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Interview MCQ Set 1

1. What is the basis for numerous spatial domain processing techniques?
a) Transformations
b) Scaling
c) Histogram
d) None of the Mentioned

View Answer

Answer: c [Reason:] Histogram is the basis for numerous spatial domain processing techniques.

2. In _______ image we notice that the components of histogram are concentrated on the low side on intensity scale.
a) bright
b) dark
c) colourful
d) All of the Mentioned

View Answer

Answer: b [Reason:] Only in dark images, we notice that the components of histogram are concentrated on the low side on intensity scale.

3. What is Histogram Equalisation also called as?
a) Histogram Matching
b) Image Enhancement
c) Histogram linearisation
d) None of the Mentioned

View Answer

Answer: c [Reason:] Histogram Linearisation is also known as Histogram Equalisation.

4. What is Histogram Matching also called as?
a) Histogram Equalisation
b) Histogram Specification
c) Histogram linearisation
d) None of the Mentioned

View Answer

Answer: b [Reason:] Histogram Specification is also known as Histogram Matching.

5. Histogram Equalisation is mainly used for ________________
a) Image enhancement
b) Blurring
c) Contrast adjustment
d) None of the Mentioned

View Answer

Answer: a [Reason:] It is mainly used for Enhancement of usually dark images.

6. To reduce computation if one utilises non-overlapping regions, it usually produces ______ effect.
a) Dimming
b) Blurred
c) Blocky
d) None of the Mentioned

View Answer

Answer: c [Reason:] Utilising non-overlapping regions usually produces “Blocky” effect.

7. What does SEM stands for?
a) Scanning Electronic Machine
b) Self Electronic Machine
c) Scanning Electron Microscope
d) Scanning Electric Machine

View Answer

Answer: c [Reason:] SEM stands for Scanning Electron Microscope.

8. The type of Histogram Processing in which pixels are modified based on the intensity distribution of the image is called _______________.
a) Intensive
b) Local
c) Global
d) Random

View Answer

Answer: c [Reason:] It is called Global Histogram Processing.

9. Which type of Histogram Processing is suited for minute detailed enhancements?
a) Intensive
b) Local
c) Global
d) Random

View Answer

Answer: b [Reason:] Local Histogram Processing is used.

10. In uniform PDF, the expansion of PDF is ________________
a) Portable Document Format
b) Post Derivation Function
c) Previously Derived Function
d) Probability Density Function

View Answer

Answer: d [Reason:] PDF stands for Probability Density Function.

Interview MCQ Set 2

1. A device driver can be thought of as a translator. Its input consists of _____ commands and output consists of _______ instructions.
a) high level, low level
b) low level, high level
c) complex, simple
d) low level, complex

View Answer

Answer: a [Reason:] None.

2. The file organization module knows about :
a) files
b) logical blocks of files
c) physical blocks of files
d) all of the mentioned

View Answer

Answer: d [Reason:] None.

3. Metadata includes :
a) all of the file system structure
b) contents of files
c) both file system structure and contents of files
d) none of the mentioned

View Answer

Answer: c [Reason:] None.

4. For each file their exists a ___________ that contains information about the file, including ownership, permissions and location of the file contents.
a) metadata
b) file control block
c) process control block
d) all of the mentioned

View Answer

Answer: b [Reason:] None.

5. For processes to request access to file contents, they need to :
a) they need to run a seperate program
b) they need special interrupts
c) implement the open and close system calls
d) none of the mentioned

View Answer

Answer: c [Reason:] None.

6. During compaction time, other normal system operations _______ be permitted.
a) can
b) cannot
c) is
d) none of the mentioned

View Answer

Answer: b [Reason:] None.

7. When in contiguous allocation the space cannot be extended easily :
a) the contents of the file have to be copied to a new space, a larger hole
b) the file gets destroyed
c) the file will get formatted and loose all its data
d) none of the mentioned

View Answer

Answer: a [Reason:] None.

8. In the linked allocation, the directory contains a pointer to the :
I. first block
II. last block
a) I only
b) II only
c) Both I and II
d) Neither I nor II

View Answer

Answer: c [Reason:] None.

9. There is no __________ with linked allocation.
a) internal fragmentation
b) external fragmentation
c) starvation
d) all of the mentioned

View Answer

Answer: b [Reason:] None.

