## ISRO-2016

Question 1 |

Which of the following is true?

√3 + √7 = √10 | |

√3 + √7 ≤ √10 | |

√3 + √7 < √10 | |

√3 + √7 > √10 |

Question 1 Explanation:

√3 + √7=1.7320508075688772935274463415059 + 2.6457513110645905905016157536393

=4.3778021186334678840290620951451

√10=3.1622776601683793319988935444327

So, √3 + √7 > √10 is true.

=4.3778021186334678840290620951451

√10=3.1622776601683793319988935444327

So, √3 + √7 > √10 is true.

Question 2 |

A given connected graph is a Euler Graph if and only if all vertices of are of

same degree | |

even degree | |

odd degree | |

different degree |

Question 2 Explanation:

A given connected graph is a Euler Graph if and only if all vertices of are of even degree.

→ Let G(V, E) be an Euler graph. Thus G contains an Euler line Z, which is a closed walk.

→ Let this walk start and end at the vertex u ∈ V. Since each visit of Z to an intermediate vertex v of Z contributes two to the degree of v and since Z traverses each edge exactly once, d(v) is even for every such vertex.

→ Each intermediate visit to u contributes two to the degree of u, and also the initial and ﬁnal edges of Z contribute one each to the degree of u. So the degree d(u) of u is also even.

**Proof:**→ Let G(V, E) be an Euler graph. Thus G contains an Euler line Z, which is a closed walk.

→ Let this walk start and end at the vertex u ∈ V. Since each visit of Z to an intermediate vertex v of Z contributes two to the degree of v and since Z traverses each edge exactly once, d(v) is even for every such vertex.

→ Each intermediate visit to u contributes two to the degree of u, and also the initial and ﬁnal edges of Z contribute one each to the degree of u. So the degree d(u) of u is also even.

Question 3 |

The maximum number of edges in a n-node undirected graph without self loops is

n ^{2} | |

n(n-1)/2 | |

n-1 | |

n(n+1)/2 |

Question 3 Explanation:

The set of vertices has size n, the number of such subsets is given by the binomial coefficient C(n,2)⋅C(n,2)=n(n-1)/2.

Question 4 |

The minimum number of NAND gates required to implement the Boolean function A + AB’+ AB’C is equal to

0 | |

1 | |

4 | |

7 |

Question 4 Explanation:

A + AB’+ AB’C = A(1+B’+B’C) = A

No GATE is required to implement the function A.

No GATE is required to implement the function A.

Question 5 |

The minimum Boolean expression for the following circuit is:

AB + AC + BC | |

A + BC | |

A + B | |

A + B + C |

Question 5 Explanation:

If the switches are in parallel then use “+”, and if they are serial then use “ ”.

A(B+C) + AB + (A+B)C = AB + AC + AB + AC + BC

= (AB+AB) + (AC+AC) + BC

= AB + AC + BC

A(B+C) + AB + (A+B)C = AB + AC + AB + AC + BC

= (AB+AB) + (AC+AC) + BC

= AB + AC + BC

Question 6 |

For a binary half-subtractor having two inputs A and B, the correct set of logical expressions for the outputs D (= A minus B) and X (=borrow) are

D = AB + A’B , X = A’B | |

D = A’B + AB’ , X = AB’ | |

D = A’B + AB’ , X = A’B | |

D = AB + A’B , X = AB’ |

Question 6 Explanation:

The function table for the Half Subtractor is as follows

A-B= D= A’B + AB’

X= A’B

A-B= D= A’B + AB’

X= A’B

Question 7 |

Consider the following gate network

Which one of the following gates is redundant?

Which one of the following gates is redundant?

Gate No. 1 | |

Gate No. 2 | |

Gate No. 3 | |

Gate No. 4 |

Question 7 Explanation:

W’ + W’Z + Z’XY = W’(1+Z) + Z’XY

= W’ + Z’XY

The term W’Z is redundant which is represented by GATE 2.

= W’ + Z’XY

The term W’Z is redundant which is represented by GATE 2.

Question 8 |

The dynamic hazard problem occurs in

combinational circuit alone | |

sequential circuit only | |

Both (a) and (b) | |

None of the above |

Question 8 Explanation:

→ A dynamic hazard is the possibility of an output changing more than once as a result of a single input change.

