A) +*!
B)#*!
C)+*?
D)-*?
Answer C.
explanation.
+ symbol for one or more time.
* for 0 or more time.
? for zero or more time.
B)#*!
C)+*?
D)-*?
Answer C.
explanation.
+ symbol for one or more time.
* for 0 or more time.
? for zero or more time.
Monday 30 November 2015
Q39,p3,j15.Which one of the follwing is correct?
A)Java applet canot be written in any programming language.
B)An applet is not a small program.
C)An applet can be run on its own.
D)Applet are embedded in another application.
Answer D
B)An applet is not a small program.
C)An applet can be run on its own.
D)Applet are embedded in another application.
Answer D
Q38,p3,J15.The servelet response interface enables a servelet to formulate a response for a client using method-------------------
a) void log(Exception e,String S).
b) void destroy()
c)int get server port()
d) void set context type(string type)
b) void destroy()
c)int get server port()
d) void set context type(string type)
Q37,p3,j15. In Java, when we implement an interface method, it must be declared as
A) Private
B)protected.
C)Public.
D)Friend
Answer C.
B)protected.
C)Public.
D)Friend
Answer C.
Q36,p3,J15. The number of nodes of height h in any n element heap is--------
A) h
B)2^h
C) ceil(n/2^h)
D)ceil(n/(2^(h+1)).
Answer D.
B)2^h
C) ceil(n/2^h)
D)ceil(n/(2^(h+1)).
Answer D.
let f(n) and g(n) be asypotically non negative functions . Which of the follwing is correct.
A) ϴ(f(n)*g(n)=min(f(n),g(n).
B)ϴ(f(n)*g(n))=max(f(n),g(n).
C)ϴ(f(n)+g(n))=min(f(n),g(n).
D)ϴ(f(n)+g(n))=max(f(n),g(n).
Answer D.
B)ϴ(f(n)*g(n))=max(f(n),g(n).
C)ϴ(f(n)+g(n))=min(f(n),g(n).
D)ϴ(f(n)+g(n))=max(f(n),g(n).
Answer D.
Sunday 29 November 2015
Q33,p3J15.Which of the following is asymptotically smaller.
A)lg(lg*n)
B) lg*(lg n)
C)lg(n!)
D) lg*( n!).
Answer A.
Explanation.
Iterated algo is very slowly growing function.
Lg*2=1
lg*4=2
Lg*16=3
lg*65536=4.
lets take n=16.
option A.
lg*16=3.
lg(3)= approx 1.7
Option B.
log*(log 16)
log*(4)=2.
Option C.
log(16!)
greater than 1.7
option D
Log*(16!) greater than 1.7.
B) lg*(lg n)
C)lg(n!)
D) lg*( n!).
Answer A.
Explanation.
Iterated algo is very slowly growing function.
Lg*2=1
lg*4=2
Lg*16=3
lg*65536=4.
lets take n=16.
option A.
lg*16=3.
lg(3)= approx 1.7
Option B.
log*(log 16)
log*(4)=2.
Option C.
log(16!)
greater than 1.7
option D
Log*(16!) greater than 1.7.
Q 32,p3,j15 The travelling salesman problem can be solved in:
Options
A) POlynomial time using dynamic programming.
B)Polynomial time using branch and bound algorithm.
C)Exponential time using dynamic programming and branch and bound algorithms.
D)Polynomial time using backtrack algorithms.
Answer C.
A) POlynomial time using dynamic programming.
B)Polynomial time using branch and bound algorithm.
C)Exponential time using dynamic programming and branch and bound algorithms.
D)Polynomial time using backtrack algorithms.
Answer C.
Q31,P3,J15. An all pair shortest paths problem is efficiently solved using.
A) Dijkstra algorithm. B)Bellman ford algorithm C)Kruskal algorithm D)Floyd Warshal algorithm.
Answer D.
Answer D.
Q30,P3,J15. The maximum payload for TCP segment is
A) 65535 B)65515 C)65495 D)65475Answer C.
Explanation.
Each segment including the header must fit in 65515 byte. header is 20 byte. so max payload is 65495
Explanation.
