Java Program 41 - Java Program to Print Alphabet Patterns - Study Viral

Java Program 41 - Java Program to Print Alphabet Patterns - Study Viral

In This Video You Can See how to Make Following
Kind of Alphabet Pattern...With two ways using integer and character data-type

EDCBA
EDCB
EDC
ED
E

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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 61-64)

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 61-64)

Q 61. In RDBMS, which type of Join returns all rows that satisfy the join condition?
 
(A) Inner Join
(B) Outer Join
(C) Semi Join
(D) Anti Join

Answer : (B) Outer Join

Explanation : Outer Join return all rows that satisfy condition from both the participating relations. An outer join extends the result of a simple join. An outer join returns all rows that satisfy the join condition and also returns some or all of those rows from one table for which no rows from the other satisfy the join condition. 

 

Q 62. Consider a relation book (title, price) which contains the titles and prices of different books. Assuming that no two books have the same price, what does the following SQL query list ?

Select title
from book as B
where (select count ( * )
    from book as T
    where T.price > B.price) < 7

(A) Titles of the six most expensive books.
(B) Title of the sixth most expensive books.
(C) Titles of the seven most expensive books.
(D) Title of the seventh most expensive books.

Answer :  (C) Titles of the seven most expensive books.

Q 63.  In a Hierarchical database, a hashing function is used to locate the ________.

(A) Collision
(B) Root
(C) Foreign Key
(D) Records

Answer :  (B) Root

Explanation :  A Data Model is the methodology used by a particular DBMS to organize and access the data.

Hierarchical, Network and Relational Model are the three popular data models. 

However, the relational model is more widely used.

Hierarchical Model :

• The hierarchical model was developed by IBM in 1968.
  
• The data is organize in a tree structure where the nodes represent the records and the branches of the tree represent the fields.
  
• Since the data is organized in a tree structure, the parent node has the links to its child nodes.
  
• If we want to search a record, we have to traverse the tree from the root through all its parent nodes to reach the specific record. Thus, searching for a record is very time consuming.
  
• The hashing function is used to locate the root.
  

Q 64. Relations produced from E – R Model will always be in ________.

(A) 1 NF
(B) 2 NF
(C) 3 NF
(D) 4 NF

Answer :  (A) 1 NF

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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 60)

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 60)

Q 60. In which of the following scheduling criteria, context switching will never take place ?

(A) ROUND ROBIN
(B) Preemptive SJF
(C) Non-preemptive SJF
(D) Preemptive priority


Answer : (C) Non-preemptive SJF


In Non – preemptive algorithms context switching will never take place because it doesn’t allow to switch the process until it is completed fully.

Except Non-Preemptive SJF , All are Preemptive.
CPU will leave the Shotest Process only after its Full execution,Once process is started to execute.

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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 59)

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 59)

Q 59. Consider the following three processes with the arrival time and CPU burst time given in milliseconds:

Process	Arrival Time	Burst Time
P1	0	7
P2	1	4
P3	2	8

The Gantt Chart for preemptive SJF scheduling algorithm is _________.

(A)

P1

P2

P3

0                                           7                      13                        21

(B)

P1

P2

P1

P

0                      1                        5                      11                    19    

(C)

P1

P2

P3

0                                           7                      13                        19

(D)

P2

P3

P1

0                      4                       12                                            19

Answer : (B)

Explanation : 
A) Not preemptive. P1 continues execution without being preempted.
B) Preemptive SJF . P1 executes for 1 second followed by execution of P2 for 4 secs. Then P1 finishes execution followed by P3.
C) Not preemptive as each of the processes carries on execution without being preempted.
D)Preemptive but P2 starts execution even before it has arrived which can't happen in reality.

P₁ arrive at 0 it will be served by CPU for 1 unit. After 1 unit P₂ arrived ans it is shortest So it will execute for 4 unit, after its execution There are 2 processes out of which P₁ is shortest so it will execute for 6 unit after its execution P₁ will execute for 8 unit.

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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 56-58)

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 56-58)

Q 56. Which of the following statements are true ?
(a) External Fragmentation exists when there is enough total memory space to satisfy a request but the available space is contiguous.
(b) Memory Fragmentation can be internal as well as external.
(c) One solution to external Fragmentation is compaction.

Code:
(A) (a) and (b) only
(B) (a) and (c) only
(C) (b) and (c) only
(D) (a), (b) and (c)

Answer : (C) (b) and (c) only

Explanation:

External Fragmentation exists when there is enough total memory space to satisfy a request but the available space is contiguous. This is a false statement because external fragmentation occurs due to non contiguous available space.

Memory Fragmentation can be internal as well as external.True

One solution to external Fragmentation is compaction or shuffle memory contents.True

Q 57. Page information in memory is also called as Page Table. The essential contents in each entry of a page table is/are _________.

(A) Page Access information
(B) Virtual Page number
(C) Page Frame number
(D) Both virtual page number and Page Frame Number

Answer : (C) Page Frame number

Explanation: Page information in memory is also called as Page Table. The essential contents in each entry of a page table is page frame number. (i.e. Frame Number is essential in page table.)


