SecondaryMemory
Question 1 
A hard disk has 63 sectors per track, 10 platters each with 2 recording surfaces and 1000 cylinders. The address of a sector is given as a triple , where c is the cylinder number, h is the surface number and s is the sector number. Thus, the 0^{th} sector is addressed as <0, 0, 0>, the 1^{st} sector as <0,0,1>, and so on.
The address of the 1039^{th} sector is
The address of the 1039^{th} sector is
〈0, 15, 31〉
 
〈0, 16, 30〉  
〈0, 16, 31〉  
〈0, 17, 31〉 
Question 1 Explanation:
We know in each track there are 63 sectors. And we know there is one track per surface.
From the given options we can calculate the sector numbers as
Option A  15*63+31 = 976
Option B  16*63+30 = 1038
Option C  16*63+31 = 1039
Option D  17*63+31 = 1102
Hence Option C is the answer.
From the given options we can calculate the sector numbers as
Option A  15*63+31 = 976
Option B  16*63+30 = 1038
Option C  16*63+31 = 1039
Option D  17*63+31 = 1102
Hence Option C is the answer.
Question 2 
Using a larger block size in a fixed block size file system leads to :
Better disk throughout but poor disk space utilization  
Better disk throughout and better disk space utilization  
Poor disk throughput but better disk space utilization  
Poor disk throughput and poor disk space utilization 
Question 2 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.
Question 3 
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?
1,280,000  
1280  
1250  
128,000 
Question 3 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
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 4 
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 3D disk address is 2000,15,55> representing cylinder, surface, sector numbers respectively.
3520  
253000  
50600  
250600 
Question 4 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
Hence, Position of sector < 2000, 15, 55>
(2000263)+(1563)+55
=252000+945+55
=253000
Question 5 
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:
256 MB, 19b  
256MB,8b  
512 MB, 20b  
64 GB, 28b 
Question 5 Explanation:
Let first find capacity of disk,
16 surfaces × 128 tracks × 256 sectors/track × 512 bytes/sector
= 2^{4} × 2^{7 }× 2^{8 }× 2^{9}
= 2^{28}
= 256 MB
Now let's find no. of bits required to specify a particular sector,
16 surfaces × 128 tracks × 256 sectors/track
= 2^{4} × 2^{7} × 2^{8} = 2^{19}
Hence 19 bits.
16 surfaces × 128 tracks × 256 sectors/track × 512 bytes/sector
= 2^{4} × 2^{7 }× 2^{8 }× 2^{9}
= 2^{28}
= 256 MB
Now let's find no. of bits required to specify a particular sector,
16 surfaces × 128 tracks × 256 sectors/track
= 2^{4} × 2^{7} × 2^{8} = 2^{19}
Hence 19 bits.
Question 6 
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?
0.0443s and 936 Kbps
 
0.0843s and 1536 Kbps
 
0.1043s and 1736 kbps  
0.0943s and 1636 kbps

Question 7 
Comprehension:
Question 9195 concerns a disk with a sector size of 512 bytes, 2000 tracks per surface. 50 sectors per track, five doublesided platters, and average seek time of 10 milliseconds.
Q91: If one track of data can be transferred per revolution, then what is the data transfer rate?
2,850 KBytes/second  
4,500 KBytes/second  
5,700 KBytes/second  
2,250 KBytes/second 
Question 7 Explanation:
It is given that the disk platters rotate at 5400 RPM.
In 60s  5400 rotations Time for one rotation = 60s/5400 = (1/90) seconds In (1/90)seconds one track size can be transferred. Each track is of size 25KB. In 1/90 seconds 25KB can be transferred. In 1 seconds the amount of data that can be transferred = 25KB*90 = 2250KB So the data transfer rate = 2250 KB/s
In 60s  5400 rotations Time for one rotation = 60s/5400 = (1/90) seconds In (1/90)seconds one track size can be transferred. Each track is of size 25KB. In 1/90 seconds 25KB can be transferred. In 1 seconds the amount of data that can be transferred = 25KB*90 = 2250KB So the data transfer rate = 2250 KB/s