(A) Inconsistency
(B) Consistency
(C) Atomicity
(D) Isolation
Ans: Inconsistency
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Q. Identify the true statement from the given statements.
(1) FIFO is non-preemptive
(2) Round Robin is non-preemptive.
(3) Multilevel Queue Scheduling is non-preemptive.
Ans: (1) FIFO is non-preemptive
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Q. Dijkstra’s algorithm is based on:
(A) Greedy Approach | (B) Dynamic Programming |
(C) Backtracking Paradigm | (D) Divide and conquer paradigm |
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Q. Identify the total number of tokens in the given statement:
printf ( “ A % B = ” , &I ) ;
(A) 7 | (B) 8 |
(C) 9 | (D) 13 |
Explanation:
1. printf
2. (
3. “A%B=”
4. ,
5. &
6. I
7. )
8. ;
Option: (B) 8 is the total number of tokens in the given statement
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Q. Identify the true statement from the given statements.
1. HTTP may use different TCP connection for different objects of a web page if non-persistent connections are used.
2. FTP uses two TCP connections, one for data and another for control.
3. TELNET and FTP can only use TWO connection at a time.
(A) 1 | (B) 1 and 2 |
(C) 2 and 3 | (D) 1, 2 and 3 |
Explanation:
HTTP may use different TCP connection for different objects of a webpage if non-persistent connections are used, because client and server need to establish one new TCP connection for each object of a webpage. But, in case of persistent connection only one TCP connection can serve the purpose.
FTP uses two TCP connections, port 20 used to sand data and port number 21 used for control.
Telnet uses port number 23 as a TCP connection.
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Q. _________________ number of queues are required to implement a stack.
(A) 1 | (B) 2 |
(C) 3 | (D) 4 |
Ans: Option (B) 2 is the correct one.
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Q. ____________________ IP address can be used in WAN.
(A) 265.0.0.1 | (B) 172.16.0.10 |
(C) 15.1.5.6 | (D) 127.256.0.1 |
Explanation: (B) 172.16.0.10 IP address can be used in WAN, (A) 265.0.0.1 and (D) 127.256.0.1 are not valid IP addresses to be used in IPv4, because an IP address in IPv4 can be represented in 32bit binary number or 4 byte, which can be further represented in doted decimal number where dot (.) being placed after each 8bit, so with 8 bit, 28=256 or 0 to 255 are the range of the IP address. In (A) 265.0.0.1 and (D) 127.256.0.1, 256 is invalid for any IP address of IPv4 addressing. And (C) 15.1.5.6 is being used for private addresses only.
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Q. ____________transport layer protocol is used to support electronic mail.
(A) SNMP | (B) IP |
(C) SMTP | (D) TCP |
Q. ______________is used to convert from recursive to iterative implementation of an algorithm.
(A) Array | (B) Tree |
(C) Stack | (D) Queue |
Q. _________________ pairs of traversals is not sufficient to build a binary tree.
(A) Preorder and Inorder | (B) Postorder and Inorder |
(C) Postorder and Preorder | (D) None of the above |
Explanation: (C) Postorder and Preorder pairs of traversals is not sufficient to build a binary tree.
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Q. Among all given option, __________________ must reside in the main memory.
(A) Assembler | (B) Compiler |
(C) Linker | (D) Loader |
Explanation: Option:(D) Loader is the correct one. Loader resides in MAIN memory to load the object program into main memory from the secondary memory, when executing any program.
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Q. Time to Live (TTL) field in the IP datagram is used _____________
(A) to optimize throughput | (B) to prevent packet looping |
(C) to reduce delays | (D) to prioritize packets |
Explanation: Option: (B) to prevent packet looping is the correct one.
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Q. Maximum number of superkeys for the relation schema
R (X, Y, Z, W) with X as the key is:
(A) 6 | (B) 8 |
(C) 9 | (D) 12 |
Ans: (B) 8
Explanation:
Maximum number of superkeys possible for any relations with n attributes can be express with the following expression: 2^(n-1)
Here, n= 4
So, the possible superkeys is =2^(4-1)= 2^3=8
Example: X, XY, XZ, XW,XYZ,XYW,XWZ, XYZW
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Q. If the period of a signal is 100ms. Then its frequency in Hertz is ___________
(A) 10 | (B) 100 |
(C) 1000 | (D) 10000 |
Explanation: As we know Time period and Frequency has inverse relation with each other. So, as the time increases frequency goes down and when the time period decreases the frequencies goes up. For example:
Time | Hertz |
1Second | 1 Hertz |
1Milisecond | 1 Kilo Hertz |
1Microsecond | 1 Mega Hertz |
1Nenosecond | 1 Giga Hertz |
Similarly:
Time | Hertz |
1 Millisecond | 1 Kilo Hertz |
10 Millisecond | 100 Hertz |
100 Millisecond | 10 Hertz |
1000Millisecond | 1 Hertz |
So, when the period of a signal is 100ms, its frequency in Hertz becomes 10 Hz.
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