ISRO2011-38

Question

A fast wide SCSI-II disk drive spins at 7200 RPM, has a sector size of 512 bytes, and holds 160 sectors per track. Estimate the sustained transfer rate of this drive

• A. 576000 Kilobytes / sec
• B. 9600 Kilobytes / sec
• C. 4800 Kilobytes / sec
• D. 19200 Kilobytes / sec

Comments

given:

spines=7200 RPM,size=512 bytes,sectors=160

i will provide the answer shortly

check the answer option is given in kilobytes and sec so you have to modify two things that is given below

size=512 bytes=>0.5kb,rotation spin is=7200 min convert to sec(ie.,1/60)

=160*0.5kb

=80kb =>80*7200 RPM=576000 kb/min

to convert it into sec =>576000/60=9600kb/sec Answer is 9600 kb/sec option is[B]

Answer 1

Disk is making 7200 revolutions in 1 minute.

For one rotation it will take $\frac{60}{7200}$ sec.

i.e. 1 revolution ---------------------------  $\frac{1}{120}$ sec

in one rotation it will read a track. i.e. 160 sectors * 512 Bytes.

In $\frac{1}{120}$ sec  it will read 160 * 512 bytes.

then in 1 sec it will read $\frac{160 * 512}{\frac{1}{120}}$ bytes

= 160 * 512 bytes * 120

= 9830400 bytes / sec

= $\frac{9830400}{1024}$ KiloBytes / sec

= 9600 Kilobytes/Sec

Answer 2

Answer is B)

7200 rounds per 60 sec

1 round takes 0.00833 sec which is the rotation speed of drive

In one rotation, it can read 512*160 bytes 

Hence , transfer rate of the drive is

= total bytes read in one rotation/ rotation speed 

= 512*160/ 0.00833

= 9603.841 kilobytes per sec 

Comments

Mine is coming B) . even you can calculate. What is the answer?

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ohhhhhhhh i m sorry 1kb=1024..i m putting 1000 thats why i m getting wrong ans

ur rt

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Option B is right @asu

Answer 3

Answer is B)

7200 rounds per 60 sec

1 round takes 0.00833 sec which is the rotation speed of drive

In one rotation, it can read 512*160 bytes 

Hence , transfer rate of the drive is

= total bytes read in one rotation/ rotation speed 

= 512*160/ 0.00833

= 9603.841 kilobytes per sec = approx 9600 kbytes /sec

Answer 4

7200  rotations in 60 sec

1 rotation = ? sec

=> 60/7200

= 1/120 sec

 

Data on 1 track = 512 x 160 bytes

i.e 1 rotation reads 512 x 160 bytes

Therefore,

In 1/120 sec , R/W head reads 512x160 bytes

In 1 sec  =  512 x 160 x 120

               = 9600 K bytes

Ans = B