UGC-NET December 2020 and June 2021 computer Science and application solutions

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Explanation:
Regular language does not have any memory to compare.
A. L={ (01)n0k | n > k, k>=0 }
Here, occurance of n is greater than occurance of k, for that we need to compare value of n with k, and regular language does not have power to compare. so,it is not regular language. It is Context Free Language.

B. L={ cnbkan+k | n >= 0, k>=0 }
Here, occurance of c is less than or equal to ocuurance of a and Finite Automata cannot do this comparison. So, it is not regular language . It is Context Free Language.

C. L={ 0n1k | n≠k }
Here, occurance of n is not equal to occurance of k, and this type of comparison cannot be done by Finite Automata.So, it is also not regular language . It is Context Free Language.

So, Option(III) is correct.

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Explanation:
Given,
pushdown automata M=({q0, q1, q2}, {a, b}, {a, b, z}, δ, q0, z, {q2})
where,
{q0, q1, q2} = Finite state of states
{a, b} = Input alphabet
{a, b, z} = Stack alphabet
δ = Transition Function
q0 = Initial state
z = Bottom of the stack
{q2} = Final state
Let’s Draw pushdown automata for a given Transition function: First, keep pushing any number of a’s and b’s in the stack , and then pop one a from stack on seeing input alphabet a and pop one b from stack on seeing input alpbabet b. Keep popping untill we reach end of string λ. when we reach end of string λ and Top of Stack is z, then go to final state.
Let’s take one string and analyze them
suppose string Language generated by above push down automata (L)= {wwR | w Є {a, b}+}
So, Option(IV) is correct.

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Explanation:
Given,
(D.C)16
Calculation,
To obtain octal equivalent for a given number , first convert into base 2 representation and then make a group of 3 bits.
why 3 bits? Because octal representation have base 8 and log28 = 3bits.
(D)16 = (13)10 = (1101)2
(C)16 = (12)10 = (1100)2
(D.C)16 = (1101.1100)2
make a pair of 3 bits
(D.C)16 = (001 101.110 000)2 = (15.60)8 = (15.6)8
So, Option(II) is correct.

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Explanation:
Given,
Block size = 1KByte = 1024 Byte
Record Pointer = 7 Byte
Key Size = 9 Byte
Block pointer = 6 Byte

Order of leaf node =?
Formula,
n*(Key size + Record Pointer) + Block Pointer <= Block size
Calculation,
=> n*(9+7) + 6 <= 1024 Byte
=> 16n <= 1018
=> n <= 1018/16
=> n <= 63.625
=> n = 63
So,Option(I) is correct.

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Explanation:
Statement A: L3 =L1 ∩ L2
L1 = { 0n1n0m | n>=1, m>=1 }
= {010, 0100, 00110, 001100……………………}
L2 = { 0n1m0m | n>=1, m>=1 }
= {010, 0010, 01100, 001100…………………. }
L3 =L1 ∩ L2
= {010, 001100, 000111000…………………….}
= {0n1n0n | n>=1}
Statement A is correct.
Statement B: L1 and L2 are context free languages but L3 is not a context free language
L1 = { 0n1n0m | n>=1, m>=1 }
let’s draw push down automata for above language: =1, m>=1 }”>
Language L2 is Context Free Language.

L3 = { 0n1n0n | n>=1}
Here, n times 0 followed by n times 1 followed by n times 0, this kind of comparison cannot be done by Push Down Automata. So, it is not Context Free Language. It is Context Sensitive Language.
Statement B is correct.

Statement C: L1 and L2 are not context free languages but L3 is a context free language
As statement B is correct, Statement C is obviously incorrect.

Statement D: L1 is a subset of L3
L1 = { 0n1n0m | n>=1, m>=1 }
= {010, 0100, 00110, 001100……………………}
L3 ={ 0n1n0n | n>=1}
= {010, 001100, 000111000…………………….}
As see, L1 is superset L3
Statement D is incorrect.
So, Option(IV) is correct.

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Explanation:
S-> XY
->0XY | 1XY | 0Y
->0XY0 | 0XY1 | 0X0 | 1XY0 | 1X0 | 0Y0 | 0Y1 | 00
->0XY0 | 0XY1 | 000 | 1XY0 | 100 | 000 | 001 | 00

Option(i): has at least one 1
This is incorrect, because one of the string of above language is 00 which don’t contain 1.

Option(ii): has no consecutive 0’s or 1’s
This is incorrect, because one of the string of above language is 100 which has consecutive 0’s.

Option(iii): should end with 0
This is incorrect, because one of the string of above language is 001 which ends with 1.

Option(iv): has at least two 0’s
This is correct. Because X and Y always end with 0, so string has atleast two 0’s.

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Answer: (Marks to all from UGC NET )

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UGC-NET December 2020 and June 2021 computer Science and application solutions

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Explanation:
The inﬁx expression form : operand1 operator operand2
The reverse Polish notation form: operand1 operand2 operator
Order of precedence: = (A + B) * (C * D – E) * F / G
= [A*{B+C*(D+E)}] / {F*(G+H)}
= (A+B) * (C * D – E) * F / G
= AB+ * (C * D – E) * F / G
= AB+ * (CD* E) * F / G
= AB+ * CD*E– * F / G
= (AB+ CD*E – *) * F / G
= AB+ CD*E – * F * / G
= A B + C D * E – * F * G /
Note: Here blue is used to highlight operands and Orange is used to highlight operator that need to be evaluated first.
So, Option(I) is correct.

