UGC NET Computer Science Computer Memory Back-up Devices Previous Year Questions (PYQs) – Page 1 of 2
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The total storage capacity of a floppy disk having $80$ tracks and storing $128$ bytes / sector is $163,840$ bytes. How many sectors does this disk have?
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Solution
Total capacity $=$ Number of tracks $\times$ Number of sectors $\times$ Bytes per sector
$163840=80\times \text{Number of sectors}\times 128$
$\text{Number of sectors}=\frac{163840}{80\times128}$
$\text{Number of sectors}=16$
$163840=80\times \text{Number of sectors}\times 128$
$\text{Number of sectors}=\frac{163840}{80\times128}$
$\text{Number of sectors}=16$
For a magnetic disk with concentric circular tracks, the seek latency is not linearly proportional to the seek distance due to:
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Solution
Seek latency is the time taken by the disk arm to move the read/write head to the required track.
The arm does not move with uniform speed throughout the seek operation.
It takes time to start moving and also takes time to stop at the required track due to inertia.
Because of this starting and stopping inertia, seek latency is not linearly proportional to the seek distance.
concern a disk with a sector size of $512$ bytes, $2000$ tracks per surface, $50$ sectors per track, five double-sided platters, and average seek time of $10$ milliseconds.
If $T$ is the capacity of a track in bytes, and $S$ is the capacity of each surface in bytes, then $(T,S)=$ _____
If $T$ is the capacity of a track in bytes, and $S$ is the capacity of each surface in bytes, then $(T,S)=$ _____
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Solution
Sector size $=512$ bytes
Number of sectors per track $=50$
So, capacity of one track is:
$T=512\times 50$
$T=25600$ bytes
Since,
$1K=1024$ bytes
So,
$T=\frac{25600}{1024}=25K$
Now, each surface has $2000$ tracks.
Capacity of each surface is:
$S=25K\times 2000$
$S=50000K$
Therefore,
$(T,S)=(25K,50000K)$
Number of sectors per track $=50$
So, capacity of one track is:
$T=512\times 50$
$T=25600$ bytes
Since,
$1K=1024$ bytes
So,
$T=\frac{25600}{1024}=25K$
Now, each surface has $2000$ tracks.
Capacity of each surface is:
$S=25K\times 2000$
$S=50000K$
Therefore,
$(T,S)=(25K,50000K)$
concern a disk with a sector size of $512$ bytes, $2000$ tracks per surface, $50$ sectors per track, five double-sided platters, and average seek time of $10$ milliseconds.
What is the capacity of the disk, in bytes?
What is the capacity of the disk, in bytes?
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Solution
Given,
Sector size $=512$ bytes
Sectors per track $=50$
Tracks per surface $=2000$
Number of double-sided platters $=5$
So, number of surfaces:
$5\times 2=10$
Capacity of one track:
$512\times 50=25600$ bytes
Since,
$1K=1024$ bytes
So,
$\frac{25600}{1024}=25K$
Capacity of one surface:
$25K\times 2000=50000K$
Capacity of total disk:
$50000K\times 10=500000K$
Sector size $=512$ bytes
Sectors per track $=50$
Tracks per surface $=2000$
Number of double-sided platters $=5$
So, number of surfaces:
$5\times 2=10$
Capacity of one track:
$512\times 50=25600$ bytes
Since,
$1K=1024$ bytes
So,
$\frac{25600}{1024}=25K$
Capacity of one surface:
$25K\times 2000=50000K$
Capacity of total disk:
$50000K\times 10=500000K$
concern a disk with a sector size of $512$ bytes, $2000$ tracks per surface, $50$ sectors per track, five double-sided platters, and average seek time of $10$ milliseconds.
Given below are two statements:
Statement I: The disk has a total number of $2000$ cylinders.
Statement II: $51200$ bytes is not a valid block size for the disk.
In the light of the above statements, choose the correct answer from the options given below:
Given below are two statements:
Statement I: The disk has a total number of $2000$ cylinders.
Statement II: $51200$ bytes is not a valid block size for the disk.
In the light of the above statements, choose the correct answer from the options given below:
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Solution
Statement I is true.
Number of cylinders is equal to the number of tracks per surface.
Given,
Tracks per surface $=2000$
So, total number of cylinders $=2000$.
Therefore, Statement I is true.
Statement II is false.
Given sector size is:
$512$ bytes
A valid block size should be a multiple of sector size.
Now,
$51200=512\times 100$
So, $51200$ bytes is a valid block size.
But Statement II says that $51200$ bytes is not a valid block size.
Therefore, Statement II is false.
Number of cylinders is equal to the number of tracks per surface.
Given,
Tracks per surface $=2000$
So, total number of cylinders $=2000$.
Therefore, Statement I is true.
Statement II is false.
