**Ans . **

D

**Explanation :**ALU of 8085 have five flip flops whose states (set/reset) are determined by the result data of other registers and accumulator. They are called as Zero, Carry, Sign, Parity and Auxiliary-Carry flags.

**Ans . **

C

**Explanation :**In 8085 microprocessor the address bus is of 16 bits.

**Ans . **

A

**Explanation :**Index Mode is used to access an array whose elements are in successive memory locations. The content of the instruction code, represents the starting address of the array and the value of the index register, and the index value of the current element. By incrementing or decrementing index register different element of the array can be accessed..

**Ans . **

B

**Explanation :**1. In Memory mapped I/O, there are no specific input or output instructions. The CPU can manipulate I/O data residing in interface registers with the same instructions that are used to manipulate memory words i.e. the same set of instructions are used for reading and writing memory can be used to input and output.3. Asynchronous serial communication is a form of serial communication in which the communicating endpoints' interfaces are not continuously synchronized by a common clock signal. 4.Synchronous communication requires that the clocks in the transmitting and receiving devices are synchronized

**Ans . **

A

**Explanation :**Cycle in precedence graph tells that schedule is not conflict serializable. DFS and BFS traversal of graph are possible even if graph contains cycle. And hence DFS and BFS are also possible for non serializable graphs. But Topological sort of any cyclic graph is not possible. Thus topological sort guarantees graph to be serializable . Option D is not valid because in a transaction with more indices might have to come before lower one. Also two non- conflicting schedule can occur simultaneously.

**Ans . **

C

**Explanation :**LXI rp, data performs the load register pain immediate operation

**Ans . **

C

**Explanation :**S2 is a conflict serializable schedule as visible from the sequence but S1 contains a loop, so it is not a conflict serializable schedule.

**Ans . **

B

**Explanation :**As we all know primary key is a subset of candidate key and candidate key is a subset of super key. because A relation can have one or more candidate keys and one out of these candidate keys can be selected as a primary key for the relation. One or more attributes which can derive all the other attributes in a relation is called a super key.

**Ans . **

C

**Explanation :**As we know for Many To One representation, we consider only the primary keys of relation having many attributes. An arrow mark from A to B represents that it is a many to one relationship between A to B.

**Ans . **

A

**Explanation :**on careful analysis of relation R, we derive that the candidate key is AB, so for the first functional dependency i.e A-->c there exists a Partial Dependency. The second Functional dependency B-->D also shows partial dependency, so the relation cannot be in 2NF.

**Ans . **

D

**Explanation :**In the given query we first select the average salary of employees of the company in the inner query, and submit to outer query which is selecting all the employees whose salary is greater than the computed average salary.

**Ans . **

D

**Explanation :**As we know Different kinds of phosphor are available for use. and various factors to differentiate include color and persistence which tell how long it will continue to emit light even after the CRT beam is removed.

**Ans . **

D

**Explanation :**Segments allow objects to be manipulated by simply referring to the segment numbrs as a result segments can be translated ,rotated , panned and zoomed

**Ans . **

A

**Explanation :**Glass contains liquid crystal and serves as a bonding surface for a conductive coating. Conductive coating acts as a conductor so that a voltage can be applied across the liquid crystal.Liquid Crystal ia a substance which will polarize light when a voltage is applied to it.Polarized film is a transparent sheet that polarizes light.

**Ans . **

D

**Explanation :**Maskable Interrupt: Those interrupts that can be delayed due to occurrence of a much higher priority interrupt that has occurred to the processor. Periodic Interrupt: If the interrupt is occurring after a fixed interval in timeline then those interrupts are called as periodic interrupts. Synchronous Interrupt: Those interrupts whose source is in phase with the system clock is called synchronous interrupt. that is those interrupts which are dependent on the system clock. Division by Zero: While executing a program if we have a condition such as a value which need to be divided by zero is called a exception.

**Ans . **

B

**Explanation :**Virtual memory is am illusion to the programmers of earlier generation that they have a very large memory available at their disposal. it is a special feature of operating system (OS) that allows a computer to execute programs of large size inspite of less physical memory by temporarily transferring pages of data from random access memory (RAM) to disk storage..

