Diode Bis reverse biased.
When diode is reverse based, the barrier height increases as the direction of applied voltage (V) and the direction of barrier potential (V₀) is same. The effective barrier height under reverse biased is (V₀ + V).
Diode C is forward biased.
As the direction of applied voltage is opposite to the barrier potential, therefore, the effective barrier height is reduced to (V₀ - V).

X: Message signal generator
Y: Modulator
Z: Power amplifier 
Function of elements
X : It converts one form of message signal into electrical energy. The variation of current or voltage changes in accordance to variation in pressure, ete. also known as transducer.
Y : Used for superimposing the low frequency modulating signals over high frequency carrier wave.
Z : Power amplifier strengthens the modulated signals for transmission.

Here, R = 10 \(k\Omega \) = 10 x 10³ \(\Omega \); C = 12 pF = 12 x 10^-12 F
v_c = 40 MHz = 40 x 10⁶ Hz = 4 x 10⁷ Hz
Time period of carrier frequency, \({ T }_{ c }=\frac { 1 }{ { v }_{ c } } =\frac { 1 }{ 4\times { 10 }^{ 7 } } =2.5\times { 10 }^{ -8 }s\)
Time constant of C-R circuit, \(\tau =RC=\left( 10\times { 10 }^{ 3 } \right) \times \left( 12\times { 10 }^{ -12 } \right) =12\times { 10 }^{ -8 }s\)
As \({ T }_{ c }<\tau ,\) therefore, circuit is good enough for detection.

\(Here, \ R=100\Omega , \ { E }_{ v }=220V, \ v=50Hz\)
\((a) \ { I }_{ v }=?, \ { I }_{ v }=\frac { { E }_{ v } }{ R } =\frac { 220 }{ 100 } =2.2A\)
\((b) \ Net \ power \ consumed \ over \ a \ full \ cycle,\)
\(P={ E }_{ v }{ I }_{ v }=220\times 2.2=484W\)

\((111001.01)_{ 2 }=(111001)_{ 2 }+(0.01)_{ 2 }\)
\( (111001)_{ 2 }=1\times2^{ 0 }+0\times2^{ 1 }+0\times2^{ 2 }+1\times2^{ 3 }+1\times2^{ 4 }+1\times2^{ 5 5 }\)
\( =1+0+0+8+16+32=(57)_{ 10 } \ (0.01)_{ 2 }=0\times2^{ -1 }+1\times2^{ -2 }=0+0.25=(0.25)_{ 10 }\)
\(\therefore \ (111001.01)_{ 2 }=(57)_{ 10 }+(0.25)_{ 10 }=(57.25)_{ 10 }\)

(a) For the cut-off or threshold frequency, the energy h v₀ of the incident radiation must be equal to work function  Φ₀, so that
\({ V }_{ 0 }=\frac { { \phi }_{ 0 } }{ h } =\frac { 2.14eV }{ 6.63\times { 10 }^{ -34 }Js }\)
\(=\frac { 2.14\times 1.6\times { 10 }^{ -19 }J }{ 6.63\times { 10 }^{ -34 }Js } =5.16\times { 10 }^{ 14 }Hz\)
Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.
(b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the potential energy eV₀ by the retarding potential V₀. Einstein’s Photoelectric equation is 
\(e{ V }_{ 0 }=\frac { hc }{ \lambda } -{ \phi }_{ 0 }\)
\(or\ \lambda =\frac { hc }{ \left( e{ V }_{ 0 }+{ \phi }_{ 0 } \right) }\)
\(or \ \lambda =\frac { \left( 6.63\times { 10 }^{ -34 }Js \right) \times \left( 3\times { 10 }^{ 8 }m/s \right) }{ \left( e\times 0.6V+2.14eV \right) } \)
\(\lambda =\frac { 19.89\times { 10 }^{ -26 }Jm }{ 2.74\times 1.6\times { 10 }^{ -19 }J } =454nm\)

Only (c) is right; the rest cannot represent electrostatic field lines.
(a) is wrong because field lines must be normal to a conductor.
(b) is wrong because lines of force cannot start from a negative charge.
(d) is wrong because lines of force cannot intersect each other.
(e) is wrong because electrostatic field lines cannot form closed loops.

