Q1: Consider a disk with the following specifications: rotation speed of 6000 RPM, average seek time of 5 milliseconds, 500 sectors/track, 512-byte sectors. A file has content stored in 3000 sectors located randomly on the disk. Assuming average rotational latency, the total time (in seconds, rounded off to 2 decimal places) to read the entire file from the disk is _______ (2024 SET-2)
(a) 12.25
(b) 29.55
(c) 32.56
(d) 42.12
Ans: (b)
Sol: Total Time(T) = No. of sectors * ( Avg time per sector for random access)
T= 3000 * ( Avg Seek Time + Avg Rotational Latency + Transfer Time)
T= 3000* ( 5ms + 10/2 ms + 1/50 ms)
T= 30.06 sec
Q2: Consider a 512 GB hard disk with 32 storage surfaces. There are 4096 sectors per track and each sector holds 1024 bytes of data. The number of cylinders in the hard disk is ____ (2024 SET-1)
(a) 1024
(b) 2048
(c) 4096
(d) 512
Ans: (c)
Sol: Given that:
- Hard disk capacity= 512GB = 29∗230 = 239 Byte
- Number of surface = 32 = 25
- Sector/track = 4096 = 212
- sector size/capacity = 1024 = 210 Byte
- Number of cylinder = x
we know that: Disk capacity = Number of surface * number of track/surface * number of sector/track * number of byte/sector
⇒ 239 = 25∗ x ∗ 212 ∗ 210
⇒ 239 = x ∗ 212+10+5
⇒ 239 = x ∗ 227
⇒ x = 239/227 = 212 = 4096
∴ the number of cylinders is 4096.
Q3: A magnetic disk has 100 cylinders, each with 10 tracks of 10 sectors. If each sector contains 128 bytes, what is the maximum capacity of the disk in kilobytes? (2020)
(a) 12,80,000
(b) 1280
(c) 1250
(d) 1,28,000
Ans: (c)
Sol: Magnetic disk capacity
= 100 x 10 x 10 x 128 = 1280000 Bytes = 12380000/1024 KB = 1250 KB
Q4: 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: (2018)
(a) 12
(b) 25
(c) 38
(d) 44
Ans: (d)
Sol: Bit String: D4FE2003:1101 0100 1111 1110 0010 0000 0000 0011
Number of vacant tracks: 18
Number of occupied tracks:14
Percentage occupied tracks: (14/32)∗100 = 43.75% ≃ 44%
Option d) is correct.
Q5: SATA is the abbreviation of (2017)
(a) Serial Advanced Technology Attachment
(b) Serial Advanced Technology Architecture
(c) Serial Advanced Technology Adapter
(d) Serial Advanced Technology Array
Ans: (a)
Q6: 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: (2016)
(a) 256 Mbyte, 19 bits
(b) 256 Mbyte, 28 bits
(c) 512 Mbyte, 20 bits
(d) 64 Gbyte, 28 bits
Ans: (a)
Sol: Answer is (A).
16 surfaces = 4 bits, 128 tracks = 7 bits, 256 sectors = 8 bits, sector size 512 bytes = 9 bits
Capacity of disk =24+7+8+9 = 228 = 256 MB
To specify a particular sector we do not need sector size, so bits required = 4 + 7 + 8 = 19.
Q7: 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 ? (2015)
(a) 300.5 ms
(b) 255.5 ms
(c) 255 ms
(d) 300 ms
Ans: (d)
Sol: Avg. time to transfer = Avg. seek time + Avg. rotational delay + Data transfer time
Avg Seek Time given that : time to move between successive tracks is 1 ms
time to move from track 1 to track 1 : 0 ms
time to move from track 1 to track 2 : 1 ms
time to move from track 1 to track 3 : 2 ms
..
..
time to move from track 1 to track 500 : 499 ms
Avg Seek time 
Avg Rotational Delay
RMP : 600
600 rotations in 60 sec
one Rotation takes 60/600 sec = 0.1 sec
Avg Rotational Delay = 0.1/2 { usually (Rotation time/2) is taken as Avg Roational Delay}
= .05 sec
= 50 ms
Data Transfer Time
One 1 Roatation we can read data on one complete track. = 100 × 500 = 50,000 B data is read in one complete rotation one complete rotation takes 0.1 s ( we seen above ) 0.1 → 50,000 bytes.
250 bytes → 0.1×250/50,000 = 0.5 ms
Avg. time to transfer = Avg. seek time + Avg. rotational delay + Data transfer time
= 249.5 + 50 + 0.5
= 300 ms
Q8: Six files F1, F2, F3, F4, F5 and F6 have 100, 200, 50, 80, 120, 150 records respectively. In what order should they be stored so as to optimize act. Assume each file is accessed with the same frequency (2015)
(a) F3, F4, F1, F5, F6, F2
(b) F2, F6, F5, F1, F4, F3
(c) F1, F2, F3, F4, F5, F6
(d) Ordering is immaterial as all files are accessed with the same frequency.
