answer : INR r
Correct Answer : registers
Correct Answer : 13 and 9 respectively
Correct Answer : 4 bits
Correct Answer : -3
MOV M, r
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
Correct Answer : underscore
Correct Answer : -2.3 E - 8
MOV A, D
RAL
MOV D, A
Correct Answer : decimal 40
Correct Answer : a string of capital alphabets A to Z, digits 0 to 9 and special characters included in question mark
Correct Answer : 8 data bits are multiplexed with 8 least significant bits of address bus
Correct Answer : ADC M
Correct Answer : Exchange bus
Correct Answer : only the sign of mantissa
Correct Answer : RC circuit
Correct Answer : 01101100 00000000 00000000 10000100
Correct Answer : 8086
Correct Answer : 360 KB and 1.2 MB respectively
Correct Answer : Shadow RAM
Correct Answer : (- 5.2) ** (4.3)
Correct Answer : UV light intensity x exposure time in seconds
Correct Answer : OFH is stored at location 02 and 0 A 5 at location 03 and address of word is 02
Correct Answer : bits
Correct Answer : 5*2
Correct Answer : only when the logical expression controlling the loop becomes true for repeat loop and false for while loop
Correct Answer : DIMENSION
Correct Answer : -3.4 e 37 to + 3.4 e 37
IMPLICIT INTEGER (A - F)
Correct Answer : all variable names which begin with A, B, C, D, E or F as being of integer type
Correct Answer : Eight 4 bit shift registers
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.
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 AXExample for output port: OUT 03H, AL; This sends data available in AL to a port whose address is 03H.
Correct Answer : multiple interrupt processing ability
Explanation : The processor if handles more devices as interrupts then it has multiple interrupt processing ability.
Correct Answer : nonmaskable interrupt
Explanation : NMI is the acronym for nonmaskable interrupt.
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.
Correct Answer : even address memory bank
Explanation : In static memory, the lower 8-bit bank is called even address memory bank.
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.
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.
Correct Answer : memory locations
Explanation : In memory-mapped scheme, the devices are viewed as memory locations and are addressed likewise.
Correct Answer : inputs to outputs
Explanation : If DIR is 1, then the direction is from A(inputs) to B(outputs).
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.
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.
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.
Correct Answer : RESET
Explanation : If reset pin is enabled then the control word register is cleared.
Explanation : Compared to other 12-bit ADCs, it is of very low cost and useful for slow practical applications.
Correct Answer : 2048 clock cycles
Explaination : The internal integrator needs 2048 clock cycles to integrate voltage difference between input low and input high.
Correct Answer : Direct addressing
READ *, (X.(I), I = 0, 4)
Correct Answer : 5 values would be read and stored in X (0), X (1), X (2), (3)(4)
Correct Answer : Machine language programming
Correct Answer : a positive number and closer to zero
Correct Answer : PARAMETER
CY = 1ACC = 10000001
Correct Answer : 1 and 00000011
MOV r1, r2
r1
r2
Correct Answer : causes contents of registers r2 to be moved to r1, and uses register addressing
Correct Answer : AB / CF * AD - E * + *
Correct Answer : 256 KB to 1 MB of main memory
Correct Answer : Power consumption and bus loading
Correct Answer : F DISK
MOV A, M
Correct Answer : Register indirect addressing
Correct Answer : Both are true
Correct Answer : one character set having 60 characters
Correct Answer : -1
Correct Answer : 60 ns
Correct Answer : Programmable CRT controller
Correct Answer : .9432 E - 4 is changed to .09432 E - 3 and .5452 E - 3 is not changed
MVIA, 00H
LXIH, OEAOFG
V LDAZ ADI 0005H
Correct Answer : Adds decimal number 5 to contents of memory
Correct Answer : 13
Correct Answer : 100 ns
Correct Answer : MN / MX
Correct Answer : IA * B
Correct Answer : MEM MAKER
Explanation : The 8257 can accomplish three types of operations and they arei) verify DMA operationii) write operationiii) read operation.
Correct Answer : HLDA
Explanation : If the HLDA signal is received by the DMA controller, it indicates that the bus is available.
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.
Correct Answer : burst transfer
Explanation : If more than one channel requests service simultaneously, then the transfer occurs as a burst or continuous transfer.
