ace test series

Question

 

i dont understand how ans come 53??? 

MY CAL

AVE ROTATIONAL DELAY =10 MS 

DATA TRANSFER RATE IS 128 KB SO 0.5 SEC = 500 MS REQUIRED FOR 64 KB 

THEN TRANSFER FROM TRACKS 32 MS

TOTAL= 10+500+32=533 MS ACCORDING TO ME!!!CORRECT ME IF M WRONG!!

Answer 1

Rotational delay = 20ms

Seek time = 32ms

---

1 rotation = 20ms

to cover a sector(512B) in a track , 1/256 rotation = 20/256ms

1 B in = 20/(256*512) ms

64*1024 B in = 20*64*1024/(256*512) = 10ms

Therefore,

Latency = 10+32+10 =52 ms.