NIMCET 2016 — Computer PYQ

Q: How do I solve NIMCET 2016 Computer questions?

Practice NIMCET 2016 Computer previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of NIMCET Computer questions?

NIMCET Computer questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are NIMCET previous year papers available for free?

Yes. Mathem Solvex provides free access to NIMCET previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

NIMCET 2016 — Computer PYQ

NIMCET | Computer | 2016

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 cyclinderwise and the addressing format is <cylinder no., surface no., sector no.>. A file of size 42797KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is cylinder number of the last sector of the file, if it is stored in a contiguous manner?

Choose the correct answer:

Explanation

Data:
File size is 42797 KB= 42797 ×210 B=85594×29 B.
recording surfaces = 16 (0 to 15)
cylinders = (0-16383)
sectors(0-63
one sector=512 B
Calculation:
The file will be stored in 85594 sectors, that is, it needs to cross 85594 sectors
starting of the file is
number of cylinders to cross = (85594÷16)× 64 = 83 cylinders
remaining sectors to cross = 85594-(83×16× 64) = 602
number of surfaces to cross = 9
so to cross 9 surfaces we need to cross on more cylinder as the file has started at surface 9 and no surface in the

cylinder is 16 so
number of the cylinder to cross=83+1=84
so cylinder number = 1200+84=1284