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, \(m=\frac{A_{m}}{A_{c}}=\frac{1}{2}=0.5\) 

The components of electric vector associated with light wave, along the direction of aligned molecules of a polaroid, get absorbed. As a result after passing through it, the components perpendicular to the direction of aligned molecules will be obtained in the form of plane polarised light.
(b) When unpolarised light is incident on the boundary between two transperent media, the reflected light is polarised, with electric vector perpendicular to the plane of incidence when the reflected and refracted light rays make a right angle, as shown in the figure below.
Since, \(\angle CBQ+\angle QBD={ 90 }^{ o }\)
(90 - i_B) + (90 - r) = 90^o
i_B + r = 90
r = 90 - i_B
Using Snell's law,
\(\mu =\frac { sin{ i }_{ B } }{ sinr } \)
\(=\frac { sin{ i }_{ B } }{ sin\left( 90-{ i }_{ B } \right) } \)
\(=\frac { sin{ i }_{ B } }{ { cosi }_{ B } } \)
\(\mu =tan{ i }_{ B }\).

(i) Average power delivered by an AC circuit is
\({ P }_{ av }={ V }_{ rms }{ I }_{ rms }cos\phi \)
where is minimum, the power delivered is minimum and hence, power dissipated will be maximum for the circuit.

While the slide is in the middle of the potentiometer only half of its resistance (R₀/2) will be between the points A and B. Hence, the total resistance between A and B, say, R₁, will be given by the following expression:
\(\frac{1}{R_{1}}=\frac{1}{R}+\frac{1}{\left(R_{0} / 2\right)}\)
\(R_{1}=\frac{R_{0} R}{R_{0}+2 R}\)
The total resistance between A and C will be sum of resistance between A and B and B and C, i.e., R₁ + R₀/2
∴ The current flowing through the potentiometer will be
\(I=\frac{V}{R_{1}+R_{0} / 2}=\frac{2 V}{2 R_{1}+R_{0}}\)
The voltage V1 taken from the potentiometer will be the product of current I and resistance R₁,
\(V_{1}=I R_{1}=\left(\frac{2 V}{2 R_{1}+R_{0}}\right) \times R_{1}\)
Substituting for R₁, we have a
\(V_{1}=\frac{2 V}{2\left(\frac{R_{0} \times R}{R_{0}+2 R}\right)+R_{0}} \times \frac{R_{0} \times R}{R_{0}+2 R}\)
\(V_{1}=\frac{2 V R}{2 R+R_{0}+2 R}\)
\(\text { or } V_{1}=\frac{2 V R}{R_{0}+4 R}\)

Given, AC = 2a, BP = r
AP = r + a and PC = r-a

The potential at P is V.
\(\therefore\) V = Potential at P value due to A+ Potential at P due to B + Potential at P due to C
\(V=\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } \left[ \frac { q }{ AP } -\frac { 2q }{ BP } +\frac { q }{ CP } \right] \ \)
\(=\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } .q\left[ \frac { 1 }{ \left( r+a \right) } -\frac { 2 }{ r } +\frac { 1 }{ \left( r-a \right) } \right] \)
\(=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{r(r-a)-2(r+a)(r-a)+r(r+a)}{r(r+a)(r-a)}\right]\)
\(=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{r^{2}-r a-2 r^{2}+2 a^{2}+r^{2}+r a}{r\left(r^{2}-a^{2}\right)}\right]\)
\(=\frac{q \cdot 2 a^{2}}{4 \pi \varepsilon_{0} r\left(r^{2}-a^{2}\right)}=\frac{q \cdot 2 a^{2}}{4 \pi \varepsilon_{0} \cdot r \cdot r^{2}\left(1-\frac{a^{2}}{r^{2}}\right)}\)
According to the question,
If r/a>> 1 Therefore, V = \(\frac { q.2{ a }^{ 2 } }{ { 4\pi \varepsilon }_{ 0 }.{ r }^{ 3 } } \Rightarrow V\propto \frac { 1 }{ { r }^{ 3 } } \)
As, we know that electric potential at a point on axial line due to an electric dipole is
\(V\propto \frac { 1 }{ { r }^{ 2 } } \)
In case of electric monopole, \(V\propto \frac { 1 }{ { r } } \) .
Then, we conclude that for larger r, the electric potential due to quadrupole is inversely proportional to the cube of the distance r, while due to an electric dipole, it is inversely proportional to the square of r and inversely proportional to the distance r for a monopole.

