In the given circuit, we have two diodes \( D_1 \) and \( D_2 \). The voltage across the diodes and the resistor is affected by the direction in which the diodes are connected and their characteristics.
- \( D_1 \) is forward-biased because its anode is connected to +5V, and the cathode is connected to the node where \( V_o \) is measured.
- \( D_2 \) is reverse-biased because its anode is connected to the node \( V_o \), and its cathode is connected to ground.
Since \( D_2 \) is reverse-biased, it will not conduct, and \( D_1 \) will conduct. Therefore, the output voltage \( V_o \) will be zero because the voltage drop across the conducting diode \( D_1 \) is almost zero in a forward-biased condition.
Thus, the output voltage is zero.
Therefore, the correct answer is (2) Zero.
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
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: