Secondary-Memory

Question 1
A particular disk unit uses a bit string to record the occupancy or vacancy of its tracks, with 0 denoting vacant and 1 for occupied. A 32-bit segment of this string has hexadecimal value D4FE2003. The percentage of occupied tracks for the corresponding part of the disk, to the nearest percentage is
A
12
B
25
C
38
D
44
       Computer-Organization       Secondary-Memory       ISRO-2018
Question 2
Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors per track. 512 bytes of data are stored in a bit serial manner in a sector. The capacity of the disk pack and the number of bits required to specify a particular sector in the disk are respectively:
A
256 Mbyte, 19 bits
B
256 Mbyte, 21 bits
C
512 Mbyte, 20 bits
D
64 GB, 28 bits
       Computer-Organization       Secondary-Memory       ISRO-2016
Question 2 Explanation: 
→ Given that the disk pack has 16 surfaces, 128 tracks per surface, 256 sectors per track and each sector size is 512 bytes.
→ So the disk pack capacity = 16*128*256*512 bytes = 256 MB
→ To specify a sector we need the information about surface number, track number and sector number within a track.
→ Surface number needs 4 bits as there are 16 surfaces(24), track number needs 7 bits as there are 128 tracks(27) within a surface, within a track the sector number needs 8 bits as there are 256 sectors (28).
→ Total number bits needed to specify a particular sector = 4+7+8 = 19 bits.
Question 3
There are 200 tracks on a disc platter and the pending requests have come in the order – 36, 69, 167, 76, 42, 51, 126, 12 and 199. Assume the arm is located at the 100 track and moving towards track 199. If the sequence of disc access is 126, 167, 199, 12, 36, 42, 51, 69 and 76 then which disc access scheduling policy is used?
A
Elevator
B
Shortest seek-time first
C
C-SCAN
D
First Come First Served
       Computer-Organization       Secondary-Memory       ISRO CS 2014
Question 3 Explanation: 
Total tracks are :200
Tracks order is 36, 69, 167, 76, 42, 51, 126, 12 and 199
Track position is at 100 and moving towards 199.
sequence of disc access is 126, 167, 199, 12, 36, 42, 51, 69 and 76
From the above sequence , we can observe that disc moving from 100 to 126 from there to moving right side direction until last disc 199
From the last disc(199) to first disc(12) and from there to next disc(36) and so on.
C-SCAN sweeps the disk from end-to-end, but as soon it reaches one of the end tracks it then moves to the other end track without servicing any requesting location. As soon as it reaches the other end track it then starts servicing and grants requests headed to its direction
Question 4

A Computer uses a memory unit with 256K word of 32 bits each. A binary instruction code is stored in one word of memory. The instruction has four parts: an indirect bit, an operation code and a register code part to specify one of 64 registers and an address part. How many bits are there in operation code, the register code part and the address part?

A
7, 7, 18
B
18, 7, 7
C
7, 6, 18
D
6, 7, 18
       Computer-Organization       Secondary-Memory       UGC-NET CS 2018 DEC Paper-2
Question 4 Explanation: 
An instruction size is given as 32-bits.
Now, the instruction is divided into four parts :
An indirect bit
Register code part : Since number of registers given as 64(26) so to identify each register uniquely 6-bits are needed.
Address part : 256K(218) word memory is mentioned so to identify each word uniquely 18-bits are needed.
Operation code:
Size of Operation code = Complete instruction size - (size of indirect bit + size of register code + size of address part)
Size of Operation code = 7-bits
Question 5
What is the average access time for a Drum rotating at 4000 revolutions per minute?
A
2.5 milliseconds
B
5.0 milliseconds
C
7.5 milliseconds
D
4.0 milliseconds
       Computer-Organization       Secondary-Memory       Nielit Scientist-B CS 22-07-2017
Question 5 Explanation: 
→ 4000 revolution per minute means 60/4000 seconds for 1 revolution.
→ Average access time is time to complete 1/2 revolution= 7.5milliseconds.
Question 6
A disk has 200 tracks(numbered 0 through 199). At a given time, it was servicing the request of reading data from track 120 and at the previous request, service was for track 90. The pending requests(in order of their arrival) are for track numbers 30 70 115 130 110 80 20 25. How many times will the head change its direction for the disk scheduling policies SSTF(Shortest Seek Time First) and FCFS(First come first serve)?
A
2 and 3
B
3 and 3
C
3 and 4
D
4 and 4
       Computer-Organization       Secondary-Memory       Nielit Scientist-B CS 22-07-2017
Question 6 Explanation: 
According to Shortest Seek Time First:
90-> 120-> 115-> 110-> 130-> 80-> 70-> 30-> 25-> 20
Change of direction(Total 3); 120->15; 110->130; 130->80
According to First Come First Serve:
90-> 120-> 30-> 70-> 115-> 130-> 110-> 80-> 20-> 25
Change of direction(Total 4); 120->30; 30->70; 130->110;20->25
Question 7
When we move from the outermost track to the innermost track in a magnetic disk, then density(bits per linear inch)
A
increases
B
decreases
C
remains the same
D
either remains constant or decreases
       Computer-Organization       Secondary-Memory       Nielit Scientist-B CS 2016 march
Question 7 Explanation: 
→ we move from the outermost track to the innermost track in a magnetic disk, then density(bits per linear inch) increases
→ The density, in bits per linear inch, increases as we move from the outermost track to the innermost track (this same phenomenon is present in a phonograph record).
Question 8
Where does the swap reside?
A
RAM
B
ROM
C
DISK
D
On-chip cache
       Computer-Organization       Secondary-Memory       Nielit Scientist-B CS 4-12-2016
Question 8 Explanation: 
●Swap space is an area on disk that temporarily holds a process memory image.
● When memory is full and process needs memory, inactive parts of process are put in swap space of disk.
Question 9
A hard disk system has the following parameters : Number of tracks = 500 Number of sectors/track = 100 Number of bytes /sector = 500 Time taken by the head to move from one track to adjacent track = 1 ms Rotation speed = 600 rpm. What is the average time taken for transferring 250 bytes from the disk ?
A
300.5 ms
B
255.5 ms
C
255 ms
D
300 ms
       Computer-Organization       Secondary-Memory       ISRO CS 2015
Question 9 Explanation: 
Avg. time to transfer = Avg. seek time + Avg. rotational delay + Data transfer time
For Avg. seek time:
Given that, time to move between successive tracks is 1ms.
Time to move from track1 to track1 = 0ms
Time to move from track1 to track2 = 1ms
Time to move from track1 to track3 = 2ms
⋮ Time to move from track1 to track500 = 499ms
∴ Avg. seek time = 0+1+2+...+499/500 = 249.5ms
Avg. rotational delay:
600 rotations in 60sec.
One rotation takes 60/600 s = 100ms ∴ Avg. rotational delay = 100/2 = 50ms

