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Andrew S. Tanenbaum (OS) Edition 4 Exercise 5 Question 17 (Page No. 431)

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Answer:

$7200\;\text{rpm} = 60\;\text{sec} \\ \Rightarrow 1 \;\text{rotation} = \frac{1000}{120}\;\text{msec}$

Now, $\dfrac{\frac{1000}{120}}{200}\;\text{sectors} = \frac{1}{24}\;\text{msec}$

$\therefore \;\text{Cylinder skew} = \;\frac{24\;\text{secotrs passed}}{ {1\;\text{msec}}} = 24$

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