Ans: (a)
Sol: This is basically optimal storage on tapes problem.Greedy apprach is used to solve this problem.The files are to be stored sequentially on tape.To read a particular file we need to start from beginning of tape..here goal is to find such order of storage that cost to access file is minimum..in order to achieve the files are stored in increasing order of length
So in above eg files will be stored in order F3 F4 F1 F5 F6 F2
Q9: Consider a typical disk that rotates at 15000 rotations per minute (RPM) and has a transfer rate of 50 x 106 bytes/sec. If the average seek time of the disk is twice the average rotational delay and the controller's transfer time is 10 times the disk transfer time, the average time (in milliseconds) to read or write a 512-byte sector of the disk is ______. (2015 SET-2)
(a) 8.2
(b) 6.1
(c) 3.5
(d) 9.5
Ans: (b)
Sol: Average time to read/write = Avg. seek time + Avg. rotational delay + Effective transfer time
Rotational delay = 60/15 = 4 ms
Avg. rotational delay = (1/2) × 4 = 2 ms
Avg. seek time = 2×2 = 4 ms
Disk transfer time = (512 Bytes/50∗106 Bytes/sec) = 0.0102 ms
Effective transfer time = 10 × disk transfer time = 0.102 ms
So, avg. time to read/write = 4 + 2 + 0.0102 + 0.102 = 6.11 ms ≈ 6.1 ms
Q10: Consider a disk pack with a seek time of 4 milliseconds and rotational speed of 10000 rotations per minute (RPM). It has 600 sectors per track and each sector can store 512 bytes of data. Consider a file stored in the disk. The file contains 2000 sectors. Assume that every sector access necessitates a seek, and the average rotational latency for accessing each sector is half of the time for one complete rotation. The total time (in milliseconds) needed to read the entire file is ____________. (2015 SET-1)
(a) 12530
(b) 14020
(c) 17564
(d) 15642
Ans: (b)
Sol: Since each sector requires a seek,
Total time = 2000 (seek time + avg. rotational latency + data transfer time)
Since data transfer rate is not given, we can take that in 1 rotation, all data in a track is read. i.e., in 60/10000 = 6ms, 600 × 512 bytes are read. So, time to read 512 bytes = 6/600ms = 0.01ms
= 2000 × (4ms + 60 × 1000/2 × 10000 + 0.01)
= 2000 × (7.01 ms)
= 14020 ms.
Q11: In case of a DVD, the speed of data transfer is mentioned in multiples of? (2013)
(a) 150 KB/s
(b) 1.38 MB/s
(c) 300 KB/s
(d) 2.40 MB/s
Ans: (b)
Sol: Modern compact discs support a writing speed of 52X and higher, with some modern DVDs supporting speeds of 16X and higher. It is important to note that the speed of writing a DVD at 1X (1,385,000 bytes per second) is approximately 9 times faster compared to writing a CD at 1X (153,600 bytes per second). However, the actual speeds depend on the type of data being written to the disc.
Q12: Consider a hard disk with 16 recording surfaces (0-15) having 16384 cylinders (0-16383) and each cylinder contains 64 sectors (0-63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is (cylinder no., surface no., sector no.). A file of size 42797 KB is stored in the disk and the starting disk location of the file is (1200, 9, 40). What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner? (2013)
(a) 1281
(b) 1282
(c) 1283
(d) 1284
Ans: (d)
Sol: First convert ⟨1200,9,40⟩ into sector address.
(1200 × 16 × 64) + (9×64) + 40 = 1229416
Number of sectors to store file = (42797 KB)/512 = 85594
Last sector to store file = 1229416 + 85594 = 1315010
Now, do reverse engineering,
1315010/(16 × 64) = 1284.189453 (1284 will be cylinder number and remaining sectors = 194)
194/64 = 3.03125 (3 is surface number and remaining sectors are 2)
∴ ⟨1284, 3, 1⟩ is last sector address.
Correct Answer: D
Q13: Which RAID level gives block level striping with double distributed parity? (2011)
(a) RAID 10
(b) RAID 2
(c) RAID 6
(d) RAID 5
Ans: (c)
Sol: RAID 0 – striping
RAID 1 – mirroring
RAID 5 – striping with parity
RAID 6 – striping with double parity
RAID 10 – combining mirroring and striping
Hence, Option (C) RAID 6 is the Answer.
Q14: A fast wide SCSI-II disk drive spins at 7200 RPM, has a sector size of 512 bytes, and holds 160 sectors per track. Estimate the sustained transfer rate of this drive (2011)
(a) 576000 Kilobytes / sec
(b) 9600 Kilobytes / sec
(c) 4800 Kilobytes / sec
(d) 4800 Kilobytes / sec
Ans: (b)
Sol: Disk is making 7200 revolutions in 1 minute.
For one rotation it will take 60/7200 sec.
i.e. 1 revolution --------------------------- 1/120 sec
in one rotation it will read a track. i.e. 160 sectors * 512 Bytes.
In 1/120 sec it will read 160 * 512 bytes.
then in 1 sec it will read (160∗512)/(1/120) bytes
= 160 * 512 bytes * 120
= 9830400 bytes / sec
= 9830400/1024 KiloBytes / sec
= 9600 Kilobytes/Sec
Q15: An application loads 100 libraries at startup. Loading each library requires exactly one disk access. The seek time of the disk to a random location is given as 10ms. Rotational speed of disk is 6000rpm. If all 100 libraries are loaded from random locations on the disk, how long does it take to load all libraries? (The time to transfer data from the disk block once the head has been positioned at the start of the block may be neglected) (2011)
(a) 0.5s
(b) 1.5s
(c) 1.25s
(d) 1s
Ans: (b)
Sol: Disk access time = Seek time + Rotational Latency + Transfer time
(given that transfer time is neglected)
Seek time = 10 ms
Rotational speed = 6000 rpm
- 60s→6000 rotations
- 1 rotation → 60/6000s
- Rotational latency = (1/2) × (60/6000s) = 5 ms
Total time to transfer one library = 10+5 = 15 ms
∴ Total time to transfer 100 libraries = 100 × 15 ms = 1.5s
Correct Answer: B
Avg Rotational Delay
