CS609 System Programming Solved MCQs + Solved Subjective Questions For Final Term Exam

Whatare the differences betweenhardware and softwareinterrupts?

Software interrupts are invoked bymeansof some software instruction or statement and hardware interrupt isinvoked bymeansof some hardware controllergenerally.The only differencebetween themisthe methodbywhich theyare invoked

Howwould aprocedure written forsoftwareinterrupt will be different from that written for hardwareinterrupt?

Write a program thatwillturn on/off thespeaker and connectitwith theintervaltimerwhenever

Ctrl+Alt+S is pressedTimerinterrupt

#include<dos.h>

#include<bios.h

void interrupt(*oldint15) ( );

void interruptnewint15(unsigned intBP, unsignedintDI, unsigned intSI, unsignedintDS, unsigned intES, unsigned intDX, unsigned intCX, unsignedintBX, unsigned intAX, unsignedintIP, unsigned intCS, unsigned intflags);

void main ( )

{

oldint15=getvect(0x15); setvect(0x15, newint15); keep (0, 1000);

}

void interruptnewint15(unsigned intBP, unsignedintDI, unsigned intSI, unsignedintDS, unsigned intES, unsigned intDX, unsigned intCX, unsignedintBX, unsigned intAX, unsignedintCS, unsigned intIP, unsigned intflags)

{

if (_AH = = 0x4F)

{

if (_AL== 0x1F)

{

outport(0x43, 0xB4); outport(0x42, 0xFF); outport(0x42, 0x21);

outport(0x61, inport(0x61) ^ 3);

}

}

else

(*oldint15)( );

}

CS609 MCQs For Papers

Whatis thedifference between a simpleprogramand aTSR program? Howcan westop multipleloading ofaTSR programinto memory?

TSR (Terminateand StayResident)program is a programwhichprogramsthe interruptnumber

65H butin this casethe new interrupt65H function remains in memoryevenafterthe terminationof theprogramand hence thevectorof int65h does notbecomea danglingpointer. ATSR need tobe loadedonce inmemory.Multipleloadingwillleaveredundantcopies in memorySo we needto havesomecheckwhichwillload theprogramonlyonce

Howinterrupts are processed? List down fivedifferences between hardware and software interrupts.

Software interrupts are invoked bymeansof some software instruction or statement and hardware interrupt isinvoked bymeansof some hardware controllergenerally.

5w.

hatisnulmlodem2marks

Ifdatais to transferredfromone computer toanotherthrough some media whichcan carry digitaldata thenthe modemcanbe eliminated andthe UARTon both computers can be interconnected. Such arrangementis called a NULLmodem

6t.ypesofinterrupts3marks

Interrupt means to breakthe continuityof some ongoing task. When wetalkof computer interrupt wemean exactlythe same in termsof the processor. When an interruptoccursthe continuityof the processorisbroken and the execution branchesto an interrupt service routine.Two types ofinterrupts are:

Software interrupts

Hardware Interrupts.

DMAstands forDirecMtemoryAcces.

REGS is a

Size of IVTis 1024bytes

SettheInterruptvector means to change thedouble word sized interruptvector within IVT.

T/F

Displaydevice (Monitor) performs memorymappedI/O. NMI stands for .

CS609 MCQs For Papers

Ifkeyboard buffer is emptythehead andtailpoints atthe samelocation. T/F

Tailof keyboard should getto get thestartof buffer.

Interrupt 65His empty.

Intervaltimeis used to divide inputfrequency.T/F

An I/O devicecannotbedirectlyconnected tothe busses so controlleris placedbetweenCPU

andI/O.

Standard PC canhave PPI.

1,4, 8, 16

The BIOS int 0x1Ahcanbeused toconfigureRTC. Keyboarduses port64Has status port.

Communicationb/w keyboard andkeyboard controlleris Asynchronous serial

Synchronouserial

Parallel

None

0xF3means typematic rate willbe sent innextbyte. Bycascadingtwo DMAs bits can betransferred.

