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Solution

Total capacity $=$ Number of tracks $\times$ Number of sectors $\times$ Bytes per sector

$163840=80\times \text{Number of sectors}\times 128$

$\text{Number of sectors}=\frac{163840}{80\times128}$

$\text{Number of sectors}=16$

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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.

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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,

$(T,S)=(25K,50000K)$

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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:

$50000K\times 10=500000K$

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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.

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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

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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

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Solution

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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

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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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Solution

DRAM stores data using capacitors.

Capacitors lose charge with time, so DRAM requires refreshing.

Statement 1:

DRAM is slower than SRAM.

So, statement 1 is true.

Statement 2:

DRAM uses fewer components per bit, so its packing density is higher than SRAM.

So, statement 2 is true.

Statement 3:

DRAM is not faster than SRAM.

So, statement 3 is false.

Statement 4:

DRAM requires data refreshing.

So, statement 4 is true.

Therefore, correct statements are:

$1,2,4$

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Solution