Correct Answer:
Cache memory
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Correct Answer:
0000H to FFFFH
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Correct Answer:
Accumulator low
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Correct Answer:
Accumulator high
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Correct Answer:
Program counter
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Correct Answer:
Non mask able interrupt
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Correct Answer:
Address data
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Correct Answer:
Address latch enable
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Correct Answer:
Source index
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Correct Answer:
Destination index
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Correct Answer:
Base pointer
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Correct Answer:
Effective address
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Correct Answer:
Segment base address
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Correct Answer:
Physical address
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Correct Answer:
Dual inline package
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Correct Answer:
40 pin
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Correct Answer:
Physical
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Correct Answer:
20 bit
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Correct Answer:
64 kb
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Correct Answer:
Both A and B
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Correct Answer:
16 bit
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Correct Answer:
Status register
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Correct Answer:
Stack segment
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Correct Answer:
AL
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Correct Answer:
AX
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Correct Answer:
All of these
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Correct Answer:
Data segment
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Correct Answer:
Code segment
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Correct Answer:
Instruction pointer
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Correct Answer:
All of these
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Correct Answer:
Both A and B
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Correct Answer:
Execution unit
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Correct Answer:
Bus interface unit
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Correct Answer:
9432 E – 4 is changed to .09432 E – 3 and .5452 E – 3 is not changed
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Correct Answer:
True
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Correct Answer:
main memory and I/O devices
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Correct Answer:
False
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Correct Answer:
234
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Correct Answer:
Both A and R are correct but R is not correct explanation of A
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integer part, fraction part along with positive or negative sign
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Correct Answer:
5
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Correct Answer:
Concentrated refresh
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Correct Answer:
a pair of cross coupled NAND
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R is asserted, S is asserted
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Correct Answer:
the same as
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Correct Answer:
log2 (m)
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Correct Answer:
halves
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Correct Answer:
RS Latch
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Correct Answer:
set of {AND,OR,NOT}
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Correct Answer:
register
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Correct Answer:
can not change
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Correct Answer:
Circuit A has more gates than circuit B
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Correct Answer:
BYTE
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Cannot be determined
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Correct Answer:
NAND-NAND
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Correct Answer:
6
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Correct Answer:
PA0 – PA7
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Correct Answer:
Synchronous DRAM
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Correct Answer:
Indirect addressing
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1000 0011 1110 1000
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Correct Answer:
Timer 2
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Correct Answer:
Both (A) and (B)
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Correct Answer:
Edge triggered and level sensitive
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Correct Answer:
Two
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Correct Answer:
15 lines
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Correct Answer:
2 – 4 ms
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Correct Answer:
Four
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Correct Answer:
TRUE
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Correct Answer:
6
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Correct Answer:
CONFIG.SYS
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Correct Answer:
Enables all channels.
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Correct Answer:
ENIAC
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Correct Answer:
33 MHz
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Correct Answer:
Windows 98 is RTOS
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Correct Answer:
HOLD and HLDA
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Correct Answer:
Input port using mode 0
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Correct Answer:
64 bit
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Correct Answer:
Three
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Correct Answer:
16 bit address bus
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Correct Answer:
USART
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Correct Answer:
A machine cycle consists of one or more instruction cycle.
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Correct Answer:
All of the above
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Correct Answer:
Low =-15 volt to –3 vol, high = +3 volt to +15 volt
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Correct Answer:
Ultraviolet rays
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Correct Answer:
3
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Correct Answer:
During interrupt servicing
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Correct Answer:
64 bit microprocessor.
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Correct Answer:
1.38 MB/s
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Correct Answer:
Synchronous DRAM
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Correct Answer:
Floating
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Correct Answer:
When hold line is a logical 1
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Correct Answer:
To introduce wait states
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Correct Answer:
String instructions.
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Correct Answer:
0A and carry flag is set
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Correct Answer:
256
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Correct Answer:
Assembler directives
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Correct Answer:
Contents can be erased using ultra violet rays
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Correct Answer:
Write
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Correct Answer:
Near.
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Correct Answer:
From I/O to memory.
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Correct Answer:
4 ns.
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Correct Answer:
T1.
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Correct Answer:
2.5 MHz.
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Correct Answer:
Associative memory
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Correct Answer:
All of these
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Power Engineering, Power Plant Engineering MCQs | Topic-wise