Given, input frequency = 50 Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
\(\therefore\) Output frequency = 50 Hz
For a full-wave rectifier, the output frequency is twice the input frequency.
\(\therefore\)  Output frequency = 2 x 50 = 100 Hz.

n-type semiconductor is obtained when pentavalent impurity added to Si or Ge. All these materials are electrically neutral, so n-type semiconductor is also neutral.

By considering the situation given in the question,
At t = 0, N_A = N₀ (maximum), while N_B = 0. As time increases, N_A decreases exponentially and the number of atoms of B increases. They become (N_B) maximum and finally drop to zero exponentially by radioactive decay law. So, graph showing the variation of number of atoms of A and B will be shown as below:

(i) By Malus law, intensity of emergent light from \({ P }_{ 2 }\) is \(I={ I }_{ \circ }cos^{ 2 }\theta \) , where \(\theta \) is the angle between \({ P }_{ 1 }and{ P }_{ 2 }\)|
When  \(\theta ={ 90 }^{ \circ }\) 
\(\\ I={ I }_{ \circ }\times 0\)
Intensity of emergent light, I = 0
(ii) Intensity of light from \({ P }_{ 3 }\)
 \(=\left( \frac { { I }_{ \circ } }{ 2 } cos^{ 2 }\beta \right) \left[ cos^{ 2 }\left( { 90 }^{ \circ }-\beta \right) \right] \)

\(=\frac { { I }_{ \circ } }{ 2 } cos^{ 2 }\beta sin^{ 2 }\beta =\frac { { I }_{ \circ } }{ 8 } sin^{ 2 }2\beta \)
\(As,\ \frac { { I }_{ \circ } }{ 8 } sin^{ 2 }2\beta =\frac { { I }_{ \circ } }{ 8 } (given)\)
\(so, \ \left( sin2\beta \right) ^{ 2 }=1 \ \Rightarrow 2\beta ={ 90 }^{ \circ } \ \Rightarrow \beta ={ 45 }^{ \circ }\)

de-Broglie matter wave equation,
\(\lambda =\frac { h }{ p } =\frac { h }{ \sqrt { 2mK } } \ \left[ \because K=\frac { { P }^{ 2 } }{ 2m } \right] \)
where K is kinetic energy and m is a mass of the particle.
\(K=\frac { { h }^{ 2 } }{ 2m{ \lambda }^{ 2 } } \)   [ for same wavelength \( \lambda] \)
\(K\propto \frac { 1 }{ m }\)
\(\Rightarrow \ { K }_{ e }:{ K }_{ \alpha }:{ K }_{ p }=\frac { 1 }{ { m }_{ e } } :\frac { 1 }{ { m }_{ \alpha } } :\frac { 1 }{ { m }_{ p } } \)
where \({ m }_{ e },{ m }_{ p } \ and \ { m }_{ \alpha }\) are masses of electron, proton and \(\alpha \) -particle, respectively.
Also, \({ K }_{ e },k_{ p } \ and \ K_{ \alpha }\) are their respective kinetic energies.
\(\because \ m_{ \alpha }>m_{ p }>m_{ e }\)
\(\\ \Rightarrow m_{ \alpha }m_{ p }>m_{ e }m_{ \alpha }>m_{ e }m_{ p }\)
\(\\ { K }_{ e }>k_{ p }>K_{ \alpha }\)
(i) \(\alpha \) particle possess minimum kinetic energy
(ii) The electron has maximum kinetic energy. The magnifying power of an electron microscope is inversely related to the wavelength of radiation used. The Smaller wavelength of the electron beam in comparison to visible light increases the magnifying power of the microscope.