Data transfer time:
In one rotation, we can read data on one complete track
= 100 × 500 = 50,000B data is read in one complete rotation
One complete rotation takes 100ms.
100ms → 50,000B
250B → 100/50000 × 250 = 0.5ms
∴ Avg. time to transfer = 249.5 × 50 + 0.5 = 300ms
Question 10
____ are used to access data on secondary, sequential access stores, such as disks and tapes.
A
Sequences
B
Arrays
C
Records
D
Registers
       Computer-Organization       Secondary-Memory       KVS DEC-2013
Question 10 Explanation: 
● Registers is not secondary storage devices. So it is not correct
● Array occupies primary memory(RAM). So it also not correct.
● Records used to access data on secondary devices record by record sequentially.
Question 11
The positioning time or random access time, consists of two parts: The time necessary to move the disk arm to the desired cylinder, called
A
Seek time
B
Rotational latency
C
Flash drives
D
Transfer rate
       Computer-Organization       Secondary-Memory       KVS DEC-2013
Question 11 Explanation: 
Seek Time :​ The amount of time required to move the read/write head from its current position to desired track.
Rotational latency :​ The amount of time to rotate the track when the read/write head comes to desired sector position. In simple disk, rotational latency is the time to rotate 1⁄2 disk to the access.
Question 12
The following data refers to a hard disk; number of tracks per side=600; number of sides=2; Number of bytes per sector=512; Storage capacity in bytes=21 504 000 Determine the number of sectors per track for this hard disk
A
35
B
40
C
45
D
50
       Computer-Organization       Secondary-Memory       KVS DEC-2013
Question 12 Explanation: 
Storage capacity in bytes= (number of tracks per side*number of sides*Number of bytes per sector*number of sectors per track)
Here, we have to find number of sectors per track.
Total storage capacity / (Number of bytes per sector*number of sides*number of tracks per side)
=21504000 / (512*2*600)
=35
Question 13
On which of the following storage media, storage of information is organized as a single continuous spiral groove?
A
CD-ROM
B
RAM
C
Hard disk
D
Floppy disk
       Computer-Organization       Secondary-Memory       KVS DEC-2017
Question 13 Explanation: 
Compact Disk: ​ original physical design for audio
Question 14
____ refers to the amount of time required to position the read write head of a hard disk on appropriate sector
A
Load Time
B
Seek time
C
Access time
D
Rotational latency
       Computer-Organization       Secondary-Memory       KVS DEC-2017
Question 14 Explanation: 
Seek time → It measures the time it takes the head assembly on the actuator arm to travel to the track of the disk where the data will be read or written. The data on the media is stored in sectors which are arranged in parallel circular tracks (concentric or spiral depending upon the device type) and there is an actuator with an arm that suspends a head that can transfer data with that media.
→ ​ Average seek time​ is the average of all possible seek times which technically is the time to do all possible seeks divided by the number of all possible seeks, but in practice it is determined by statistical methods or simply approximated as the time of a seek over one third of the number of tracks.
Rotational latency (sometimes called rotational delay or just latency​ :
It is the delay waiting for the rotation of the disk to bring the required disk sector under the read-write head. It depends on the rotational speed of a disk (or spindle motor), measured in revolutions per minute (RPM).
→ Maximum latency = 60/rpm
→ Average latency = 0.5*Latency Time
Transfer Time Transfer time is the time to transfer the data. It depends on the rotating speed of the disk and number of bytes to be transferred. Disk Access Time Disk Access Time = Seek Time + Rotational Latency + Transfer Time
Question 15
The maximum amount of information that is available in one portion of the disk access arm for a removal disk pack (without further movement of the arm with multiple heads)
A
a plate of data
B
a cylinder of data
C
a track of data
D
a block of data
       Computer-Organization       Secondary-Memory       UGC NET CS 2011 Dec-Paper-2
Question 15 Explanation: 
A cylinder of data having maximum amount of information that is available in one portion of the disk access arm for a removal disk pack (without further movement of the arm with multiple heads).
Question 16