**Ans . **

B

**Explanation :**Before 8085,8086 the CPU of intel corporation was 8080A which were requiring +5V,-5V and +12V for functioning. With the advancement in technology, it was possible to run the microprocessor with single +5V.Hence the LAST DIGIT were taken as 5.

**Ans . **

A

**Explanation :**Digital Differential Analyzer (DDA) algorithm is the simple line generation algorithm which is explained step by step here. Step 1 - Get the input of two end points (X0,Y0) and (X1,Y1). Step 2 - Calculate the difference between two end points. Step 3 - Based on the calculated difference in step-2, you need to identify the number of steps to put pixel. If dx > dy, then you need more steps in x coordinate; otherwise in y coordinate. Step 4 - Calculate the increment in x coordinate and y coordinate. Step 5 - Put the pixel by successfully incrementing x and y coordinates accordingly and complete the drawing of the line.

**Ans . **

C

**Explanation :**Language which contains strings of even length of a's is a regular language. Apart from that, no other language given in options is a regular one, because they require computations.

**Ans . **

D

**Explanation :**L1* = {e} L2* = 1* L1* U L2* L1* = 1* Concatenation has higher priority than Union.

**Ans . **

A

**Explanation :**simply read the word and try to find out the meaning of the word.

**Ans . **

D

**Explanation :**Solve the composition of hog(x) as per rules of function h0g(x)= h(g(x)) = h(1/(x2+1) ) = (x2+1)-4 h0g0f(x) = hog(f(x)) = h0g(x3-4x) = [(x3-4x)2+1]-4

**Ans . **

C

**Explanation :**To find out the number of multiples from 0 to 100, we need to divide 100 by 6(integer division). We get 16. Since the range starts from 0, So the Total numbers will be 17. From -6 to 34, there will be 7 multiples.

**Ans . **

D

**Explanation :**A Hamiltonian path is a path in graph that visits each vertex exactly once. A Hamiltonian cycle is a Hamiltonian path that is a cycle. Since, Hamiltonian cycle satisfies all the properties given in the options.

**Ans . **

D

**Explanation :**Given that (P-->Q)?(R-->S) (P?R) here if P then Q and if R then S now P V R means either Q is true or S is true. so Y will be Q V S

**Ans . **

B

**Explanation :**ECL is the fastest logic family of all logic families. (High speeds are possible in ECL because the transistors are used in difference amplifier configuration, in which they are never driven into saturation and thereby the storage time is eliminated.

**Ans . **

D

**Explanation :**JK flip-flop is in holding mode and toggle mode when the JK inputs are 00 and 11 respectively. If JK inputs are 01, JK flip-flop is in reset mode, while the inputs are 10, JK flip-flop is in set mode. It behaves almost like SR flip-flop but JK flip-flop has toggle mode. J0=K0=1 J1=K1=Q'0 J2=K2=Q'1Q'0

**Ans . **

A

**Explanation :**0.40518 = 0?80+4?8-1+0?8-2+5?8-3+1?8-4 = 0+0.5+0+0.009765625+0.000244140625 = 0.51000976562510

**Ans . **

C

**Explanation :**Writing down the octal 2357 in binary form as follows (2357)8 = (010 011 101 111)2 now grouping them in pair of 4 bits 0100 1110 1111 --- so it will be 4EF.

**Ans . **

D

**Explanation :**We cannot pass Header files to functions in C++ as there is no matching type construct available for this purpose.

**Ans . **

A

**Explanation :**In C++, Whenever we call an overloaded function or operator, the compiler determines the most appropriate definition by comparing the no of arguments and argument types, you used to call the function or operator with the parameter types specified in the definitions. so option A is correct

**Ans . **

B

**Explanation :**IExtern storage class is also called as class with the global visibility.

**Ans . **

D

**Explanation :**Structure reference operator cannot be overloaded and all other can be overloaded..

**Ans . **

B

**Explanation :**A number A which is a power of 2 will have 1 at MSB and all other bits 0(e.g. we consider 8 then its binary will be 1000). In that way, A-1, will have all its bit 0 except for its MSB which is 0(e.g. we consider 8-1 i.e. 7 then its binary will be 0111). If we take bitwise AND of these two numbers, it will be all 0's.