Given q₁ = 2 \(\times\) 10^-7 C, q₂ = 3 \(\times\) 10^-7 C   r = 30 cm = 0.3 m
Therefore, Force of repulsion, F = 9 \(\times\) 10⁹ \(\times\) \(\frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \)
= \(9\times { 10 }^{ 9 }\times \frac { 2\times { 1 }0^{ -7 }\times 3\times { 10 }^{ -7 } }{ { (0.3) }^{ 2 } } =\frac { 54\times { 10 }^{ -5 } }{ 9\times { 10 }^{ -2 } } \)
 \(=\ 6\times { 10 }^{ -3 }N\)

Potentiometer does not draw any (net) current front-the cell. Voltmeter draws some current from cell, when connected across at, hence measures terminal voltage.

Work done in bringing the charge q₁ from infinity to position r₁
W₁ = q₁ V(r₁)
Work done in bringing charge q₂ to the position r₂
W₂ = q₂ V(r₂) + \(\frac{q_1q_2}{4\pi\epsilon_0r_{12}}\)
Hence, total work done in assembling the two charges
W = W₁ +W₂
= q₁V(r₁) + q₂V(r₂) + \(\frac{q_1q_2}{4\pi\epsilon_0r_{12}}\)

(i) According to Gauss' theorem,
\(\phi=\frac { \Sigma q }{ { \varepsilon }_{ 0 }{ \varepsilon }_{ r } } \propto \Sigma q\)

\(\therefore\)  \(\frac { { \phi }_{ { s }_{ 1 } } }{ { \phi }_{ { s }_{ 2 } } } =\frac { 2Q }{ 2Q+4Q } \)
\(=\frac { 2Q }{ 6Q } =\frac { 1 }{ 3 } \)                                         
(ii) If a medium of dielectric constant \({ \varepsilon }_{ r }\) is introduced in the space inside S ₁ in place of air, then
\({ \phi }_{ { s }_{ 1 } }=\frac { \Sigma q }{ { \varepsilon }_{ 0 }{ \varepsilon }_{ r } } =\frac { 2Q }{ { \varepsilon }_{ 0 }{ \varepsilon }_{ r } } \) 

During landing or take off, the tyres of aircraft's get charged due to the friction between tyres and ground. In case, the tyres are slightly conducting, the charge developed on the tyres will not stay on them and it finds its way to the earth.

Here the hollow cylinder is positively charged. We know that the electric field lines appear to come out from the conductor. thus the field lines for a uniformely positive charged hollow cyclinder is shown in the figure.

Towards the earth.

The cause of charging is actual transfer of electrons from one body to the other.

Given,
The area of plate of the capacitor, A = 6 x 10^-3 m²
Distances between the plates, d = 3mm = 3 x 10^-3 m
Voltage supplied, V = 100V
Capacitance of a parallel plate capacitor is given by, \(C=\frac{\epsilon \times A}{d}\) 
Here,
ε = permittivity of free space = 8.854 x10^-12 N^-1 m ^-2 C^-2
\(C=\frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}=17.81 \times 10^{-12} \mathrm{~F}=17.71 \mathrm{pF}\)
Therefore, each plate of the capacitor is having a charge of
q = VC = 100 x 17.81 x 10^-12 C = 1.771 x 10^-9 C

The first refracting ABC forms the image I₁ of the object O. The image I₁ acts as virtual object for the second refracting surface ADC, which forms the real image I as shown in the diagram
For refraction at ABC 
\(\frac { { n }_{ 2 } }{ { v }_{ 1 } } -\frac { { n }_{ 1 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 1 } } \)
For refraction at ADC
\(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ { v }_{ 1 } } =\frac { { n }_{ 1 }-{ n }_{ 2 } }{ { R }_{ 2 } } \)
Adding equation (i) and equation (ii)
 \(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
We know, If  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\) 

Given, the equation of carrier wave,
 c(t) = 1 sin (8\(\pi \) t)        ..........(i)
(i) According to the figure,
Amplitude of modulating signal,
A_m = 1 V
Amplitude of carrier wave,
A_C =2
T_m = 1 s
From Eq.(i), we get
 \({ \omega }_{ m }=\frac { 2\pi }{ { T }_{ m } } =\frac { 2\pi }{ 1 } =2\pi \ rad/s\)    ....(ii)
 c(t) = 2 sin (8 \(\pi \) t)
So,  \({ \omega }_{ c }=4{ \omega }m_{ }\)
From Eq. (ii) , we get
So,   \({ \omega }_{ c }=4{ \omega }_{ m }\)
Amplitude of modulated wave,
A = A_m + A_c
 = 1 + 2= 3 V
The sketch of the amplitude modulated waveform is shown below:

