McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Computer Science Engineering (CSE) .
| 601. |
The virtual memory basically stores the next segment of data to be executed on the |
| A. | secondary storage |
| B. | disks |
| C. | ram |
| D. | rom |
| Answer» A. secondary storage | |
| Explanation: none. | |
| 602. |
The associatively mapped virtual memory makes use of |
| A. | tlb |
| B. | page table |
| C. | frame table |
| D. | none of the mentioned |
| Answer» A. tlb | |
| Explanation: tlb stands for translation look-aside buffer. | |
| 603. |
The main reason for the discontinuation of semi conductor based storage devices for providing large storage space is |
| A. | lack of sufficient resources |
| B. | high cost per bit value |
| C. | lack of speed of operation |
| D. | none of the mentioned |
| Answer» B. high cost per bit value | |
| Explanation: 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. | |
| 604. |
The digital information is stored on the hard disk by |
| A. | applying a suitable electric pulse |
| B. | applying a suitable magnetic field |
| C. | applying a suitable nuclear field |
| D. | by using optic waves |
| Answer» A. applying a suitable electric pulse | |
| Explanation: the digital data is sorted on the magnetized discs by magnetizing the areas. | |
| 605. |
For the synchronization of the read head, we make use of a |
| A. | framing bit |
| B. | synchronization bit |
| C. | clock |
| D. | dirty bit |
| Answer» C. clock | |
| Explanation: the clock makes it easy to distinguish between different values red by a head. | |
| 606. |
One of the most widely used schemes of encoding used is |
| A. | nrz-polar |
| B. | rz-polar |
| C. | manchester |
| D. | block encoding |
| Answer» C. manchester | |
| Explanation: the manchester encoding used is also called as phase encoding and it is used to encode both clock and data. | |
| 607. |
The drawback of Manchester encoding is |
| A. | the cost of the encoding scheme |
| B. | the speed of encoding the data |
| C. | the latency offered |
| D. | the low bit storage density provided |
| Answer» D. the low bit storage density provided | |
| Explanation: the space required to represent each bit must be large enough to accommodate two changes in magnetization. | |
| 608. |
The read/write heads must be near to disk surfaces for better storage. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: by maintaining the heads near to the surface greater bit densities can be achieved. | |
| 609. |
pushes the heads away from the surface as they rotate at their standard rates. |
| A. | magnetic tension |
| B. | electric force |
| C. | air pressure |
| D. | none of the mentioned |
| Answer» C. air pressure | |
| Explanation: due to the speed of rotation of the discs air pressure develops in the hard disk. | |
| 610. |
The air pressure can be countered by putting in the head-disc surface arrangement. |
| A. | air filter |
| B. | spring mechanism |
| C. | coolant |
| D. | none of the mentioned |
| Answer» B. spring mechanism | |
| Explanation: the spring mechanism | |
| 611. |
The method of placing the heads and the discs in an air tight environment is also called as |
| A. | raid arrays |
| B. | atp tech |
| C. | winchester technology |
| D. | fleming reduction |
| Answer» C. winchester technology | |
| Explanation: the disks and the heads operate faster due to the absence of the dust particles. | |
| 612. |
A hard disk with 20 surfaces will have heads. |
| A. | 10 |
| B. | 5 |
| C. | 1 |
| D. | 20 |
| Answer» D. 20 | |
| Explanation: each surface will have its own head to perform read/write operation. | |
| 613. |
The set of corresponding tracks on all surfaces of a stack of disks form a |
| A. | cluster |
| B. | cylinder |
| C. | group |
| D. | set |
| Answer» B. cylinder | |
| Explanation: the data is stored in these sections called as cylinders. | |
| 614. |
The data can be accessed from the disk using |
| A. | surface number |
| B. | sector number |
| C. | track number |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: none. | |
| 615. |
The read and write operations usually start at of the sector. |
| A. | center |
| B. | middle |
| C. | from the last used point |
| D. | boundaries |
| Answer» D. boundaries | |
| Explanation: the heads read and write data from the ends to the center. | |
| 616. |
To distinguish between two sectors we make use of |
| A. | inter sector gap |