**Ans . **

B

**Explanation :**In oracle datatype, varchar(size) is the size of maximum accomodoation, here the size of the string is decided by the actual size of the string not by its allocated space. So here, "Ram" will be of size 3, but in case of char(size) is the allocated size irrespective of its content, so the specified space is 20..

**Ans . **

C

**Explanation :**As per the definition of the standard integrity constraints.

**Ans . **

D

**Explanation :**You need three tables: For R1: Since it is one : many, you need just the tables for E1 and E2. E2 will have a foreign key referencing E1. Thus multiple tuples in E2 will reference the same tuple in E1

**Ans . **

A

**Explanation :**here clearly the dependencies are preserved but they are not lossless as both R1 and R2 cannot be formed back again as they donot have a common attribute which is a key.

**Ans . **

A

**Explanation :**data pointer = child pointer + search field value = 14+7 = 21B so max number of children it can have = 1KB/21B = 1024/21 = 48.76 = 48(approx) All other options are greater than the 48 can go with option 1 as asnwer

**Ans . **

A

**Explanation :**As we know the array is sorted. so in worst case, last item insert Insertion-O(n), in worst case, last item delete Deletion-O(n), maximum and minimum is always at end , we know the positions. so the Maximum-O(1) and Minimum-O(1).

**Ans . **

B

**Explanation :**push G F E D C B A pop 5 times insert popped in queue: (rear) E D C B A (front) delete 2 from queue: (rear) E D C (front) push in stack POP 1.

**Ans . **

A

**Explanation :**Since number of keys n is less than the size of table m so the Expected number of collisions is less than 1.

**Ans . **

B

**Explanation :**(a) True. ref:http://www.geeksforgeeks.org/dynamic-programming-set-24-optimal-binary-search-tree/ (b) False. ref: https://www.quora.com/Why-is-DFS-preferred-for-finding-connected-components-in-directed-graphs (c) True. because: inorder is needed (d) True. ref: https://www.quora.com/Why-is-DFS-preferred-for-finding-connected-components-in-directed-graphs

**Ans . **

A

**Explanation :**a. Application of layer iv. HTTP b. Transport layer i. TCP c. Network layer ii. IP ??????? d. Datalink layer iii. PPP

**Ans . **

X

**Explanation :**Application layer can send any size of data. There is no limit defined by standards. The lower layers divides the data if needed. so all the given options are not correct.

**Ans . **

D

**Explanation :**default gateway. A default gateway serves as an access point or IP router that a networked computer uses to send information to a computer in another network or the Internet.Default simply means that this gateway is used by default, unless an application specifies another gateway.

**Ans . **

C

**Explanation :**A distance-vector routing protocol requires that a router inform its neighbors about its topology changes periodically.

**Ans . **

B

**Explanation :**Link state protocols, sometimes called shortest path first or distributed database protocols, are built around a well-known algorithm from graph theory, E. W. Dijkstra's shortest path algorithm, which is is a graph search algorithm that solves the single-source shortest path problem for a graph with non-negative edge path costs, producing a shortest path tree. This algorithm is often used in routing and as a subroutine in other graph algorithms.

**Ans . **

D

**Explanation :**given expression is p+[3-5]*[xyz] Since, p has to be present as the first character of each string, so III string is not valid also p occurs only once, so VI becomes invalid. Since 6 is not a part of the set, II string is also invalid. so the question can be seen as I. p443y -> matched Il. p6y -> not matched III. 3xyz -> not matched IV. p35z -> matched V. p353535x -> matched Vl. ppp5 -> not matched

**Ans . **

C

Explanation :

in assembly AX=AH+BL AL is the lower 8 bits AH is the higher bits 8 HERE AL and Ah both contains 15 or F so AX contains 0F0F or 0000 1111 0000 1111 XOR AL AL will return 0 in AL so AX becomes 0000 1111 0000 0000 MOV CL,3 will store 3 in CL shr ax,cl will shift right ax by 3 so 0000 1111 0000 0000 will become 0000 0001 1110 0000 or 01E0 in hexadecimal i.e C 01E0 H