How to calculate the capacity of disk pack in the below question ?

Question

A disk pack has 19 surfaces and storage area on each surface has an outer diameter of 33cm and inner diameter of 22cm .The maximum recording storage density on any track is 200 bits/cm and minimum spacing between tracks is 0.25 mm , then how to calculate the capacity of disk pack ?

Answer 1

Disk capacity = total number of surfaces * no. of tracks per surface * amount of data per track 
----- Now here our main task is to calculate no of tracks on each surface. 
----- recording width =(outer radius -inner radius)=(16.5-11)=5.5 cm
{*** we cant also use (outer diameter -inner diameter ) but in whis way we are calculating no of tracks twice so divide by 2 is reauired here} 
----now if w=total recording width on surface =5.5 cm =55 mm 
          x= inter/inner track distance=0.25 mm 
     y= track recording width  ;; that is not given in our problem and very small in comparison to (inter/inner track distance) neglect it  
   now no of tracks =(recording width / ( inter/inner track distance +  track recording width)) 
                             =(w/x+y) ≈(w/x)=55/0.25=220 tracks  
--------- now we calculate  amount of data per track 
that is = parimeter of a inner track * storage density of a track=(22/7)*22 * 200 bits (here i m not calculate it completely )
so disk capacity = 19 * 220 *200 *(22/7)*22 = 57803428.57 =57803429 bits
----- to calculate track capacity we are using (parimeter of a inner track)
bcoz when disk containing same no of sectors per track then inner track is more dense as compare to outer track.

Comments

what is the meaning of  minimum spacing it means it could more than 0.25 mm

Answer 2

Disk capacity = total number of surfaces * no. of tracks per surface * amount of data per track

Total number of tracks per surface = (outer radius - inner radius)/ inter track gap  = (33cm/2 - 22cm/2 )/0.25mm = 220

perimeter of inner track = 2 * pi * inner radius = 2 * (22/7 )*11 cm

density = 200 bits/ cm

amount of data present in single track = perimeter * density = 13829 bits (approximately)

disk capacity= 19 * 220 * 13829  bits = 6.89 MB

Comments

hi thanks . ya i did it wrong still answer is 6.89 u also recalculate.

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Hi thanks, edited. i consider K as 1000 ( approximately )  rather 1024 .  anyway thanks :)

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what is the meaning of  minimum spacing it means it could more than 0.25 mm