(i) Shyam is responsible, caring and intelligent.
(ii) It is the phenomena of redistribution of light energy in a medium due to the superposition of two coherent light waves.
(iii) For constructive interference or bright fringe, the phase difference between the two coherent light waves should be \(2n\pi \) 
i.e  \(\phi =2n\pi (n=0,1,2,3)\)

(a) Speed of an electron, v = 5.20 x 10⁶ m/s
Magnetic field experienced by the electron, B = 1.30 x 10⁻⁴ T
Specific charge of an electron, ec
e = Charge on the electron = 1.6 x 10⁻¹⁹ C
m = Mass of the electron = 9.1 x 10⁻³¹ kg⁻¹
The force exerted on the electron is given as:
\(F=e|\vec{v} \times \vec{B}|\)
\(=e v B \sin \theta\)
θ = Angle between the magnetic field and the beam velocity
The magnetic field is normal to the direction of beam.
∴ θ = 90^∘
F = eVB   ...(1)
The beam traces a circular path of radius, r. It is the magnetic field, due to its bending nature, that provides the centripetal force \(\left(F=\frac{m v^{2}}{r}\right)\)for the beam.
Hence, equation (1) reduces to:
\(e v B=\frac{m v^{2}}{r}\)
\(\therefore r=m \frac{v}{e B}=\frac{v}{\left(\frac{e}{m}\right) B}\)
\(=\frac{5.20 \times 10^{6}}{1.76 \times 10^{11}} \times 1.30 \times 10^{-4}=0.227 m=22.7 \mathrm{~cm}\)
Therefore, the radius of the circular path is 22.7 cm.
Energy of the electron beam, E = 20 MeV = 20 x 10⁶ x 1.6 x 10^-19J
The energy of the electron is given as:
\(E=\frac{1}{2} m v^{2}\)
\(\therefore v=\left(2 \frac{E}{m}\right)^{\frac{1}{2}}\)
\(=\sqrt{\frac{2 \times 20 \times 10^{6} \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}}=2.652 \times 10^{9} \mathrm{~m} / \mathrm{s}\)
This result is incorrect because nothing can move faster than light. In the above formula, the expression (mv²/2) for energy can only be used in the non-relativistic limit, i.e., for v << c
When very high speeds are concerned, the relativistic domain comes into consideration.
In the relativistic domain, mass is given as:
\(m=m_{0}\left[1-\frac{v^{2}}{c^{2}}\right]^{\frac{1}{2}}\)
Where,
m₀ = Mass of the particle at rest
Hence, the radius of the circular path is given as:
\(r=m \frac{v}{e} B\)
\(=\frac{m_{0} v}{e B \sqrt{\frac{c^{2}-v^{2}}{c^{2}}}}\)

Here, slit width,d = 0.28 mm = \(0.28\times { 10 }^{ -3 }m\)
Distance between slit and screen, D = 1.4m
 \(y=1.2cm=1.2\times { 10 }^{ -2 }m,n=4,\lambda =?\)
For constructive interference,
 \(y=\eta \lambda \frac { D }{ d } or\lambda =\frac { yd }{ nD }\)
\( \\ =\frac { 1.2\times { 10 }^{ -2 }\times 0.28\times { 10 }^{ -3 } }{ 4\times 1.4 } =6\times { 10 }^{ -7 }m\)
= 600 nm
Hence, the wavelength of the light is 600 nm.

(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.

(i) Priyanshu's father is responsible, makes his child understood the concepts of solenoid and toroid.
(ii) A solenoid has magnetic field straight within the turns and in toroid, it is in form of concentric circles.
(iii) The magnitude of magnetic field in solenoid is given by \(B={ { \mu } }_{ \circ }nI\)
(where, n = number of turns/length, I = current in coil)

(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.

\(We \ know \ that,{ E }_{ axial }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { 2\rho r }{ { ({ r }^{ 2 }-{ r }^{ 2 }) }^{ 2 } } \)
\(CaseI,\)
\( When \ r=10cm=0.1m\)
\(\\ { E }_{ axial }=3.75\times { 10 }^{ 5 }N/C\)
\(\\ 3.75\times { 10 }^{ 5 }=9\times { 10 }^{ 5 }\times \frac { 2\rho \times 0.1 }{ { { [(0.1) }^{ 2 }-{ a }^{ 2 }] }^{ 2 } } \ .....(i)\)
\(Case \ II,\)
\( When \ r=20cm=0.2m\)
\(\\ { E }_{ axial }=3\times { 10 }^{ 4 }\times \frac { 2\rho \times 0.2 }{ { { [(0.2) }^{ 2 }-{ a }^{ 2 }] }^{ 2 } } \ .......(ii)\)
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

There are two protons p₁ and p₂ with an electron e.
Distance between two protons id given by
r₁ = 1.5\(\overset { o }{ A } \) = 1.5 x 10^-10 m

Distance between proton p1 and electron e is given by
r2 = 1 \(\overset { o }{ A } \)
 = 1 x 10-10 m
Distance between proton p2 and electron e is given by
r3 = 1\(\overset { o }{ A } \)
=1 x10-10 m
The total potential energy of the system,
\(U=\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } .\left[ \frac { { q }_{ p1 }{ q }_{ p2 } }{ { r }_{ 1 } } +\frac { { q }_{ p1 }{ q }_{ e } }{ { r }_{ 2 } } +\frac { { q }_{ p2 }{ q }_{ e } }{ { r }_{ 13 } } \right] \)......(i)
Given qp1 = qp2
= 1.6 x10-19 C
and  qe= -1.6 x10-19 C
Putting these values in Eq.(i), we get
\(U=9 \times 10^{9}\left[\frac{1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{1.5 \times 10^{-10} \cdots}\right.\) \(+\frac{\left(1.6 \times 10^{-19}\right) \times\left(-1.6 \times 10^{-19}\right)}{10^{-10}}\) \(\left.+\frac{1.6 \times 10^{-19} \times\left(-1.6 \times 10^{-19}\right)}{10^{-10}}\right]\) 
\(=\frac{9 \times 10^{9} \times 1.6 \times 1.6 \times 10^{-38}}{10^{-10}}\left[\frac{1}{1.5}-1-1\right]\)
= -30.72 x 10-19 J
\(=\frac{-30.72 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}=-19.2 \mathrm{eV}\)
Here, we use that potential energy at infinity is zero.