Each segment including the header must fit in 65515 byte. header is 20 byte. so max payload is 65495
Q29,p3,J15.Suppose a digitized voice channel is made by digitizing 8KHZ bandwith Analog voice signal .It is required to sample the signal at twice the highest frequency (two samples per hertz)..What is the rate required ,if it is assumed that each sample requires 8 bits.
Options
A) 32kbps B)64 KBPS C)128 KBPS D)256 KBPS.
Answer C.
Bandwidth of analog cahnnel is 8K/Sec.
Required sample 2/hz. =2s.
one sample is 8 bit.
so bit rate = (8k)*2*(8 bit)=128 kbps.
A) 32kbps B)64 KBPS C)128 KBPS D)256 KBPS.
Answer C.
Bandwidth of analog cahnnel is 8K/Sec.
Required sample 2/hz. =2s.
one sample is 8 bit.
so bit rate = (8k)*2*(8 bit)=128 kbps.
Q28,p3,j15. A message 'COMPUTER NETWORK'encrypted (ignore quotes) using columnar transposition cipher with a key "layer" the encrypted message is?
A) CTTOEWMROPNRUEK
B)MROUEKCTTPNROEW
C)OEWPNRCTTUEKMRO
D)UEKPNRMROOEWCTT
Answer A.
B)MROUEKCTTPNROEW
C)OEWPNRCTTUEKMRO
D)UEKPNRMROOEWCTT
Answer A.
|
C
|
O
|
M
|
P
|
U
|
|
T
|
E
|
R
|
N
|
E
|
|
T
|
W
|
O
|
R
|
K
|
Layer is of 5 length hence computer network is written in 5
columns horizontally. Then it is
read vertically as. CTTOEWMROPNRUEK
Q27.p3,j15. What are two modes of IP security?
A) Transport and certifiates
B) Transport and tunnel.
C) Certificate and tunnel.
D)Pre shared and transport.
Answer B
B) Transport and tunnel.
C) Certificate and tunnel.
D)Pre shared and transport.
Answer B
Q26,p3,j15. Which of the control field in TCP header is used to specify whether the sender has no more data to transmit.?
A)Fin B)RST C)SYN D) PSH.
Answer A.
Answer A.
Q25,p3,j15. Which transmission technique guarantees that data packets will be recieved by the receiver in the same order in which they were sent by the sender.
A) broadcasting B)unicasting C)Packet switching D)Circuit Switching.
Answer D.
Explanation.
Circuit switching is a methodology of implementing a telecommunications network in which two network nodes establish a dedicated communications channel or circuit through the network before the nodes may communicate.
Answer D.
Explanation.
Circuit switching is a methodology of implementing a telecommunications network in which two network nodes establish a dedicated communications channel or circuit through the network before the nodes may communicate.
Friday 27 November 2015
Q24,p3,j15. Which one of the follwing is true?
Options
A) The resolvent of two horn clause is not a horn clause.
B)The resolvent of two horn clause is a horn clause.
C) If we resolve a negated goal G against a fact or rule A to get clause C Then C has positive literal or non null goal.
D) if we resolve a negated goal G against a fact or rule A to get clause C then C has positive literals or null goal.
Answer B.
A) The resolvent of two horn clause is not a horn clause.
B)The resolvent of two horn clause is a horn clause.
C) If we resolve a negated goal G against a fact or rule A to get clause C Then C has positive literal or non null goal.
D) if we resolve a negated goal G against a fact or rule A to get clause C then C has positive literals or null goal.
Answer B.
Q23,p3,J15. Which of the following is false for the programming language Prolog.?
a)
A Prolog variable can be assigned a value once.
b)
Prolog is a strongly typed language.
c)
The scope of a variable in Prolog is a single
clause or a rule.
d)
The scope of a variable in Prolog is a single query.
Answer B.
Explanation.
Prolog is a statically typed language .
Q22,p3,J15. The clausal form of the disjunctive normal form ~Av~Bv~C v D is
Options
A)
A^B^C->D B)AvBVCVD->True. C)A^B^C^D->true D)A^B^C^D->false.
Answer A.
Explanation
~Av(~Bv~C)vD.
~AV ~(B^C)VD.
~(A^B^C)VD.
We know A-> B = ~AV B.
