NET JUNE 2015 PAPER -3 SOLVED PART-1 22


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  1. For the 8-bit word 00111001, the check bits stored with it would be 0111. Suppose when the word is read from memory, the check bits are calculated to be 1101. What is the data word that was read from memory?
    (A) 10011001 (B) 00011001
    (C) 00111000 (D) 11000110
    Answer: B

Explanation .
Given 8 bit word 00111001. encoded in d8=0 d7=0 d6=1 d5=1 d4=1 d3=0 d2=0 d1=1.
parity is calculated as 01111  which are written as p8p4p2p1.
p1=1
p2=1
p4=1
p8=0.

encode the string as 
d8 d7 d6 d5 p8 d4 d3 d2 p4 d1 p2 p1.UGC net study material
001101001111.
check bit at other end is 1101.
c1=1
c2=0
c4=1
c8=1. 
xor(1101,0111)=1010 . its decimal conversion is 10. so 10th bit is changed.
 10th bit is d6.  so we flip it. and we get
00011001


2.   Consider a 32 – bit microprocessor, with a 16 – bit external data bus, driven by an 8 MHz input clock. Assume that the microprocessor has a bus cycle whose minimum duration equals four input clock cycles. What is the maximum data transfer rate for this microprocessor?
(A) 8×106 bytes/sec (B) 4×106 bytes/sec
(C) 16×106 bytes/sec (D) 4×109 bytes/sec
Answer: B

Explanation.
Minimum bus cycle duration = 4 clock cycle.
maximum bus cycle rate= 8 MHZ/4= 2M/s where 1 MHZ= 1000000/sec.
maximum data transferred per  bus cycle rate= 2bytes.
data transferred rate per second = data transferred  per bus cycle* bus cycle rate.= 2bytes*2M/s


3 The RST 7 instruction in 8085 microprocessor is equivalent to:
(A) CALL 0010 H (B) CALL 0034 H
(C) CALL 0038 H (D) CALL 003C H
Answer: C

Explanation.

Interrupt Vector Address
RST 0 0000H
RST 1 0008H
RST2 0010H
RST3 0018H
RST4 0020H
RST5 0028H
RST6 0030H
RST7 0038H

4 The equivalent hexadecimal notation for octal number 2550276 is:
(A) FADED (B) AEOBE
(C) ADOBE (D) ACABE
Answer: C

010 for 2.
101 for 5
and similarly.

so we organize every thing as.
010101101000010111110.
Now this binary representation is grouped in to 4 to get the Hexadecimal notation.

0 1010 1101  0000   1011   1110 so after converting every group in HEX. we get

ADOBE.


5 The CPU of a system having 1 MIPS execution rate needs 4 machine cycles on an average for executing an instruction. The fifty percent of the cycles use memory bus. A memory read/write employs one machine cycle. For execution of the programs, the system utilizes 90 percent of the CPU time. For block data transfer, an IO device is attached to the system while CPU executes the background programs continuously. What is the maximum IO data transfer rate if programmed IO data transfer technique is used?
(A) 500 Kbytes/sec (B) 2.2 Mbytes/sec
(C) 125 Kbytes/sec (D) 250 Kbytes/sec
Answer: D

CPU speed 1MIPS=10^6 instruction per sec

1 Cpu instruction =4 machine cycles(Avg)

1 memory read/access = 1 machine cycle

CPU utilization =90%

Programmed I/O

each byte transfer requires 4 cycles (instructions)

in status

check status

Branch 

Read/Write data

max data transfer rate CPU speed /4 = 10^6/4=250 kbytes/sec


6. The number of flip-flops required to design a modulo-272 counter is:
(A) 8 (B) 9
(C) 27 (D) 11
Answer: B

Explanation.
 The number of flipflop required to design a Modulo 272 counter is Log2 272.
  8 flips flops  do the work up to 256. hence we need 9 flip flops to achieve the task.


7. Let E1 and E2 be two entities in E-R diagram with simple single valued attributes. R1 and R2 are two relationships between E1 and E2 where R1 is one-many and R2 is many-many. R1 and R2 do not have any attributes of their own. How many minimum number of tables are required to represent this situation in the Relational Model?
(A) 4 (B) 3
(C) 2 (D) 1
Answer: B

Explanation.