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.
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.
Correct Answer : rotating priority scheme
Explanation : In this scheme, the priorities assigned to the channels are not fixed.
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.
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.
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.
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.
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.
Correct Answer : 16 MB
Explanation : The 80286 with its 24-bit address bus is able to address 16 Mbytes of physical memory.
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.
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.
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.
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.
Correct Answer : None of the above
Explanation : The three blocks of an internal architecture of 80287 are:1. bus control logic2. data interface and control unit3. floating point unit.
Explanation : The data interface and control unit contains status and controls words, TAG words and error pointers.
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.
Correct Answer : infinity control bits
Explanation : The infinity control bit is initialized to zero after reset.
Correct Answer : barrel shifter
Explanation : The barrel shifter arranges and presents the data to be shifted successively, whenever required for the execution.
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.
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.
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.
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.
Correct Answer : GA, GB
Explanation : GA register is used as source and GB as destination pointers during DMA operations.
Correct Answer : DRQ and EXT
Explanation : The DRQ and EXT are used for data transfer control and operation termination signals during DMA operations.
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.
Explanation : The 80386DX is a 32-bit processor that supports, 8-bit/16-bit/32-bit data operands.
Correct Answer : 32 address lines
Explanation : The 80386, with its 32-bit address bus, can address up to 4 GB of physical memory.
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.
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).
Correct Answer : 20 MHz and 33 MHz
Explanation : The operating frequency of 80386DX is 20MHz and 33 MHz.
Explanation : The 80386 has on-chip address translation cache, and the instruction set is upward compatible with all its predecessors.
Correct Answer : 132 pin
Explanation : The 80386DX is available in a 132-pin grid array package.
Correct Answer : E
Explanation : A 32 bit register, known as an extended register, is represented by the register name with a prefix of E.
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.
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.
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.
Correct Answer : RF is set
Explanation : If RF (resume flag) is set, any debug fault is ignored during the instruction cycle.
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.
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.
Correct Answer : level 0
Explanation : The real mode programs are executed at the highest privilege level i.e. level 0.
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.
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).
Correct Answer : 48-bit, 48-bit
Explanation : 80387 consists of two 48-bit registers, known as instruction and data pointer registers.
Explanation : Several resource sharing strategies have been investigated by the developers. Some of these are1. Partitioned resources2. Threshold sharing3. Full sharing.
Correct Answer : HALT
Explanation : An optimization may require the use of HALT instruction, when either of the two logical processors is idle.
Correct Answer : operating system
Explanation : The HALT instruction is a privileged instruction that can be only used by operating system.
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.
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).
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.
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.
Correct Answer : Accumulator and B register
Explanation : In some instructions, the Accumulator and B register are used to store the operands.
Correct Answer : 8051
Explanation : The 8051 stack is not a top-down data structure, like other Intel processors.
Correct Answer : parallel-in serial-out register
Explanation : The transmit buffer of serial data buffer is a parallel-in serial-out 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.
Correct Answer : PCON and SCON
Explanation : The registers, PCON and SCON contain control and status information about serial port.
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.
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.
Correct Answer : two machine cycles
Explanation : For an interrupt to be guaranteed served it should have duration of two machine cycles.
Correct Answer : write operation
Explanation : The control word register can only be written and cannot be read.
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.
Correct Answer : read/load LSB first and then MSB
Explanation : To access 16 bit, first LSB is loaded first, and then MSB.
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 countersiv) loading of the counter registers.
Correct Answer : 1.4 E - 45 to 3.4 E + 38
Correct Answer : A = B
Correct Answer : I/O devices have 16 bit addresses
Correct Answer : 6809
Correct Answer : integer constants or integer variable names or integer expressions
Correct Answer : operation code, low order byte of data/address and high order byte of data / address
Correct Answer : 32 bit register and 16 bit data bus
Correct Answer : S = 0, Z = 1, AC = 1, P = 0 and CY=1
Correct Answer : byte, short, int and long
Correct Answer : RST 9.5
Correct Answer : 111
Correct Answer : TRAP and RST lines are not high
Correct Answer : Shift
Correct Answer : logic group
Correct Answer : Solid state device
Correct Answer : 125000