Q 58. Consider a virtual page reference string 1, 2, 3, 2, 4, 2, 5, 2, 3, 4. Suppose LRU page replacement algorithm is implemented with 3 page frames in main memory. Then the number of page faults are_________.

(A) 5
(B) 7
(C) 9
(D) 10


Answer : (B) 7

Explanation: Page replacement diagram for LRU:

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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 54-55)

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 54-55)

Q 54. Normally user programs are prevented from handling I/O directly by I/O instructions in them. For CPU's having explicit I/O instructions, such I/O protection is ensured by having the I/O instructions privileged. In a CPU with memory mapped I/O, there is no explicit I/O instruction. Which one of the following is true for a CPU with memory mapped I/O ?

(A) I/O protection is ensured by operating system routines.
(B) I/O protection is ensured by a hardware trap.
(C) I/O protection is ensured during system configuration.
(D) I/O protection is not possible.

Answer : (A) I/O protection is ensured by operating system routines.

Explanation: Memory mapped I/O means, accessing I/O via general memory access as opposed to specialized IO instructions.
The programmer can directly access any memory location directly. To prevent such an access, the OS (kernel) will divide the address space into kernel space and user space. An user application can easily access user application. To access kernel space, we need system calls (traps).


Q 55. Which UNIX/Linux command is used to make all files and sub-directories in the directory “progs” executable by all users ?

(A) chmod− R a+x progs
(B) chmod −R 222 progs
(C) chmod−X a+x progs
(D) chmod −X 222 progs

Answer : (A) chmod− R a+x progs

Solution : Option A will make all files and sub-directories in the directory (progs) executable by all users.
-R is for recurssive calls, so that sub-directories within the directories can be accessed.
a is for all users
+x is for making things executable

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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 51-53)

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 51-55)

Q 51. At a particular time of computation, the value of a counting semaphore is 10. Then 12 P operations and “x” V operations were performed on this semaphore. If the final value of semaphore is 7, x will be:

(A) 8
(B) 9
(C) 10
(D) 11

Answer :  (B) 9
Solution : P operation : Decrements the value of semaphore by 1
V operation : Increments the value of semaphore by 1
Initially, value of semaphore = 10
After 12 P operations, value of semaphore = 10 – 12 = – 2
Now, after x V operations, value of semaphore = 7

– 2 + xV = 7
xV = 2 + 7 = 9 

Q 52. In a paged memory, the page hit ratio is 0.40. The time required to access a page in secondary memory is equal to 120 ns. The time required to access a page in primary memory is 15 ns. The average time required to access a page is ________.

(A) 105
(B) 68
(C) 75
(D) 78

Answer :  (D) 78

Solution : Average access time = hit ratio * primary memory access time + (1 – hit ratio) * secondary memory access time

Average access time = 0.4 * 15 + 0.6 * 120←
Average access time = 6 + 72
Average access time = 78.
So, option (D) is correct.


Q 53. In a multi-user operating system, 30 requests are made to use a particular resource per hour, on an average. The probability that no requests are made in 40 minutes, when arrival pattern is a poisson distribution, is _________.
 
(A) e^-15
(B) 1 – e^-15
(C) 1 – e^-20
(D) e^-20


Answer : (D) e^-20

Solution : In One Hour = 30 Requests Sent
 in 40 minutes =   30 x 40   =  20  Requests Sent

                                        ^{60 
 Here 20 is Number of request Sent in 40 minutes. (Let 𝛌 is number of request in 40 minutes)
Therefore}  𝛌 = 20

^{But according to given statement in question Actual requests are made in 40 minutes is ZERO.
Let x is actual requests in 40 minutes. (According to Question)
i.e. x = 0} 

  

e−t/θe−t/θ
^e−t/θ

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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 46-50)

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 46-50)

Q 46. Which of the following statements are true ?

(a) Three broad categories of Networks are

     (i) Circuit Switched Networks

    (ii) Packet Switched Networks

    (iii) Message Switched Networks

(b) Circuit Switched Network resources need not be reserved during the set up phase.

(c) In packet switching there is no resource allocation for packets.

Code :

1. (a) and (b) only
2. (b) and (c) only
3. (a) and (c) only
4. (a), (b) and (c)

Answer: 3). (a) and (c) only

Q 47. In Challenge-Response authentication the claimant ________.

1. Proves that she knows the secret without revealing it
2. Proves that she doesn’t know the secret
3. Reveals the secret
4. Gives a challenge

Answer: 1). Proves that she knows the secret without revealing it

Reason: Challenge-Response authentication is a computer security protocol in which one party ask a question and another party must provide a valid answer. I.e. she knows the secret without revealing it or will provide the secret when it is asked.

Q 48. Decrypt the message “WTAAD” using the Caesar Cipher with key=15.

1. LIPPS
2. HELLO
3. OLLEH
4. DAATW

Answer: 2). HELLO
Reason: In Ceaser cipher algorithm encryption and decryption takes place. During encryption right shift takes place depending upon the key and in decryption left shift takes place depending on key. In the given question key value is 15 and decryption is asked:
Given message is WTAAD
Left shift 15 on W is H
Left shift 15 on T is E
Left shift 15 on A is L
Left shift 15 on A is L
Left shift 15 on D is O
The decrypted message will HELLO.

