DECEMBER 2011 - PAPER II

28. On receiving an interrupt from an I/O device, the CPU

(A) halts for predetermined time
(B) branches off to the interrupt service routine after completion of the current instruction
(C) branches off to the interrupt service routine immediately.
(D) hands over control of address bus and data bus to the interrupting device.

Ans:-B

---

29. The maximum amount of information that is available in one portion of the disk access arm for a removal disk pack (without further movement of the arm with multiple heads)

(A) a plate of data
(B) a cylinder of data
(C) a track of data
(D) a block of data

Ans:-B

---

30. Consider a logical address space of 8 pages of 1024 words mapped with memory of 32 frames. How many bits are there in the physical address ?

(A) 9 bits
(B) 11 bits
(C) 13 bits
(D) 15 bits

Ans:-D

Explanation:-
The question asks about the number of bits in the physical address. The word size is 1024. So, the number of bits required to calculate the offset into each page = 2¹⁰=1024=10 bits.
So for 32 frames(2⁵) of 1024 words each,we would require a total of 5+10 bits = 15 bits. Therefore the option is D.

---

31. CPU does not perform the operation

(A) data transfer
(B) logic operation
(C) arithmetic operation
(D) all of the above

Ans:-D

---

32. A chip having 150 gates will be classified as

(A) SSI
(B) MSI
(C) LSI
(D) VLSI

Ans:-B

Explanation:- The number of gates for the different classification are as follows.
Small Scale Integration(SSI) - <10
Medium Scale Integration(MSI) - between 10 to 1000
Large Scale Integration(LSI) - >1000
Very Large Scale Integration(VLSI) - >100000
So the answer is option B.

---

33.If an integer needs two bytes of storage, then the maximum value of unsigned integer is

(A) 2¹⁶ –1
(B) 2¹⁵ –1
(C) 2¹⁶
(D) 2¹⁵

Ans:-A

Explanation:- The same question is asked in June 2012,paper -II but there they have asked for the maximum value of signed integer.The link to that answer is JUNE 2012 - PAPER II. Here it is the maximum value of unsigned integer and the answer is option A.

---

34. Negative numbers cannot be represented in

(A) signed magnitude form
(B) '1’ s complement form
(C) '2’ s complement form
(D) none of the above

Ans:-D

Explanation:-
Negative numbers can be represented in signed magnitude form, or '1' complement form or '2''s complement form. So the option is D.

---

35. The cellular frequency reuse factor for the cluster size N is

(A) N
(B) N²
(C) 1/N
(D) 1/N²

Ans:-C

---