Monday, 16 January 2017

UGC Net Computer Science Paper II Dec 13 Page 4 Solved

UGC Net Computer Science Paper II Dec 13 Page 4 Solved

UGC Net Computer Science Paper II Dec 13 Page 4 Solved

31. The dual of a Boolean expression is obtained by interchanging
(A) Boolean sums and Boolean products
(B) Boolean sums and Boolean products or interchanging 0’s and 1’s
(C) Boolean sums and Boolean products and interchanging 0’s & 1’s
(D) Interchanging 0’s and 1’s
Answer C
Explanation :-
Dual of a Boolean expression
The duality principle ensures that "if we exchange every symbol by its dual in a formula, we get the dual result".
(a) 0 . 1 = 0: is a true statement asserting that "false and true evaluates to false"

(b) 1 + 0 = 1: is the dual of (a): it is a true statement asserting that "true or false evaluates true."

(c) 1 . 1 = 1: it is a true statement asserting that "true and true evaluates to true".

(d) 0 + 0 = 0: (d) is the dual of (c): it is a true statement asserting, correctly, that "false or false 

More example
F1 = x'yz' + x' y' z
The dual of F1 is : ( x' + y + z' ) ( x' + y' + z )
Complement each literal : F1' = ( x + y' + z ) ( x + y + z' )
F2 = x ( y' z' + yz )
The dual of F2 is x + ( y' + z' ) ( y + z )
Complement each literal: F2' = x' + ( y + z ) ( y' + z' )  

Product of Sums (POS)

A boolean expression consisting purely of Maxterms (sum terms) is said to be in canonical product of sums form.
Example
Lets say, we have a boolean function F defined on two variables A and B. So, A and B are the inputs for F and lets say, output of F is true i.e., F = 1 when only one of the input is true or 1.
now we draw the truth table for F


Now we will create a column for the maxterm using the variables A and B. If input is 1 we take the complement of the variable and if input is 0 we take the variable as is.


To get the desired canonical POS expression we will multiply the maxterms (sum terms) for which the output is 0.
F = (A+B) . (A’+B’)

UGC Net Computer Science Paper II Dec 13 Page 3 Solved

 UGC Net Computer Science Paper II Dec 13 Page 3 Solved

UGC Net Computer Science Paper II Dec 13 Page 3 Solved

21. What is the value of the postfix expression ?
a b c d + – ∗ (where a = 8, b = 4,
c = 2 and d = 5)
(A) –38
(B) –83
(C) 24 
(D) –24
Answer D
Explanation :-
Algorithm for evaluation postfix expressions.

1) Create a stack to store operands (or values).
2) Scan the given expression and do following for every scanned element.
a) If the element is a number, push it into the stack
b) If the element is a operator, pop operands for the operator from stack.
           ​Evaluate the operator and push the result back to the stack 
3) When the expression is ended, the number in the stack is the final answer

 Postfix Expression=a b c d + - *

UGC Net Computer Science Paper II Dec 13 Page 2

UGC Net Computer Science Paper II Dec 13 Page 2

UGC Net Computer Science Paper II Dec 13 Page 2 Solved

11. The student marks should not be greater than 100. This is
(A) Integrity constraint
(B) Referential constraint
(C) Over-defined constraint
(D) Feasible constraint
Answer A
Explanation :-
Before one can start to implement the database tables, one must define the integrity constraints. Intergrity means something like 'be right' and consistent. The data in a database must be right and in good condition.

There are the domain integrity, the entity integrity, the referential integrity and the foreign key integrity constraints.

Domain Integrity

Domain integrity means the definition of a valid set of values for an attribute. You define 
data type, 
  • lenght or size
  • is null value allowed
  • is the value unique or not
  • for an attribute.

Sunday, 15 January 2017

UGC Net Computer Science Paper II Dec 13 Page 1

UGC Net Computer Science Paper II Dec 12 Page 1

UGC Net Computer Science Paper II Dec 13 Page 1

1. When data and acknowledgement are sent in the same frame, this is called as
(A) Piggy packing
(B) Piggy backing
(C) Back packing
(D) Good packing
Answer B
Explanation :-
he usual purpose of piggybacking is simply to gain free network access rather than any malicious intent, but it can slow down data transfer for legitimate users of the network. Furthermore, a network that is vulnerable to piggybacking for network access is equally vulnerable when the purpose is data theft, dissemination of viruses, or some other illicit activity.


It's quite simple to access an unsecured wireless network: All you have to do is get into the range of a Wi-Fi hotspot's signal and select your chosen network from the options presented. However, unauthorized network access, even to free Wi-Fi, may be illegal. People have been fined for accessing hot spots from outside businesses, such as coffee shops, that provide free Wi-Fi for customers' use. (http://whatis.techtarget.com/)

2. Encryption and Decryption is