GATE Technical: Linked Answer Questions questions with solutions

Q1. Statement for Linked Answer Questions 52 and 53: Relation R has eight attributes A B C D E F G H. Fields of R contain only atomic values. F = {C H → G, A → BC, B → C F H, E → A, F → E G} is a set of functional dependencies (FDs) so that F+ is exactly the set of FDs that hold for R. Q.52 How many candidate keys does the relation R have?

1. 3
2. 4
3. 5
4. 6

Answer: 4

Attribute D never appears on any FD right-hand side, so it must belong to every candidate key. Testing minimal supersets, the closures of AD, BD, DE, and DF each yield all eight attributes, and these are the only minimal keys. There are 4 candidate keys (option index 1), not the stored 6.

Q2. Statement for Linked Answer Questions 54 and 55: A computer uses 46-bit virtual address, 32-bit physical address, and a three-level paged page table organization. The page table base register stores the base address of the first-level table (T1), which occupies exactly one page. Each entry of T1 stores the base address of a page of the second-level table (T2). Each entry of T2 stores the base address of a page of the third-level table (T3). Each entry of T3 stores a page table entry (PTE). The PTE is 32 bits in size. The processor used in the computer has a 1 MB 16-way set associative virtually indexed physically tagged cache. The cache block size is 64 bytes. Q.54 What is the size of a page in KB in this computer?

1. 2
2. 4
3. 8
4. 16

Answer: 8

With page size 2^p bytes and 4-byte (32-bit) entries, each table page holds 2^(p-2) entries, giving (p-2) index bits per level. The 46-bit virtual address splits as 46 = p + 3*(p-2) => 4p-6=46 => p=13, so page size = 2^13 = 8 KB (option C), not the stored 4 KB.

Q3. Statement for Linked Answer Questions 57 and 58: The figure above shows coils 1 and 2, with dot markings as shown, having 4000 and 6000 turns respectively. Both the coils have a rated current of 25 A. Coil 1 is excited with single phase, 400 V, 50 Hz supply. The coils are to be connected to obtain a single phase, 400/1000 V, auto-transformer to drive a load of 10 kVA. Which of the options given should be exercised to realize the required auto-transformer?

1. Connect A and D; Common B
2. Connect B and D; Common C
3. Connect A and C; Common B
4. Connect A and C; Common D

Answer: Connect A and C; Common B

Connecting A and C with Common B allows for the proper transformation ratio needed to achieve the desired output voltage of 1000 V from the 400 V input, utilizing the turns ratio effectively between the two coils.

Q4. Statement for Linked Answer Questions 52 and 53: A massless beam has a loading pattern as shown in the figure. The beam is of rectangular cross-section with a width of 30 mm and height of 100 mm. The beam is simply supported at A and C, with span AC = 4000 mm. A uniformly distributed load of intensity 3000 N m⁻¹ acts over the right half BC, where AB = 2000 mm and BC = 2000 mm. The maximum bending moment occurs at

1. Location B
2. 2675 mm to the right of A
3. 2500 mm to the right of A
4. 3225 mm to the right of A

Answer: 2500 mm to the right of A

Total UDL load = 6000 N acting at x=3000 mm, giving RA=1500 N and RC=4500 N. Shear stays 1500 N up to B (2000 mm) then 1500-3000(x-2000)=0 gives x=2500 mm, where the maximum bending moment occurs.

Q5. Statement for Linked Answer Questions 52 and 53: In orthogonal turning of a bar of 100 mm diameter with a feed of 0.25 mm/rev, depth of cut of 4 mm and cutting velocity of 90 m/min, it is observed that the main (tangential) cutting force is perpendicular to the friction force acting at the chip-tool interface. The main (tangential) cutting force is 1500 N. Q.52 The orthogonal rake angle of the cutting tool in degree is

1. zero
2. 3.58
3. 5
4. 7.16

Answer: zero

The orthogonal rake angle is zero because the main cutting force is perpendicular to the friction force, indicating that the cutting action is purely orthogonal with no inclination of the cutting edge.

Q6. Statement for Linked Answer Questions 54 and 55: In a simple Brayton cycle, the pressure ratio is 8 and temperatures at the entrance of compressor and turbine are 300 K and 1400 K, respectively. Both compressor and gas turbine have isentropic efficiencies equal to 0.8. For the gas, assume a constant value of specific heat at constant pressure (cp) equal to 1 kJ/kgK and ratio of specific heats as 1.4. Neglect changes in kinetic and potential energies. Q.54 The power required by the compressor in kW/kg of gas flow rate is

1. 194.7
2. 243.4
3. 304.3
4. 378.5

Answer: 304.3

T2s = 300*8^(0.4/1.4) = 300*1.811 = 543.4 K, so ideal work = 1*(543.4-300) = 243.4 kJ/kg. Actual compressor work = 243.4/0.8 = 304.3 kW per kg/s, which is the third option.