Consider a hard disk with 16 recording surfaces (0-15) having 16384 cylinders (0-16383) and each cylinder contains 64 sectors (0-63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is $\langle\text{cylinder no., surface no., sector no.}\rangle$. A file of size 42797 KB is stored in the disk and the starting disk location of the file is $\langle 1200, 9, 40 \rangle$. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?

Given parameters:

• Number of recording surfaces per cylinder = $16$

• Number of sectors per track/surface = $64$

• Capacity per sector = $512\text{ bytes} = 0.5\text{ KB}$

• Total sectors per cylinder = $16 \times 64 = 1024\text{ sectors}$

Total sectors required for the file of size $42797\text{ KB}$:

Starting address is $\langle 1200, 9, 40 \rangle$:

• The number of sectors preceding this address within cylinder 1200 is:

  $$\text{Offset} = 9 \times 64 + 40 = 576 + 40 = 616\text{ sectors}$$

• The number of sectors available in cylinder 1200 from the starting sector onward:

  $$1024 - 616 = 408\text{ sectors}$$

Remaining sectors to be allocated after cylinder 1200:

Number of full cylinders occupied:

These correspond to cylinders $1201$ through $1200 + 83 = 1283$.

Sectors allocated in the full cylinders:

Remaining sectors to be placed in the next cylinder:

These remaining 194 sectors will reside in cylinder $1284$.

Therefore, the cylinder number of the last sector of the file is 1284 (Option D).