(ii) We know, Lorentz force, F = Bqv sin\(\theta \)
Where \(\theta \) = angle between velocity of particle and magnetic field=90⁰
So,Lorentz force, F = Bqv    [sin 90⁰ = 1]
When a charged particle enters in a magnetic field in a direction normal to the field, then in this condition,
Lorentz force = Centripetal force
\(Bqv=\frac { mv^{ 2 } }{ r } \Rightarrow r=\frac { mv }{ Bq } \)

(i) Post office box and meter bridge are two electrical appliances based on the principal of Wheatstone bridge.
(ii) The post office box is used practically in post and telegraph department to locate the snapping of telephone line.
The broken telephone line will touch the ground. Using post office box, resistance S of broken line is determined.
As resistance per unit length of line is known, the length of broken line can be calculated. Therefore, snapping of line is located.

Current sensitivity, \({ I }_{ s }=\frac { NAB }{ k } \) and
Voltage sensitivity, \(V_{ s }=\frac { NAB }{ kR } \)
Since, the resistance of the coil may vary, it implies an increase in current sensitivity may not necessarily increase voltage sensitivity.
Thus, the trajectory of both the particles will be same.

(i)When As is implanted in Si-crystal, n-type wafer is created. The no of majority carrier electrons due to doping of As is
\(n_{ e }={ N }_{ D }=\frac { 1 }{ 10^{ 6 } } \times5\times10^{ 28 }=5\times10^{ 22 }/m^{ -3 }\)
No of minority carriers (holes) in \(n-\)type wafer is 
\({ n }_{ h }=\frac { { n }_{ 1 }^{ 2 } }{ n_{ e } } =\frac { (1.5\times10^{ 16 })^{ 2 } }{ 5\times10^{ 22 } } =0.45\times10^{ 10 }/m^{ 3 }\)
when B is implanted in Si-crystal p-type wafer is created with no of holes
\({ n }_{ h }=N_{ A }=\frac { 200 }{ 10^{ 6 } } \times(5\times10^{ 28 })=1\times10^{ 25 }/m^{ 3 }\)
Minority carriers (electrons) created in the p-type wafer is
\({ n }_{ e }=\frac { { n }_{ 1 }^{ 2 } }{ n_{ h } } =\frac { (1.5\times10^{ 16 })^{ 2 } }{ 1\times10^{ 25 } } =2.25\times10^{ 10 }/m^{ 3 }\)
When \(p-n\)a junction is reverse biased, the minority carrier holes of \(n-\)region wafer \((n_{ h }=0.45\times10^{ 10 }/m^{ 3 })\) would contributte  more  to the reverse saturation current than minority carrier than minority carrier electrons \((n_{ e }=2.25\times10^{ 7 }/m^{ 3 })\)

As the nucleus is massive, recoil momentum of the atom may be neglected and the entire energy of the transition may be considered transferred to the Auger electron. As there is single valence electron in Cr, the energy states may be thought of as given by the Bohr model.
The energy of the nth state
\(E_n=Z^2R{1\over n^2}\) where R is the Rydberg constant and Z = 24
In transition from n = 2 to n = 1,
Energy released
\(\Delta E=-RZ^2\left[{1\over4}-1\right]={3\over4}Z^2R\)
The energy required to eject a n = 4 electron is 
\(E_4=Z^2R{1\over16}={Z^2R\over16}\)
So K.E. of Auger electron is given by
K.E. \(=Z^2R\left({3\over4}-{1\over10}\right)\)
\(={11\over16}\times24\times24\times13.6eV\)
= 5385.6eV