(i) The total charge of the capacitor remains conserved on introduction of dielectric slab. Also, the capacitance of capacitor increases to K times of original values. 
\(\therefore \\ \)  CV = C' V '= (KC) V' \(\Rightarrow V'=\frac { V }{ K } \) 
\(\therefore \\ \) New electric field,
\(E'=\frac { V' }{ d } =\left( { \frac { { V }/{ K } }{ d } } \right) =\left( \frac { V }{ d } \right) \frac { 1 }{ k } =\frac { E }{ K } \) 
​​\(\therefore \\ \)  On introduction of dielectric medium, new electric field E' becomes \(\frac { 1 }{ K } \) times of its original value (decrease).
(ii) The thickness of dieletric slab is \(\frac{d}{2}, \text { i.e. } t=\frac{d}{2}\)
The capacitance of a capacitor due to dielectric slab,
\(\begin{aligned} C & =\frac{\varepsilon_0 A}{d-t+\frac{t}{K}} \end{aligned}\)
\(\begin{aligned} =\frac{\varepsilon_0 A}{d-\frac{d}{2}+\frac{d}{2 K}}=\frac{2 \varepsilon_0 A}{d\left(1+\frac{1}{K}\right)} \end{aligned}\)

Let the distance between the plates be increased by a very small distance\(\triangle x\) = Force x Increased distance
= F.\(\Delta x\)     ...(i)  

Increase in volume of capacitor
= Area of plates x Increased distance
= A.\(\Delta x\)
u = Energy density = \(\frac { Energy }{ Volume } \)
Increase in energy = u x voulme = u.A.\(\Delta x\)   .....(ii)
As, Energy = Work done
F.\(\Delta x\) = u.A.\(\Delta x\) \([\because From \ Eqs.(i) \ and \ (ii)]\) 
 \(\Rightarrow\) F = u.A
\(=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }^{ 2 }.A\quad [\because u=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }^{ 2 } \ and \ E=\frac { V }{ d } ]\)
\(=\frac { 1 }{ 2 } \varepsilon _{ 0 }.\frac { { V }^{ 2 } }{ { d }^{ 2 } } .A=\left( \frac { { \varepsilon }_{ 0 }A }{ d } .A \right) \frac { V }{ d } \times \frac { 1 }{ 2 } \)
\(=\frac { 1 }{ 2 } .E.C.V=\frac { 1 }{ 2 } QE \ [\because C=\frac { { \varepsilon }_{ 0 }A }{ d } ,CV=Q]\)

Given, velocity of electromagnetic wave in vacum, c = 3 X 10⁸ m / s
Relative electric permeability,\(\varepsilon _{ r }\) = 2
and magnetic permeability, \(\mu _{ r }\) = 1
Since, velocity of electromagnetic wave in a medium can be calculated by
v = \(\frac { 1 }{ \sqrt { \varepsilon _{ 0 }\varepsilon _{ r }\mu _{ 0 }\mu _{ r } } } =\frac { 1 }{ \sqrt { \varepsilon _{ 0 }\mu _{ 0 } } \quad x \quad \sqrt { \mu _{ r }\varepsilon _{ r } } } \)
Where,
\(\frac { 1 }{ \sqrt { \varepsilon _{ 0 }\mu _{ 0 } } } =c \ \Rightarrow \ v \ = \ \frac { c }{ \sqrt { \mu _{ r }\varepsilon _{ r } } } \)     ...(i)
Therefore, v \(=\frac { 3x10^{ 8 } }{ \sqrt { 2x1 } } \Rightarrow \) \(v=\frac { 3 }{ \sqrt { 2 } } \times10^{ 8 }\) m/s.

The amplitude modulated radiowaves having frequency 530 kHz to 1710 kHz (or wavelength between 175 m to 566 m) which are travelling directly following the surface of earth are known as ground waves. The short wave communication over long distance is only possible via sky waves. It is not possible via ground waves because the ground waves can bend round the corners of the objects on earth and hence, their intensity falls with distance. Moreover the ground wave transmission becomes weaker as frequency increases.

