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OS : 2011 GATE Questions

Questions : 4


QID : 251GATE-2011, 1M

A thread is usually defined as a ‘light weight process’ because an operating system (OS) maintains smaller data structures for a thread than for a process. In relation to this, which of the followings is TRUE?

Options:
A) On per-thread basis, the OS maintains only CPU register state
B) The OS does not maintain a separate stack for each thread
C) On per-thread basis, the OS does not maintain virtual memory state
D) On per thread basis, the OS maintains only scheduling and accounting information
Ans : Option A
Solution :
QID : 253GATE-2011, 1M

Let the time taken to switch between user and kernel modes of execution be t1 while the time taken to switch between two processes be t2. Which of the following is TRUE?

Options:
A) t1 > t2
B) t1 = t2
C) t1 < t2
D) Nothing can be said about the relation between t1 and t2
Ans : Option C
Solution :
Process switching also involves mode changing
QID : 255GATE-2011, 2M

Consider the following table of arrival time and burst time for three processes P0, P1 and P2.

The pre-emptive shortest job first scheduling algorithm is used. Scheduling is carried out only at arrival or completion of processes. What is the average waiting time for the three processes?

Options:
A) 5.0ms
B) 4.33ms
C) 6.33ms
D) 7.33ms
Ans : Option A
Solution :

QID : 254GATE-2011, 2M

An application loads 100 libraries at startup. Loading each library requires exactly one disk access. The seek time of the disk to a random location is given as 10ms. Rotational speed of disk is 6000rpm. If all 100 libraries are loaded from random locations on the disk, how long does it take to load all libraries? (The time to transfer data from the disk block once the head has been  positioned at the start of the block may be neglected)

Options:
A) 0.50s
B) 1.50s
C) 1.25s
D) 1.00s
Ans : Option B
Solution :
6000 rotations ________ 60 sec
Thus, 1 rotation ________10 ms

Therefore, Rotational latency (RL) = 5ms (Avg is the half of one rotation time which is required to set the head at the start of sector)

Time for 1 library = Seek Time + RL + Transfer Time
= 10ms + 5ms + zero (its to be neglected as given)
= 15ms

Time to load all libraries = 15 ×100 = 1500ms = 1.5sec
 

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