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
Consider the following logic circuit.
The output is Y = 0 when :
The logic gate equivalent to the combination of logic gates shown in the figure is
The output (Y) of the given logic implementation is similar to the output of an/a …………. gate.
A bead of mass \( m \) slides without friction on the wall of a vertical circular hoop of radius \( R \) as shown in figure. The bead moves under the combined action of gravity and a massless spring \( k \) attached to the bottom of the hoop. The equilibrium length of the spring is \( R \). If the bead is released from the top of the hoop with (negligible) zero initial speed, the velocity of the bead, when the length of spring becomes \( R \), would be (spring constant is \( k \), \( g \) is acceleration due to gravity):
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: