TEXAS 1999

Here is Texas paper for you.
in this paper there was 20 questions as follows in 60 minutes .
second part consists of 36 que. in 30 minutes all questions are
diagramatical.(figurs)..

1. if a 5-stage pipe-line is flushed and then we have to execute 5 and
12
instructions respectively then no. of cycles will be
a. 5 and 12
b. 6 and 13
c. 9 and 16
d.none

2. k-map

ab
----------
c 1 x 0 0
1 x 0 x

solve it

a. A.B
B. ~A
C. ~B
D. A+B

3.CHAR A[10][15] AND INT B[10][15] IS DEFINED
WHAT'S THE ADDRESS OF A[3][4] AND B[3][4]
IF ADDRESS OD A IS OX1000 AND B IS 0X2000

A. 0X1030 AND 0X20C3
B. OX1031 AND OX20C4
AND SOME OTHERS..

4. int f(int *a)
{
int b=5;
a=&b;
}

main()
{
int i;
printf("\n %d",i);
f(&i);
printf("\n %d",i);
}

what's the output .

1.10,5
2,10,10
c.5,5
d. none

5. main()
{
int i;
fork();
fork();
fork();
printf("----");
}

how many times the printf will be executed .
a.3
b. 6
c.5
d. 8

6.
void f(int i)
{
int j;
for (j=0;j<16;j++)
{
if (i & (0x8000>>j))
printf("1");
else
printf("0");
}
}
what's the purpose of the program

a. its output is hex representation of i
b. bcd
c. binary
d. decimal

7.#define f(a,b) a+b
#define g(a,b) a*b

main()
{

int m;
m=2*f(3,g(4,5));
printf("\n m is %d",m);
}

what's the value of m
a.70
b.50
c.26
d. 69

8.
main()
{
char a[10];
strcpy(a,"\0");
if (a==NULL)
printf("\a is null");
else
printf("\n a is not null");}

what happens with it .
a. compile time error.
b. run-time error.
c. a is null
d. a is not null.

9. char a[5]="hello"

a. in array we can't do the operation .
b. size of a is too large
c. size of a is too small
d. nothing wrong with it .

10. local variables can be store by compiler
a. in register or heap
b. in register or stack
c .in stack or heap .
d. global memory.

11. average and worst time complexity in a sorted binary tree is

12. a tree is given and ask to find its meaning (parse-tree)
(expression tree)
ans. ((a+b)-(c*d)) ( not confirmed)
13. convert 40.xxxx into binary .

14. global variable conflicts due to multiple file occurance
is resolved during
a. compile-time
b. run-time
c. link-time
d. load-time

15.
two program is given of factorial.
one with recursion and one without recursion .
question was which program won't run for very big no. input because
of stack overfow .
a. i only (ans.)
b. ii only
c. i& ii both .
c. none

16.
struct a
{
int a;
char b;
int c;
}

union b
{
char a;
int b;
int c;
};
which is correct .
a. size of a is always diff. form size of b.(ans.)
b. size of a is always same form size of b.
c. we can't say anything because of not-homogeneous (not in ordered)
d. size of a can be same if ...

bye..
p.sreenivasa rao

______________________________________________________

TEXAS 1999

----------------------------------------------------------------------------

>^M
> THIS IS TI 1999 jadavpur for ECE students.for cs another paper is ^M
>given^M
>^M
>1.two transistors are connected Vbe is 0.7volts .this is simple ckt.one ^M
>transistor is diode equivalent. & asked the o/p across the 2 nd transistor.^M
>2.simple k map ans is Bbar.^M
>3.^M
>^M
> Emitter^M
>---R-------transistorbase| --^M
> | ---^M
> collector^M
> in above capacitor is connected parallel with resistance ^M
>r.capacitor is not shown^M
> in fig.capacitor is used for in this ckt:^M
>^M
>^M
> ans:a.speedupb.active bypass c.decoupling^M
> 4.^M
>^M
> -----R------I----------o/p^M
> |___R____ |^M
> in above r is resistence.I is cmos inverter.^M
> then ckt is used for:^M
>^M
>^M
> a.schmitt trigger b.latch c.inverter ^M
>d.amplifier^M
>^M
>^M
> 5.simple amplifier ckt openloop gain of amplifier is 4.V in ^M
>=1v.asked for V x?^M
> amplifdier + is connected to base. - is connected to i/p in between ^M
>5k is connected.^M
> from o/p feedback connected to - of amplifier with 15k.this is ckt.^M
>^M
>^M
> 6.resistence inductot cap are serially connected to ac voltage 5 ^M
>volts.voltage across^M
> inductor is given.R I C values are given & asked for^M
> voltages across resistence & capacitor.^M
> 7.^M
> ___ R_____^M
> | |^M
> ---R------OPAMP ----------^M

> |---^M
> R1 R1 is for wjhat i mean what is the purpose of R1.^M
> |^M
>^M
> ground^M
>^M
>^M
> 8.asked for Vo at the o/p.it is like simple cmos realization that is n ^M
>block is above^M
> & p block is below.Vdd is 3 volts at supply.V threshold 5 volts.^M
> 9.2 d ffs are connected in asyncro manner .clock 10 MEGAHZ.gate delay ^M
>is 1 nanosec.^M
> A B are the two given D FFs.asked for AB output is:^M
>^M
>^M
> a.updown^M
> b.up c. updown glitching like that (take care abt glitching word)^M
>^M
> 10.^M
>^M
>^M
> ----------------| subtractor|---------o/p^M
> |___HPF____|^M

>^M
> the ckt is LPF ,HPF or APF ?^M
>^M
> 11.in a queue at the no of elements removed is proportional to no of ^M
>elements in^M
> the queue.then no of elements in the queue:^M
> a.increases decreases exp or linearly(so these are the 4 options given ^M
>choose 1 option)^M
> 12.with 2 i/p AND gates u have to form a 8 i/p AND gate.which is the ^M
>fastest in the^M
> following implementations.^M
> ans we think ((AB)(CD))((EF)(GH))^M
> 13.with howmany 2:1 MUX u can for 8:1 MUX.answer is 7.^M
> 14. there are n states then ffs used are log n.^M
> 15.cube each side has r units resistence then the resistence across ^M
>diagonal of cube.^M
> 16.op amp connections asked for o/p^M
> the answer is (1+1/n)(v2-v1).check it out.practise this type of model.^M
> 17.^M
> _____________ supply^M
> ---|__ ___|^M
> Ii >________ |___ Tranistot^M
> > _______Vo^M
> > _______Vo^M
> |^M
> |^M
> R |^M
> | | Io^M
> ground.^M
>^M
>^M
>^M
>^M
> asked for Io/Ii=? transistor gain is beta.^M
>^M
>^M
> a.(1+beta)square b.1+beta c. beta^M
>^M
>^M
> 18.y=kxsquare. this is transfer function of a block with i/p x & o/p ^M
>y.if i/p is^M
> sum of a & b then o/p is :--^M
>^M
> a. AM b.FM c. PM^M
> 19.^M
> ------MULTIPLIER--- |^M

> | |^M
> _____R__|__OPAMP______________________Vo^M
> ---^M
> |^M
> ground.^M
> v in = -Ez then o/p Vo =?^M
> answer is squareroot of -Ez.multiplier i/ps are a & b then ^M
>its o/p^M
> is a.b;^M