| B. | splitting bit |
| C. | numbering bit |
| D. | none of the mentioned |
| Answer» A. inter sector gap | |
| Explanation: this means that we leave a little gap between each sector to differentiate between them. | |
| 617. |
The process divides the disk into sectors and tracks. |
| A. | creation |
| B. | initiation |
| C. | formatting |
| D. | modification |
| Answer» C. formatting | |
| Explanation: the formatting process deletes the data present and does the creation of sectors and tracks. | |
| 618. |
The access time is composed of |
| A. | seek time |
| B. | rotational delay |
| C. | latency |
| D. | both seek time and rotational delay |
| Answer» D. both seek time and rotational delay | |
| Explanation: the seek time refers to the time required to move the head to the required disk. | |
| 619. |
The disk drive is connected to the system by using the |
| A. | pci bus |
| B. | scsi bus |
| C. | hdmi |
| D. | isa |
| Answer» B. scsi bus | |
| Explanation: none. | |
| 620. |
is used to deal with the difference in the transfer rates between the drive and the bus. |
| A. | data repeaters |
| B. | enhancers |
| C. | data buffers |
| D. | none of the mentioned |
| Answer» C. data buffers | |
| Explanation: the buffers are added to store the data from the fast device and to send it to the slower device at its rate. | |
| 621. |
is used to detect and correct the errors that may occur during data transfers. |
| A. | ecc |
| B. | crc |
| C. | checksum |
| D. | none of the mentioned |
| Answer» A. ecc | |
| Explanation: ecc stands for error correcting code. | |
| 622. |
______ has been developed specifically for pipelined systems. |
| A. | Utility software |
| B. | Speed up utilities |
| C. | Optimizing compilers |
| D. | None of the mentioned |
| Answer» C. Optimizing compilers | |
| 623. |
The fetch and execution cycles are interleaved with the help of ________ |
| A. | Modification in processor architecture |
| B. | Clock |
| C. | Special unit |
| D. | Control unit |
| Answer» B. Clock | |
| 624. |
Each stage in pipelining should be completed within ____ cycle. |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 4 |
| Answer» A. 1 | |
| 625. |
To increase the speed of memory access in pipelining, we make use of _______ |
| A. | Special memory locations |
| B. | Special purpose registers |
| C. | Cache |
| D. | Buffers |
| Answer» C. Cache | |
| 626. |
The situation wherein the data of operands are not available is called ______ |
| A. | Data hazard |
| B. | Stock |
| C. | Deadlock |
| D. | Structural hazard |
| Answer» A. Data hazard | |
| 627. |
The time lost due to the branch instruction is often referred to as _____ |
| A. | Latency |
| B. | Delay |
| C. | Branch penalty |
| D. | None of the mentioned |
| Answer» C. Branch penalty | |
| 628. |
The algorithm followed in most of the systems to perform out of order execution is ______ |
| A. | Tomasulo algorithm |
| B. | Score carding |
| C. | Reader-writer algorithm |
| D. | None of the mentioned |
| Answer» A. Tomasulo algorithm | |
| 629. |
The logic operations are implemented using _______ circuits. |
| A. | Bridge |
| B. | Logical |
| C. | Combinatorial |
| D. | Gate |
| Answer» C. Combinatorial | |
| 630. |
The carry generation function: ci + 1 = yici + xici + xiyi, is implemented in ____________ |
| A. | Half adders |
| B. | Full adders |
| C. | Ripple adders |
| D. | Fast adders |
| Answer» B. Full adders | |
| 631. |
Which option is true regarding the carry in the ripple adders? |
| A. | Are generated at the beginning only |
| B. | Must travel through the configuration |
| C. | Is generated at the end of each operation |
| D. | None of the mentioned |
| Answer» B. Must travel through the configuration | |
| 632. |
In full adders the sum circuit is implemented using ________ |
| A. | And & or gates |
| B. | NAND gate |
| C. | XOR |
| D. | XNOR |
| Answer» C. XOR | |
| 633. |
The usual implementation of the carry circuit involves _________ |
| A. | And & or gates |
| B. | XOR |
| C. | NAND |
| D. | XNOR |
| Answer» B. XOR | |
| 634. |
The advantage of I/O mapped devices to memory mapped is ___________ |
| A. | The former offers faster transfer of data |
| B. | The devices connected using I/O mapping have a bigger buffer space |
| C. | The devices have to deal with fewer address lines |
| D. | No advantage as such |
| Answer» C. The devices have to deal with fewer address lines | |
