Question

A disc drive has a average seek time of 10ms, 32 sectors on each track and 512 bytes per sector. if the average time to read 8kbytes of continuously stored data is 20ms, what is the rotational speed of the disc drive? 3600 rpm 6000 rpm 3000 rpm 2400 rpm

Answer 1

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ANSWER : 3000 rpm

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SOLUTION:
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8192 bytes / 512 bytes per sector = 16 sectors

3000 rpm / 60 sec = 50 rps

50 rps * 32 sectors = 1600 sectors / sec

time to read 16 sectors is 16 / 1600 = 0.01 sec

0.01 sec to ms = 10 

adding 10 ms for seek = 20 ms

Answer: 3000 rpm

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Answer 2

Answer:

6000 rpm

Explanation:

average time required  = seek time + rotational latency + transfer time

seek time = 10 ms

rotational latency: suppose x rotations are made in 1 min(60 sec) i.e. x rpm

1 rotation time is 60/x

We know, rotational latency = 1/2 * time for 1 rotation

Therefore, rotational latency = 1/2 * 60/x = 30/x sec  

transfer time: tracks required to transfer 8 KB data  = (8 * 1024) / (32 * 512) = 1/2

1 track transfer is equivalent 1 rotation

So, transfer time = 1/2 * 60/x = 30/x sec

average time required = 20 ms (given) = 10 ms + 30/x sec + 30/x sec

Solving it will give,

10 ms  = 60/x sec

which implies, x = (60 * 1000 ms) / 10 ms = 6000 rpm