4,8, 16, 32

Timerinterruptoccurs 18.2timesin a second.

PPI interconnection bits is cleared toindicate low nibble is beingsent. D1, D2, D3D, 4

WriteTSR thatuse 17h to ignorespaces. 5 marks

#include dos.h>

void interrupt (*old)( ); void interrupt newfunc ( ); main( )

{ old=getvect(0x17); setvect(0x17,newfunc); keep(0,1000);

CS609 MCQs For Papers

}

void interrupt newfunc( )

{

if(_AH==0)

{

if( _AL!=‘ ‘ ) (*old)();

}

}

WriteTSR to displaynamein center withinterrupt8 provided 1990 is center index. 10 marks

voidinterrupt (*1990)();

voidmain()

{ old=getvect(0x08); setvect(0x08,newint); keep(0,1000);

}

voidinterrupt newint ()

{

… (*1990)();

}

If CPUID instructionis not present then the processorcan be a

?486processor

?386 processor

?286 processor

?Allof the above

QuestionNo. 2

On the execution of IRETinstruction,numberof bytes popped fromstack is

?4 bytes

?6 bytes

?8bytes

?10 bytes

QuestionNo. 3

Keyboard Status Byteis located at the address

?0040:0000H

?0040:0013H

?0040:0015H

?0040:0017H

QuestionNo. 4

If weuse keep (0, 1000) in aTSR program,the memoryallocated to it is

?64000 bytes

?32000 bytes

?16000 bytes

?80000 bytes

QuestionNo. 5

CS609 MCQs For Papers

Maximumnumberofinterrupts in a standard PC is

?64

?128

?256

?512

InbootblockBIOS parameterblock startsfrom

03H

05H

08H

0BH

File canbeviewedas collectionof clusters or blocks.

Physically

Logically

Both physicallyandlogically

None

Whatwillbethe valueof DLRegister when we areaccessing C driveusing undocumented service21H/32H?

0

1

2

3

Operating systemnamecontains bytesin bootblock.

3

5

8

11

WhenLSN is equalto zero(0) it means Firstblock of the disk

FirstblockofthelogicaldriveFirstblock of hidden blocks Noneof the given

The sizeof DPB data structureisbytes

16

32

64

128

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Cluster numbercan also be referredas block number. True

False

Using theentryand the FATwe can access the contents of file. Reserved blocks

RootDirectoryNumber of FATcopies Noneof the given

Control information in files is maintained using

BPB DPB FCBFPB

Wecan access blocks for FATusing BPB

DPB

FCB

Both BPB and DPB

File systemusedin CD’s isfilesystem

ContiguousChained Indexed None

Disadvantageof FAT32 isLarge disk canbe managed inFAT32

Cluster sizeis reduced

Internalfragmentation is reduced

Verylargetable

Practically entries are there in FAT32

2^26

2^28

2^30

2^32

NTFS volumecanbe accesseddirectlyin DOS.

True

False

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Whatwillhappenif NTFS volumeis accessed in DOS? Convertit to FATvolume

Nothing willhappen ErrorofinvalidmediaNoneof the given

Advantages of FAT32 is/areLarge disk sizecan bemanagedin FAT32

Cluster sizeis reduced

Internalfragmentation is reduced

Allofthegiven

FATbased file systemcanstore file namein form

ASCII UNICODE

BothASCIIandUNICODE

None

How manybytes can beusedto store a file nameinNTFS?

128

255

510

1024

IN NTFS, FATandroot directoryisreplaced by

FCB

MFT

Hidden blocks

Boot sector

LSNofFSInfoblockisavailable at

BPB

FAT

Root Directory

None ofthe given

DOSdevice drivers donot understandthe_data structures. FAT12

FAT16

FAT32

NTFS

For supportinglongfile names, fragments canbe supported.

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12

20

26

32

If a file ishavingmore than one clusterthenit will bemanagedby

FAT BPB DPB None

File systemkeeps the backup of its bootblock. FAT12

FAT16

FAT32

None

If a file size is 12K andthe size oftheclusteris 4Kthen clusters are usedfor the file.

