(i)  It is the branch of physics incorporated with technology towards the design of circuits using transistors and microchips.
(ii) It depicts the behavior and movement of electrons in a semiconductor, vacuum, or gas.
(iii) Electronics deals with electrical circuits that involve active components such as transistors, diodes, integrated circuits, and sensors, associated with the passive components like resistors, inductors, capacitors, and transformers.

(i) For a given frequency of incident light the number of photoelectrons emitted is directly proportional to the intensity of the incident light. The saturation current is also directly proportional to the intensity of incident light.
(ii) Maximum kinetic energy of the photo electrons is independent of intensity 0 the incident light.
(iii) Maximum kinetic energy of the photo electrons from a given metal is directly proportional to the frequency of incident light.
(iv) For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.
(v) There is no time lag between incidence of light and ejection of photo electrons.

According to law of reflection,
(i) The incident ray, reflected ray and normal to the reflecting surface all are coplanar (ie. lie in the same plane).
(ii) The angle of incidence i is equal to the angle of reflection r.
i = r

(i) The electromagnetic oscillations of LC system can be compared with the mechanical oscillations of a spring-mass system.
(ii) There are two forms of energy involved in LC oscillations. One is electrical energy of the charged capacitor; the other magnetic energy of the inductor carrying current.
(iii) Likewise, the mechanical energy of the spring-mass system exists in two forms; the potential energy of the compressed or extended spring and the kinetic energy of the mass. The Table lists these two pairs of energy.
(iv) By examining, the analogies between the various quantities can be understood and these correspondences.
(v) The angular frequency of oscillations of a spring-mass is given by equation
\(\omega =\sqrt { \frac { k }{ m } } \)
k ⟶ \(\frac { 1 }{ C } \) and m ⟶ L. Therefore, the angular frequency of LC oscillations is given by
ω = \(\frac { 1 }{ \sqrt { LC } } \)

To calculate the total electrostatic potential energy, we use the following procedure. We bring all the charges one by one and arrange them according to the configuration as shown in Figure.
a) Bringing a charge q₁ from infinity to the point A requires no work, because there are no other charges already present in the vicinity of charge q₁.
b) To bring the second charge q₂ to the point B, work must be done against the electric field created by the charge q₁. So the work done on the charge q₂ is W = q₂ V_1B. Here V_1B is the electrostatic potential due to the charge q₁ at point B.
\(U_I=\frac{1}{4\piε_o}\frac{q_1q_2}{r_{12}}\)   .....(1)
Note that the expression is same when q₂ is brought first and then q₁ later.
c) Similarly to bring the charge q₃ to the point C, work has to be done against the total electric field due to both the charges q₁ and q₂. So the work done to bring the charge q₃ = q₃ (V_1C + V_2C). Here V_1C is the electrostatic potential due to charge q₁ at point C and V_2C is the electrostatic potential due to charge q₂ at point C.
The electrostatic potential is
\(U_{II}=\frac{1}{4\piε_o}(\frac{q_1q_2}{r_{13}}+\frac{q_2q_3}{r_{23}})\)    .....(2)
d) Adding equations (1) and (3), the total electrostatic potential energy for the system of three charges q₁, q₂ and q₃ is
U = U_I + U_II
\(U=\frac{1}{4\piε_o}(\frac{q_1q_2}{r_{12}}+\frac{q_2q_3}{r_{13}}+\frac{q_2q_3}{r_{23}})\)  ....(3)

According to the superposition principle, the total electrostatic force on charge q₁ is the vector sum of the forces due to the other charges,
\(\vec { { F }_{ 1 }^{ tot } } =\bar { { F }_{ 12 } } +\bar { { F }_{ 13 } } +\bar { F_{ 14 } } \)
The following diagram shows the direction of each force on the charge q₁.

