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This section includes 154 Mcqs, each offering curated multiple-choice questions to sharpen your Memory Management knowledge and support exam preparation. Choose a topic below to get started.
51. |
In fixed sized partition, the degree of multiprogramming is bounded by ___________. |
A. | the number of partitions |
B. | the CPU utilization |
C. | the memory size |
D. | All of these |
Answer» B. the CPU utilization | |
52. |
When memory is divided into several fixed sized partitions, each partition may contain ________. |
A. | exactly one process |
B. | atleast one process |
C. | multiple processes at once |
D. | None of these |
Answer» B. atleast one process | |
53. |
External fragmentation will not occur when : |
A. | first fit is used |
B. | best fit is used |
C. | worst fit is used |
D. | no matter which algorithm is used, it will always occur |
Answer» E. | |
54. |
External fragmentation exists when : |
A. | enough total memory exists to satisfy a request but it is not contiguous |
B. | the total memory is insufficient to satisfy a request |
C. | a request cannot be satisfied even when the total memory is free |
D. | None of these |
Answer» B. the total memory is insufficient to satisfy a request | |
55. |
The disadvantage of moving all process to one end of memory and all holes to the other direction, producing one large hole of available memory is : |
A. | the cost incurred |
B. | the memory used |
C. | the CPU used |
D. | All of these |
Answer» B. the memory used | |
56. |
__________ is generally faster than _________ and _________. |
A. | first fit, best fit, worst fit |
B. | best fit, first fit, worst fit |
C. | worst fit, best fit, first fit |
D. | None of these |
Answer» B. best fit, first fit, worst fit | |
57. |
Another solution to the problem of external fragmentation problem is to : |
A. | permit the logical address space of a process to be noncontiguous |
B. | permit smaller processes to be allocated memory at last |
C. | permit larger processes to be allocated memory at last |
D. | All of these |
Answer» B. permit smaller processes to be allocated memory at last | |
58. |
If relocation is static and is done at assembly or load time, compaction _________. |
A. | cannot be done |
B. | must be done |
C. | must not be done |
D. | can be done |
Answer» B. must be done | |
59. |
A solution to the problem of external fragmentation is : |
A. | compaction |
B. | larger memory space |
C. | smaller memory space |
D. | None of these |
Answer» B. larger memory space | |
60. |
In internal fragmentation, memory is internal to a partition and : |
A. | is being used |
B. | is not being used |
C. | is always used |
D. | None of these |
Answer» C. is always used | |
61. |
A multilevel page table is preferred in comparison to a single level page table for translating virtual address to physical address because : |
A. | it reduces the memory access time to read or write a memory location |
B. | it helps to reduce the size of page table needed to implement the virtual address space of a process |
C. | it is required by the translation look aside buffer |
D. | it helps to reduce the number of page faults in page replacement algorithms |
Answer» C. it is required by the translation look aside buffer | |
62. |
The segment base contains the : |
A. | starting logical address of the process |
B. | starting physical address of the segment in memory |
C. | segment length |
D. | none of the mentioned |
Answer» C. segment length | |
63. |
If there are 32 segments, each of size 1Kb, then the logical address should have : |
A. | 13 bits |
B. | 14 bits |
C. | 15 bits |
D. | 16 bits |
Answer» B. 14 bits | |
64. |
The protection bit is 0/1 based on : |
A. | write only |
B. | read only |
C. | read write |
D. | none of the mentioned |
Answer» D. none of the mentioned | |
65. |
If the offset is legal : |
A. | it is used as a physical memory address itself |
B. | it is subtracted from the segment base to produce the physical memory address |
C. | it is added to the segment base to produce the physical memory address |
D. | none of the mentioned |
Answer» B. it is subtracted from the segment base to produce the physical memory address | |
66. |
When the entries in the segment tables of two different processes point to the same physical location : |
A. | the segments are invalid |
B. | the processes get blocked |
C. | segments are shared |
D. | all of the mentioned |
Answer» D. all of the mentioned | |
67. |
The segment limit contains the : |
A. | starting logical address of the process |
B. | starting physical address of the segment in memory |
C. | segment length |
D. | none of the mentioned |
Answer» D. none of the mentioned | |
68. |
The offset d of the logical address must be : |
A. | greater than segment limit |
B. | between 0 and segment limit |
C. | between 0 and the segment number |
D. | greater than the segment number |
Answer» C. between 0 and the segment number | |
69. |
Consider a computer with 8 Mbytes of main memory and a 128 K cache. The cache block size is 4 K. It uses a direct mapping scheme for cache management. How many different main memory blocks can map onto a given physical cache block ? |
A. | 2048 |
B. | 256 |
C. | 64 |
D. | 8 |
Answer» D. 8 | |
70. |
Each entry in a segment table has a : |
A. | segment base |
B. | segment peak |
C. | segment value |
D. | none of the mentioned |
Answer» B. segment peak | |
71. |
A memory buffer used to accommodate a speed differential is called: |
A. | stack pointer |
B. | cache |
C. | accumulator |
D. | disk buffer |
Answer» C. accumulator | |
72. |
CPU fetches the instruction from memory according to the value of: |
A. | program counter |
B. | status register |
C. | instruction register |
D. | program status word |
Answer» B. status register | |
73. |
When divisions of decimals are involved in program, these numbers are stored in |