2

3

4

5

is the first logical sector ofNTFS partition. DPB

MFT

Bootsector

None

Afile has2clusters andthe size ofclusteris 4K.What will bethe size offile?

2K

8K

16K

32K

InFAT32root directoryentriesarethere

128

256

512

None

Block#2is thesafest blocktostore the backup of boot block. True

False

CS609 MCQs For Papers

1. Thereare two main types of interrupts namely HardwareandSoftwareinterrupts

2. To set theinterruptvectormeans is to change thedouble word sizedinterruptvector within the IVT. (True /False)

3. The service numberis usuallyplaced inthe register. (AL,CL,AH,AX)

4. The keyboardmakes use of interruptnumber for its inputoperations. (9,10,11,12)

5. The service

15H/FFH )

is calledthe keyboardhook service. (15H/2FH,15H/4FH,

6. The BIOS interrupt canbe usedto configureRTC. (1AH, 2AH, 3AH, 4AH)

7. The intervaltimer canoperatein modes. (Five,Seven, Four,Six)

8. is Disk interrupt.(10H,11H,13H,14H)

9. PPI stands forPeripheralProgrammableInterface (PPI)

10.Int is used to control theprintervia theBIOS. (17H, 18H, 20H, 21H)

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11.Counter register canbe used to divide clocksignal. (TrueFalse) Note:itisalsousedtodividefrequency(See#69)

12.DCEstands for _DataCommunicationEquipment

13.InterruptIDRegister_is used toidentifythe causeof interrupt.

14.The bit# of thecoprocessor controlword is theinterruptenable flag. (7,8,9, 6)

15.Thereare kinds of serialcommunication. (2,3,4,5)

16. store thebase address for LPT1. (40:00H , 40:02H ,40:08H, 40:1AH)

17.The amountof memoryaboveconventional memory(extended memory)canbe determinedusingthe service . (15H/88H, 16H/88H, 17H/88H, 21H/88H)

18.The outputon themonitoris controller bya controllercalled withinthePC. (Videocontroller, Bus controller, Ramcontroller, None of thegiven)

19.The keyboardinputcharacter scan codeis received at port. (60H,61H,62H,63H,64H )

20. is LED controlbyte.(0xFD,0xED, 0xFF, 0xEE)

21.In RS232C these abbreviations stands for[3 marks]

DTRDTR (dataterminalready)

DSRDSR (data setready)

CTSCTS (clearto send) RTSRTS (Request tosend)

22.Forwhat purpose these servicesare used?[3 marks]

INT1AH/01SetClockCounter

INT1AH/02ReadTime INT1AH/03SetTime INT1AH/04Read Date

CS609 MCQs For Papers

INT1AH/05SetDate

23.Howcommunication between modemcan performed (in terms ofdata transfer). [5marks]

Modem is generallyused tosend /receive data to/froman analog telephone.Had the telephone line beenpurelydigitaltherewouldhave beenno need of a modem inthis form. If datais to transferred fromonecomputerto anotherthrough somemediawhich can carrydigitaldata thenthe modemcanbe eliminated andthe UARTon both computerscan be interconnected.Sucharrangement is calleda NULLmodem.

BYMISHII

Function totogglespeaker betweenon and off

The port61h is usedto controlthe speakeronlytheleastsignificant2 bits are important.Bit0 is usedto connect theinterval timer tothe speakerand thebit#1 is used toturn the speakeron off. Restof thebits are used byotherdevices.

Parallelcommunication

PPI is used toperformparallelcommunication. Devices likeprinterare generallybasedon parallelcommunication. It’s calledparallelbecause a number of bits aretransferredfromone point toanotherparallelon variouslines simultaneously

Communicationwith paritycheck

The line controlregister contains important information aboutthe behaviour of thelinethrough which thedata willbe transferred. In itvarious bits signifythe word size, length

ofstop bits, paritycheck,paritytype andalso the acontrolbit toload thedivisor value.

