

MCQOPTIONS
Saved Bookmarks
This section includes 239 Mcqs, each offering curated multiple-choice questions to sharpen your Operating System knowledge and support exam preparation. Choose a topic below to get started.
151. |
Type of magnetic tape named reel is known to be of size |
A. | 1/2 inch |
B. | 1/3 inch |
C. | 1/4 inch |
D. | 1/5 inch |
Answer» B. 1/3 inch | |
152. |
Storage which has a smallest storage capacity is |
A. | floppy disk |
B. | zip disk |
C. | hard disk |
D. | compact disk |
Answer» B. zip disk | |
153. |
What is the storage capacity of a dual layered Blue Ray disc? |
A. | 80 GB |
B. | 50 GB |
C. | 10 GB |
D. | 25 GB |
Answer» C. 10 GB | |
154. |
DVD technology uses an optical media to store the digital data. DVD is an acronym for |
A. | Digital Vector Disc |
B. | Digital Volume Disc |
C. | Digital Versatile Disc |
D. | Digital Visualization Disc |
Answer» D. Digital Visualization Disc | |
155. |
What type of devices are CDs or DVDs ? |
A. | Input |
B. | Output |
C. | Software |
D. | Storage |
Answer» E. | |
156. |
Which of the following has the smallest storage capacity ? |
A. | zip disk |
B. | hard disk |
C. | floppy disk |
D. | data cartridge |
Answer» D. data cartridge | |
157. |
What is the storage capacity of a single layered Blue Ray disc? |
A. | 80 GB |
B. | 50 GB |
C. | 10 GB |
D. | 25 GB |
Answer» E. | |
158. |
When you save to this, your data will remain intact even when the computer is turned off. |
A. | RAM |
B. | Motherboard |
C. | Secondary storage device |
D. | Primary storage device |
Answer» D. Primary storage device | |
159. |
What is the main purpose of the secondary storage device? |
A. | To Store Data |
B. | To Networking |
C. | To increase the speed of Computer |
D. | To Install Operating System |
Answer» B. To Networking | |
160. |
Storage capacity of secondary storage is said to be of |
A. | Virtually unlimited capacity |
B. | Virtually limited capacity |
C. | Virtually volatile capacity |
D. | Virtually non-volatile capacity |
Answer» B. Virtually limited capacity | |
161. |
Process of dividing the disk into tracks and sectors is called |
A. | tracking |
B. | allotting |
C. | crashing |
D. | formatting |
Answer» E. | |
162. |
CD-ROMs are pre-stamped by a process named to be |
A. | Stamping |
B. | Recording |
C. | Mastering |
D. | Manufacturing |
Answer» D. Manufacturing | |
163. |
Term which is used for the disk content that is recorded at the time of manufacturing and cannot be changed or erased by user is |
A. | read only |
B. | write only |
C. | run only |
D. | save only |
Answer» B. write only | |
164. |
Example of serial access store is |
A. | floppy disc |
B. | program tape |
C. | magnetic tape |
D. | magnetic disc |
Answer» D. magnetic disc | |
165. |
Storage devices falling in direct access form of secondary storage is/are of |
A. | One type |
B. | Two types |
C. | Three types |
D. | Four types |
Answer» C. Three types | |
166. |
Encoding scheme used by optical disk drives is named as |
A. | BCD |
B. | EBCDIC |
C. | ASCII |
D. | CLV |
Answer» E. | |
167. |
Which of the following correctly represents the track pattern of an optical disk ? |
A. | a |
B. | b |
C. | c |
D. | d |
Answer» B. b | |
168. |
Optical disk storage of direct access drive includes CD-ROM and |
A. | Winchester disk |
B. | Floppy disk |
C. | Zip disk |
D. | WORM disk |
Answer» E. | |
169. |
Warping may end up as twisted tape ribbon, which may cause |
A. | Loss of data |
B. | Deletion of data |
C. | Garbage values in data |
D. | Converted form of data |
Answer» B. Deletion of data | |
170. |
For positioning of read and write heads over desired track requires time, defined to be |
A. | Latency |
B. | Throughput time |
C. | Access time |
D. | Seek time |
Answer» E. | |
171. |
All of the followings are included in removable media except …………. |
A. | CD-ROMs |
B. | Diskette |
C. | DVDs |
D. | Hard Disk Drive |
Answer» E. | |
172. |
In streamer tape, number of tracks can vary depending upon the |
A. | Tape drive |
B. | Tape controller |
C. | Tape word length |
D. | Tape streaming |
Answer» B. Tape controller | |
173. |
A magnetic disk's surface is divided into invisible concentric circles known to be |
A. | Cells |
B. | Intervals |
C. | Tracks |
D. | Records |
Answer» D. Records | |
174. |
Reading data is performed in magnetic disk by |
A. | Lower surface |
B. | Read/write leads |
C. | Sectors |
D. | Track |
Answer» C. Sectors | |
175. |
To overcome limitations of primary storage, we can use |
