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25 feb my Exam
MCQS for Moaz File Mostly..
1.how can we read / write the disk block when LSN is given
2.why it is not feasiable to calculate FS infor block information again and again FAT32
3.how descriptor describes a memory segment and what are the attribute of memory segment
4.Where reserved sector = 1 and sector per FAT = 9. Use appropriate assumption where needed
5.Write three Data Structures for Memory DOS use
6.Write down the purpose of interrupt 12H and interrupt 15H/88H
7.Define Head, Sector, Tracks
8.How a NTFS volume can be accessed in DOS
9. GDT,LDT and IDT entries
1. calculate the sector#for the following cluster 1CH
We have following information
Block per cluster = 8
First user block number = 20
Pray 4 me JazakAllah
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Please share karein apney apney papers.....
please students share your cs609 current paper
Cs 609 8/3/2017 my today ppr object was new but subjective was from last 5 chaptrz Jremember in dua
QNO:13:Find the root directory sector. Where reserved sector = 1 and sector per
FAT = 9. Use appropriate assumption where needed? 5 Marks
Ans: Ans: Root directory sector can find by this formula Reserved Sectors + 2 * (size of FAT) So, Root DIR
Sector = Reserved Sectors + 2 * (size of FAT) = 1 + 2 * 9 = 19
ram checks the extended memory size using int 15H/88H and reports its size.
QNO36: Explain the bits 0,1,2 of 64 status port ? [ 3 marks]
Ans: Port 64H Status Register • Bits 0 = (1 = Output Buffer full) • Bits 1 = (1 = Input Buffer full)
• Bits 2 = (Null)
d from the Partition Table + Boot Block
QNO51: Write down the anatomy of NTFS based file system?
Ans: The FAT and root directory has been replaced by the MFT. It will generally have two copies the
other copy will be a mirror image of the original. Rests of the blocks are reserved for user data. In the
middle of the volume is a copy of the first 16 MTF record which are very important to the system.
Number of reserved blocks=2 Number of blocks in FAT= 7 Number of blocks in root directory=32 Find the number of systems blocks. make the appropriate assumptions when needed.
Answer: Now in the example sector per FAT is unknown.
No. of System Area Blocks = Reserved Block + Sector per FAT * No. of FAT’s + No. of entries *
Root dir =reserved sector+2*sector per FAT 2*sector per FAT= Root dir - reserved sector=32-2=30
Sector per FAT=30/2=15
No. of System Area Blocks = 2 + 15 * 7 + 32=151
Defragmentation Software reserves space for each file in contiguous block by moving the data in clusters and Re-adjusting.
• As a result of defragmentation the FAT entries will change and data will move from one cluster to other localized cluster to reduce seek time.
• Defragmentation has high computation cost and thus cannot be performe d frequently