| 635. |
The system is notified of a read or write operation by ___________ |
| A. | Appending an extra bit of the address |
| B. | Enabling the read or write bits of the devices |
| C. | Raising an appropriate interrupt signal |
| D. | Sending a special signal along the BUS |
| Answer» D. Sending a special signal along the BUS | |
| 636. |
To overcome the lag in the operating speeds of the I/O device and the processor we use ___________ |
| A. | Buffer spaces |
| B. | Status flags |
| C. | Interrupt signals |
| D. | Exceptions |
| Answer» B. Status flags | |
| 637. |
The method which offers higher speeds of I/O transfers is ___________ |
| A. | Interrupts |
| B. | Memory mapping |
| C. | Program-controlled I/O |
| D. | DMA |
| Answer» D. DMA | |
| 638. |
The instruction, Add #45, R1 does _______ |
| A. | Adds the value of 45 to the address of R1 and stores 45 in that address |
| B. | Adds 45 to the value of R1 and stores it in R1 |
| C. | Finds the memory location 45 and adds that content to that of R1 |
| D. | None of the mentioned |
| Answer» B. Adds 45 to the value of R1 and stores it in R1 | |
| 639. |
In the case of, Zero-address instruction method the operands are stored in _____ |
| A. | Registers |
| B. | Accumulators |
| C. | Push down stack |
| D. | Cache |
| Answer» C. Push down stack | |
| 640. |
The addressing mode which makes use of in-direction pointers is ______ |
| A. | Indirect addressing mode |
| B. | Index addressing mode |
| C. | Relative addressing mode |
| D. | Offset addressing mode |
| Answer» A. Indirect addressing mode | |
| 641. |
The addressing mode/s, which uses the PC instead of a general purpose register is ______ |
| A. | Indexed with offset |
| B. | Relative |
| C. | direct |
| D. | both Indexed with offset and direct |
| Answer» B. Relative | |
| 642. |
_____ addressing mode is most suitable to change the normal sequence of execution of instructions. |
| A. | Relative |
| B. | Indirect |
| C. | Index with Offset |
| D. | Immediate |
| Answer» A. Relative | |
| 643. |
The reason for the implementation of the cache memory is ________ |
| A. | To increase the internal memory of the system |
| B. | The difference in speeds of operation of the processor and memory |
| C. | To reduce the memory access and cycle time |
| D. | All of the mentioned |
| Answer» B. The difference in speeds of operation of the processor and memory | |
| 644. |
The effectiveness of the cache memory is based on the property of ________ |
| A. | Locality of reference |
| B. | Memory localisation |
| C. | Memory size |
| D. | None of the mentioned |
| Answer» A. Locality of reference | |
| 645. |
The spatial aspect of the locality of reference means ________ |
| A. | That the recently executed instruction is executed again next |
| B. | That the recently executed won’t be executed again |
| C. | That the instruction executed will be executed at a later time |
| D. | That the instruction in close proximity of the instruction executed will be executed in future |
| Answer» D. That the instruction in close proximity of the instruction executed will be executed in future | |
| 646. |
The correspondence between the main memory blocks and those in the cache is given by _________ |
| A. | Hash function |
| B. | Mapping function |
| C. | Locale function |
| D. | Assign function |
| Answer» B. Mapping function | |
| 647. |
The copy-back protocol is used ________ |
| A. | To copy the contents of the memory onto the cache |
| B. | To update the contents of the memory from the cache |
| C. | To remove the contents of the cache and push it on to the memory |
| D. | None of the mentioned |
| Answer» B. To update the contents of the memory from the cache | |
| 648. |
The address space is 22 bits the memory is 32 bit word addressable what is the memory size |
| A. | 16MB |
| B. | 512KB |
| C. | 4MB |
| D. | 1GB |
| Answer» A. 16MB | |
| 649. |
In which cycle the memory is read and the contents of memory at the address containedin the PC register are loaded into in to IR. |
| A. | Execution Cycle |
| B. | Memory Cycle |
| C. | Fetch Cycle |
| D. | Decode Cycle |
| Answer» C. Fetch Cycle | |
| 650. |
The part of machine level instruction, which tells the central processor what has to be done, is |
| A. | Operation code |
| B. | Address |
| C. | Locator |
| D. | Flip-Flop |
| Answer» A. Operation code | |
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