Q: 1 What is the maximum size of the disk for the following details?
No. of Cylinders : 1024
No. of Heads : 16 (two tracks per head)
Sector per Track : 63
Each Sector : 512 bytes
128 MB
528 MB
1056 MB
428 MB
[ Option C ]
To calculate the maximum disk size, we use the formula:
Disk Size = Cylinders × Heads × Sectors per Track × Bytes per Sector.
Here, the important detail is that each head has two tracks, so the effective number of heads becomes 16*2 = 32. This increases the total storage capacity accordingly.
Substituting the values, we get, 1024*32*63*512 = 1,056,964,608 bytes. This represents the total number of bytes that can be stored on the disk based on the given configuration.
Now converting bytes into megabytes using decimal notation (1 MB = 106 bytes), we get approximately 1056 MB.
Q: 2
Match the following—
| List – I | List – II |
|---|---|
| A. Disk Scheduling | 1. Round Robin |
| B. Batch Processing | 2. Scan |
| C. Time Sharing | 3. LIFO |
| D. Interrupt Processing | 4. FIFO |
A-3, B-4, C-2, D-1
A-4, B-3, C-2, D-1
A-2, B-4, C-1, D-3
A-3, B-4, C-1, D-2
[ Option C ]
Disk scheduling algorithms manage the order in which disk I/O requests are processed. Common methods are FCFS (FIFO), SSTF, SCAN, LOOK, etc. The SCAN (also called Elevator algorithm) which moves the disk arm across the disk servicing requests in one direction.
In Batch Processing, jobs are processed in the order they arrive or according to some simple sequence. A common scheduling method is FIFO (First In First Out), where the first job to arrive is the first to be processed.
Time-sharing systems use Round Robin Scheduling to give each process a fair share of CPU.
Interrupts are handled in the reverse order of occurrence, often following a LIFO (Last In First Out) scheme where the most recent interrupt is serviced first.
Q: 3 The smallest amount of information that can be read from or written to the disk is __________.
Block
Sector
Track
Cylinder
[ Option B ]
In disk storage, data is organized into structures like tracks, sectors, and cylinders. The Sector is the smallest unit of data that can be physically read from or written to a disk.
Q: 4 Consider a disk queue with requests for Input/Output to blocks on cylinders in the given order. The disk head is initially at cylinder 53.
98, 183, 37, 122, 14, 124, 65, 67
What will be the number of total head movement for SSTF and SCAN algorithms respectively?
236, 208
208, 236
208, 208
236, 236
[ Option A ]
Disk scheduling algorithms decide the order in which I/O requests are serviced to minimize disk head movement. Two common algorithms are:
SSTF (Shortest Seek Time First):
SCAN (Elevator Algorithm):
Q: 5 Which of the following is NOT a characteristic of RAID?
RAID is a set of physical disk drives viewed by the OS as a single logical drive.
Data are distributed across physical drives of an array in a scheme known as striping.
Redundant disk capacity is used to stores parity information, which guaranteeing data recoverability in case of disk failure.
RAID is primarily used to compress and encrypt data across multiple disks.
[ Option D ]
RAID (Redundant Array of Independent Disks) is a storage technology that combines multiple physical disks into a single logical unit to improve performance and reliability (fault tolerance). Key characteristics of RAID include:
Q: 6 If there are nine 1 TB drives and one 500GB drive, what is the size of RAID5?
1TB
500GB
4.5GB
4.5TB
[ Option D ]
In RAID 5, data and parity information are distributed across all drives. The usable storage capacity is calculated using the formula:
RAID 5 Capacity = (N-1)*Size of Smallest Drive
In the given question, there are a total of 10 drives (Nine 1 TB drives and one 500 GB drive), and since RAID 5 always considers the smallest disk size, all drives are effectively treated as 500 GB each for capacity calculation.
So,
Q: 7 A hard disk has a rotational speed of 6000 RMP. Its average latency time is?
5*10-3 sec
0.05 sec
1 sec
0.5 sec
[ Option A ]
The rotational speed of the hard disk is 6000 RPM, which means the disk completes 6000 rotations in one minute. To find the time for one complete rotation, we divide 60 seconds by 6000, giving 60/6000=0.01 seconds per rotation.
The average rotational latency is the time the system waits for the required sector to come under the read/write head. On average, this waiting time is half of one full rotation.
So, the average latency time is 0.01/2=0.005 seconds, which can also be written as 5*10-3 seconds.
Q: 8 Consider a disk queue with request for I/O to blocks on cylinders 98, 183, 37, 122, 14, 124, 65, 67. Consider the Shortest Seek Time First (SSTF) algorithm for disk scheduling. If the initial head position is at 53, the total head movement in cylinders is
640
236
208
450
[ Option B ]
In SSTF (Shortest Seek Time First), at every step we choose the request which is closest to the current head position.
Given:
At each step, keep track of:
Step 1: Head at 53
Pending: {98, 183, 37, 122, 14, 124, 65, 67}
Distances from 53:
Closest: 65 (Distance 12).
Movement: 53 → 65
Remaining: 98, 183, 37, 122, 14, 124, 67
Step 2: Head at 65
Pending: {98, 183, 37, 122, 14, 124, 67}
Distances from 65:
Closest: 67 (Distance 2).
Movement: 65 → 67
Remaining: 98, 183, 37, 122, 14, 124
Step 3: Head at 67
Pending: {98, 183, 37, 122, 14, 124}
Distances from 67:
Closest: 37 (Distance 30).