So it turns out to be A^B^C->D.
Q21,p3,j15. The transition function for the language L={w|na(w) and nb(w) are both odd} is given by.
Δ(q0,a)=q1.; Δ(q0,b)=q2.
Δ(q1,a)=q0. Δ(q1,b)=q3.
Δ(q2,a)=q3. Δ(q2,b)=q0.
Δ(q3,a)=q2. Δ(q3,b)=q1.
The initial and final states are.
Options.
A)
Q0 and q0
B)
Q0 and q1.
C)
Q0 and q2
D)
Q0 and q3.
Answer D.
Explanation.
Accepted words are { ab, ba, aaab,abaa, aaba,bbba-----}
A and b are of odd length. Hence from diagram from the above
transitions. option D is correct.
Q20,P3,j15. The regular expression corresponding to the language L
where L={xε {0,1}* |x ends with 1 and does not
contain substring 00} is:
Options
A.
(1+01)*(10+01)
B.
(1+01)*01
C.
(1+01)*(1+01)
D.
(10+01)01.
Answer C.
Explanation.
Option A can yield string terminating with 0 hence it fails.
Option D an yield
string 1001. Hence it also fails.
Option C. generates all string with length1,2,3,----- terminating
with 1 and does not contain 00 substring.
Q19,p3,j15. Minimal deterministic finite automation for the language L={0n, | n>=0, n|= 4} will have:
Options
A.)One final state among 5 states.
B.) 4 final states among 5 states.
C.) 1 final state among 6 states.
D.) 5. Final states among 6 states.
Answer D.
Explanation.
Accepted words are {Epsilon,o,oo,ooo,ooooo,oooooo---}
Q18,p3,j15. Match the following.
|
List 1
|
List 2
|
|
A. Flood
Gun
|
1. An
electron gun designed to flood the entire screen with electrons.
|
|
B. Collector
|
2. Partly
energized by flooding gun, stores the charge generated by the writing gun.
|
|
C. Ground
|
3. Used
to discharge the collector.
|
|
D. Phosphorus
|
4. Used
in memory tube display and similar to those used in standard CRT.
|
|
E. Writing
gun system
|
5. Used
in memory tube display , basically the
same as electron gun used in conventional CRT.
|
Codes
A B C D E
A. i. ii. Iii. Iv. V.
B. i. iii iv v i
C. III I ii. V. iv.
D. iv. V. i. ii. Iii.
Answer A.
Thursday 26 November 2015
Q17. P3,j15. Which of the following statements are incorrect?
1.
Mapping the coordinates of the points and lines
that form the pictures into the appropriate coordinates on the devices or
workstations is known as viewing
transformation.
2.
The right handed Cartesian coordinate system in whose coordinate we describe the
picture is known as world coordinate system.
3.
The coordinate system that corresponds to the
device or workstation where the image is to be displayed is known as physical device coordinate system.
4.
Left hand coordinate system In which the
displayed area of the virtual device corresponds to unit |x| square whose left
hand corner is at the origin of the coordinate system, is known as normalized
device coordinate system.
Codes
A)
I only. B)1
and 2. C) 3 only D) 4
only.
Answer D.
Explanation.
In normalized device coordinate system lower left corner
corresponds to (0,0) and upper right corner corresponds to (1,1).
Wednesday 25 November 2015
Q16,p3,j15. which of the following is no the basic primitive of the graphics kernel system.
Options.
A) polyline. B)polydraw. c) Fillarea D)Polymarker.
Answer B.
Explanation.
Basic primitive of GKS are Polyline,polymarker, fillarea,text.
A) polyline. B)polydraw. c) Fillarea D)Polymarker.
Answer B.
Explanation.
Basic primitive of GKS are Polyline,polymarker, fillarea,text.
Q15,p3,j15. The process of dividing an analog signal into a string of discrete outputs,each of constant amplitude is called.
Options
A)
Strobing B)Amplification C)Conditioning D)Quantization.
Answer D.
Q13,p3,j15. Give the number of principal vanishing point along with their direction for the standard perspective transformation.
A) Only one in direction K.
B)
Two in direction I,J.
C)
Three in direction I,J,k.
D)
Only two in direction j,K
Answer A.