10. The major disadvantage with linked allocation is that :
a) internal fragmentation
b) external fragmentation
c) there is no sequential access
d) there is only sequential access

View Answer

Answer: d [Reason:] None.

11. If a pointer is lost or damaged in a linked allocation :
a) the entire file could get damaged
b) only a part of the file would be affected
c) there would not be any problems
d) none of the mentioned

View Answer

Answer: a [Reason:] None.

12. FAT stands for :
a) File Attribute Transport
b) File Allocation Table
c) Fork At Time
d) None of the mentioned

View Answer

Answer: b [Reason:] None.

13. By using FAT, random access time is __________
a) the same
b) increased
c) decreased
d) not affected

View Answer

Answer: c [Reason:] None.

Interview MCQ Set 3

1. A large number of disks in a system improves the rate at which data can be read or written :
a) if the disks are operated on sequentially
b) if the disks are operated on selectively
c) if the disks are operated in parallel
d) all of the mentioned

View Answer

Answer: c [Reason:] None.

2. RAID stands for :
a) Redundant Allocation of Inexpensive Disks
b) Redundant Array of Important Disks
c) Redundant Allocation of Independent Disks
d) Redundant Array of Independent Disks

View Answer

Answer: d [Reason:] None.

3. If the mean time to failure of a single disk is 100,000 hours, then the mean time to failure of some disk in an array of 100 disks will be :
a) 100 hours
b) 10 days
c) 10 hours
d) 1000 hours

View Answer

Answer: d [Reason:] None.

4. The solution to the problem of reliability is the introduction of __________
a) aging
b) scheduling
c) redundancy
d) disks

View Answer

Answer: c [Reason:] None.

5. The technique of duplicating every disk is known as :
a) mirroring
b) shadowing
c) redundancy
d) all of the mentioned

View Answer

Answer: a [Reason:] None.

6. The mean time to failure of a mirrored disk depends on :
I) the mean time to failure of individual disks
II) the mean time to repair
a) Only I
b) Only II
c) Both I and II
d) Neither I nor II

View Answer

Answer: c [Reason:] None.

7. RAID level ________ refers to disk arrays with striping at the level of blocks, but without any redundancy.
a) 0
b) 1
c) 2
d) 3

View Answer

Answer: a [Reason:] None.

8. RAID level _______ refers to disk mirroring.
a) 0
b) 1
c) 2
d) 3

View Answer

Answer: b [Reason:] None.

9. RAID level ______ is also known as bit interleaved parity organisation.
a) 0
b) 1
c) 2
d) 3

View Answer

Answer: d [Reason:] None.

10. A single parity bit can be used for :
a) detection
b) multiple error corrections
c) few error corrections
d) all of the mentioned

View Answer

Answer: a [Reason:] None.

11. RAID level ______ is also known as memory style error correcting code(ECC) organization.
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] None.

12. RAID level 3 does not have _________ as in RAID level 1.
a) efficiency
b) enough storage space for data
c) storage overhead
d) time consumption overhead

View Answer

Answer: c [Reason:] There is one mirror disk for every disk in level 1.

Interview MCQ Set 4

1. A large number of disks in a system improves the rate at which data can be read or written :
a) if the disks are operated on sequentially
b) if the disks are operated on selectively
c) if the disks are operated in parallel
d) all of the mentioned

View Answer

Answer: c [Reason:] None.

2. RAID stands for :
a) Redundant Allocation of Inexpensive Disks
b) Redundant Array of Important Disks
c) Redundant Allocation of Independent Disks
d) Redundant Array of Independent Disks

View Answer

Answer: d [Reason:] None.

3. If the mean time to failure of a single disk is 100,000 hours, then the mean time to failure of some disk in an array of 100 disks will be :
a) 100 hours
b) 10 days
c) 10 hours
d) 1000 hours

View Answer

Answer: d [Reason:] None.

4. The solution to the problem of reliability is the introduction of __________
a) aging
b) scheduling
c) redundancy
d) disks

View Answer

Answer: c [Reason:] None.

5. The technique of duplicating every disk is known as :
a) mirroring
b) shadowing
c) redundancy
d) all of the mentioned

View Answer

Answer: a [Reason:] None.

