In this circuit, we have two logic gates: a NAND gate and a NOT gate.
The first gate is a NAND gate, which gives a low output (0) only when both inputs are high (1). Since one of the inputs is 1, the other input should be 0 for the output to be 0. So, \( P = 0 \).
The second gate is a NOT gate that inverts the input. The input to this NOT gate is 0, so its output will be 1.
Thus, \( Q = 1 \). Therefore, \( P = 0 \) and \( Q = 1 \).
The correct answer is (B) : P = 0, Q = 1.
The circuit consists of two logic gates: a NOT gate and an AND gate. Let's analyze the circuit step by step:
1. The input to the NOT gate is 1, so the output of the NOT gate will be 0. Thus, the value of \( P \) is 0.
2. The output of the NOT gate (which is 0) is the input to the AND gate. The other input to the AND gate is also 0. Since an AND gate gives an output of 1 only when both inputs are 1, the output of the AND gate is 0.
Thus, the values of \( P \) and \( Q \) are:
- \( P = 0 \)
- \( Q = 1 \)
Therefore, the correct answer is \( P = 0, Q = 1 \).
Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is