Explanation.
As things get further away, from us, they seems smaller and closer together. When they get far enough away, distances become ever tinier and so form a single point.Q12,p3,j15. In the indexed scheme of blocks to a file, the maximum possible size of the file depends on.
Options
A)
The number of blocks used for index and size of
index.
B)
Size of blocks and size of address.
C)
Size of index.
D)
Size of blocks.
Answer B.
Tuesday 24 November 2015
Q11,p3,j15. The relation schema R1 and R2 form a lossless join decomposition of R if and only if.
1. R1 intersection R2 -->>(R1-R2)
2. R1->R2.
3. R1 intersection R2 ->>(R2-R1).
4.R2->R1 intersection R2.
codes
A) 1 and 2 happens. B)1 and 4 happens. C) 1 and 3 happens D) 2 and 3 happens.
Answer C.
Explanation.
for lossless decomposition .Atleast one of the following dependencies should be in F+.
R1 intersction R2->R1. Eq. 1.
or
R1 intersection R2->R2. Eq2.
when a decompostion satisfies eq 1. given above then that will solve the R1 intersection R2 -->>(R1-R2).
when a decomposition satisfies eq2 given above then that will solve the R1 intersection R2 -->>(R2-R1).
2. R1->R2.
3. R1 intersection R2 ->>(R2-R1).
4.R2->R1 intersection R2.
codes
A) 1 and 2 happens. B)1 and 4 happens. C) 1 and 3 happens D) 2 and 3 happens.
Answer C.
Explanation.
for lossless decomposition .Atleast one of the following dependencies should be in F+.
R1 intersction R2->R1. Eq. 1.
or
R1 intersection R2->R2. Eq2.
when a decompostion satisfies eq 1. given above then that will solve the R1 intersection R2 -->>(R1-R2).
when a decomposition satisfies eq2 given above then that will solve the R1 intersection R2 -->>(R2-R1).
Q10,P3,J15. The Relation
vendor order(v_no,v_ord_no,v_name,qty_sup,unit_price) is in 2nf because:
options
A) Non key attribute v_name is dependent on v_no ,which is part of composite key.
B)Non key attribute v_name is dependent on qty_sup.
C)key attribute qty_sup is dependent on primary key unit_price.
D)Key attribute v_ord_no is dependent on primary key unit_price.
Answer A.
options
A) Non key attribute v_name is dependent on v_no ,which is part of composite key.
B)Non key attribute v_name is dependent on qty_sup.
C)key attribute qty_sup is dependent on primary key unit_price.
D)Key attribute v_ord_no is dependent on primary key unit_price.
Answer A.
Monday 23 November 2015
Q9,p3,J15.select the False statement from the following statements about Normal forms.
A) Loss less preserving decomposition in to 3nf is always possible.
B)Loss less preserving decomposition in to BCNF is always possible.
C)Any relation with two attributes is in BCNF.
D)BCNF is stronger than 3NF.
Answer B.
explanation.
Table 1(A,B,C) with FD AB->C and C->B cannot be preserved when decomposition is done.
B)Loss less preserving decomposition in to BCNF is always possible.
C)Any relation with two attributes is in BCNF.
D)BCNF is stronger than 3NF.
Answer B.
explanation.
Table 1(A,B,C) with FD AB->C and C->B cannot be preserved when decomposition is done.
Q8,p-3,j15.The student information in a university is stored in the relation
STUDENT(Name,sex,marks,Dept_name)
Consider the following SQL query "Select Dept_name from student where sex='M' group by dept_name having avg(marks)>select avg(marks)from student . It returns the name of department for which:
Options
A)The average marks of male students is more than the average marks of the same department.
B)The average marks of male students is more than the average marks of students in the university.
C)The average marks of male students is more than the average marks of male students in university.
d) The average marks of students is more than the average marks of male students in the university.
Answer B.
Consider the following SQL query "Select Dept_name from student where sex='M' group by dept_name having avg(marks)>select avg(marks)from student . It returns the name of department for which:
Options
A)The average marks of male students is more than the average marks of the same department.
B)The average marks of male students is more than the average marks of students in the university.