The bit7 if setindicates thatthebase+0 andbase +1 willactas the divisor registerotherwise if cleared willindicatethatbase+0 is thedataregister.

RS232Cwork flow?

RS232Cis a standard for physicaldimension of theconnectorinterconnecting a DTE(Data terminalequipment) and DCE(Datacommunication equipment).

Data is receivedthrough theRxD line. Datais send through theTxD line. DTR (data terminal ready)indicates thatthe data terminalis liveand kicking.DSR(data setready)indicates that the dataset is live.Wheneverthe sender cansend dataitsends the signalRTS( Request tosend) if as a resultthe receiveris free and canreceive dataitsend the sender anacknowledgethrough

CTS( clearto send) indicating thatits clear tosend now.

1AH services

Clock Counter1AH/00

SetClockCounter 1AH/01

ReadTime1AH/02

SetTime1AH/03

Read Date1AH/04

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SetDate1AH/05

SetAlarm1AH/06

DisableAlarm1AH/07

ReadAlarm1AH/09

Write function in relationto COM portand modem

The initialize () function initializes the COM portwhose numberis passedas parameter using BIOS services.The recievechar()functionuses theCOM portnumber toreceivea bytefrom the COM portusing BIOS services. thesendchar() function sends a characterto theCOM portusing BIOS servicewhosenumber is passed as parameter.And thegetcomstatus() functionretrieves

the statusof the COM portwhose number has been specifiedand returns themodemand line status in anunsignedint.

Coprocesser

To access theblock withinclusterusingBIOS servicesthe cluster number should beconvertedinto .

Selectcorrectoption: CHS

LBA

LSN

Noneof the given

Question # 2 of 10 ( Starttime:12:21:55PM )TotalMarks:1

The practicallimitof blocks percluster is . Selectcorrectoption:

32blocks per cluster

64blocks percluster

128blocks per cluster

256blocks per cluster

Question # 3 of 10 ( Starttime:12:23:05PM )TotalMarks:1

Maximumpossibleentries in FAT12 are . Selectcorrectoption:

1024

2048

4096

65536

Question # 4 of 10 ( Starttime:12:24:20PM )TotalMarks:1

Whenwe talk aboutFATbased file system, inuserdata area firstcluster numberis . Selectcorrectoption:

0

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1

2

Noneof the given

Question # 5 of 10 ( Starttime:12:25:04PM )TotalMarks:1

Ifwe know theclusternumber,we canaccess theblocks withinthe clusterusingBIOS services directly.

Selectcorrectoption:

True

False

Question # 6 of 10 ( Starttime:12:26:16PM )TotalMarks:1

Jumpcode partcontains bytes in bootblock. Selectcorrectoption:

3

5

8

11

Question # 7 of 10 ( Starttime:12:27:22PM )TotalMarks:1

Indos we havelimitof . Selectcorrectoption:

128blocks percluster

256blocks per cluster

32blocks per cluster

64blocks per cluster

Question # 8 of 10 ( Starttime:12:28:01PM )TotalMarks:1

Drive parameterblock is derived from . Selectcorrectoption:

FCB FAT BPB CPB

Question # 9 of 10 ( Starttime:12:29:26PM )TotalMarks:1

The directorystructure of DOS is like . Selectcorrectoption:

Array Tree Linkedlist

Noneof the given

CS609 MCQs For Papers

Question # 10 of 10 ( Starttime:12:30:19PM )TotalMarks:1

is the firstblock on disk. Selectcorrectoption:

LSN =0

LBA=0

LBA=1

Both LBA=0 and LSN=0

Question # 1 of 10 ( Starttime:12:20:29PM ) TotalMarks:1

To access theblock withinclusterusingBIOS servicesthe clusternumbershouldbe converted into .