A. | Flash memory |
B. | Cache memory |
C. | Auxiliary memory |
D. | Virtual memory |
Answer» D. Virtual memory | |
176. |
A normal CD-ROM usually can store up to __________data? |
A. | 680 KB |
B. | 680 Bytes |
C. | 680 MB |
D. | 680 GB |
Answer» D. 680 GB | |
177. |
As compared to diskettes, the hard disks are |
A. | Slowly accessed |
B. | Less rigid |
C. | More expensive |
D. | More portable |
Answer» D. More portable | |
178. |
Number of tracks possessed by a streamer tape is defined to be |
A. | 2-20 tracks |
B. | 3-30 tracks |
C. | 4-40 tracks |
D. | 5-50 tracks |
Answer» D. 5-50 tracks | |
179. |
A Digital Versatile Disk (DVD) can store______ Gigabyte of information. |
A. | 4.6 |
B. | 5.6 |
C. | 6.6 |
D. | 7.6 |
Answer» B. 5.6 | |
180. |
During the recovery cycle from a faliure one |
A. | each pair of physical block is examined |
B. | specified pair of physical block is examined |
C. | first pair of physical block is examined |
D. | last pair of physical block is examined |
Answer» B. specified pair of physical block is examined | |
181. |
.............. Gigabytes = 1 Terabyte. |
A. | 1204 |
B. | 1104 |
C. | 1024 |
D. | None of the above |
Answer» D. None of the above | |
182. |
In a computer system, related files are generally stored in the same |
A. | track |
B. | table |
C. | folder |
D. | wallet |
Answer» D. wallet | |
183. |
CD can store up to ___________ MB of data. |
A. | 550 |
B. | 1000 |
C. | 750 |
D. | 1500 |
Answer» D. 1500 | |
184. |
Which of following can not be accessed randomly |
A. | DRAM |
B. | SRAM |
C. | ROM |
D. | Magnetic tape |
Answer» E. | |
185. |
1 gigabytes is equal to |
A. | 1024 bytes |
B. | 1024 kilobytes |
C. | 1024 megabytes |
D. | 1024 bits |
Answer» D. 1024 bits | |
186. |
The size of a floppy disks varies from |
A. | 250KB to 360 KB |
B. | 360 KB to 1 GB |
C. | 360 KB to 2 GB |
D. | 360 KB to 2.88 MB |
Answer» E. | |
187. |
Time a device takes to read actual data is classified as |
A. | transfer time |
B. | seek time |
C. | seek delay |
D. | access delay |
Answer» B. seek time | |
188. |
Backing store that have short access time and consists no moving parts is classified as |
A. | timed discs |
B. | bubble memory |
C. | exchangeable discs |
D. | non exchangeable discs |
Answer» C. exchangeable discs | |
189. |
A piece of time taken by disc to rotate it and read data from right place is classified as |
A. | rotational delay |
B. | access delay |
C. | seek time delay |
D. | reversal delay |
Answer» B. access delay | |
190. |
Type of backup storage in which data is read in a sequence is classified as |
A. | permanent access |
B. | storage access |
C. | direct access |
D. | serial access |
Answer» E. | |
191. |
In fixed head discs, rotational delay plus transfer time is equal to |
A. | access time |
B. | delay time |
C. | processing time |
D. | storage time |
Answer» B. delay time | |
192. |
Secondary storage memory type is |
A. | volatile memory |
B. | non volatile memory |
C. | impact memory |
D. | virtual memory |
Answer» C. impact memory | |
193. |
Which access method is used to access cassette tape? |
A. | Direct |
B. | Both Sequential and Direct |
C. | Sequential |
D. | None of the above |
Answer» D. None of the above | |
194. |
For recording data in magnetic disks, a standard binary code is used of size |
A. | 4 bits |
B. | 8 bits |
C. | 16 bits |
D. | 32 bits |
Answer» C. 16 bits | |
195. |
Large length of blank spaces to allow acceleration and deceleration are called |
A. | streamer gaps |
B. | cartridge gaps |
C. | block gaps |
D. | inter block gaps |
Answer» E. | |
196. |
Which device is used to backup the data? |
A. | Floppy Disk |
B. | Tape |
C. | Network Drive |
D. | All of the above |
Answer» E. | |
197. |
Storage location of information being accessed determines variation in |
A. | Data storage |
B. | Location |
C. | Access time |
D. | Processing time |
Answer» D. Processing time | |
198. |
CD-ROM stands for |
A. | Compactable Read Only Memory |
B. | Compactable Disk Read Only Memory |
C. | Compact Disk Read Only Memory |
D. | Compact Data Read Only Memory |
Answer» D. Compact Data Read Only Memory | |
199. |
In sequential access device, data can be retrieved in same way as |
A. | Operated |
B. | Accessed |
C. | Updated |
D. | Stored |
Answer» E. | |
200. |
The time required to locate and retrieve the data from the storage unit is |
A. | Access time |
B. | Access mode |
C. | Storage capacity |
D. | Storage cost |
Answer» B. Access mode | |