UGC NET CS 2015 Jun – III

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?

• A. 10011001
• B. 00011001
• C. 00111000
• D. 11000110

Answer: (B)

UGC NET CS 2015 Jun - III Microprocessor
Discuss it

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. 10_th 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?

• A. 8 x 10⁶ bytes/sec
• B. 4 x 10⁶ bytes/sec
• C. 16 x 10⁶ bytes/sec
• D. 4 x 10⁹ bytes/sec

Answer: (B)

UGC NET CS 2015 Jun - III Microprocessor
Discuss it

In a 32 - bit microprocessor, with a 16 - bit external data bus, data transfered 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:

• A. CALL 0010 H
• B. CALL 0034 H
• C. CALL 0038 H
• D. CALL 003C H

Answer: (C)

UGC NET CS 2015 Jun - III Microprocessor
Discuss it

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:

• A. FADED
• B. AEOBE
• C. ADOBE
• D. ACABE

Answer: (C)

UGC NET CS 2015 Jun - III
Discuss it

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?

• A. 500 Kbytes/sec
• B. 2.2 Mbytes/sec
• C. 125 Kbytes/sec
• D. 250 Kbytes/sec

Answer: (D)

UGC NET CS 2015 Jun - III Computer Organization and Architecture
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Question 6

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

• A. 8
• B. 9
• C. 27
• D. 11

Answer: (B)

UGC NET CS 2015 Jun - III Digital Logic & Number representation Sequential circuits
Discuss it

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?

• A. 4
• B. 3
• C. 2
• D. 1

Answer: (B)

UGC NET CS 2015 Jun - III
Discuss it

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:

• 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 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)

UGC NET CS 2015 Jun - III SQL
Discuss it

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:

• 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)

UGC NET CS 2015 Jun - III Database Design(Normal Forms)
Discuss it

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:

• 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)

UGC NET CS 2015 Jun - III Database Design(Normal Forms) 50 DBMS MCQs with Answers
Discuss it

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.