Movement: 67 → 37
Remaining: 98, 183, 122, 14, 124
Step 4: Head at 37
Pending: {98, 183, 122, 14, 124}
Distances from 37:
Closest: 14 (Distance 23).
Movement: 37 → 14
Remaining: 98, 183, 122, 124
Step 5: Head at 14
Pending: {98, 183, 122, 124}
Distances from 14:
Closest: 98 (Distance 84).
Movement: 14 → 98
Remaining: 183, 122, 124
Step 6: Head at 98
Pending: {183, 122, 124}
Distances from 98:
Closest: 122 (Distance 24).
Movement: 98 → 122
Remaining: 183, 124
Step 7: Head at 122
Pending: {183, 124}
Distances from 122:
Closest: 124 (Distance 2).
Movement: 122 → 124
Remaining: 183
Step 8: Head at 124
Pending: {183}
Distances from 124:
Movement: 124 → 183
Total Head Movement:
12+2+30+23+84+24+2+59 = 236 Cylinders.
Q: 9 Consider a hard disk with 16 recording surface (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?
1284
1282
1286
1288
[ Option A ]
Each sector of the disk stores 512 bytes, which means 1 KB = 2 sectors. Therefore, a file of size 42797 KB requires 42797*2 = 85594 sectors in total.
The disk has 16 recording surfaces and 64 sectors per surface, so each cylinder contains 16*64 = 1024 sectors. This means that once 1024 sectors are filled, the storage moves to the next cylinder.
The file starts at disk location <1200, 9, 40>. The starting sector number within cylinder 1200 is calculated as (9*64)+40 = 616. Hence, the number of free sectors remaining in cylinder 1200 is 1024-616 = 408 sectors.
After filling these 408 sectors, the remaining sectors to be stored are 85594-408 = 85186 sectors. These remaining sectors occupy 83 complete cylinders because 83*1024 = 84992 sectors.
Since some sectors are still left after filling these 83 cylinders, they go into the next cylinder, which is cylinder number 1284.
Q: 10 Which of the following is a disk scheduling policy ensuring selection of the disk I/O request that requires the least movement of the disk arm from its current position?
Shortest-service-time-first (SSTF)
First In First Out (FIFO)
SCAN algorithm
Back and forth over disk
[ Option A ]
In an operating system, multiple processes often request disk I/O operations at the same time. To manage these requests efficiently, the OS uses Disk Scheduling Algorithms, which decide the order in which the disk head (arm) moves to service requests. The main goal is to reduce Seek Time, which is the time taken by the disk arm to move to the track where data is located.
Among the given options, the Shortest-Service-Time-First (SSTF) algorithm selects the disk I/O request that is closest to the current position of the disk arm. This means the arm moves the least possible distance, which minimizes seek time and improves performance.
| ALGORITHM | DESCRIPTION |
|---|---|
| FIFO (First In First Out) | Serves requests in the order they arrive. Simple but may cause long seek times. |
| SSTF (Shortest-Service-Time-First) | Chooses the request nearest to the current head position. It reduces seek time but may cause starvation. |
| SCAN (Elevator Algorithm) | Moves arm in one direction, serving requests, then reverses. It provides fairness and reduces variance. |
Q: 11 A wrong sentence related to FAT 32 and NTFS file system is?
FAT 32 has lower disk utilization compared to NTFS
Read and write speeds of NTFS are faster than that of FAT 32
FAT 32 store individual files of size up to 32 GB
NTFS stands for New Technology File System
[ Option C ]
Q: 12 The time required to move the read/write head to the desired track of a disk is called
Latency
Access Time
Seek Time
Page Fault
[ Option C ]
In a disk system, data is stored on circular tracks, and a read/write head is used to access this data. When a request is made, the head must first move to the correct track where the data is located. The time taken for this movement is an important performance factor.
Seek Time is defined as the time required to move the read/write head to the desired track.
| TERM | EXPLANATION |
|---|---|
| Latency | Time taken by the disk to rotate and bring the required data under the read/write head. |
| Access Time | Total time to access data. It includes seek time, latency, and transfer time. |
| Page Fault | Occurs in virtual memory when required data is not found in RAM. |
Q: 13 Match RAID levels in Column I with their description in Column II:
| Column – I | Column – II |
|---|---|
| a. RAID 2 | i. Bit-interleaved parity |
| b. RAID 3 | ii. Mirrored |
| c. RAID 5 | iii. Block-interleaved distributed parity |
| d. RAID 1 | iv. Redundant via Hamming Code |
Choose the correct option:
a – iii, b – i, c – ii, d – iv
a – ii, b – iv, c – iii, d – i
a – iv, b – i, c – iii, d – ii
a – iv, b – iii, c – i, d – ii
[ Option C ]
RAID (Redundant Array of Independent Disks) is a storage technology that combines multiple physical disks into one logical unit for better performance, reliability, or both. Each RAID level has a specific method for distributing and storing data and parity across disks.
| RAID LEVEL | DESCRIPTION | KEY FEATURE |
|---|---|---|
| RAID 0 | Disk Striping, splits data across all disks. | No redundancy, highest performance. |
| RAID 1 | Mirrored. | Data is duplicated on two disks for fault tolerance. |
| RAID 2 | Redundant via Hamming Code. | Uses error-correcting (Hamming) code for fault tolerance. |
| RAID 3 | Bit-interleaved parity. | Data is striped at the byte level with a dedicated parity disk. |
| RAID 5 | Block-interleaved distributed parity. | Data and parity information are distributed across all disks. |
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