The charges q₂ and q₄ are equi-distant from q₁. As a result the strengths (magnitude) of the forces \(\vec { { F }_{ 12 } } \) and \(\vec { { F }_{ 14 } } \) are the same even though their directions are different. Therefore the vectors representing these two forces are drawn with equal lengths. But the charge q₃ is located farther compared to q₂ and q₄. Since the strength of the electrostatic force decreases as distance increases, the strength of the force \(\vec { { F }_{ 13 } } \) is lesser than that of forces \(\vec { { F }_{ 12 } } \) and \(\vec { { F }_{ 14 } } \). Hence the vector representing the force \(\vec { { F }_{ 13 } } \) is drawn with smaller length compared to that for forces \(\vec { { F }_{ 12 } } \) and \(\vec { { F }_{ 14 } } \).
From the figure, r₂₁ =\(\sqrt { 2 } \) m = r₄₁ and r₃₁ = 2m
The magnitudes of the forces are given by
F₁₃ = \(\frac { kq^{ 2 } }{ r_{ 31 }^{ 2 } } =\frac { 9\times 10^{ 9 }\times 10^{ -12 } }{ 4 } \)
F₁₃  = 2.25 x 10^-3 N
F₁₂ = \(\frac { kq^{ 2 } }{ r_{ 31 }^{ 2 } } ={ F }_{ 14 }=\frac { 9\times 10^{ 9 }\times 10^{ -12 } }{ 2 } \)
= 4.5 x 10^-3N
From the figure, the angle θ = 45⁰. In terms of the components, we have
\(\vec { { F }_{ 12 } } ={ F }_{ 12 }cos\theta \hat { i } -{ F }_{ 12 }sin\theta \hat { j } \)
= 4.5 x 10^-3 x \(\frac { 1 }{ \sqrt { 2 } } \hat { i-4.5\times { 10 }^{ -3 }\times \frac { 1 }{ \sqrt { 2 } } \hat { j } } \)
 \(\vec { { F }_{ 13 } } =F_{ 13 }\hat { i } \) = 2.25 x 10^-3 N\(\hat { i } \)
\(\vec { { F }_{ 14 } } ={ F }_{ 14 }cos\theta \hat { i } +{ F }_{ 14 }sin\theta \hat { j } \)
= 4.5 x 10^-3 x \(\frac { 1 }{ \sqrt { 2 } } \hat { i+4.5\times { 10 }^{ -3 }\times \frac { 1 }{ \sqrt { 2 } } \hat { j } } \)
Then the total force on q₁ is,
\(\vec { { F }_{ 1 }^{ tot } }={ (F }_{ 12 }cos\theta \hat { i } -{ F }_{ 12 }sin\theta \hat { j } )+{ F }_{ 13 }\hat { i } +{ (F }_{ 14 }cos\theta \hat { i } +{ F }_{ 14 }sin\theta \hat { j } )\)
\(\vec { { F }_{ 1 }^{ tot } } =({ F }_{ 12 }cos\theta +F_{13}+{ F }_{ 14 }cos\theta )\hat { i } +(-{ F }_{ 12 }sin\theta +{ F }_{ 14 }sin\theta )\)\(\hat { j } \)
Since F₁₂ = F₁₄, the j ^th component is zero.
Hence we have
\(\vec { { F }_{ 1 }^{ tot } } =({ F }_{ 12 }cos\theta +F_{13}+{ F }_{ 14 }cos\theta )\hat { i } \)
substituting the values in the above equation,
\(\left( \frac { 4.5 }{ \sqrt { 2 } } +2.25+\frac { 4.5 }{ \sqrt { 2 } } \right) \times10^{-3}\hat { i }=(4.5\sqrt { 2 } +2.25)\times 10^{-3}\hat { i } \)
\(\vec { { F }_{ 1 }^{ tot } } \) = 8.61 x 10^-3 N\(\hat { i } \)
The resultant force is along the positive x-axis.

Directions : Question numbers 53, 54, 55 are based on following passage.
PASSAGE : ANPN transistor is used in common emitter made in an amplifier circuit. A change of 40 HA in the base current changes the output current by 2 mA and 0.04 V in input voltage.

\(\begin{array}{l} \alpha=\frac{I_c}{I_{\mathrm{c}}} ; \mathrm{I}_{\mathrm{c}}=\alpha \mathrm{I}_{\mathrm{t}}=9 \mathrm{~mA} \\ \mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{t}}-\mathrm{I}_{\mathrm{c}}=10-9=1 \mathrm{~mA} \end{array}\)

(i) Consider a parallel beam of light, incident on a reflecting plane surface such as a plane mirror XY as shown in Figure.