Q 49. To guarantee correction of upto t errors, the minimum Hamming distance dmin in a block code must be ________.

1. t+1
2. t−2
3. 2t−1
4. 2t+1

Answer: 4). 2t+1

Q 50. Encrypt the Message “HELLO MY DEARZ” using Transposition Cipher with

1. HLLEO YM AEDRZ
2. EHOLL ZYM RAED
3. ELHL MDOY AZER
4. ELHL DOMY ZAER

Answer: 3). ELHL MDOY AZER 
Reason:
In question encryption of given message using Transposition Cipher with 

Given message is “HELLO MY DEARZ” Now arrange them in group of 4
I.e. HELL OMYD EARZ
i.e. replace the following...

• second character in above format to the first place
• fourth character to the second place
• first character to the third place
• third character to the fourth place.

Encrypted message will be ELHL MDOY AZER.

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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 41-45)

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 41-45)

Q 41. A slotted ALOHA network transmits 200-bit frames using a shared channel with a 200 Kbps bandwidth. Find the throughput of the system, if the system (all stations put together) produces 250 frames per second :

1. 49
2. 368
3. 149
4. 151

Answer: 1). 49


Solution : Here G is 1/4, Therefore S = G x e^-G = 0.195 (19.5), 
this means 250 x 0.195 = 49. Only 49 frames out of 250 will probably survive.

Q 42. The period of a signal is 100 ms. Its frequency is _________.

1. 100³ Hertz
2. 10⁻² KHz
3. 10⁻³ KHz
4. 10⁵ Hertz

Answer: 2). 10⁻² KHz
Solution: Explanation: We know that:
frequency(f) = 1 / Time period(T)
Given time period(T) = 100 ms
f = 1 / 100ms
f = 1000 / 100000 ms
We know that 1000 ms = 1 sec and we can write 1000 as 1 k, i.e. f = 1k / 100s
f = 10^-2 kHz (1 / s = Hz) So, option (B) is correct.

Q 43. The dotted-decimal notation of the following IPV4 address in binary notation is _________.

10000001 00001011 00001011 11101111

1. 111.56.45.239
2. 129.11.10.238
3. 129.11.11.239
4. 111.56.11.239

Answer: 3). 129.11.11.239

Explanation:  IPV4 address = 10000001 00001011 00001011 11101111
10000001 = 129,       00001011 = 11,     00001011 = 11,     11101111 = 239

IPV4 address 10000001 00001011 00001011 11101111 is equivalent to 129.11.11.239
So, option (C) is correct.

Q 44. Which of the following statements are true ?

(a) Advanced Mobile Phone System (AMPS) is a second generation cellular phone system.

(b) IS – 95 is a second generation cellular phone system based on CDMA and DSSS.

(c) The Third generation cellular phone system will provide universal personnel communication.

Code :

1. (a) and (b) only
2. (b) and (c) only
3. (a), (b) and (c)
4. (a) and (c) only

Answer: 2). (b) and (c) only
Reason:

• Advanced Mobile Phone System (AMPS) is a First generation cellular phone system.False
• IS – 95 is a second generation cellular phone system based on CDMA and DSSS.True 
• The Third generation cellular phone system will provide universal personnel communication.True

SO, option (B) is correct.

45. Match the following symmetric block ciphers with corresponding block and key sizes :

Answer: 2). (ii)     (iv)       (i)      (iii)

Reason:

Data Encryption Standard(DES) block cipher uses block size 64 and key size 64

International Data Encryption Algorithm (IDEA) uses block size 64 and key size 12.

BLOW FISH is the relacement for DES or IDEA, it uses block size 64 and key size ranges between 32 and 448

Advanced Encryption Standard (AES) uses block size 128 and key sizes 128, 192, 256

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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 37-40)

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 37-40)

Q 37. Context sensitive language can be recognized by a :

(1) Finite state machine

(2) Deterministic finite automata

(3) Non-deterministic finite automata

(4) Linear bounded automata

Answer:  (4) Linear bounded automata

Q 38. The set A={ 0ⁿ 1ⁿ 2ⁿ | n=1, 2, 3, ……… } is an example of a grammar that is :

(1) Context sensitive

(2) Context free

(3) Regular

(4) None of the above

 

Answer:  (1) Context sensitive

Q 39. A bottom-up parser generates :

(1) Left-most derivation in reverse

(2) Right-most derivation in reverse

(3) Left-most derivation

(4) Right-most derivation

 

Answer:  (2) Right-most derivation in reverse

Q 40. Consider the following statements( ) :

S1 : There exists no algorithm for deciding if any two Turing machines M1 and M2 accept the same language.

S2 : The problem of determining whether a Turing machine halts on any input is undecidable.

Which of the following options is correct ?

(1) Both S1 and S2 are correct

(2) Both S1 and S2 are not correct

(3) Only S1 is correct

(4) Only S2 is correct

 

Answer:  (1) Both S1 and S2 are correct

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