A. | real numbers |
B. | rational numbers |
C. | original numbers |
D. | irrational numbers |
Answer» B. rational numbers | |
74. |
Whole numbers from '0' and all negative numbers are classified as |
A. | integers |
B. | real numbers |
C. | prime numbers |
D. | composite numbers |
Answer» B. real numbers | |
75. |
Function which is used in stepwise refinement of program is classified |
A. | procedure |
B. | library format |
C. | routine format |
D. | subroutine format |
Answer» B. library format | |
76. |
Data types are differed on basis of |
A. | way of storage |
B. | type of operations |
C. | both (a) and (b) |
D. | type of operators used |
Answer» D. type of operators used | |
77. |
Statement "130 num(subs) = 2*subs-1" is an example of |
A. | array in BASIC |
B. | array in PASCAL |
C. | array in COMAL |
D. | array in COBOL |
Answer» B. array in PASCAL | |
78. |
In high-level language COMAL, area is calculated as |
A. | Area: = Width*Length |
B. | 100 Area = Width*Length |
C. | 100 Area: = Width*Length |
D. | length 100: area*20width |
Answer» D. length 100: area*20width | |
79. |
In line "110 DIM num(10)" in BASIC language, 'Line 110' declares |
A. | one-dimensional array |
B. | two-dimensional array |
C. | three-dimension array |
D. | multi-dimension array |
Answer» B. two-dimensional array | |
80. |
To obtain better memory utilization, dynamic loading is used. With dynamic loading, a routine is not loaded until it is called. For implementing dynamic loading, |
A. | special support from hardware is required |
B. | special support from operating system is essential |
C. | special support from both hardware and operating system is essential |
D. | user programs can implement dynamic loading without any special support from hardware or operating system |
Answer» E. | |
81. |
In a paged memory, the page hit ratio is 0.35. The required to access a page in secondary memory is equal to 100 ns. The time required to access a page in primary memory is 10 ns. The average time required to access a page is : |
A. | 3.0 ns |
B. | 68.0 ns |
C. | 68.5 ns |
D. | 78.5 ns |
Answer» D. 78.5 ns | |
82. |
When there is a large logical address space, the best way of paging would be : |
A. | not to page |
B. | a two level paging algorithm |
C. | the page table itself |
D. | all of the mentioned |
Answer» C. the page table itself | |
83. |
Illegal addresses are trapped using the _____ bit. |
A. | error |
B. | protection |
C. | valid invalid |
D. | access |
Answer» D. access | |
84. |
When the valid invalid bit is set to valid, it means that the associated page : |
A. | is in the TLB |
B. | has data in it |
C. | is in the process s logical address space |
D. | is the system s physical address space |
Answer» D. is the system s physical address space | |
85. |
Memory protection in a paged environment is accomplished by : |
A. | protection algorithm with each page |
B. | restricted access rights to users |
C. | restriction on page visibility |
D. | protection bit with each page |
Answer» E. | |
86. |
The percentage of times a page number is found in the TLB is known as : |
A. | miss ratio |
B. | hit ratio |
C. | miss percent |
D. | None of the mentioned |
Answer» C. miss percent | |
87. |
In high-level language Pascal, area is calculated as |
A. | Area: = Width*Length; |
B. | length 100: area*20width |
C. | 100 Area = Width*Length |
D. | 100 Area = Width*Length |
Answer» B. length 100: area*20width | |
88. |
In line "550 INPUT NUM: PRINT 6.9*NUM*NUM+27" of a BASIC program, 'PRINT and INPUT' is |
A. | reserved words |
B. | reserved variable |
C. | reserved constant |
D. | reserved exponent |
Answer» B. reserved variable | |
89. |
In line "100 INPUT NUM" of a BASIC program, NUM is if chosen by programmer is classified as |
A. | identifier |
B. | exponents |
C. | real constants |
D. | reserved variable |
Answer» B. exponents | |
90. |
A special quantity named in a program and its value can be changed is called |
A. | variables |
B. | constant |
C. | mantissa |
D. | exponent |
Answer» B. constant | |
91. |
Loop statement which is repeated to some given number of times is classified as |
A. | GO loop |
B. | FOR loop |
C. | REPEAT loop |
D. | GO REPEAT loop |
Answer» C. REPEAT loop | |
92. |
Set of data whose items are organized together is classified as |
A. | data structure |
B. | string structure |
C. | positive structure |
D. | variable structure |
Answer» B. string structure | |
93. |
Comments used to make program easier are also called as |
A. | narrative |
B. | unmarked stringe |
C. | marked variable |
D. | unmarked variable |
Answer» B. unmarked stringe | |
94. |
An assembler translates |
A. | routine into subroutine |
B. | processing time into manual time |
C. | assembly code into machine code |
D. | machine code into assembly code |
Answer» D. machine code into assembly code | |
95. |
Name given by a programmer to some data is classified as |
A. | mantissa |
B. | identifier |
C. | exponent |
D. | identification |
Answer» C. exponent | |
96. |
Functions used in programs that are defined by programmers are called |
A. | built-in functions |
B. | program layout |
C. | program procedure |
D. | user-defined function |
Answer» E. | |
97. |
A program which interprets each line of high level program at time of execution is called |
A. | executor |
B. | instructor |
C. | translator |
D. | interpreter |
Answer» E. | |
98. |
In COMAL language programs, variables in program procedure are declared local using |
A. | START function |
B. | OPEN function |
C. | END function |
D. | CLOSED function |
Answer» E. | |
99. |
An instruction which tells assembler how to deal with whole program is classified as |
A. | director |
B. | directive |
C. | compiler |
D. | direction |
Answer» C. compiler | |
100. |
In statement 110 IF Num = INT(Num) THEN PRINT "Whole number' "of BASIC language, built-in function is |
A. | IF |
B. | INT |
C. | Num |
D. | |
Answer» C. Num | |