C)The average marks of male students is more than the average marks of male students in university.
d) The average marks of students is more than the average marks of male students in the university.
Answer B.
Q7,p3,J15. Let E1 and E2 be two entities in ER diagram with simple single valued attributes. R1,R2 are two relationships between E1 and E2. where R1 is one to many and R2 is many to many. R1 and R2 do not have any attributes of their own.How many minimum number of tables are required to represent this situation in relational model?
Options A)4 B)3 C)2 D)1.
Answer B.
Explanation.
Let relation R1 is composed of table emp(ssn,name, Superssn). A query like this can be made from single table like. Find the employee ssn, name, super visor ssn, supervisor name from emp table. An ssn will map to a single supervisor and a supervisor can map to many employee.
In many to many relationship like "a student can take many subjects and many students can take a subject". it needs two table.eg. student(rollno,subject,name) where pk= roll no and subject. and fk = subject.
Foreign key map to pk of subjet (Subjetid, Subjname, Roomno).
Hence it required 3 tables
Answer B.
Explanation.
Let relation R1 is composed of table emp(ssn,name, Superssn). A query like this can be made from single table like. Find the employee ssn, name, super visor ssn, supervisor name from emp table. An ssn will map to a single supervisor and a supervisor can map to many employee.
In many to many relationship like "a student can take many subjects and many students can take a subject". it needs two table.eg. student(rollno,subject,name) where pk= roll no and subject. and fk = subject.
Foreign key map to pk of subjet (Subjetid, Subjname, Roomno).
Hence it required 3 tables
Sunday 22 November 2015
Q6,p3,j15 , The number of flip flops required to design a modulo 272 counter is
Options A)8 B)9 C)27. D.)11.
Answer B
Explanation.
The number of flipflop required to design a Modulo 272 counter is Log2 272.
8 flips flops do the work up to 256. hence we need 9 flip flops to achieve the task.
Answer B
Explanation.
The number of flipflop required to design a Modulo 272 counter is Log2 272.
8 flips flops do the work up to 256. hence we need 9 flip flops to achieve the task.
Q4,p3,j15. The equivalent hexadecimal notation for octal number 2550276 is.
Options A)FADED B)AEOBE C)ADOBE D)ACABE
Answer C.
Explanation .
We write binary equivalent of every no in a group of 3.
So 2550276 is written as below.
010 for 2.
101 for 5
and similarly.
so we organize every thing as.
010101101000010111110.
Now this binary representation is grouped in to 4 to get the Hexadecimal notation.
Answer C.
Explanation .
We write binary equivalent of every no in a group of 3.
So 2550276 is written as below.
010 for 2.
101 for 5
and similarly.
so we organize every thing as.
010101101000010111110.
Now this binary representation is grouped in to 4 to get the Hexadecimal notation.
0
1010 1101 0000 1011
1110 so after converting every group in HEX. we get
ADOBE.
Q3,p3,j15. RST 7 instruction in 8085 microprocessor is equivalent to.
Option A)call 0010 H B) Call 0034 H C)CALL 0038 H D)CALL 003CH
Answer C.
Explanation.
Answer C.
Explanation.
|
Interrupt
|
Vector Address
|
|
RST 0
|
0000H
|
|
RST 1
|
0008H
|
|
RST2
|
0010H
|
|
RST3
|
0018H
|
|
RST4
|
0020H
|
|
RST5
|
0028H
|
|
RST6
|
0030H
|
|
RST7
|
0038H
|
Q2,p3,j15. Consider a 32 bit microprocessor with a 16 bit external data bus,driven by 8 Mhz input clock.Assume that microprocessor has a bus cycle whose minimum duration equals to four input clock cycle. What is the maximum data transfer rate for this micro processor.
Options A) 8x10 6bytes /sec B) 4x10 6 bytes/sec c) 16x 10 6 bytes/sec
Answer. BExplanation.
Minimum bus cycle duration = 4 clock cycle.
maximum bus cycle rate= 8 MHZ/4= 2M/s where 1 MHZ= 1000000/sec.
maximum data transferred per bus cycle rate= 2bytes.
data transferred rate per second = data transferred per bus cycle* bus cycle rate.= 2bytes*2M/s
4* 106 bytes/sec
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