For carrier signal, \({ \omega }\) = 8 \(\pi \) 
\(T=\frac { 2\pi }{ \omega } =\frac { 2\pi }{ 8\pi } =\frac { 1 }{ 4 } =0.25s\)
(ii) Modulation index, \(m=\frac{A_{m}}{A_{c}}=\frac{1}{2}=0.5\) 

\(We \ know \ that,{ E }_{ axial }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { 2\rho r }{ { ({ r }^{ 2 }-{ r }^{ 2 }) }^{ 2 } } \)
\(CaseI,\)
\( When \ r=10cm=0.1m\)
\(\\ { E }_{ axial }=3.75\times { 10 }^{ 5 }N/C\)
\(\\ 3.75\times { 10 }^{ 5 }=9\times { 10 }^{ 5 }\times \frac { 2\rho \times 0.1 }{ { { [(0.1) }^{ 2 }-{ a }^{ 2 }] }^{ 2 } } \ .....(i)\)
\(Case \ II,\)
\( When \ r=20cm=0.2m\)
\(\\ { E }_{ axial }=3\times { 10 }^{ 4 }\times \frac { 2\rho \times 0.2 }{ { { [(0.2) }^{ 2 }-{ a }^{ 2 }] }^{ 2 } } \ .......(ii)\)
Solving the Eqs. (i) and (ii), we get 
a = 0.05 m
Therefore, lemgth of the dipole is 2z.
So, 2a = 2 x 0.05
or  2a = 0.1 m

As \(\tau =nIBA \ sin \ \theta ,\ therefore,\ (i) \ \tau =0, \ when \ sin\theta =0 \ or \ \theta =0°, \ i.e.\)when the plane of coil is perpendicular to the direction of magnetic field. (ii) \(\tau =\) maximum, when \(sin \ \theta \)=maximum=1 or \(\theta =90°\)
\({ \tau }_{ max }=nIBA\times 1=nIBA\)
It will be so when the plane of coil is parallel to the direction of magnetic field.
(ii) Here, n = 200; r = 0.05m; I = 2.0 A; B = 0.20 T; \(\theta =90°-60°\)=\(30°\)
\( \tau =nIBA \ sin \ \theta ,\ therefore,\ (i) \ \tau =0, \ when \ sin\theta =0 \ or \ \theta =30°-60°, \ i.e.,\)
\( \\ { \tau }_{ max }=nIBA\times 1=nIBA\)
\( \tau =nIBA \ sin \ \theta =nIB({ \pi r }^{ 2 })sin\theta =200\times 2.0\times 0.20\left[ (22/7)\times { \left( 0.05 \right) }^{ 2 } \right] \times sin30°\)
\(=0.314\quad N-m=0.31\ Nm\)
(iii) From the above study, we find that when potential energy of the coil is minimum, both force and torque acting on the coil are zero. The same is true in real life. a person who is humble and boasts of nothing, would be a happy person, with no pulls and pressure of life.

Using superposition principle, we may write electric potential at the origin (x = 0) due to various charges as
\(V={1\over 4\pi\epsilon_o}[{q\over 1}+{q\over 2}+{q\over 4}+{q\over 8}_....]\)
\(V={q\over 4\pi\epsilon_o}[{1\over1}+{1\over2}+{1\over2^2}+{1\over2^3}+....]\)
As sum of infinite G.P. series, S = \({a\over 1-r}\)
Where a is first term and r is common ratio.
\(V={q\over 4\pi\epsilon_o}\){\(1\over (1-1/2)\)}=\(2q\over 4\pi\epsilon_o\)

From Kirchhoff's junction rule, we have
\( { I }_{ 1 }={ I }+{ I }_{ 2 }\)   ...........(i)
Applying Kirchhoff's loop rule to outer loop containing 10V cell, we get
\(10=IR+{ 10I }_{ 1 }\) ............(ii)
Applying Kirchhoff's loop rule to outer loop containing 2V cell, we get
\(2={ 5I }_{ 2 }-RI\)
\(2=5\left( { I }_{ 1 }-I \right) -RI\)
\(4={ 10I }_{ 1 }-10I-2RI\)
Subtracting eq (ii) from eq (i), we get
\(6=3RI+10I\)
\(2=I\left( R+\frac { 10 }{ 3 } \right) \)
From Ohm's law, we have
\(V=I\left( R+{ R }_{ off } \right) \)
Comparing eq (iii) and (iv), we get
\({ R }_{ ef }=\frac { 10 }{ 3 } \Omega \)
If \({ E }_{ eff }\) and \({ R }_{ eff }\) are the effective voltage and effective internal resistance of the combination, then the equivalent circuit is shown.