\(Here,\beta =2mm=2\times { 10 }^{ -3 }m;D=1.2m \ and \ d=0.24\times { 10 }^{ -3 }m; \ { \phi }_{ 0 }=2.2eV,{ V }_{ 0 }=?\)
\(As \ \beta =\frac { D }{ d } \lambda \ or \ \lambda =\beta \frac { d }{ D } =\frac { \left( 2\times { 10 }^{ -3 } \right) \times \left( 0.24\times { 10 }^{ -3 } \right) }{ 1.2 } =4\times { 10 }^{ -7 }m\)
\({ eV }_{ 0 }=\frac { hc }{ \lambda } -{ \phi }_{ 0 } \ or \ { V }_{ 0 }=\frac { hc }{ e\lambda } -\frac { { \phi }_{ 0 } }{ e } =\frac { \left( 6.63\times { 10 }^{ -34 } \right) \times \left( 3\times { 10 }^{ 8 } \right) }{ 1.6\times { 10 }^{ -19 }\times \left( 4\times { 10 }^{ -7 } \right) } -\frac { 2.2eV }{ e } =3.1-2.2=0.9V\)

\({ N }_{ 1 }={ N }_{ 0 }{ e }^{ -{ \lambda }_{ 1 }t },\quad { N }_{ 2 }={ N }_{ 0 }{ e }^{ -{ \lambda }_{ 2 }t }\)
\(\\ \frac { { N }_{ 1 } }{ { N }_{ 2 } } ={ e }^{ \left( { \lambda }_{ 2 }-{ \lambda }_{ 1 } \right) t }\quad or\quad t=\frac { \log _{ e }{ \left( { N }_{ 1 }{ /N }_{ 2 } \right) } }{ { \lambda }_{ 2 }-{ \lambda }_{ 1 } } =\frac { \log _{ e }{ 140/1 } }{ \log _{ 2 }{ 2\left[ \frac { 1 }{ { T }_{ 2 } } -\frac { 1 }{ { T }_{ 1 } } \right] } } =\frac { \log _{ e }{ 140 } }{ \log _{ 10 }{ 2 } } \left( \frac { { T }_{ 2 }{ T }_{ 1 } }{ { T }_{ 1 }-{ T }_{ 2 } } \right) \)
On putting the values, we get \(t=6\times { 10 }^{ 9 }\ years\)

(i) Electric field at x = 0
\(E={1\over 4\pi \epsilon_o}[{q\over 1^2}+{q\over 2^2}+{q\over 4^2}+....]\)
\(={q\over 4\pi\epsilon_o}[{1\over 1-1/4}]={q\over 3\pi\epsilon_o}\)
(ii) If consecutive charges have opposite signs, then at x = 0, 
Electric field, \(E'={q\over 4\pi\epsilon_o}[{1\over 1^2}-{1\over 4}+{1\over 16}-{1\over 64}+....]^\times\)
\(E'={q\over 4\pi\epsilon_o}[{1\over 1-(-1/4)}]={q\over 5\pi\epsilon_o}\) 

In LR circuit, the impedance is given by
\({ Z }_{ L }=\sqrt { { R }^{ 2 }+{ X }_{ L }^{ 2 } } \quad ...(i)\)
In LCR circuit, the impedance is given by
\(Z=\sqrt { { R }^{ 2 }+{ \left( { X }_{ L }-{ X }_{ C } \right) }^{ 2 } } \quad ...(ii)\)
From equations, (i) and (ii), we find that
\(Z<{ Z }_{ L }\)
\(since \ I=\frac { E }{ Z } \)
As Z decreases on connecting a  capacitor in series with LR circuit, hence the current I in the circuit increases.

If \({ i }_{ a }\)and \({ i }_{ b }\) are the 3 antiparallel currents flowing in two parallel wires separated by a distance R, then the repulsive force experienced by unit length (1m) of either wire is
\(F=\frac { { \mu }_{ 0 }{ i }_{ a }{ i }_{ b } }{ 2\pi R } \)
Weight per meter length of above wire
\(=0.073{ Nm }^{ -1 }\)
According to problem, the weight of lower wire is supported by upward magnetic repulsion, therefore
\(\frac { { \mu }_{ 0 }{ i }_{ a }{ i }_{ b } }{ 2\pi R } \)\(=0.073\)
\(R=\frac { { \mu }_{ 0 }{ i }_{ a }{ i }_{ b } }{ 2\pi \times 0.073 } \) 
\(=\frac { 4\pi \times { 10 }^{ -7 }\times 100\times 20 }{ 2\pi \times 0.073 } \)
\(=4.4\times { 10 }^{ -3 }m = 4.48mm\)