The circumference of the two concentric disks are divided into 100 sections each. For the outer disk, 100 of the sections are painted red and 100 of the sections are painted blue.For the inner disk, the sections are painted red and blue in an arbitrary manner. It is possible to align the two disks so that____of the sections on the inner disks have their colours matched with the corresponding section on outer disk.
A
100 or more
B
125 or more
C
150 or more
D
175 or more
       Computer-Organization       Secondary-Memory       UGC NET CS 2011 June-Paper-2
Question 16 Explanation: 
→ We fix the larger disk first, then place the smaller disk on the top of the larger disk so that the centers and sectors coincide. There are 200 ways to place the smaller disk in such a manner.
→ For each such alignment, some sectors of the two disks may have the same color. Since each sector of the smaller disk will match the same color sector of the larger disk 100 times among all the 200 ways and there are 200 sectors in the smaller disk, the total number of matched color sectors among the 200 ways is 100 × 200 = 20,000.
→ Note that there are only 200 ways. Then there is at least one way that the number of matched color sectors is 20,000 / 200 = 100 or more.
Question 17
Optical storage is a
A
high-speed direct access storage device.
B
low-speed direct access storage device.
C
medium-speed direct access storage device.
D
high-speed sequential access storage device.
       Computer-Organization       Secondary-Memory       UGC NET CS 2010 Dec-Paper-2
Question 17 Explanation: 
Optical storage is a low-speed direct access storage device.
Question 18
For a magnetic disk with concentric circular tracks, the seek latency is not linearly proportional to the seek distance due to
A
Non-Uniform distribution of requests
B
arm starting or stopping inertia
C
higher capacity of tracks on the periphery of the plate
D
use of unfair arm scheduling policies
       Computer-Organization       Secondary-Memory       UGC NET June-2019 CS Paper-2
Question 18 Explanation: 
For a magnetic disk with concentric circular tracks, the seek latency is not linearly proportional to the seek distance due to arm starting or stopping inertia.
Question 19
A magnetic disk has 100 cylinders, each with 10 tracks of 10 sector. If each sector contains 128 bytes, what is the maximum capacity of the disk in kilobytes?
A
1,280,000
B
1280
C
1250
D
128,000
       Computer-Organization       Secondary-Memory       ISRO CS 2020
Question 19 Explanation: 
There are 100 cylinders, each with 10 tracks, and 10 sectors. Each sector is 128 bytes.
Maximum capacity of the disk = 100*10*10*128 bytes
= 10^4 * 128 bytes
= (128*10^4 / 1024) KB
= (10^4/ 8) = 1250 KB
Question 20
Consider a hard disk containing 1000cylinders, 10 platters each with 2 recording surfaces and 63 sectors per track. What is the position of the sector whose 3-D disk address is 2000,15,55> representing cylinder, surface, sector numbers respectively.
A
3520
B
253000
C
50600
D
250600
       Computer-Organization       Secondary-Memory       APPSC-2016-DL-CS
Question 20 Explanation: 
Here each cylinder has 2 recording surfaces and each surface has 63 sectors. Now 2000 cylinders means required sector have (2000*2*63) sectors + 15 surface means in addition to previous sectors required sector have (15* 63) sectors +55 sectors before itself.
Hence, Position of sector < 2000, 15, 55>
(2000263)+(1563)+55
=252000+945+55
=253000
Question 21

Consider the disk which has average seek time of 32 ns and rotational rate of 360 rpm (round per minute), each track of the disk has 512 sectors, each of size 512 bytes.
What is the time taken to read four continuous sectors? And what is the data transfer rate?

A
0.0443s and 936 Kbps
B
0.0843s and 1536 Kbps
C
0.1043s and 1736 kbps
D
0.0943s and 1636 kbps
       Computer-Organization       Secondary-Memory       CIL 2020
Question 22
Using a larger block size in a fixed block size file system leads to :
A
Better disk throughout but poor disk space utilization
B
Better disk throughout and better disk space utilization
C
Poor disk throughput but better disk space utilization
D
Poor disk throughput and poor disk space utilization
       Computer-Organization       Secondary-Memory       TNPSC-2017-Polytechnic-CS
Question 22 Explanation: 
Larger block size means less number of blocks to fetch and hence better throughput. But larger block size also means space is wasted when only small size is required.
There are 22 questions to complete.
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