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A disk unit has 24 recording surfaces. It has a total of 14000 cylinders. There is an average of 400 sectors per track. Each sector contains 512 bytes of data. What is the data transfer rate at a rotational speed of 7200 r.p.m?
1. 68.80 × 106 bytes/s
2. 24.58 × 106 bytes/s
3. 98.80 × 103 bytes/s
4. 24.58 × 103 bytes/s

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Correct Answer - Option 2 : 24.58 × 106 bytes/s

Concept:

In secondary memory, the magnetic disk is used to store data on the disk surface and the data transfer rate can be calculated at a rotational speed is:

Rd = ST × BS × RS

Where,

Rd = data transfer rate (bits per sec) on track

ST = total number of sectors per track

BS = total data per sector

RS = rotational speed of disk in revolution per sec

Calculation:

Given that:

ST = 400 sectors / track

BS = 512 bytes per sector

RS = 7200 r.p.m

\({R_d} = 7200 \times \frac{1}{{60}}\) 

\({R_d} = 120\;r.p.s\) 

∴ Rd = ST × BS × RS

= 400 × 512 × 120

= 24.58 × 106 bytes per sec

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