Question 1 
A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch. The logic of switching of the bulb resembles
an AND gate  
an OR gate  
an XOR gate  
a NAND gate 
Question 1 Explanation:
Truth table of XOR gate
\begin{array}{ccc} A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}
So, from the XOR gate truth table it is clear that the bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch.
\begin{array}{ccc} A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}
So, from the XOR gate truth table it is clear that the bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch.
Question 2 
Consider a vector field {\overrightarrow{A}}( \overrightarrow{r}). The closed loop line integral \int \overrightarrow{A}\cdot \overrightarrow{di} can be expressed as
\int \int ( \overline{V}\times \overrightarrow{A} )\cdot \overrightarrow{ds} over the closed surface bounded by the loop  
\int \int \int ( \overline{V}\cdot \overrightarrow{A} )dv over the closed volume bounded by the loop
 
\int \int \int ( \overline{V}\cdot \overrightarrow{A} )dv over the open volume bounded by the loop
 
\int \int ( \overline{V}\times \overrightarrow{A} )\cdot \overrightarrow{ds} over the open surface bounded by the loop 
Question 2 Explanation:
According to Stoke's theorem
\oint_{c} \vec{A} \cdot \vec{d} i=\iint_{s}(\nabla \times \vec{A}) \cdot \overrightarrow{d s}
\oint_{c} \vec{A} \cdot \vec{d} i=\iint_{s}(\nabla \times \vec{A}) \cdot \overrightarrow{d s}
Question 3 
Two systems with impulse responses h_{1}\left ( t \right ) and h_{2}\left ( t \right ) are connected in cascade. Then the overall impulse response of the cascaded system is given by
product of h_{1}\left ( t \right ) and h_{2}\left ( t \right )
 
sum of h_{1}\left ( t \right ) and h_{2}\left ( t \right )
 
convolution of h_{1}\left ( t \right ) and h_{2}\left ( t \right )
 
subtraction of h_{2}\left ( t \right ) from h_{1}\left ( t \right )

Question 3 Explanation:
The overall impulse response h(t) of the cascade
system is given by:
h(t)=h_{1}(t) * h_{2}(t)
h(t)=h_{1}(t) * h_{2}(t)
Question 4 
In a forward biased pn junction diode, the sequence of events that best describes the mechanism of current flow is
injection, and subsequent diffusion and recombination of minority carriers  
injection, and subsequent drift and generation of minority carriers  
extraction, and subsequent diffusion and generation of minority carriers  
extraction, and subsequent drift and recombination of minority carriers 
Question 4 Explanation:
In a forward biased pnjunction diode, the current
flow is due to diffusion of majority carriers and
recombination of minority carriers.
Question 5 
In IC technology, dry oxidation (using dry oxygen) as compared to wet oxidation (using steam or water vapor) produces
superior quality oxide with a higher growth rate  
inferior quality oxide with a higher growth rate  
inferior quality oxide with a lower growth rate  
superior quality oxide with a lower growth rate 
Question 5 Explanation:
Dry oxidation has better quality over wet oxidation.
Dry oxidation is slower over wet oxidation.
Dry oxidation is slower over wet oxidation.
There are 5 questions to complete.