The truth table given below is for: (A and B are the inputs, Y is the output)
 

A	B	Y
0	0	1
0	1	1
1	0	1
1	1	0

1.  NOR
2.  AND
3.  XOR
4.  NAND

In the following circuit, the output \(Y\) for all possible inputs \(A\) and \(B\) is expressed by the truth table: 
    

The circuit is equivalent to: 
     

1. AND gate
2. NAND gate
3. NOR gate
4. OR gate

A p-n photodiode is made of a material with a bandgap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly:

1. $10\times {10}^{14}$ $\mathrm{Hz}$

2. $5\times {10}^{14}$ $\mathrm{Hz}$

3. $1\times {10}^{14}$ $\mathrm{Hz}$

4. $20\times {10}^{14}$ $\mathrm{Hz}$

To get an output Y = 1 from the circuit shown below, the input must be:

1. A=0 B=1 C=0

2. A=0 B=0 C=1

3. A=1 B=0 C=1

4. A=1 B=0 C=0

Transfer characteristics [output voltage (${\mathrm{V}}_{\mathrm{o}}$) vs input voltage (${\mathrm{V}}_{\mathrm{i}}$)] for a base biased transistor in CE configurations are as shown in the figure. For using the transistor as a switch, it is used:

1. In region III

2. Both in the region (I) and (III)

3. In region II

4. In region I