Selectcorrectoption:

CHS LBA LSN

Noneof the given

Question # 2 of 10 ( Starttime:12:21:55PM ) TotalMarks:1

The practicallimitof blocks percluster is . Selectcorrectoption:

32blocks per cluster

64blocks per cluster

128blocks per cluster

256blocks per cluster

Question # 3 of 10 ( Starttime:12:23:05PM ) TotalMarks:1

IfFATentryis between FFF0H toFFF6H in FAT16 then . Selectcorrectoption:

Cluster is available

Itis a Reservedcluster

Itis nextfile cluster

Itis a lastfilecluster

Question # 4 of 10 ( Starttime:12:24:20PM ) TotalMarks:1

Whatwillbethe valueof theword locatedat1Fh in DPB whennumber of freeclusters on drive is notknown?

Selectcorrectoption:

0000H

1111H FFFFH

CS609 MCQs For Papers

Noneof the given.

Question # 5 of 10 ( Starttime:12:25:04PM ) TotalMarks:1

Ifwe know theclusternumber,we canaccess theblocks withinthe cluster using BIOS services directly.

Selectcorrectoption:

True

False

Question # 6 of 10 ( Starttime:12:26:16PM ) TotalMarks:1

Jumpcode partcontains bytes in bootblock. Selectcorrectoption:

3

5

8

11

Question # 7 of 10 ( Starttime:12:27:22PM ) TotalMarks:1

Indos we havelimitof . Selectcorrectoption:

128blocks per cluster

256blocks per cluster

32blocks per cluster

64blocks per cluster

Question # 8 of 10 ( Starttime:12:28:01PM ) TotalMarks:1

Wecan access the contents of Fileby usingthe rootdirectoryentryand . Selectcorrectoption:

Reserved Blocks

Number of FATcopies

FileAllocationTable(FAT) Noneof the given

Question # 9 of 10 ( Starttime:12:29:26PM ) TotalMarks:1

The directorystructure of DOS is like . Selectcorrectoption:

Array

CS609 MCQs For Papers

Tree

Linkedlist

Noneof the given

Question # 10 of 10 ( Starttime:12:30:19PM ) TotalMarks:1

is the firstblock on disk. Selectcorrectoption:

LSN =0

LBA=0

LBA=1

Both LBA=0 and LSN=0

Question # 1 of 10 ( Starttime:02:32:19PM ) TotalMarks:1

Disadvantageof FAT32 is . Selectcorrectoption:

Large disk sizecan bemanagedin FAT32

Cluster sizeis reduced

Internalfragmentation is reduced

Verylargetable (not sure)

Question # 2 of 10 ( Starttime:02:33:47PM ) TotalMarks:1

InFAT32, lower bits are used. Selectcorrectoption:

26

28

30

32

Question # 3 of 10 ( Starttime:02:34:23PM ) TotalMarks:1

File systemusedin CD’s is filesystem. Selectcorrectoption:

Contiguous

Chained

Indexed(guess) Noneof the given

CS609 MCQs For PapersQuestion # 4 of 10 ( Starttime:02:35:53PM ) TotalMarks:1

Ifa filesize is 12K and thesize of the clusteris 4K then clustersare used for thefile. Selectcorrectoption:

2

3

4

5

Question # 5 of 10 ( Starttime:02:37:23PM ) TotalMarks:1

To store a UNICODEcharacter is/areneeded. Selectcorrectoption:

Nibble

Byte

2Bytes

4Bytes

Cluster sizeis reducedin . Selectcorrectoption:

FAT12

FAT16

FAT32

Noneof the given

Question # 7 of 10 ( Starttime:02:40:18PM ) TotalMarks:1

Forsupportinglong file names, fragments canbe supported. Selectcorrectoption:

12

20

26

32

Question # 8 of 10 ( Starttime:02:41:28PM ) TotalMarks:1

Afilehas 2 clusters and the sizeof cluster is 4K.Whatwillbethe sizeof file? Selectcorrectoption:

2K

8K

16K

32K

CS609 MCQs For Papers

InNTFS, Backupof bootblockis stored atblock # . Selectcorrectoption:

2

6

8

10

Question # 10 of 10 ( Starttime:02:44:18PM ) TotalMarks:1

Whenwe marka fileas deleted byplacing 0xE5then thechainof clusters inFATis also replaced by .

