Numerical -Magnetic disk

Question

A magnetic surface has $128$ tracks($T_{0}$ to $T_{127}$), One track has $256$ sectors and each sector capacity is $512B$ ($0.5KB$) motor operated with $3000RPM$. Amount of time required to move the head from current trak to it’s next successive track is $1$m.sec. Current position of head is $4$^th track$(T_{4})$ ans sector $25$ $(S_{25})$. The amount of time required to read $64KB$ data from (starting address) sector $153$ $(S_{153})$ of track 36 $(T_{36})$ in millisec is _______________ 

Comments

@srestha what i mean to say is while we are moving from track $4$ to $36$ the motor would also be rotating (don't know if it is the case, just my assumption, if this is not true then this is not true).

So, in $20\,ms$ we can make a full rotation, so as the time to reach track $36$ is $32\,ms$ so we would be at initial position i.e sector $25$ again before reaching track $36$. 

So now we are again at sector $25$ and time till we reach track $36$ is $12\,ms$ remaining.

Now, time to reach sector $153$ from where we will start reading, will be $10\,ms$ as calculated above. But, as the time to reach track $36$ is $12\,ms$ then it would have gone past sector $153$, so when we reach track $36$, we will let it rotate to reach $153$ again so, we can start reading.

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@Shobhit Joshi

my doubt is we take that the motor is not rotating while we moved from track 4 to 32, does this happen ?

yes that motor is not rotating.

Seek Time = Go to the particular Track on the surface where sector is preserved !

Read/write header is not rotating, it will move in Horizontal direction only, but Sectors should be preserved even disk in circle shape.

So, to reach 36th Track from 4th track ===> you need 32 ms

Now you are at, S$_{25}$  of T$_{36}$, Now you have to move around the track

Rotational speed = Go to the particular Sector on the Track  ===> Now the disk is rotating !

128 sectors you have to pass,

 

Given that 3000 rotation ------ 60 sec

∴ 1 rotation --- 20 m.sec

with in one rotation, we will complete one rounding of track

∴ to complete rounding one track you need 20ms

1 track --- 20 ms

256 sectors ----- 20 ms

1 sector = $\frac{20ms}{256}$

128 sectors = $\frac{128*20ms}{256}$ = 10ms

 

remaining you have to transfer the 64 KB

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@Shaik Masthan thanks for the confirmation !