UPPCL AE 2018:33

Question

Consider a disk pack with $32$ surfaces, $128$ tracks per surface, and $256$ sectors per track, $512$ bytes of data are stored in a bit serial manner in a sector. The capacity of the disk pack and the number of bits required to address a particular sector in the disk are respectively:

• A. $512 \; \text{MB}, 20$
• B. $256 \; \text{MB}, 20$
• C. $128 \; \text{MB}, 18$
• D. $1024 \; \text{MB}, 21$

Answer 1

Number of surfaces in disk = $2^5$

Number of tracks per surface = $2^7$

Number of sectors per track = $2^8$

Data stored in a sector = $2^9\text{B}$

$\text{Capacity of disk} = 2^5 \times 2^7 \times 2^8 \times 2^9 = 2^{29}B = 512\text{MB}$

$\text{Number of sectors in disk}$ = $2^5 \times 2^7 \times 2^8 = 2^{20}$

Hence, $\text{20 bits}$ required for addressing sectors.