Two four-bit operands supply eight address bits, requiring 2⁸=256 ROM words. Their product needs eight result bits since 15×15=225. Capacity is 256×8=2048 bits, or 2 Kbits, D.
More Combinational Circuits questions
GATE 2013 CS (CS1) Q15– Inputs D₀,D₁,D₂,D₃ and outputs X₀,X₁,V have rows: 0000→xx0; 1000→001; x100→011; xx10→101;…
GATE 2013 CS (CS1) Q56– A RAM chip stores 1024 eight-bit words. How many 2-to-4 decoders with enable are needed to…
GATE 2010 CS (CS1) Q19– The shown 4:1 multiplexer has select inputs P (MSB),Q (LSB). What is f?
GATE 2015 CS (CS2) Q65– A half adder uses XOR and AND gates. A full adder uses two half adders and one OR gate.…
GATE 2021 CS (CS2) Q15– Which 4-to-1 multiplexer implements f(x,y,z)=Σm(0,1,3,4,5,6), with select lines s1=y,s0=z?
GATE 2026 CS (CS2) Q59– Consider the digital circuit shown below with two input lines A and B, two select lines S…