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Explanation:
Some Boolean algebra laws:
i) Identity Law
x + 0 = x , x .1 = x
ii) Absorption Law
x + xy = x , x.(x + y) = x
iii) Idempotent Law
x + x = x , x . x = x
iv) Domination Law
x + 1 = 1 , x . 0 = 0
v) Commutative Law
x + y = y + x , x . y = y . x
vi) Associative Law
x + (y + z) = (x + y) + z , x .(y . z) = (x . y). z
vii) Distributive Law
x.(y + z) = x.y + x.z , x + (y.z) = (x + y).(x + z)
viii) Complementation Law
x + x’ = 1 , x . x’ = 0
ix) DeMorgan’s Law
(x . y)’ = x’ + y’ , (x + y)’ = x’ . y’

So, Option(III) is correct.

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Explanation:
ODD Function: X-OR is an ODD Function Because it gives High output for Odd number of High inputs.
Le us check for two input A and B
Logical equation of X-OR gate.
Y = A B = AB’ + A’B Universal Gate: The NAND Gate and NOR Gate are said to be Universal Gate because any Logic circuit can be implemented with it.
Self Complementing Code: In Self Complementing code, the code of a digit and the code of 9’s complement of the digit are 1’s complement to each other.
2421 code
let’s take decimal number 2,
9’s complement of the digit 2 = 7
the 2421 code of a digit 2 = 0010 ………….Code(1)
the 2421 code of a digit 7 = 1101 ………….Code(2)
Code (1) and code(2) are 1’s complement to each other. That’s why 2421 code is self complement code.
Buffer: Amplification
So, Option(III) is correct.

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Explanation:
Combinational Circuits
• Combinational circuits are defined as time independent circuits which do not depends upon previous inputs to generate any output.
• Output depends upon only present input.
• As combinational circuit don’t have clock, they don’t require triggering. A. Output at any time is function of inputs at that time. (True)
Outputs depend on present inputs only.
B. Contains memory elements. (False)
• Combinational Circuits does not have any memory elements because it depends on present inputs only.
C. Do not have feedback paths. (True)
D. Clock is used to trigger the circuits to obtain outputs. (False)
So, Option(II) is correct.

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Explanation:
The inﬁx expression form : operand1 operator operand2
The reverse Polish notation form: operand1 operand2 operator
Order of precedence: = [A*{B+C*(D+E)}] / {F*(G+H)}
= [A*{B+C*DE+}] / {F*(G+H)}
= [A*{B+CDE+*}] / {F*(G+H)}
= [A*BCDE+*+] / {F*(G+H)}
= [ABCDE+*+*] / {F*(G+H)}
= [ABCDE+*+*] / {F*GH+}
= [ABCDE+*+*] / {F*GH+}
= [ABCDE++] / {FGH+*}
= ABCDE+*+* FGH+*/
Note: Here blue is used to highlight operands and Orange is used to highlight operator that need to be evaluated first.
So, Option(IV) is correct.

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Explanation:
Class B : Out of 32 bits, Class B have
Network id = 16 bits and
Host id = 16 bits
Calculation
Given subnet bits = 6 bits
Maximum number of the subnets = 26 – 2 = 62
Note that 2 is substracted because
* all 0’s is used to represent Network and
Maximum number of hosts in each subnet
Host id bits = 32 – (Network id bit + subnet id bits) = 32 – (16+ 6) = 10
Maximum number of hosts in each subnet = 210– 2 = 1022
Note that 2 is substracted because
* all 0’s is used to represent Subnet and
So, Option(I) is correct.

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Explanation:
Given,
Number of registers = 8
Number of bits in each register = 16bits
Formula,
Number of multiplexer in the bus = number of bits in each register
Size of multiplexer : number of register X 1
Calculation,
Number of multiplexer in the bus:
Number of Multiplexer = number of bits in each register
Here, number of bits in each register = 16
Therefore, Number of register = 16
Size of multiplexer:
Size of multiplexer : number of register X 1
Here, number of register = 8
Therefore, Size of multiplexer = 8×1
So, Option(II) is correct.

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Explanation:
Formula,
Conditional Probability Calculation,
The probability of a toothache, given evidence of a cavity,  P(toothache Ո cavity ) = 0.108 + 0.012 = 0.12 P(cavity ) = 0.108 + 0.012 + 0.072 + 0.008 = 0.2  So,Option(IV) is correct.

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Explanation:
Formula,
P(A U B) = P(A) + P(B) – P(A Ո B)
Calculation,
P(cavity U toothache) = P(cavity) + P(toothache) – P(cavity Ո toothache) P(cavity ) = 0.108 + 0.012 + 0.072 + 0.008 = 0.2 P(toothache) = 0.108 + 0.012 + 0.016 + 0.064 = 0.2 P(cavity Ո toothache) = 0.108 + 0.012 = 0.12

P(cavity U toothache) = P(cavity) + P(toothache) – P(cavity Ո toothache)
P(cavity U toothache) = 0.2 + 0.2 – 0.12 = 0.28
So, Option(III) is correct.

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Explanation:
Formula, Calculation,  P(cavity Ո (toothache U catch)) = 0.108 + 0.012 + 0.072 = 0.192 P(toothache U cavity) = 0.108 + 0.012 + 0.072 + 0.016 + 0.064 + 0.144 = 0.416  = 0.4615
So, Option(I) is correct.

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Explanation:
Calculation, P(cavity ) = 0.108 + 0.012 + 0.072 + 0.008 = 0.2

So, Option(III) is correct.

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Explanation:
Formula,
Conditional Probability Calculation,
The probability of a cavity, given evidence of a toothache,  P(cavity Ո toothache ) = 0.108 + 0.012 = 0.12 P(toothache) = 0.108 + 0.012 + 0.016 + 0.064 = 0.2  = 0.6
So,Option(IV) is correct.

UGC-NET December 2020 and June 2021 computer Science and application solutions

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