Computer Architecture Set 25

Computer Science Engineering (CSE)

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This set of Computer Architecture Multiple Choice Questions & Answers (MCQs) focuses on Computer Architecture Set 25

Q1 |                       is used to implement virtual memory organisation.
  • page table
  • frame table
  • mmu
  • none of the mentioned
Answer: mmu
Explaination:
the mmu stands for memory management unit.
Q2 |               translates the logical address into a physical address.
  • mmu
  • translator
  • compiler
  • linker
Answer: mmu
Explaination:
the mmu translates the logical address into a physical address by adding an offset.
Q3 | The DMA doesn’t make use of the MMU for bulk data transfers.
  • true
  • false
Answer: false
Explaination:
the dma stands for direct memory access, in which a block of data gets directly transferred from the memory.
Q4 | The virtual memory basically stores the next segment of data to be executed on the                     
  • secondary storage
  • disks
  • ram
  • rom
Answer: secondary storage
Explaination:
none.
Q5 | The associatively mapped virtual memory makes use of                 
  • tlb
  • page table
  • frame table
  • none of the mentioned
Answer: tlb
Explaination:
tlb stands for translation look-aside buffer.
Q6 | The main reason for the discontinuation of semi conductor based storage devices for providing large storage space is                     
  • lack of sufficient resources
  • high cost per bit value
  • lack of speed of operation
  • none of the mentioned
Answer: high cost per bit value
Explaination:
in the case of semi conductor based memory technology, we get speed but the increase in the integration of various devices the cost is high.
Q7 | The digital information is stored on the hard disk by                           
  • applying a suitable electric pulse
  • applying a suitable magnetic field
  • applying a suitable nuclear field
  • by using optic waves
Answer: applying a suitable electric pulse
Explaination:
the digital data is sorted on the magnetized discs by magnetizing the areas.
Q8 | For the synchronization of the read head, we make use of a                 
  • framing bit
  • synchronization bit
  • clock
  • dirty bit
Answer: clock
Explaination:
the clock makes it easy to distinguish between different values red by a head.
Q9 | One of the most widely used schemes of encoding used is                     
  • nrz-polar
  • rz-polar
  • manchester
  • block encoding
Answer: manchester
Explaination:
the manchester encoding used is also called as phase encoding and it is used to encode both clock and data.
Q10 | The drawback of Manchester encoding is                     
  • the cost of the encoding scheme
  • the speed of encoding the data
  • the latency offered
  • the low bit storage density provided
Answer: the low bit storage density provided
Explaination:
the space required to represent each bit must be large enough to accommodate two changes in magnetization.
Q11 | The read/write heads must be near to disk surfaces for better storage.
  • true
  • false
Answer: true
Explaination:
by maintaining the heads near to the surface greater bit densities can be achieved.
Q12 |            pushes the heads away from the surface as they rotate at their standard rates.
  • magnetic tension
  • electric force
  • air pressure
  • none of the mentioned
Answer: air pressure
Explaination:
due to the speed of rotation of the discs air pressure develops in the hard disk.
Q13 | The air pressure can be countered by putting               in the head-disc surface arrangement.
  • air filter
  • spring mechanism
  • coolant
  • none of the mentioned
Answer: spring mechanism
Explaination:
the spring mechanism
Q14 | The method of placing the heads and the discs in an air tight environment is also called as               
  • raid arrays
  • atp tech
  • winchester technology
  • fleming reduction
Answer: winchester technology
Explaination:
the disks and the heads operate faster due to the absence of the dust particles.
Q15 | A hard disk with 20 surfaces will have            heads.
  • 10
  • 5
  • 1
  • 20
Answer: 20
Explaination:
each surface will have its own head to perform read/write operation.
Q16 | The set of corresponding tracks on all surfaces of a stack of disks form a
  • cluster
  • cylinder
  • group
  • set
Answer: cylinder
Explaination:
the data is stored in these sections called as cylinders.
Q17 | The data can be accessed from the disk using                     
  • surface number
  • sector number
  • track number
  • all of the mentioned
Answer: all of the mentioned
Explaination:
none.
Q18 | The read and write operations usually start at               of the sector.
  • center
  • middle
  • from the last used point
  • boundaries
Answer: boundaries
Explaination:
the heads read and write data from the ends to the center.
Q19 | To distinguish between two sectors we make use of                   
  • inter sector gap
  • splitting bit
  • numbering bit
  • none of the mentioned
Answer: inter sector gap
Explaination:
this means that we leave a little gap between each sector to differentiate between them.
Q20 | The            process divides the disk into sectors and tracks.
  • creation
  • initiation
  • formatting
  • modification
Answer: formatting
Explaination:
the formatting process deletes the data present and does the creation of sectors and tracks.
Q21 | The access time is composed of
  • seek time
  • rotational delay
  • latency
  • both seek time and rotational delay
Answer: both seek time and rotational delay
Explaination:
the seek time refers to the time required to move the head to the required disk.
Q22 | The disk drive is connected to the system by using the            
  • pci bus
  • scsi bus
  • hdmi
  • isa
Answer: scsi bus
Explaination:
none.
Q23 |                 is used to deal with the difference in the transfer rates between the drive and the bus.
  • data repeaters
  • enhancers
  • data buffers
  • none of the mentioned
Answer: data buffers
Explaination:
the buffers are added to store the data from the fast device and to send it to the slower device at its rate.
Q24 |                 is used to detect and correct the errors that may occur during data transfers.
  • ecc
  • crc
  • checksum
  • none of the mentioned
Answer: ecc
Explaination:
ecc stands for error correcting code.
Q25 | ______ has been developed specifically for pipelined systems.
  • Utility software
  • Speed up utilities
  • Optimizing compilers
  • None of the mentioned
Answer: Optimizing compilers
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