RRB JE Stage II exam is scheduled to be conducted from 28th August to 1st September 2019. The overall merit for the selection will be made on the basis of the marks scored in stage II, so candidates need to prepare thoroughly for the second stage. To help the aspirants, Adda247 has come up with a study plan in which we will provide you daily quizzes of all technical subjects. The below quiz will be helpful for the candidates who opted the exam group Computer Science & IT in RRB JE stage II.
Q1.Which of the following are two types of basic adder circuits?
(a) Sum and Carry
(b) Asynchronous and Synchronous
(c) Half-adder and Full-adder
(d) One and Two’s Complement
Q2.Which of the following is the correct statement to select all record from table STUDENT where subject is COMPUTER?
(a) Select from STUDENT where subject=’COMPUTER’;
(b) Select * from STUDENT where subject=”COMPUTER”;
(c) Select all from STUDENT where subject=’computer’;
(d) Select * from STUDENT where subject=’Computer’;
Sol. The SELECT statement is used to select data from a database, WHERE clause is used to filter records and (*) is used to select all records from table.
SELECT * FROM table_name WHERE condition;
Q3.A symbol on the screen that represents a disk, document or program that you can select _________.
S3. Ans. (c)
Sol. Icon is a symbol on screen that may represent many things.
Q4.Which among the following is the latest version of Microsoft Windows?
(a) Microsoft Office 365
(b) Windows 9.1
(c) Windows 10
(d) Windows 8.1
S4. Ans. (c)
Sol. Windows 10 is the latest version of Microsoft’s Windows operating system.
Q5.Changing desktop wallpaper is a simple task which can be accessed through ________ section of all control panel items.
S5. Ans. (a)
Sol. Wallpapers can be changed by the personalize section of all control panel items.
Q6.Different icons of application software can be found in which bar in the latest version of Microsoft Windows?
(a) Start Menu
(d) Control Pannel
S6. Ans. (a)
Sol. Microsoft re-introduced Start Menu bar and you can access various application software like paint or Word.
Q7. In the blocked state :
(a) the processes waiting for I/O are found
(b) the process which is running is found
(c) the processes waiting for the processor are found
(d) All of the above
Sol. A process that is blocked on some event (such as I/O operation completion or a signal). A process may be blocked due to various reasons such as when a particular various reasons such as when a particular process has exhausted the CPU time allocated to it or it is waiting for an event to occur.
Q8.A method, which transfers the entire block of data from its own buffer to main memory takes place without intervention by CPU.
(a) Programmed input/output
(b) Interrupt driven input/output
(c) Direct Memory Access (DMA)
(d) Resident monitor
Sol. Direct Memory Access (DMA) allows certain hardware subsystems within the computer to access system memory independently of the central processing unit (CPU).
Q9.Which one is true for FCFS scheduling?
(a) It is simplest CPU scheduling
(b) Average waiting time under the FCFS policy, however is often quite long
(c) FCFS scheduling algorithm is non preemptive
(d) All of above
Sol. The simplest CPU-scheduling algorithm is the first come first serve (FCFS) scheduling algorithm. With this algorithm, processes are assigned the CPU in the order they request it. The average waiting time under an FCFS policy is generally not minimal and may very substantially if the process’s CPU burst times vary greatly. Also FCFS scheduling algorithm is non-preemptive. That means, once the CPU has been allocated to a process, that process keeps the CPU until it release the CPU, either by terminating or by requesting I/O.
Q10.Wasting of memory within a partition, due to a difference in size of a partition and of the object resident within it, is called —
(a) External fragmentation
(b) Internal fragmentation
Sol. Internal fragmentation is the space wasted inside of allocated memory blocks because of restrictions on the allowed sizes of allocated blocks. Allocated memory may be slightly larger than requested memory; this size difference is memory internal to a partition, but not being used.
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