→ Dynamic hazards often occur in larger logic circuits where there are different routes to the output (from the input).

→ If each route has a different delay, then it quickly becomes clear that there is the potential for changing output values that differ from the required / expected output. e.g.

→ A logic circuit is meant to change output state from 1 to 0, but instead changes from 1 to 0 then 1 and finally rests at the correct value 0. This is a dynamic hazard.

→ As a rule, dynamic hazards are more complex to resolve, but note that if all static hazards have been eliminated from a circuit, then dynamic hazards cannot occur.

→ Dynamic hazards often occur in larger logic circuits where there are different routes to the output (from the input).

→ If each route has a different delay, then it quickly becomes clear that there is the potential for changing output values that differ from the required / expected output. e.g.

→ A logic circuit is meant to change output state from 1 to 0, but instead changes from 1 to 0 then 1 and finally rests at the correct value 0. This is a dynamic hazard.

→ As a rule, dynamic hazards are more complex to resolve, but note that if all static hazards have been eliminated from a circuit, then dynamic hazards cannot occur.

Question 9 |

The logic circuit given below converts a binary code y1,y2,y3 into

Excess-3 code | |

Gray code | |

BCD code | |

Hamming Code |

Question 9 Explanation:

X1= Y1

X2= Y1⊕ Y2

X3= Y2 ⊕ Y3

X2= Y1⊕ Y2

X3= Y2 ⊕ Y3

Question 10 |

The circuit given below in the figure below is

An oscillating circuit and its output is a square wave | |

The one whose output remains stable in ‘1’ state | |

The one having output remains stable in ‘0’ state | |

has a single pulse of three times propagation delay |

Question 10 Explanation:

The square wave has alternating amplitudes(0 and 1) with duty cycle 1.

An odd number of cascaded NOT gates produce a square wave.

Note: Duty cycle= Ratio of durations in which the circuit is ON and OFF in a cycle.

An odd number of cascaded NOT gates produce a square wave.

Note: Duty cycle= Ratio of durations in which the circuit is ON and OFF in a cycle.

Question 11 |

If 12A7C

_{16}= X_{8}, then the value of X is224174 | |

425174 | |

6173 | |

225174 |

Question 11 Explanation:

Given, (12A7C)

MAke blocks of 3 bits each from LSB to MSB.

(Note: In the last block append zeros (as MSBs) if number bits is not three)

(000 010 010 101 001 111 100)

Each of the above blocks represents a digit in base 8 and they can be converted to base 8 as shown below.

= (0 2 2 5 1 7 4)

_{16}= (0001 0010 1010 0111 1100)_{2}MAke blocks of 3 bits each from LSB to MSB.

(Note: In the last block append zeros (as MSBs) if number bits is not three)

(000 010 010 101 001 111 100)

Each of the above blocks represents a digit in base 8 and they can be converted to base 8 as shown below.

= (0 2 2 5 1 7 4)

_{8}Question 12 |

The Excess-3 code is also called

Cyclic Redundancy Code | |

Weighted Code | |

Self-Complementing Code | |

Algebraic Code |

Question 12 Explanation:

Excess-3 code is also called Self-Complementing Code. Because 1’s complement of an excess-3 number is equivalent to 9’s complement of the corresponding decimal digit.

→ In excess-3 code, each of the 4-bit numbers represents decimal digit which is 3 less than the actual decimal digit. So the bits have no fixed weight.

Excess-3 code is neither CRC nor Algebraic Code which is used for error detection and/or correction.

→ In excess-3 code, each of the 4-bit numbers represents decimal digit which is 3 less than the actual decimal digit. So the bits have no fixed weight.

Excess-3 code is neither CRC nor Algebraic Code which is used for error detection and/or correction.

Question 13 |

The simplified SOP (Sum of Product) form the Boolean expression (P + Q’ + R’)(P + Q’ + R)(P + Q + R’)

(P’Q + R) | |

(P + Q’R’) | |

(P Q’ + R ) | |

(PQ + R) |

Question 13 Explanation:

Question 14 |

Which of the following binary number is the same as its 2’s complement?