Selectcorrectoption:

E5

1

0

N

Quiz: CS609Total Marks10

Instructions:

1) Upload this Quizafter solvingwith in 24 hours No ExtraTimewillbe given or Quiz will notbe acceptedviaemail.

2) Do No markmore thenone choices.

3) It is better youhighlightthe choice withRead like a)correctchoice

1) The intervaltimercan operatein modes.

a) Three b) Four c) Five d) Six

2) Highest capacityphysicalcapacityof the disk according to the IDE interfaceis

. a) 127GB b) 100 GB

c) 80 GB

d) 300 GB

3) PartitionTablecan be read using theextended Services.

a) 13H

CS609 MCQs For Papers

b) 14 H

c) 15 H

d) Noneof given

4) File controlblock(FCB) is bytelong.

a) / 32
b) / 64
c) / 16
d) / 128

5) On the execution of IRETinstruction,numberof bytes popped fromstack is

a) 4 bytes b) 6bytes c) 8 bytes d) 10 bytes

6) If CPUID instructionis not present then the processorcan be a

a) 486 processor b) 386 processor c) 286 processor

d) Allofthegivenchoices

7) Extended memory is available if the processoris of the type

a) AT

b) XT

c) Allof thegiven choices d) Noneof them

8) The builtin mechanism within the UARTforerrordetection is

a) hamming code b) parity

c) CRC16(cyclic redundancycheck16 bit)

d) CRC32(cyclic redundancycheck32 bit)

9)In Protected Mode, the segment registers are used as

a) Descriptor b) Selector

c) Allof thegiven choices

d) Noneof the givenchoices

10)If threeProgrammableinterrupt controllers are cascaded then how many interrupt driven hardware IOdevices can be attached

CS609 MCQs For Papers

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First block on logicaldrive.

LBA

3/2

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FALSE

2^28

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26

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DPB

CS609 MCQs For Papers

Therearetwomaintypesofinterrupts,namely

PCbasedandWindowbased

HardwarebasedandKernalbased

Hardwareinterrupts andSoftware interrupts

Noneofthegiven

QuestionNo:2(Marks:1)-Pleasechooseone

StandardPCoperatesintwomodesintermsofmemorywhichare

RealmodeandExtendedMode BasemodeandMemoryMode Noneofthegiven

Realmodeandprotectedmode

QuestionNo:3(Marks:1)-Pleasechooseone

BPBstandsfor .

BIOSparameterblock BIOSprocessingblock Baseprocessingblock BIOSpartitionblock

QuestionNo:4(Marks:1)-Pleasechooseone

TheFunctionofI/Ocontrolleristoprovide .

I/O controlsignals

Buffering

ErrorCorrectionandDetection

Allofgiven

QuestionNo:5(Marks:1)-Pleasechooseone

IVTisatablecontaining byteentries each ofwhichisafaraddressofan interruptserviceroutine.

2

4

8

16

QuestionNo:6(Marks:1)-Pleasechooseone

Eachparagraphin keep functionis bytesinsize.

CS609 MCQs For Papers

8

16

32

QuestionNo:7(Marks:1)-Pleasechooseone

Interrupt9usuallyreadsthe fromkeyboard.

ASCIIcode

Scancode

BothASCIIandScancode

Noneoftheabove

QuestionNo:8(Marks:1)-Pleasechooseone

AsoftwareinterruptdoesnotrequireEOI(Endofinterrupt).

True

False

QuestionNo:9(Marks:1)-Pleasechooseone

Tostoreeachcharacterinkeyboardbuffer bytesarerequired.

2

4

6

8

QuestionNo:10(Marks:1)-Pleasechooseone

Interrupt isempty;wecanuseitsvectorasaflag.

