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?
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.
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?
8 x 106 bytes/sec
4 x 106 bytes/sec
16 x 106 bytes/sec
4 x 109 bytes/sec
Question 2 Explanation:
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.
The RST 7 instruction in 8085 microprocessor is equivalent to:
CALL 0010 H
CALL 0034 H
CALL 0038 H
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 HSo, option (C) is correct.
The equivalent hexadecimal notation for octal number 2550276 is:
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.
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?
The number of flip-flops required to design a modulo - 272 counter is:
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.
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?
Question 7 Explanation:
Refer:GATE | GATE-CS-2005 | Question 90 Option (B) is correct.
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:
The Average marks of Male students is more than the average marks of students in the same Department
The average marks of male students is more than the average marks of students in the University
The average marks of male students is more than the average marks of male students in the University
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.
Select the 'False' statement from the following statements about Normal Forms:
Lossless preserving decomposition into 3NF is always possible
Lossless preserving decomposition into BCNF is always possible
Any Relation with two attributes is in BCNF
BCNF is stronger than 3NF
Question 9 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
Non_key attribute V_name is dependent on Qty_sup
Key attribute Qty_sup is dependent on primary_key unit price
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.
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