(ii) The incident wavefront is AB and the reflected wavefront is A'B' in the same medium. These wavefronts are perpendicular to the incident rays L, M and reflected rays L', M' respectively.
(iii) By the time point A of the incident wavefront touches the reflecting surface, point B is yet to travel a distance BB' to touch the reflecting surface a B'.
(iv) When point B falls on the reflecting surface at H', point A would have reached A.
(v) This is applicable to all the points on the wavefront. Thus, the reflected wavefront A'B' emanates as a plane wavefront. The two normals Nand N' are considered at the points where the rays Land Mfallon the reflecting surface.
(vi) As reflection happens in the same medium, the speed of light is the same before and after the reflection.
(vii) Hence, the time is taken for the ray to travel from B to B' is the same as the time taken for the ray to travel from A to A'.
(viii) Thus, the distance BB' is equal to the distance AA'; (A~A' = BB').
(a) The incident rays, the reflected rays, and the normal are in the same plane.
(b) Angle of incidence,\(\angle i=\angle NAL={ 90 }^{ o }-\angle NAB=\angle BAB'\) 
Angle of reflection,
∠r= ∠N' B' M' = 900 - ∠N' B' A' = A' B' A' 
(ix) For the two right-angle triangles, ΔABB' and ΔB' A' A', the right angles, ∠B and ∠A' are equal, (∠B and∠A = 900); the two sides, ∠A' and ∠B' are equal, (AA'= BB'); the side AB' is common.
(x) Thus, the two triangles are congruent. As per the property of congruency, the two angles, ∠BAB' and A' B' A' must also be equal.
i = r
Hence, the laws of reflection are proved.

V_BE =0.7V
Input junction is a forward biased.
Since,
V_BE = 0.7V
V_CE = V_BE+ V_CB
V_CB = V_CE-V_BE
To determine V_CB we find I_C
\({ I }_{ C }\cong { I }_{ C }\frac { 2-{ V }_{ BE } }{ { R }_{ 2 } } =\frac { 2-0.7 }{ 1k\Omega } \)
I_C = 1.3mA
V_CE = V_CC - I_C (R₁ + R₂)
= 10 - 1.3mA (10K + 1K)
V_CE = - 4.3 V
V_CE =-4.3V-0.7
V_CB = - 5V

i) AND gate
a) Circuit Symbol:
The circuit symbol of a two input AND gate is shown in Figure (a). A and B are inputs and Y is the output. It is a logic gate and hence A, B, and Y can have the value of either 1 or 0

Two input AND gate

Inputs		outputs
A	B	Y = A + B
0	0	0
0	1	0
1	0	0
1	1	1

Truth table
b) Boolean equation:
Y = A.B
It performs logical multiplication and is different from arithmetic multiplication.
c) Logic operation:
The output of AND gate is high only when all the inputs are high. In the rest of the cases, the output is low. It is represented in the truth table (Figure (b).
ii) OR gate
a) Circuit Symbol:
The circuit symbol of a two input OR gate is shown in Figure (a). A and B are inputs and Y is the output. 

The input OR gate

Inputs		outputs
A	B	Y = A + B
0	0	0
0	1	1
1	0	1
1	1	1

Truth table
a) Boolean equation:
A + B = Y
It performs logical addition and is different from arithmetic addition.
b) Logic operation:
The output of OR gate is high (logic 1 state) when either of the inputs or both are high. The truth table of OR gate is shown in Figure (a).
iii) NOT gate
a) Circuit Symbol:
The circuit symbol of NOT gate is shown in Figure (a). A and B are inputs and Y is the output. 

NOT gate

Inputs	Output
A	Y = Ā
0	1
1	0

Truth table
a) Boolean equation:
Y = Ā
b) Logic operation:
The output is the complement of the input. It is represented with an overbar. It is also called as inverter. The truth table infers that the output Y is I when input A is 0 and vice versa. The truth table of NOT is shown in Figure (b).
iv) NAND gate
a) Circuit Symbol:
The circuit symbol of NAND gate is shown in Figure (a). A and B are inputs and Y is the output.

Two input NAND gate

Inputs		Output (AND)	outputs (NAND)
A	B	Z = A.B	Y = \(\overline { A.B } \)
0	0	0	1
0	1	0	1
1	0	0	1
1	1	1	0

Truth table
b) Boolean equation:
Y = \(\overline { A.B } \)
Logic operation:
The output Y equals, the complement of AND operation. The circuit is an AND gate followed by a NOT gate. Therefore, it is summarized as NAND. The output is at logic zero only when all the inputs are high. The rest of the cases, the output is high (Logic I state). The truth table of NAND gate is shown in Figure (b).
v) NOR gate
a) Circuit Symbol:
The circuit symbol of NOR gate is shown in Figure (a). A and B are inputs and Y is the output.