6. The mean time to failure of a mirrored disk depends on :
I) the mean time to failure of individual disks
II) the mean time to repair
a) Only I
b) Only II
c) Both I and II
d) Neither I nor II

View Answer

Answer: c [Reason:] None.

7. RAID level ________ refers to disk arrays with striping at the level of blocks, but without any redundancy.
a) 0
b) 1
c) 2
d) 3

View Answer

Answer: a [Reason:] None.

8. RAID level _______ refers to disk mirroring.
a) 0
b) 1
c) 2
d) 3

View Answer

Answer: b [Reason:] None.

9. RAID level ______ is also known as bit interleaved parity organisation.
a) 0
b) 1
c) 2
d) 3

View Answer

Answer: d [Reason:] None.

10. A single parity bit can be used for :
a) detection
b) multiple error corrections
c) few error corrections
d) all of the mentioned

View Answer

Answer: a [Reason:] None.

11. RAID level ______ is also known as memory style error correcting code(ECC) organization.
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] None.

12. RAID level 3 does not have _________ as in RAID level 1.
a) efficiency
b) enough storage space for data
c) storage overhead
d) time consumption overhead

View Answer

Answer: c [Reason:] There is one mirror disk for every disk in level 1.

Interview MCQ Set 5

1. What is incorrect methods of revocation of access rights ?
a) Immediate/Delayed
b) Selective/General
c) Partial/total
d) Crucial

View Answer

Answer: d [Reason:] None.

2. Why is it difficult to revoke capabilities ?
a) They are too many
b) They are not defined precicely
c) They are distributed throughout the system
d) None of the mentioned

View Answer

Answer: c [Reason:] None.

3. What is reaquisition scheme to revoke capability ?
a) When a process’s capability is revoked then it wont be able to reacquire it
b) Pointers are maintained for each object which can be used to revoke
c) Indirect pointing is done to revoke object’s capabilities
d) Master key can be used compare and revoke.

View Answer

Answer: a [Reason:] None.

4. What is false regarding Back-Pointers scheme to revoke capability ?
a) List of pointers is maintained with each object
b) When revocation is required these pointers are followed
c) This scheme is not adopted in MULTICS system
d) These point to all capabilities associated with that object

View Answer

Answer: c [Reason:] None.

5. What is true about Indirection to revoke capability ?
a) Capabilities point indirectly to the objects
b) Each capability will not have a unique entry in global
c) Table entries cannot be reused for other capabilities
d) This system was adopted in MULTICS system

View Answer

Answer: a [Reason:] None.

6. How can Keys be defined or replaced ?
a) create [keyname] [bits].
b) set-key
c) Key
d) MAKE [Key Name].

View Answer

Answer: b [Reason:] None.

7. What are characteristics of Hydra system ?
a) It consist of known access rights and interpreted by the system
b) A user can of protection system can declare other rights
c) Hydra system is not flexible
d) Hydra doesn’t provide rights amplification

View Answer

Answer: a [Reason:] None.

8. What are characteristics of rights amplification in Hydra ?
a) This scheme allows a procedure to be certified as trustworthy
b) Amplification of rights cannot be stated explicitly in declaration
c) It includes kernel rights such as read
d) All of the mentioned

View Answer

Answer: a [Reason:] None.

9. What is the problem of mutually suspicious subsystem ?
a) Service program can steal users data
b) Service program can malfunction and retain some rights of data provided by user
c) Calling program can get access to restricted portion from service program
d) Calling program gets unrestricted access

View Answer

Answer: b [Reason:] Both calling program and service program are vulnerable to access each others private data/rights.

10. What are characteristics of Cambridge CAP system as compared to Hydra system ?
a) It is simpler and less powerful than hydra system
b) It is more powerful than hydra system
c) It is powerful than hydra system
d) It is not as secure as Hydra system

View Answer

Answer: a [Reason:] None.

11. What are two capabilities defined in CAP system ?
a) data & software capability
b) address & data capability
c) hardware & software capability
d) software capability

View Answer

Answer: a [Reason:] None.

Interview MCQ Set 6

1. In a real time system the computer results :
a) must be produced withing a specific deadline period
b) may be produced at any time
c) may be correct
d) all of the mentioned

View Answer

Answer: a [Reason:] None.

