UGC NET CS 2015 Jun – III

  • Last Updated : 25 Oct, 2021

Question 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?
Cross
10011001
Tick
00011001
Cross
00111000
Cross
11000110


Question 1-Explanation: 
8 - bit word = 00111001, Check bits = 0111. There are four bits and it's position will be 20 = 1; i.e. p1 = 1 21 = 2; i.e. p2 = 1 22 = 4; i.e. p4 = 1 23 = 8; i.e. p8 = 0. Encoded string will be: d12 = 0; d11 = 0; d10 = 1; d9 = 1; d8 = p8 = 1; d7 = 1; d6 = 0; d5 = 0; d4 = p4 = 1; d3 = 1; d2 = p2 = 1; d1 = p1 = 1. i.e. d12 = 0; d11 = 0; d10 = 1; d9 = 1; d8 = 1; d7 = 1; d6 = 0; d5 = 0; d4 = 1; d3 = 1; d2 = 1; d1 = 1. Check bits at other end = 1101. XOR(0111, 1101) = 1010. 10th bit be change. So, new encoded string will be: d12 = 0; d11 = 0; d10 = 0; d9 = 1; d8 = 1; d7 = 1; d6 = 0 d5 = 0; d4 = 1; d3 = 1; d2 = 1; d1 = 1. And the data word that was read from memory will be - 00011001. So, option (B) is correct.
Question 2

Consider a 32 - bit microprocessor, with a 16 - bit external data bus, driven by an 8 MHz input clock. Assume that this microprocessor has a bus cycle whose minimum duration equals four input clock cycles. What is the maximum data transfer rate for this microprocessor?

Cross

8 x 106 bytes/sec

Tick

4 x 106 bytes/sec

Cross

16 x 106 bytes/sec

Cross

4 x 109 bytes/sec



Question 2-Explanation: 

In a 32 - bit microprocessor, with a 16 - bit external data bus, data transferred per bus cycle = 2B. 
Bus clock = 8 MHz. 
Minimum bus cycle duration = 4 clock cycles 
Maximum bus cycle rate = 8 MHz / 4 = 2 M/sec 
Data transfer rate = bus cycle rate * data per bus cycle = 2 M * 2 = 4 * 106 B/sec. 

So, option (B) is correct.

Question 3
The RST 7 instruction in 8085 microprocessor is equivalent to:
Cross
CALL 0010 H
Cross
CALL 0034 H
Tick
CALL 0038 H
Cross
CALL 003C H


Question 3-Explanation: 
In 8085 microprocessor interrupt and vector address are as follows:
RST 7  CALL 0038 H 
RST 6  CALL 0030 H
RST 4  CALL 0028 H
RST 3  CALL 0020 H
RST 2  CALL 0018 H
RST 1  CALL 0010 H
RST 0  CALL 0008 H 
So, option (C) is correct.
Question 4
The equivalent hexadecimal notation for octal number 2550276 is:
Cross
FADED
Cross
AEOBE
Tick
ADOBE
Cross
ACABE


Question 4-Explanation: 
Binary equivalent of octal no. 2550276 is 010 101 101 000 010 111 110. group them in 0 1010 1101 0000 1011 1110 1010 - A 1101 - D 0000 - 0 1011 - B 1110 - E So, option (C) is correct.
Question 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?
Cross
500 Kbytes/sec
Cross
2.2 Mbytes/sec
Cross
125 Kbytes/sec
Tick
250 Kbytes/sec


Question 6

The number of flip-flops required to design a modulo - 272 counter is:

Cross

8

Tick

9

Cross

27

Cross

11



Question 6-Explanation: 

With n flip - flops we can design upto Modulo 2n counter. 272 is greater than 256 i.e. 28 so we need little bit more; 29 = 512 which is greater than 272. So 9 flip - flops are sufficient to design Modulo 272 counter. Option (B) is correct.

Question 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 attribute of their own. How many minimum number of tables are required to represent this situation in the Relational Model?
Cross
4
Tick
3
Cross
2
Cross
1


Question 7-Explanation: 
Refer:GATE | GATE-CS-2005 | Question 90 Option (B) is correct.
Question 8
The STUDENT information in a university 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:
Cross
The Average marks of Male students is more than the average marks of students in the same Department
Tick
The average marks of male students is more than the average marks of students in the University
Cross
The average marks of male students is more than the average marks of male students in the University
Cross
The average marks of students is more than the average marks of male students in the University


Question 8-Explanation: 
The SQL query: SELECT DEPT_Name from STUDENT where SEX = 'M' group by DEPT_Name having avg (Marks)>SELECT avg (Marks) from STUDENT. will return The average marks of male students is more than the average marks of students in the University. So, option (B) is correct.
Question 9
Select the 'False' statement from the following statements about Normal Forms:
Cross
Lossless preserving decomposition into 3NF is always possible
Tick
Lossless preserving decomposition into BCNF is always possible
Cross
Any Relation with two attributes is in BCNF
Cross
BCNF is stronger than 3NF


Question 9-Explanation: 
  • Lossless preserving decomposition into 3NF is always possible. True
  • Lossless preserving decomposition into BCNF is always possible. False Not always possible.
  • Any Relation with two attributes is in BCNF. True
  • BCNF is stronger than 3NF. True
  • For more information on Normal form Refer:Database Normalization | Normal Forms Option (B) is correct.
    Question 10
    The Relation Vendor Order (V_no, V_ord_no, V_name, Qty_sup, unit_price) is in 2NF because:
    Tick
    Non_key attribute V_name is dependent on V_no which is part of composite key
    Cross
    Non_key attribute V_name is dependent on Qty_sup
    Cross
    Key attribute Qty_sup is dependent on primary_key unit price
    Cross
    Key attribute V_ord_no is dependent on primary_key unit price


    Question 10-Explanation: 
    The Relation Vendor Order (V_no, V_ord_no, V_name, Qty_sup, unit_price) is in 2NF because: Non_key attribute V_name is dependent on V_no which is part of composite key. For more information on Normal forms Refer:Database Normalization | Normal Forms Option (A) is correct.
    There are 75 questions to complete.