1010 | |

0101 | |

1000 | |

1001 |

Question 14 Explanation:

Hint: Number of bits=4

(Decimal value of maximum 4-bit number +1 )/2= (15+1)/2=8

(Decimal value of maximum 4-bit number +1 )/2= (15+1)/2=8

Question 15 |

The functional difference between SR flip-flop and JK flip-flop is that

JK Flip-flop is faster than SR flip-flop | |

JK flip-flop has a feedback path | |

JK flip-flop accepts both inputs 1 | |

None of them |

Question 15 Explanation:

-> JK flip flop accepts input J=K=1. When J=K=1, the state of the flip-flop gets complimented. But it's not a valid input in SR flip-flop.

-> JK flip flop doesn’t have a feedback path.

-> JK flip flop doesn’t have a feedback path.

Question 16 |

Consider a non-pipelined processor with a clock rate of 2.5 gigahertz and average cycles per instruction of four. The same processor is upgraded to a pipelined processor with five stages; but due to the internal pipeline delay, the clock speed is reduced to 2 gigahertz. Assume that there are no stalls in the pipeline. The speedup achieved in this pipelined processor is

3.2 | |

3.0 | |

2.2 | |

2.0 |

Question 16 Explanation:

→ Given that the processor clock rate = 2.5 GHz, the processor takes 2.5 G cycles in one second.

→ Time taken to complete one cycle = (1 / 2.5 G) seconds

→ Since it is given that average number of cycles per instruction = 4, the time taken for completing one instruction=(4/2.5 G) = 1.6 ns

→ In the pipelined case we know in the ideal case CPI = 1, and the clock speed = 2 GHz.

→ Time taken for one instruction in the pipelined case = (1 / 2 G) = 0.5 ns

→ Speedup = 1.6/0.5 = 3.2

→ Time taken to complete one cycle = (1 / 2.5 G) seconds

→ Since it is given that average number of cycles per instruction = 4, the time taken for completing one instruction=(4/2.5 G) = 1.6 ns

→ In the pipelined case we know in the ideal case CPI = 1, and the clock speed = 2 GHz.

→ Time taken for one instruction in the pipelined case = (1 / 2 G) = 0.5 ns

→ Speedup = 1.6/0.5 = 3.2

Question 17 |

What is the output of this C code?

5 5 5 | |

5 5 junk | |

5 junk junk | |

Compile time error |

Question 17 Explanation:

If we are given p instead of *p then it prints the address of k but we are printing the value of k.

The same rule applies in **m.

Output is 5 5 5

The same rule applies in **m.

Output is 5 5 5

Question 18 |

Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors per track. 512 bytes of data are stored in a bit serial manner in a sector. The capacity of the disk pack and the number of bits required to specify a particular sector in the disk are respectively:

256 Mbyte, 19 bits | |

256 Mbyte, 21 bits | |

512 Mbyte, 20 bits | |

64 GB, 28 bits |

Question 18 Explanation:

→ Given that the disk pack has 16 surfaces, 128 tracks per surface, 256 sectors per track and each sector size is 512 bytes.

→ So the disk pack capacity = 16*128*256*512 bytes = 256 MB

→ To specify a sector we need the information about surface number, track number and sector number within a track.

→ Surface number needs 4 bits as there are 16 surfaces(2

→ Total number bits needed to specify a particular sector = 4+7+8 = 19 bits.

→ So the disk pack capacity = 16*128*256*512 bytes = 256 MB

→ To specify a sector we need the information about surface number, track number and sector number within a track.

→ Surface number needs 4 bits as there are 16 surfaces(2

^{4}), track number needs 7 bits as there are 128 tracks(2^{7}) within a surface, within a track the sector number needs 8 bits as there are 256 sectors (2^{8}).→ Total number bits needed to specify a particular sector = 4+7+8 = 19 bits.

Question 19 |

Let the page fault service time be 10 ms in a computer with average memory access time being 20 ns. If the one-page fault is generated for every 106 memory accesses, what is the effective access time for the memory?

21.4 ns | |

29.9 ns | |

23.5 ns | |

35.1 ns |

Question 19 Explanation:

Question 20 |

Register renaming is done in pipelined processors

as an alternative to register allocation at compile time | |

for efficient access to function parameters and local variables | |

to handle certain kinds of hazards | |

as part of address translation |

Question 20 Explanation:

→ Register renaming is used to eliminate hazards that arise due to WAR (Write After Read) and WAW(Write After Write) dependencies.