The Boolean expression $\mathrm{Y}=\mathrm{A} \overline{\mathrm{B}} \mathrm{C}+\overline{\mathrm{AC}}$ can be realised with which of the following gate configurations.
A. One 3-input AND gate, 3 NOT gates and one 2-input OR gate, One 2-input AND gate
B. One 3-input AND gate, 1 NOT gate, One 2-input NOR gate and one 2-input OR gate
C. 3-input OR gate, 3 NOT gates and one 2-input AND gate
Choose the correct answer from the options given below:
1. Boolean expression: \[ Y = A \overline{B} C + \overline{AC} \]

2. Gate configurations:
- A. One 3-input AND gate, 3 NOT gates and one 2-input OR gate, One 2-input AND gate
- B. One 3-input AND gate, 1 NOT gate, One 2-input NOR gate and one 2-input OR gate
Therefore, the correct answer is (2) A,B Only.
In the digital circuit shown in the figure, for the given inputs the P and Q values are:

The truth table corresponding to the circuit given below is 
In the circuit shown, assuming the threshold voltage of the diode is negligibly small, then the voltage \( V_{AB} \) is correctly represented by:
The value of current \( I \) in the electrical circuit as given below, when the potential at \( A \) is equal to the potential at \( B \), will be _____ A. 
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
An organic compound (X) with molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$ is not readily oxidised. On reduction it gives $\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}(\mathrm{Y})\right.$ which reacts with HBr to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis give 2,3-dimethylbutan-2-ol. Compounds (X), (Y) and (Z) respectively are: