The truth table shows that the output Y is 1 only when either A is 1 and B is 0, or both A and B are 1.
This can be represented as Y = (A AND NOT B) OR (A AND B).
Logic circuit (2) contains an AND gate, and an OR gate, and a NOT gate which matches to the truth table.
So the answer is (2).
An \( \alpha \) particle is scattered from an Au target at rest as shown in the figure. \( D_1 \) and \( D_2 \) are the detectors to detect the scattered \( \alpha \) particle at an angle \( \theta \) and along the beam direction, respectively, as shown. The signals from \( D_1 \) and \( D_2 \) are converted to logic signals and fed to logic gates. When a particle is detected, the signal is 1 and is 0 otherwise. Which one of the following circuits detects the particle scattered at the angle \( \theta \) only?
A logic gate circuit is shown in the figure below. The correct combination for the input \( (P, Q) \) for which the output \( T = 1 \) is:
For the circuit shown above, the equivalent gate is:
The total number of structural isomers possible for the substituted benzene derivatives with the molecular formula $C_7H_{12}$ is __
Four capacitors each of capacitance $16\,\mu F$ are connected as shown in the figure. The capacitance between points A and B is __ (in $\mu F$)
Among, Sc, Mn, Co and Cu, identify the element with highest enthalpy of atomisation. The spin only magnetic moment value of that element in its +2 oxidation state is _______BM (in nearest integer).
X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene. X =_____ g. (nearest integer) [Given : molar mass (in g mol\(^{-1}\)) C : 12, H : 1, O : 16, N : 14]
A perfect gas (0.1 mol) having \( \bar{C}_V = 1.50 \) R (independent of temperature) undergoes the above transformation from point 1 to point 4. If each step is reversible, the total work done (w) while going from point 1 to point 4 is ____ J (nearest integer) [Given : R = 0.082 L atm K\(^{-1}\)]