Two input NANS gate

Inputs		Output (OR)	outputs (NOR)
A	B	Z = A + B	Y = \(\overline { A+B } \)
0	0	0	1
0	1	1	0
1	0	1	0
1	1	1	0

Truth table
Boolean equation:
Y =  \(\overline { A+B } \)
Logic operation:
The output Y equals the complement of OR operation (A OR B). The circuit is an OR gate followed by a NOT gate and is summarized as NOR. The output is high when all the inputs are low. The output is low for all other combinations of inputs. The truth table of NOR gate is shown in Figure (b).
vi) Ex-OR gate
a) Circuit Symbol:
The circuit symbol of Ex-OR gate is shown in Figure (a). A and B are inputs and Y is the output. The Ex-OR operation is denoted as ⊕

Ex-OR gate

Inputs		outputs (Ex-OR)
A	B	Y = A ⊕ B
0	0	0
0	1	1
1	0	1
1	1	0

Truth table
b) Boolean equation
Y =  \(A.\overline { B } \) + \(\overline { A }.B \)
Y = A ⊕ B
Logic operation: 
The output is high only when either of the two inputs is high. In the case of an Ex-OR gate with more than two inputs, the output will be high when odd number of inputs are high. The truth table of Ex-OR gate is shown in Figure (b).

Electrical system	Mechanical system
Charge q	Displacement x
Current i = \(\frac { dq }{ dt } \)	Velocity v = \(\frac { dx }{ dt } \)
Inductance L	Mass m
Reciprocal of capacitance \(\frac { 1 }{ C } \)	Force constant k
Electrical energy  = \(\frac { 1 }{ 2 } \left( \frac { 1 }{ C } \right) { q }^{ 2 }\)	Potential energy = \(\frac { 1 }{ 2 } k{ x }^{ 2 }\)
Magnetic energy = \(\frac { 1 }{ 2 } \) Li2	Kinetic energy = \(\frac { 1 }{ 2 } \) mv2
Electromagnetic energy = \(U=\frac { 1 }{ 2 } \left( \frac { 1 }{ C } \right) { q }^{ 2 }\) + \(\frac12\)Li2	Mechanic energy E = \(\frac { 1 }{ 2 } k{ x }^{ 2 }\) + \(\frac { 1 }{ 2 } \) mv2

The key components of a robot are Power conversion unit, Actuators, Electric motors, Pneumatic Air Muscles, Muscle wires, Piezo Motors and Ultrasonic Motors, Sensors, and Robot locomotion.

The output at the 1^st AND gate : A\(\overline { B } \)
The output at the 2^nd AND gate : ĀB
The output at the OR gate: Y = A. \(\overline { B } \) + Ā .B

The P.D. between A and B is given by
\(V =\left[V_{\mathrm{A}}-V_{\mathrm{B}}\right]-V_{\mathrm{b}}(\mathrm{Si}) \)
\(=[3.3-(-7.4)]-0.7 \)
\(=10.7-0.7=10 \mathrm{~V} \)
The value of current flowing through AB can be obtained by using Ohm’s law
\(I=\frac { V }{ R } =\frac { 10}{ 1\times { 10 }^{ 3 } } ={ 10 }^{ -2 }A=10mA\)

(i) De Morgan's First theorem states that the complement of the sum of two inputs is equal to the product of its complements.
\(\overline{\mathrm{A}+\mathrm{B}}=\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\)
(ii) The second theorem states that the complement of the product of two inputs is equal to the sum of its complements.
\(\overline{\mathrm{A} \cdot \mathrm{B}}=\overline{\mathrm{A}}+\overline{\mathrm{B}}\)

(i) If the portion of the output fed to the input is in phase with the input, then the magnitude of the input signal increases
(ii) It is necessary for sustained oscillations.

When a PN junction diode is forward biased, the depletion region decreases and the diode conduct once after the barrier potential is crossed, when it is reverse biased the depletion region increases and the diode does not conduct sci it is called as unidirectional device.