Given sector size is:
$512$ bytes
A valid block size should be a multiple of sector size.
Now,
$51200=512\times 100$
So, $51200$ bytes is a valid block size.
But Statement II says that $51200$ bytes is not a valid block size.
Therefore, Statement II is false.
concern a disk with a sector size of $512$ bytes, $2000$ tracks per surface, $50$ sectors per track, five double-sided platters, and average seek time of $10$ milliseconds.
If the disk platters rotate at $5400$ rpm, then approximately what is the maximum rotational delay?
If the disk platters rotate at $5400$ rpm, then approximately what is the maximum rotational delay?
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Solution
Given,
Rotational speed $=5400$ rpm
It means:
$5400$ revolutions in $60$ seconds
So, revolutions per second:
$\frac{5400}{60}=90$
Time taken for one revolution:
$\frac{1}{90}$ seconds
$=0.0111$ seconds
Maximum rotational delay is equal to time for one full revolution.
So, maximum rotational delay is approximately:
$0.011$ seconds
Rotational speed $=5400$ rpm
It means:
$5400$ revolutions in $60$ seconds
So, revolutions per second:
$\frac{5400}{60}=90$
Time taken for one revolution:
$\frac{1}{90}$ seconds
$=0.0111$ seconds
Maximum rotational delay is equal to time for one full revolution.
So, maximum rotational delay is approximately:
$0.011$ seconds
concern a disk with a sector size of $512$ bytes, $2000$ tracks per surface, $50$ sectors per track, five double-sided platters, and average seek time of $10$ milliseconds.
If one track of data can be transferred per revolution, then what is the data transfer rate?
If one track of data can be transferred per revolution, then what is the data transfer rate?
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Solution
From previous calculation, capacity of one track is:
$25K$ bytes
Given rotational speed:
$5400$ rpm
Revolutions per second:
$\frac{5400}{60}=90$
If one track of data is transferred per revolution, then data transferred per second is:
$25K\times 90$
$=2250K$ bytes/second
So, data transfer rate is:
$2250$ KBytes/second
$25K$ bytes
Given rotational speed:
$5400$ rpm
Revolutions per second:
$\frac{5400}{60}=90$
If one track of data is transferred per revolution, then data transferred per second is:
$25K\times 90$
$=2250K$ bytes/second
So, data transfer rate is:
$2250$ KBytes/second
Which RAID level is appropriate for storing the log file of a ticket booking system where a huge database update is required?
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Solution
For log files, reliability is very important because logs help in recovery after failure.
RAID Level $1$ uses mirroring, so the same data is stored on another disk also.
Therefore, RAID Level $1$ is suitable for log files.
A magnetic tape drive has a transport speed of $200$ inches per second and a recording density of $1600$ bytes per inch. The time required to write $600000$ of data grouped in $100$ characters records with a blocking factor of $10$ is
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Solution
Given,
Tape transport speed $=200$ inches/second
Recording density $=1600$ bytes/inch
Total data $=600000$ bytes
First, find the data transfer rate.
Data transfer rate $=$ Transport speed $\times$ Recording density
Data transfer rate $=200\times1600$
Data transfer rate $=320000$ bytes/second
Now, time required to write data is:
$Time=\frac{\text{Total data}}{\text{Data transfer rate}}$
$Time=\frac{600000}{320000}$
$Time=1.875$ seconds
The records are grouped in $100$ characters with blocking factor $10$, so one block has:
$100\times10=1000$ characters
This grouping affects blocking arrangement, but the total amount of data to be written remains $600000$ bytes.
Therefore, required time is:
$1.875$ sec
Tape transport speed $=200$ inches/second
Recording density $=1600$ bytes/inch
Total data $=600000$ bytes
First, find the data transfer rate.
Data transfer rate $=$ Transport speed $\times$ Recording density
Data transfer rate $=200\times1600$
Data transfer rate $=320000$ bytes/second
Now, time required to write data is:
$Time=\frac{\text{Total data}}{\text{Data transfer rate}}$
$Time=\frac{600000}{320000}$
$Time=1.875$ seconds
The records are grouped in $100$ characters with blocking factor $10$, so one block has:
$100\times10=1000$ characters
This grouping affects blocking arrangement, but the total amount of data to be written remains $600000$ bytes.
Therefore, required time is:
$1.875$ sec
Consider a disk pack with $32$ surfaces, $64$ tracks and $512$ sectors per track. $256$ bytes of data are stored in a bit serial manner in a sector. The number of bits required to specify a particular sector in the disk is:
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Solution
Number of surfaces is
$32=2^5$
So, surface number needs $5$ bits.
Number of tracks is
$64=2^6$
So, track number needs $6$ bits.
Number of sectors is
$512=2^9$
So, sector number needs $9$ bits.
Therefore, total bits required are
$5+6+9=20$
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