MCQOPTIONS
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MCQOPTIONS
This section includes 950 Mcqs, each offering curated multiple-choice questions to sharpen your Computer Science Engineering (CSE) knowledge and support exam preparation. Choose a topic below to get started.
| 651. |
The contention for the usage of a hardware device is called |
| A. | structural hazard |
| B. | stalk |
| C. | deadlock |
| D. | none of the mentioned |
| Answer» B. stalk | |
| 652. |
The periods of time when the unit is idle is called as |
| A. | stalls |
| B. | bubbles |
| C. | hazards |
| D. | both stalls and bubbles |
| Answer» E. | |
| 653. |
In pipelining the task which requires the least time is performed first. |
| A. | true |
| B. | false |
| Answer» C. | |
| 654. |
Each stage in pipelining should be completed within                         cycle. |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 4 |
| Answer» B. 2 | |
| 655. |
To increase the speed of memory access in pipelining, we make use of |
| A. | modification in processor architecture |
| B. | clock |
| C. | special unit |
| D. | control unit |
| Answer» C. special unit | |
| 656. |
In 32 bit representation the scale factor as a range of |
| A. | -128 to 127 |
| B. | -256 to 255 |
| C. | 0 to 255 |
| D. | none of the mentioned |
| Answer» B. -256 to 255 | |
| 657. |
The pipelining process is also called as |
| A. | superscalar operation |
| B. | assembly line operation |
| C. | von neumann cycle |
| D. | none of the mentioned |
| Answer» C. von neumann cycle | |
| 658. |
              have been developed specifically for pipelined systems. |
| A. | utility software |
| B. | speed up utilities |
| C. | optimizing compilers |
| D. | none of the mentioned |
| Answer» D. none of the mentioned | |
| 659. |
In double precision format, the size of the mantissa is |
| A. | 32 bit |
| B. | 52 bit |
| C. | 64 bit |
| D. | 72 bit |
| Answer» C. 64 bit | |
| 660. |
The 32 bit representation of the decimal number is called as |
| A. | double-precision |
| B. | single-precision |
| C. | extended format |
| D. | none of the mentioned |
| Answer» C. extended format | |
| 661. |
The normalized representation of 0.0010110 * 2 9 is |
| A. | 0 10001000 0010110 |
| B. | 0 10000101 0110 |
| C. | 0 10101010 1110 |
| D. | 0 11110100 11100 |
| Answer» C. 0 10101010 1110 | |
| 662. |
The numbers written to the power of 10 in the representation of decimal numbers are called as |
| A. | height factors |
| B. | size factors |
| C. | scale factors |
| D. | none of the mentioned |
| Answer» D. none of the mentioned | |
| 663. |
The sign followed by the string of digits is called as |
| A. | significant |
| B. | determinant |
| C. | mantissa |
| D. | exponent |
| Answer» D. exponent | |
| 664. |
                  constitute the representation of the floating number. |
| A. | sign |
| B. | significant digits |
| C. | scale factor |
| D. | all of the mentioned |
| Answer» E. | |
| 665. |
If the decimal point is placed to the right of the first significant digit, then the number is called |
| A. | orthogonal |
| B. | normalized |
| C. | determinate |
| D. | none of the mentioned |
| Answer» C. determinate | |
| 666. |
CSA stands for? |
| A. | computer speed addition |
| B. | carry save addition |
| C. | computer service architecture |
| D. | none of the mentioned |
| Answer» B. carry save addition | |
| 667. |
The multiplier -6(11010) is recorded as |
| A. | 0-1-2 |
| B. | 0-1+1-10 |
| C. | -2-10 |
| D. | none of the mentioned |
| Answer» B. 0-1+1-10 | |
| 668. |
The method used to reduce the maximum number of summands by half is |
| A. | fast multiplication |
| B. | bit-pair recording |
| C. | quick multiplication |
| D. | none of the mentioned |
| Answer» C. quick multiplication | |
| 669. |
The decimal numbers represented in the computer are called as floating point numbers, as the decimal point floats through the number. |
| A. | true |
| B. | false |
| Answer» B. false | |
| 670. |
The multiplicand and the control signals are passed through to the n-bit adder via |
| A. | mux |
| B. | demux |
| C. | encoder |
| D. | decoder |
| Answer» B. demux | |
| 671. |
The               is used to coordinate the operation of the multiplier. |
| A. | controller |
| B. | coordinator |
| C. | control sequencer |
| D. | none of the mentioned |
| Answer» D. none of the mentioned | |
| 672. |
The product of 1101 & 1011 is |
| A. | 10001111 |
| B. | 10101010 |
| C. | 11110000 |
