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This section includes 1690 Mcqs, each offering curated multiple-choice questions to sharpen your Technical Programming knowledge and support exam preparation. Choose a topic below to get started.
| 1351. |
When device A has a cable that plugs into device B, and device B has a cable that plugs into device C and device C plugs into a port on the computer, this arrangement is called a _________ |
| A. | port |
| B. | daisy chain |
| C. | bus |
| D. | cable |
| Answer» C. bus | |
| 1352. |
A ____ a set of wires and a rigidly defined protocol that specifies a set of messages that can be sent on the wires. |
| A. | port |
| B. | node |
| C. | bus |
| D. | none of the mentioned |
| Answer» D. none of the mentioned | |
| 1353. |
If one or more devices use a common set of wires to communicate with the computer system, the connection is called ______ |
| A. | CPU |
| B. | Monitor |
| C. | Wirefull |
| D. | Bus |
| Answer» E. | |
| 1354. |
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 lookaside buffer |
| D. | it helps to reduce the number of page faults in page replacement algorithms |
| Answer» C. it is required by the translation lookaside buffer | |
| 1355. |
Consider a computer with 8 Mbytes of main memory and a 128K 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 | |
| 1356. |
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 | |
| 1357. |
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 | |
| 1358. |
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 | |
| 1359. |
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 | |
| 1360. |
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 | |
| 1361. |
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 | |
| 1362. |
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 | |
| 1363. |
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 | |
| 1364. |
In paging the user provides only ________ which is partitioned by the hardware into ________ and ______ |
| A. | one address, page number, offset |
| B. | one offset, page number, address |
| C. | page number, offset, address |
| D. | none of the mentioned |
| Answer» B. one offset, page number, address | |
| 1365. |
In segmentation, each address is specified by : |
| A. | a segment number & offset |
| B. | an offset & value |
| C. | a value & segment number |
| D. | a key & value |
| Answer» B. an offset & value | |
| 1366. |
In paged memory systems, if the page size is increased, then the internal fragmentation generally : |
| A. | becomes less |
| B. | becomes more |
| C. | remains constant |
| D. | none of the mentioned |
| Answer» C. remains constant | |
| 1367. |
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. | |
| 1368. |
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 | |
| 1369. |
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 | |
| 1370. |
Illegal addresses are trapped using the _____ bit. |
| A. | error |
| B. | protection |
| C. | valid – invalid |
| D. | access |
| Answer» D. access | |
| 1371. |
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 | |
| 1372. |
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. | |
| 1373. |
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 | |
| 1374. |
An ______ uniquely identifies processes and is used to provide address space protection for that process. |
| A. | address space locator |
| B. | address space identifier |
| C. | address process identifier |
| D. | None of the mentioned |
| Answer» C. address process identifier | |
| 1375. |
If a page number is not found in the TLB, then it is known as a : |
| A. | TLB miss |
| B. | Buffer miss |
| C. | TLB hit |
| D. | All of the mentioned |
| Answer» B. Buffer miss | |
| 1376. |
Each entry in a Translation lookaside buffer (TLB) consists of : |
| A. | key |
| B. | value |
| C. | bit value |
| D. | constant |
| Answer» B. value | |
| 1377. |
Time taken in memory access through PTBR is : |
| A. | extended by a factor of 3 |
| B. | extended by a factor of 2 |
| C. | slowed by a factor of 3 |
| D. | slowed by a factor of 2 |
| Answer» E. | |
| 1378. |
For every process there is a __________ |
| A. | page table |
| B. | copy of page table |
| C. | pointer to page table |
| D. | all of the mentioned |
| Answer» B. copy of page table | |
| 1379. |
For larger page tables, they are kept in main memory and a __________ points to the page table. |
| A. | page table base register |
| B. | page table base pointer |
| C. | page table register pointer |
| D. | page table base |
| Answer» B. page table base pointer | |
| 1380. |
The page table registers should be built with _______ |
| A. | very low speed logic |
| B. | very high speed logic |
| C. | a large memory space |
| D. | none of the mentioned |
| Answer» C. a large memory space | |
| 1381. |
Smaller page tables are implemented as a set of _______ |
| A. | queues |
| B. | stacks |
| C. | counters |
| D. | registers |
| Answer» E. | |
| 1382. |
Paging increases the ______ time. |
| A. | waiting |
| B. | execution |
| C. | context – switch |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| 1383. |
The operating system maintains a ______ table that keeps track of how many frames have been allocated, how many are there, and how many are available. |
| A. | page |
| B. | mapping |
| C. | frame |
| D. | memory |
| Answer» D. memory | |
| 1384. |
With paging there is no ________ fragmentation. |
| A. | internal |
| B. | external |
| C. | either type of |
| D. | none of the mentioned |
| Answer» C. either type of | |
| 1385. |
If the size of logical address space is 2 to the power of m, and a page size is 2 to the power of n addressing units, then the high order _____ bits of a logical address designate the page number, and the ____ low order bits designate the page offset. |
| A. | m, n |
| B. | n, m |
| C. | m – n, m |
| D. | m – n, n |
| Answer» E. | |
| 1386. |
The size of a page is typically : |
| A. | varied |
| B. | power of 2 |
| C. | power of 4 |
| D. | none of the mentioned |
| Answer» C. power of 4 | |
| 1387. |
The _____ table contains the base address of each page in physical memory. |
| A. | process |
| B. | memory |
| C. | page |
| D. | frame |
| Answer» D. frame | |
| 1388. |
The __________ is used as an index into the page table. |
| A. | frame bit |
| B. | page number |
| C. | page offset |
| D. | frame offset |
| Answer» C. page offset | |
| 1389. |
Every address generated by the CPU is divided into two parts : |
| A. | frame bit & page number |
| B. | page number & page offset |
| C. | page offset & frame bit |
| D. | frame offset & page offset |
| Answer» C. page offset & frame bit | |
| 1390. |
Logical memory is broken into blocks of the same size called _________ |
| A. | frames |
| B. | pages |
| C. | backing store |
| D. | none of the mentioned |
| Answer» C. backing store | |
| 1391. |
Physical memory is broken into fixed-sized blocks called ________ |
| A. | frames |
| B. | pages |
| C. | backing store |
| D. | none of the mentioned |
| Answer» B. pages | |
| 1392. |
When the memory allocated to a process is slightly larger than the process, then : |
| A. | internal fragmentation occurs |
| B. | external fragmentation occurs |
| C. | both internal and external fragmentation occurs |
| D. | neither internal nor external fragmentation occurs |
| Answer» B. external fragmentation occurs | |
| 1393. |
Sometimes the overhead of keeping track of a hole might be : |
| A. | larger than the memory |
| B. | larger than the hole itself |
| C. | very small |
| D. | all of the mentioned |
| Answer» C. very small | |
| 1394. |
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. | |
| 1395. |
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 the mentioned |
| Answer» B. the total memory is insufficient to satisfy a request | |
| 1396. |
__________ 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 the mentioned |
| Answer» B. best fit, first fit, worst fit | |
| 1397. |
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 the mentioned |
| Answer» B. the memory used | |
| 1398. |
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 | |
| 1399. |
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 the mentioned |
| Answer» B. permit smaller processes to be allocated memory at last | |
| 1400. |
A solution to the problem of external fragmentation is : |
| A. | compaction |
| B. | larger memory space |
| C. | smaller memory space |
| D. | none of the mentioned |
| Answer» B. larger memory space | |