GATE CSE 1993 | Question: 6.7

Question

A certain moving arm disk storage, with one head, has the following specifications:

• Number of tracks/recording surface $= 200$
• Disk rotation speed $= 2400$ rpm
• Track storage capacity $= 62,500$ bits

The average latency of this device is $\text{P}$ ms and the data transfer rate is $\text{Q}$ bits/sec. Write the values of $\text{P}$ and $\text{Q}$.

Comments

https://gateoverflow.in/2289/gate1993-6-7

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After the syllabus change in 2016 is this topic currently within the syllabus?

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$\\ 1.\ Avg \ lateny:2400R\rightarrow 60sec\\ \\ 1R\rightarrow \dfrac{1}{40}sec\\ \\ \dfrac{1}{2}R\rightarrow \dfrac{1}{80}sec=12.5ms\\ \\ \\ $

$2.\ Data \ transfer\ rate: \\ 1R\rightarrow \dfrac{1}{40}sec\\ \\ \dfrac{1}{40}sec\rightarrow 62,500\ bits\\ \\ 1sec\rightarrow 40 \times 62,500 bits \\ \\ means\ 2500000\ bits/sec\\$

Answer 1

1.
 In 60 seconds, the disk rotates 2400 times. So it takes 25ms to rotate it one time. There are 200 tracks. On average, 100 tracks need to be traveresed to reach a specific track.So 25ms*100 tracks= 2.5s time is taken on average to reach a particular track on avg.

2.
In 25 ms, one rotation is completed and all the 62500 bits are transferred. So in 1ms, 2500 bits are transferred. Accordingly, in 1s, 2.5Mb are transferred. Thats why the speed is 2.5Mb/s