9H

13H

15H

65H

QuestionNo:11(Marks:1)-Pleasechooseone

Command registeris an bitregister

4

8

16

32

QuestionNo:12(Marks:1)-Pleasechooseone

Theintervaltimercanoperatein modes.

Three Four Five Six

QuestionNo:13(Marks:1)-Pleasechooseone

Thefollowingcommand“outportb(0x61,inportb(0x61)|3);”will .

Turnonthespeaker Turnoffthespeaker Togglethespeaker Noneoftheabove

QuestionNo:14(Marks:1)-Pleasechooseone

ThePPIactsasaninterfacebetweentheCPUandaparallel .

I/Odevice

CPU BUS

NoneofGiven

QuestionNo:15(Marks:1)-Pleasechooseone

DTEis .

Data terminalequipment Data transmittingequipment Dual terminal equipment Noneofthegiven.

QuestionNo:16(Marks:1)-Pleasechooseone

DSRstandsfor .

Datasetready Dataserviceready Datastockready Noneofthegiven

QuestionNo:17(Marks:1)-Pleasechooseone

InselftestmodetheoutputoftheUARTisroutedtoitsinput.

True

False

QuestionNo:18(Marks:1)-Pleasechooseone

Interrupt isusedtogetorsetthetime.

0AH

1AH

2AH

3AH

QuestionNo:19(Marks:1)-Pleasechooseone

isusedtosettime.

1A/02H

1A/03H

1A/04H

1A/05H

QuestionNo:20(Marks:1)-Pleasechooseone

Communicationbetweenkeyboardandkeyboardcontrolleris .

Asynchronousserial Synchronousserial Parallelcommunication Noneofthegiven

)

QuestionNo:1(Marks:1)-Pleasechooseone

Followingisnotamethod ofI/O

ProgrammedI/O InputdrivenI/O HardwareBasedI/O Noneofgiven

QuestionNo:2(Marks:1)-Pleasechooseone

TheFunctionofI/Ocontrolleristoprovide .

I/Ocontrolsignals

Buffering

ErrorCorrectionandDetection

Allofgiven

QuestionNo:3(Marks:1)-Pleasechooseone

WhichofthefollowingaretypesofISR .

BIOS(BasicI/Oservice)ISR DOSISR

ISRprovidedbythird partydevicedrivers

Allofthegiven

QuestionNo:4(Marks:1)-Pleasechooseone

Interruptservicenumberisusuallyplacedin register.

CH CL AH AL

QuestionNo:5(Marks:1)-Pleasechooseone

NMIStandfor

NonMaskableInterrupt NonMultitudeInterrupt NonMaskableInstruction NoneofGiven

QuestionNo:6(Marks:1)-Pleasechooseone

Asingleinterrupt controllercanarbitrateamong differentdevices.

4

6

8

10

QuestionNo:7(Marks:1)-Pleasechooseone

HardwareInterruptsare .

Preemptive

Non-Preemptive

BothPreemptiveandNon-Preemptive

NoneofGiven

QuestionNo:8(Marks:1)-Pleasechooseone

Themicroprocessorpackagehas manysignalsfordata. Below aresomeinCorrect priorityorder (HighertoLower).

Reset,Hold,NMI,INTR

NMI,INTR,Hold,Reset INTR,NMI,Reset,Hold NoneoftheGiven

QuestionNo:9(Marks:1)-Pleasechooseone

Theintervaltimercanoperatein modes.

Three Four Five Six

QuestionNo:10(Marks:1)-Pleasechooseone

PPIstandsfor .

ParallelProgrammable interface PeripheralProgrammableinterface PortProgrammableinterface

Noneofthegiven

QuestionNo:11(Marks:1)-Pleasechooseone

Thefollowingcommand“outportb(0x61,inportb(0x61)0xFC);”will

Turnonthespeaker Turnoffthespeaker Togglethespeaker Noneofthegiven

QuestionNo:12(Marks:1)-Pleasechooseone

ThePPIactsasaninterfacebetweentheCPUandaparallel .