2. In a safety critical system, incorrect operation :
a) does not affect much
b) causes minor problems
c) causes major and serious problems
d) none of the mentioned

View Answer

Answer: c [Reason:] None.

3. Antilock brake systems, flight management systems, pacemakers are examples of :
a) safety critical system
b) hard real time system
c) soft real time system
d) safety critical system and hard real time system

View Answer

Answer: d [Reason:] None.

4. In a ______ real time system, it is guaranteed that critical real time tasks will be completed within their deadlines.
a) soft
b) hard
c) critical
d) none of the mentioned

View Answer

Answer: b [Reason:] None.

5. Some of the properties of real time systems include :
a) single purpose
b) inexpensively mass produced
c) small size
d) all of the mentioned

View Answer

Answer: d [Reason:] None.

6. The amount of memory in a real time system is generally :
a) less compared to PCs
b) high compared to PCs
c) same as in PCs
d) they do not have any memory

View Answer

Answer: a [Reason:] None.

7. The priority of a real time task :
a) must degrade over time
b) must not degrade over time
c) may degrade over time
d) none of the mentioned

View Answer

Answer: b [Reason:] None.

8. Memory management units :
a) increase the cost of the system
b) increase the power consumption of the system
c) increase the time required to complete an operation
d) all of the mentioned

View Answer

Answer: d [Reason:] None.

9. The technique in which the CPU generates physical addresses directly is known as :
a) relocation register method
b) real addressing
c) virtual addressing
d) none of the mentioned

View Answer

Answer: b [Reason:] None.

Interview MCQ Set 7

1. Which of the following is not a block of architecture of 80287?
a) bus control logic
b) data interface and control unit
c) floating point unit
d) none of the mentioned

View Answer

Answer: d [Reason:] The three blocks of internal architecture of 80287 are: 1. bus control logic 2. data interface and control unit 3. floating point unit.

2. The unit that provides and controls the interface, between the internal 80287 bus and 80286 bus via data buffer is
a) bus control logic
b) data interface and control unit
c) floating point unit
d) none of the mentioned

View Answer

Answer: a [Reason:] The bus control logic provides and controls the interface, between the internal 80287 bus and 80286 bus via data buffer.

3. The data interface and control unit consists of
a) status and control words
b) tag words and error pointers
c) instruction decoders
d) all of the mentioned

View Answer

Answer: d [Reason:] The data interface and control unit contains status and control words, TAG words and error pointers.

4. The word that optimizes the NDP performance, by maintaining a record of empty and non-empty register locations is
a) Status and control words
b) TAG words
c) Error pointers
d) All of the mentioned

View Answer

Answer: b [Reason:] The TAG word optimizes the NDP performance by maintaining a record of empty and non-empty register locations. It helps the exception handler to identify special values in the contents of the stack locations.

5. The part of data interface and control unit, that points to the source of exception generated is
a) Status and control words
b) TAG words
c) Error pointers
d) None of the mentioned

View Answer

Answer: c [Reason:] The error pointers point to the source of exception (address of the instruction that generated the exception) generated.

6. The data bus in floating point unit is of
a) 16 bits
b) 32 bits
c) 64 bits
d) 84 bits

View Answer

Answer: d [Reason:] The data bus in floating point unit is of 84-bits. Out of this 84-bits, the lower 68 bits are significant (mantissa) data bit, the next 16 bits are used for exponent.

7. The arrangement of data that is to be shifted successively, whenever required for the execution, is done by
a) error pointer
b) data buffer
c) barrel shifter
d) none of the mentioned

View Answer

Answer: c [Reason:] The barrel shifter arranges and presents the data to be shifted successively, whenever required for the execution.

8. The word that is used to select one of the processing options, among the provided ones is
a) status word
b) control word
c) status and control words
d) none of the mentioned

View Answer

Answer: b [Reason:] The control word is used to select one of the processing options, among the ones provided by 80287.

9. After reset of 80287, the control bit that is initialized to zero is
a) masking bits
b) precision control bits
c) rounding control bits
d) infinity control bits

View Answer

Answer: d [Reason:] The infinity control bit is initialized to zero after reset.

10. The bits that are modified depending upon the result of the execution of arithmetic instructions are
a) masking bits
b) rounding control bits
c) condition code bits
d) error summary bits

View Answer

Answer: c [Reason:] 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.

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