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

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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)

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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

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

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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.

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

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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:

ProcessArrival TimeBurst Time
P107
P214
P328

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.

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

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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:


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

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    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

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

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    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 
      

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