SET 2016 Previous year papers of GATE, NET, SET

Q 01

The most important advantage of integrated circuit is its:

Easy replacement in case of circuit failure

Extremely high reliability

Reduced cost

Low power consumption

Ans .

B/C/D

Explanation : The function of an IC (integrated circuit) chip is to replace many separate electronic components which could possibly have been used to build a particular electronic circuit. Most of those separate components are replaced by just one tiny IC chip that has been manufactured ("fabricated" is the correct technical word) to include extremely miniature circuits which imitate the behavior of all those separate components.

Q 02

Which of the follwing is used to store critical data duriing subroutines and interrupt ?

Stack

Queue

Accumulator

Data Register

Ans .

Explanation : Stack is used to store critical data duriing subroutines and interrupt.

Q 03

CS stands for

Cost Segment

Counter Segment

Code Segment

Coot Segment

Ans .

Explanation : CS is Code Segment.

Q 04

......... is the most important segment and it contains the actual assembly language instructions to be executed by the microprocessor.

Data Segment

Code Segment

Stack Segment

Extra Segment

Ans .

Explanation : The code segment register is used for addressing a memory location in the code segment of memory, where executable program is stored.

Q 05

An integrated circuit is :

a complicated circuit

an integrated device

much costlier than a single transistor

available in the form of a silicon chip

Ans .

Explanation : An integrated circuit or monolithic integrated circuit (also referred to as an IC, a chip, or a microchip) is a set of electronic circuits on one small flat piece (or "chip") of semiconductor material, normally silicon.

Q 06

In order to maintain the consistency during transactions database provides :

Commit

Atomic

FlashBack

Retain

Ans .

Explanation : By atomic , either all the effects of the transaction are reflected in the database, or none are (after rollback).

Q 07

It is an abstraction through which relationships are treated as higher level entities :

Generalization

Specialization

Aggregation

Inheritance

Ans .

Explanation : It is an abstraction through which relationships are treated as higher level entities Aggregation. (In ER diagram, aggregation is used to represent a relationship as an entity set.

Q 08

In multiple granularity of locks SIX lock is compatible with :

SIX

Ans .

Explanation : IS – Intent to get S lock(s) at finer granularity.

Q 09

Let E1 and E2 be 2 entities in an ER diagram with simple single-valued attributes. R1 and R2 are 2 relationships between E1 and E2 , where R1 is one-to-many and R2 is many-to-many. R1 and R2 do not have any attributes of their own. What is the minimum number of tables required to represent the situation in the relational model ?

Ans .

Explanation : Strong entities E1 and E2 are represented as separate tables. In addition to that many-to-many relationships(R2) must be converted as seperate table by having primary keys of E1 and E2 as foreign keys. One-to-many relaionship (R1) must be transferred to ‘many’ side table(i.e. E2) by having primary key of one side(E1) as foreign key( this way we need not to make a seperate table for R1). Let relation schema be E1(a1,a2) and E2( b1,b2). Relation E1( a1 is the key) Relation E2( b1 is the key, a1 is the foreign key, hence R1(one-many) relationship set satisfy here ) Relation R2 ( {a1, b1} combined is the key here , representing many-many relationship R2 )

Q 10

In domain relational calculus we create variable for every :

Row

Column

Table

None of these

Ans .

Explanation : In domain relational calculus we create variable for every column.

Q 11

Which of the following is the tightest upper-bound that represents the time complexity of inserting an object into an binary search tree of 'n' nodes

O(1)

O(log n)

O(n)

O(n log n)

Ans .

Explanation : To insert an element, we need to search for its place first. The search operation may take O(n) for a skewed tree.

Q 12

A layer-4 firewall (a device that can look at all protocols headers up-to the transport layer) can not :

Block entire HTTP traffic during 9: 00 pm and 5:00 am

Block all ICMP traffic

Stop incoming traffic from a specific IP address but allow outgoing traffic to same IP address

Block TCP traffic from a specific user on a multi-user system during 9:00 pm and 5:00 am

Ans .

Explanation : Since,it is a layer 4 firewall,it cannot block application layer protocol like HTTP. Transport layer has port number and port number of http is 80 and thus firewall can block http traffic

Q 13

The lexical analysis for a modern language such as JAVA needs the power of which one of the following machine models in a necessary and sufficient sense ?

