UGC NET CS 2015 Jun – III

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.

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.

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.

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.

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.

Refer:GATE | GATE-CS-2005 | Question 90 Option (B) is correct.

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.

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.

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.