The given circuit has two inputs, A and B, each passing through a NOT gate, making the inputs Ā and B̄. These modified inputs are then fed into a NAND gate, which gives the output Y as follows:
1. **Output of NAND Gate:**
Y = (Ā ⋅ B̄).
2. **Apply De Morgan’s Law:**
Using De Morgan’s law:
Y = (Ā ⋅ B̄) = A + B.
Thus, the output Y behaves as an OR gate.
Answer: OR gate
For the circuit shown above, the equivalent gate is:
Which of the following circuits has the same output as that of the given circuit?
Consider the following logic circuit.
The output is Y = 0 when :
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
Two vessels A and B are connected via stopcock. Vessel A is filled with a gas at a certain pressure. The entire assembly is immersed in water and allowed to come to thermal equilibrium with water. After opening the stopcock the gas from vessel A expands into vessel B and no change in temperature is observed in the thermometer. Which of the following statement is true?
Choose the correct nuclear process from the below options:
\( [ p : \text{proton}, n : \text{neutron}, e^- : \text{electron}, e^+ : \text{positron}, \nu : \text{neutrino}, \bar{\nu} : \text{antineutrino} ] \)
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: