Test: File Systems-2 - Computer Science Engineering (CSE) MCQ

In C-SCAN disk scheduling algorithm, the disk head serve the request from ine end to other end but when it reaches to other end it immediately returns to the starting of the disk without serving any request. Refer:Disk Scheduling Algorithms So option (D) is correct.

For access time optimization sort the files in incresing order of file size. So, option (B) is correct.

Access to 50 pages will cause 50 page faults. When these pages are accessed in reverse order, the first four accesses will not cause page fault. All other access to pages will cause page faults. So total number of page faults will be 50 + 46 = 96. Option (A) is correct.

First In First Out Page Replacement Algorithms: This is the simplest page replacement algorithm. In this algorithm, operating system keeps track of all pages in the memory in a queue, oldest page is in the front of the queue. When a page needs to be replaced page in the front of the queue is selected for removal.

FIFO Page replacement algorithms suffers from Belady’s anomaly :

Belady’s anomaly states that it is possible to have more page faults when increasing the number of page frames.

4, 34, 10, 7, 19, 73, 2, 15, 6, 20 Since shortest seek time first policy is used, head will first move to 34. This move will cause 16*2 ms. After 34, head will move to 20 which will cause 14*1 ms. And so on. So cylinders are accessed in following order 34, 20, 19, 15, 10, 7, 6, 4, 2, 73 and total time will be (16 + 14 + 1 + 4 + 15 + 3 + 1 + 1 + 1 + 71)*2 = 238 ms. So option (B) is correct.

1. Device controller performs data transfer.
2. Device driver process I/O request.
3. Inturrupt handler Extracts information from the controller register and store it in data buffer
4. Kernel I/O subsystem performs I/O scheduling./li> So, option (A) is correct.

The algorithm looks similar to C- Scan.

Option (C) is correct.

According to question file control block and the index block is already in memory. So, we don't need to transfer them from anywhere else for further operation. If a block is added at the end (and the block information to be added is stored in memory), this is a single operation. So we need only one I/O operation required for indexed (single-level) allocation strategy. So, option (A) is correct.

In case of a DVD, the speed of data transfer in a DVD is typically 1,385,000 bytes per second i.e. equal to 1.38 MB / sec Option (B) is correct.

An operating system has three main functions: (1) manage the computer's resources, such as the central processing unit, memory, and other input - output sources

(2) establish a user interface, and

(3) execute and provide services for applications software. OS provides an interface between the user and the hardware and thus making the computer easy to use for the user but the primary function of OS is to manage the hardware in the most efficient way. So, option (A) is correct.

Capacity of the disk = 16 surfaces X 128 tracks X 256 sectors X 512 bytes = 256 Mbytes. To calculate number of bits required to access a sector, we need to know total number of sectors. Total number of sectors = 16 surfaces X 128 tracks X 256 sectors = 2^19 So the number of bits required to access a sector is 19. Option (A) is correct.

Seek time is the time taken by the head to move to the correct track and Rotational latency is the time taken by the disk to rotate until the head is over the desired block to read So, the total time to prepare a disk drive mechanism for a block of data to be read from is the sum of both the seek time and the latency. Option (C) is correct.

Spooler is a software which works on creating a typical request queue where data, instructions and processes from multiple sources are accumulated for execution later on. Generally, it is maintained on computer’s physical memory, buffers or the I/O device-specific interrupts. The spool is processed in FIFO manner i.e. whatever first instruction is there in the queue will be popped and executed. Option (B) is correct.

Spooling works like a typical request queue where data, instructions and processes from multiple sources are accumulated for execution later on. Generally, it is maintained on computer’s physical memory, buffers or the I/O device-specific interrupts. The spool is processed in FIFO manner i.e. whatever first instruction is there in the queue will be popped and executed. Example: printer Option (A) is correct.

The arm is at cylinder 20, so the service order = 18, 25, 35, 39, 8, 5, 3. Seek time = (20−18) + (25−18) + (35−25) + (39−35) + (39−8) + (8−5) + (5−3) = 2 + 7 + 10 + 4 + 31 + 3 + 2 = 59 Total seek time = 59 * 5 = 295 Option (B) is correct.

The read and write functionality depends on the type of microcontroller peripheral. Generally, the status bits have a read only status and can be modified by peripherals only. So, read bit can be read by CPU and written by the peripheral. Option (D) is correct.

D4FE2003 = 1101 0100 1111 1110 0010 0000 0000 0011 Total bits = 32 Total bits which are occupied, i.e. with value 1 = 14 Percentage of occupied tracks = 14/32 * 100 = 43.75 = 44 So, option (D) is correct.

FCFS Total seek time in FCFS Scheduling when the disk drive is reading from cylinder 20 for cylinders in the order 10, 22, 20, 2, 40, 6, and 38 :
= (10 + 12 + 2 + 18 + 38 + 34 + 32)*6
= 146*6
= 876 ms
As no other option matches the answer, option (D) is correct.

Main memory = 1000K Job 1 requiring 200 K arrives Job 2 requiring 350 K arrives Job 3 requiring 300 K arrives and assuming continuous allocation: Free memory = 1000 − 850(200 + 350 + 300) = 150 K (till these jobs first fit and best fit are same) Since, job 1 finishes, Free memory = 200 K and 150 K Case 1: First fit Job 4 requiring 120 K arrives Since 200 K will be the first slot, so Job 4 will acquire this slot only. Remaining memory = 200 - 120 = 80 K Job 5 requiring 150 K arrives It will acquire 150 K slot Job 6 requiring 80 K arrives It will occupy 80 K slot, so, all jobs will be allocated successfully. Case 2: Best fit Job 4 requiring 120 K arrives It will occupy best fit slot which is 150 K. So, remaining memory = 150 − 120 = 30 K Job 5 requiring 150 K arrives It will occupy 200 K slot. So, free space = 200 − 150 = 50 K Job 6 requiring 80 K arrives There is no continuous 80 K memory free. So, it will not be able to allocate. So, first fit is better. Option (A) is correct.

According to Shortest Seek Time First (SSTF) is a disk scheduling algorithm:

Therefore, option (B) is False.