Let relation R1 is composed of table emp(ssn,name, Superssn). A query like this can be made from single table like. Find the employee ssn, name, super visor ssn, supervisor name from emp table. An ssn will map to a single supervisor and a supervisor can map to many employee.
In many to many relationship like  “a student can take many subjects and many students can take a subject”. it needs two table.eg. student(rollno,subject,name) where pk= roll no and subject. and fk = subject.
Foreign key map to pk of subjet (Subjetid, Subjname, Roomno).
Hence it required 3 tables


8. The STUDENT information in a university is stored in the relation STUDENT (Name, Sex, Marks, DEPT_Name)
Consider the following SQL Query SELECT DEPT_Name from STUDENT where SEX=’M’ group by DEPT_Name having avg (Marks)>(SELECT avg (Marks) from STUDENT). It returns the Name of the Department for which:
(A) The Average marks of Male students is more than the average marks of students in the same Department
(B) The average marks of male students is more than the average marks of the students in the University
(C) The average marks of male students is more than the average marks of male students in the University
(D) The average marks of students is more than the average marks of male students in the University
Answer: B


9. Select the ‘False’ statement from the following statements about Normal Forms:
(A) Lossless preserving decomposition into 3NF is always possible
(B) Lossless preserving decomposition into BCNF is always possible
(C) Any relation with two attributes is in BCNF
(D) BCNF is stronger than 3NF
Answer: B

Explanation.
Table 1(A,B,C) with FD  AB->C and C->B  cannot be preserved  when decomposition is done


10. The Relation
Vendor Order (V_no, V_ord_no, V_name, Qty_sup, unit_price)
is in 2NF because :
(A) Non key attribute V_name is dependent on V_no which is part of composite key
(B) Non key attribute V_name is dependent on Qty_sup
(C) key attribute Qty_sup is dependent on primary_key unit price
(D) key attribute V_ord_no is dependent on primary_key unit price
Answer: A

Explanation.
for lossless decomposition .Atleast one of the following dependencies should be in F+.

R1 intersction R2->R1.                     Eq. 1.
or
R1 intersection R2->R2.                    Eq2.

when a decompostion satisfies  eq 1. given above then that will solve the  R1 intersection R2 –>>(R1-R2).
when a  decomposition satisfies eq2 given above then that will solve the  R1 intersection R2 –>>(R2-R1).


11. The relation schemas R1 and R2 form a Lossless join decomposition of R if and only if:
(a) R1⋂R2→(R1-R2)
(b) R1→R2
(c) R1⋂R2→(R2-R1)
(d) R2→R1⋂R2
(A) (a) and (b) happens (B) (a) and (d) happens
(C) (a) and (c) happens (D) (b) and (c) happens
Answer: C


12. In the indexed scheme of blocks to a file, the maximum possible size of the file depends on:
(A) The number of blocks used for index, and the size of index
(B) Size of Blocks and size of Address
(C) Size of Index
(D) Size of Block
Answer: A


13. Give the number of principal vanishing point(s) along with their direction for the standard perspective transformation:
(A) Only one in the direction K
(B) Two in the directions I and J
(C) Three in the directions I, J and K
(D) Only two in the directions J and K
Answer: A


14. Consider a triangle A(0, 0), B(1, 1), C(5, 2). The triangle has to be rotated by an angle of 450 about the point P(-1, -1). What shall be the coordinates of the new triangle?
(A) A’=(1, √2-1), B’=(-1, 2√2-1), C’=(3√2-1, (9/2)√2-1)
(B) A’=(1, √2-1), B’=(2√2-1, -1), C’=(3√2-1, (9/2)√2-1)
(C) A’=(-1, √2-1), B’=(-1, 2√2-1), C’=(3√2-1, (9/2)√2-1)
(D) A’=(√2-1, -1), B’=(-1, 2√2-1), C’=(3√2-1, (9/2)√2-1)
Answer: C


15. The process of dividing an analog signal into a string of discrete outputs, each of constant amplitude, is called:
(A) Strobing (B) Amplification
(C) Conditioning (D) Quantization
Answer: D


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