| D. | 11001100 |
| Answer» B. 10101010 | |
| 673. |
The delay reduced to in the carry look ahead adder is |
| A. | 5 |
| B. | 8 |
| C. | 10 |
| D. | 2n |
| Answer» B. 8 | |
| 674. |
A Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â gate is used to detect the occurrence of an overflow. |
| A. | nand |
| B. | xor |
| C. | xnor |
| D. | and |
| Answer» C. xnor | |
| 675. |
The usual implementation of the carry circuit involves |
| A. | and & or gates |
| B. | xor |
| C. | nand |
| D. | xnor |
| Answer» C. nand | |
| 676. |
In full adders the sum circuit is implemented using |
| A. | and & or gates |
| B. | nand gate |
| C. | xor |
| D. | xnor |
| Answer» D. xnor | |
| 677. |
The logic operations are implemented using                 circuits. |
| A. | bridge |
| B. | logical |
| C. | combinatorial |
| D. | gate |
| Answer» D. gate | |
| 678. |
The logic operations are simpler to implement using logic circuits. |
| A. | true |
| B. | false |
| Answer» B. false | |
| 679. |
The SOF is transmitted every |
| A. | 1s |
| B. | 5s |
| C. | 1ms |
| D. | 1us |
| Answer» D. 1us | |
| 680. |
The data packets can contain data upto |
| A. | 512 bytes |
| B. | 256 bytes |
| C. | 1024 bytes |
| D. | 2 kb |
| Answer» D. 2 kb | |
| 681. |
The            signal is used to indicate the beginning of a new frame. |
| A. | start |
| B. | sof |
| C. | beg |
| D. | none of the mentioned |
| Answer» C. beg | |
| 682. |
The CRC bits are computed based on the values of the |
| A. | pid |
| B. | addr |
| C. | endp |
| D. | both addr and endp |
| Answer» E. | |
| 683. |
The last field in the packet is |
| A. | pid |
| B. | addr |
| C. | endp |
| D. | crc |
| Answer» E. | |
| 684. |
The first field of any packet is |
| A. | pid |
| B. | addr |
| C. | endp |
| D. | crc16 |
| Answer» B. addr | |
| 685. |
The type/s of packets sent by the USB is/are |
| A. | data |
| B. | address |
| C. | control |
| D. | both data and control |
| Answer» E. | |
| 686. |
The 4 bit PID’s are transmitted twice. |
| A. | true |
| B. | false |
| Answer» B. false | |
| 687. |
Locations in the device to or from which data transfers can take place is called |
| A. | end points |
| B. | hosts |
| C. | source |
| D. | none of the mentioned |
| Answer» B. hosts | |
| 688. |
When the USB is connected to a system, its root hub is connected to the |
| A. | pci bus |
| B. | scsi bus |
| C. | processor bus |
| D. | ide |
| Answer» D. ide | |
| 689. |
The USB address space can be shared by the user’s memory space. |
| A. | true |
| B. | false |
| Answer» C. | |
| 690. |
The devices connected to USB is assigned a          address. |
| A. | 9 bit |
| B. | 16 bit |
| C. | 4 bit |
| D. | 7 bit |
| Answer» E. | |
| 691. |
A USB pipe is a               channel. |
| A. | simplex |
| B. | half-duplex |
| C. | full-duplex |
| D. | both simplex and full-duplex |
| Answer» D. both simplex and full-duplex | |
| 692. |
The device can send a message to the host by taking part in            for the communication path. |
| A. | arbitration |
| B. | polling |
| C. | prioritizing |
| D. | none of the mentioned |
| Answer» C. prioritizing | |
| 693. |
In USB the devices can communicate with each other. |
| A. | true |
| B. | false |
| Answer» C. | |
| 694. |
USB is a parallel mode of transmission of data and this enables for the fast speeds of data transfers. |
| A. | true |
| B. | false |
| Answer» C. | |
| 695. |
The I/O devices form the            of the tree structure. |
| A. | leaves |
| B. | subordinate roots |
| C. | left subtrees |
| D. | right subtrees |
| Answer» B. subordinate roots | |
| 696. |
The sampling process in speaker output is a                   process. |
| A. | asynchronous |
| B. | synchronous |
| C. | isochronous |
| D. | none of the mentioned |
| Answer» D. none of the mentioned | |
| 697. |
The high speed mode of operation of the USB was introduced by |
| A. | isa |
| B. | usb 3.0 |
| C. | usb 2.0 |
| D. | ansi |
| Answer» D. ansi | |
| 698. |
The data is stored on the disk in the form of blocks called |
| A. | pages |
| B. | frames |
| C. | sectors |
| D. | tables |
| Answer» D. tables | |
| 699. |
HVD stands for |
| A. | high voltage differential |
| B. | high voltage density |
| C. | high video definition |
| D. | none of the mentioned |
| Answer» B. high voltage density | |
| 700. |
The mode of data transfer used by the controller is |
| A. | interrupt |
| B. | dma |
| C. | asynchronous |
| D. | synchronous |
| Answer» C. asynchronous | |