Carry look ahead adder is frequently used for addition because it is faster.

Correct Option: A

Carry look ahead adder is frequently used for addition because it is faster.

By Forming truth table we get

From the input and output table we see that outputs are in the form of gray code.
For example: Binary 0 0 0 → gray code 0 0 0 (first bit as it then add according to arrow shown)
0 1 1→ gray code 0 1 0 etc.
Hence alternative (B) is the correct answer.

Correct Option: B

By Forming truth table we get

From the input and output table we see that outputs are in the form of gray code.
For example: Binary 0 0 0 → gray code 0 0 0 (first bit as it then add according to arrow shown)
0 1 1→ gray code 0 1 0 etc.
Hence alternative (B) is the correct answer.

Apply hit and trial method
(A) A + AC + AB + BC = A + BC: i.e., False
(B) AC + AB + BC: i.e., False
(C) AC + AB + BC = AC + AB = BC + AA = A (A + B) + C (A + B) = (A + C) (A + B)
which is ≠ LHS i.e., False
(D) A + AC + AB + BC
= A (1 + C) + AB + BC
= A + BC
= L.H.S. True
Hence alternative (D) is the correct answer.

Correct Option: D

Apply hit and trial method
(A) A + AC + AB + BC = A + BC: i.e., False
(B) AC + AB + BC: i.e., False
(C) AC + AB + BC = AC + AB = BC + AA = A (A + B) + C (A + B) = (A + C) (A + B)
which is ≠ LHS i.e., False
(D) A + AC + AB + BC
= A (1 + C) + AB + BC
= A + BC
= L.H.S. True
Hence alternative (D) is the correct answer.

The gate period is made equal to main supply period, 1/50 = 0·02 s or 20 ms. for 60 Hz, it is 16·67 ms.

Correct Option: B

The gate period is made equal to main supply period, 1/50 = 0·02 s or 20 ms. for 60 Hz, it is 16·67 ms.

Each bit require one line so that for n lines, the size of memory in the address bus is 2ⁿ
for 16 bit, 2¹⁶ = 65, 536 or 64 K.

Correct Option: A

Each bit require one line so that for n lines, the size of memory in the address bus is 2ⁿ
for 16 bit, 2¹⁶ = 65, 536 or 64 K.