I/Odevice

CPU BUS

NoneofGiven

QuestionNo:13(Marks:1)-Pleasechooseone

BIOSDONOTsupport .

LPT1

LPT2

LPT3

LPT4

QuestionNo:14(Marks:1)-Pleasechooseone

bitisclearedtoindicatethelownibbleisbeingsent.

D1

D2

D3

D4

QuestionNo:15(Marks:1)-Pleasechooseone

Thebit ofLinecontrolregisterinUART,ifclearedwill indicatethatDLLis the dataregister.

1

3

5

7

QuestionNo:17(Marks:1)-Pleasechooseone

usedtodeterminetheamountofconventionalmemoryinterfacedwiththeprocessorin kilobytes.

INT10H INT11H INT12H INT13H

QuestionNo:18(Marks:1)-Pleasechooseone

Bitnumber ofcoprocessorcontrolwordis theInterruptEnableFlag.

7

8

9

10

QuestionNo:19(Marks:1)-Pleasechooseone

Todistinguish486withPentiumCPUIDTestisused.

True

False

QuestionNo:20(Marks:1)-Pleasechooseone

is LEDcontrolbyte.

0xF3

0xED

0xE5

0xFF

2.Intservice0 can beusedto set thelineparameter of theUARTor COM port.

14H

15H

13H

Noneof the givenoption

3.In case of synchronous communication a timingsignal is required toidentifythe startand end of a bit.

True

False

4.In self testmode theoutputof the UARTis routedto itsinput

True

False

6.Thefuction uses the COM portnumberto receive a bytefromtheCOM portusing

BIOS services. recievebyte() receive() recievechar()

10.The function initializethe COM portwhosenumber is passed as parameter usingBIOS services.

Initializecom() Initialize()Recievechar()

Noneof these option

11. XONwhenever receivedindicatesthe start of communicationand XOFFwhenever receivedindicates atemporary pause inthe communication.

14. is a device incorporatedinto thePC to updatetimeeven ifthe computeris off.

Clock counter

ROM Clock

Realtimeclock

Whatis thelocationoftimercount inBIOSdata area?

Sol.

0040:006C

QuestionNo:22(Marks:3)

Writeafunctionwhichfillsthewholescreenwithblanks(space). Sol.

unsignedchar far*scr=(unsignedchar far*)0xb8000000 voidmain()

{

inti;

for(i=0;i<2000;i++)

{

*scr=0x20;

*(scr+1)=0x07;

scr=scr+2;

}

}

QuestionNo:23(Marks:5)

Forwhatpurposetheport61Hisused? Sol

Theport61his usedtocontrolthe speakeronlythe least significant 2 bits are important. Bit 0is usedtoconnect the interval timertothe speakerandthe bit #1isusedtoturnthe speaker on off. Restof the bits are used by other devices.

QuestionNo:24(Marks:10)

WritedownaCprogramthatwillprotectthe bootblocktobewrittenbyother application.

Hint.

Useinterrupt13foraccessingthebootblockinformation. Sol.

#pragmainline

#include<dos.h>

#include<bios.h>

voidinterrupt (*oldtsr) ( );

voidinterrupt newtsr (unsigned int BP, …, flags);

//mustprovide all the arguments

voidmain ( )

{

oldtsr= getvect (0x13); setvect(0x13, newtsr); //corrected keep (0, 1000);

}

voidinterrupt newtsr(unsigned int BP, unsigned int DI,

unsignedint SI, unsigned int DS, unsigned int ES, unsigned

intDX, unsigned int CX, unsigned int BX, unsigned int AX, unsigned int IP, unsigned int CS,

unsignedint flags) //corrected

{

if( _AH == 0x03)

if((_DH == 1 & _CH == 0 & _CL == 1)& _DL >= 0x80)

{

asmclc;

asmpushf;

asmpop flags;

return;

}

_ES = ES; _DX = DX;

_CX= CX; _BX = BX;