Finite Automata State

Deterministic Pushdown Automata

Non-deterministic Pushdown Automata

Turing Machine

Ans .

Explanation : Lexical analysis is the first step in compilation. In lexical analysis, program is divided into tokens. Lexical analyzers are typically based on finite state automata.

Q 14

An undirected graph G(V,E) contains n(n>2) nodes named V1,V2,......,Vn. Two nodes Vi, Vj are connected if and only if 0 < |i-j| < = 2. Each edge (Vi, Vj) is assigned a weight i+j . A simple graph with n=4 is shown below
Maharashtra NET 2016
What will be the cost of the minimum Spanning Tree (MST) of such a graph with 'n' nodes ?

1/12 (11n^2 - 5n)

n^2 - n + 1

6n - 11

2n + 1

Ans .

Explanation : Minimum spanning tree for 2 nodes would be (v1) _ (v2) Total weight 3 Minimum spanning tree for 3 nodes would be (v1) _ (v2) | (v3) Total weight= 3 + 4 = 7 Minimum spanning tree for 4 nodes would be (v1) _ (v2) _ (v4) | (v3) Total weight= 3 + 4 + 6 = 13 Minimum spanning tree for 5 nodes would be (v1) _ (v2) _ (v4) | (v3) | (v5) Total weight= 3 + 4 + 6 + 8 = 21 Minimum spanning tree for 6 nodes would be (v1) _ (v2) _ (v4) _ (v6) | (v3) | (v5) Total weight= 3 + 4 + 6 + 8 + 10 = 31 We can observe from above examples that when we add kth node, the weight of spanning tree increases by 2k-2. Let T(n) be the weight of minimum spanning tree. T(n) can be written as T(n) = T(n-1) + (2n-2) for n > 2 T(1) = 0, T(2) = 0 and T(2) = 3 The recurrence can be written as sum of series (2n – 2) + (2n-4) + (2n-6) + (2n-8) + …. 3 and solution of this recurrence is n^2 – n + 1.

Q 15

Let G be a simple undirected planar graph on 10 vertices with 15 edges. If G is a connected graph , then the number of bounded faces in any embedding on the plane is equal to :

Ans .

Explanation : If the graph is planar, then it must follow below Euler’s Formula for planar graphs v - e + f = 2 v is number of vertices e is number of edges f is number of faces including bounded and unbounded 10 - 15 + f = 2 f = 7 There is always one unbounded face, so the number of bounded faces = 6

Q 16 Let G be a complete undirected graph on 6 vertices. If vertices of G are labeled , then the number of distinct cycles of length 4 in G is equal to :

360

Ans .

Explanation : There can be total 6C4 ways to pick 4 vertices from 6. The value of 6C4 is 15. Note that the given graph is complete so any 4 vertices can form a cycle. There can be 6 different cycle with 4 vertices. For example, consider 4 vertices as a, b, c and d. The three distinct cycles are cycles should be like this (a, b, c, d,a) (a, b, d, c,a) (a, c, b, d,a) (a, c, d, b,a) (a, d, b, c,a) (a, d, c, b,a) and (a, b, c, d,a) and (a, d, c, b,a) (a, b, d, c,a) and (a, c, d, b,a) (a, c, b, d,a) and (a, d, b, c,a) are same cycles. So total number of distinct cycles is (15*3) = 45.

Q 17

The transform at the heart of JPEG compression standard for digital images is :

fOURIER TRANSFORM

LOG TRANSFROM

dESCRETE COSINE TRANSFORM

LAPLACE TRANSFORM

Ans .

Explanation : The transform at the heart of JPEG compression standard for digital images is descrete cosine transform.

Q 18

In a network of LANs connected by bridges , packets are sent from one LAN To another through intermediate bridges . Since more than one path may exist between two LANs , packets may have to be Routed through multiple bridges. Why is the spanning tree algorithm is used for bridge routing ?

For shortest path Routing between LANs

For avoiding Loops in the routing paths

For fault tolerance

For minimizing collisions

Ans .

Explanation : The main idea for using Spanning Trees is to avoid loops.

Q 19

The maximum window size for data transmission using the selective reject protocol with n -bit frame sequence number is :

2^ n

2^(n-1)

2^n - 1

2^(n-2)

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