(i) Power has to be transported at high voltage to reduce the energy loses and also the cost of transmission.
(ii) As P = E_vI_v cos\(\phi \) . To transmit a given power P at a given E_v, if cos \(\phi \) is small, I_v has to be increases. Therefore. Power loss = \({ I }_{ v }^{ 2 }R\) will increase.
(iii) Ajit understood the necessity of high tension tower which did not affect his land adversely. His uncle made practical use of his knowledge by inviting Ajit that high tension tower are essential and they cause no harm to the farmland.

Here, v=2.18 x 10⁶ ms^-1,
r=5.3 x 10^-11 m, e=1.6 x 10^-19 C.
(a) Time period of revolution of electron is given by, 
\(T=\frac { 2\pi r }{ v } =\frac { 2\pi \times 5.3\times { 10 }^{ -11 } }{ 2.18\times { 10 }^{ 6 } } =1.528\times { 10 }^{ -16 }s\) 
Equivalent current, \(I=\frac { charge }{ time } =\frac { e }{ T } \) 
\(=\frac { 1.6\times { 10 }^{ -19 } }{ 1.528\times { 10 }^{ -16 } } =1.05\times { 10 }^{ -3 }A\) 
\((b)B=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2\pi I }{ r } =\frac { { 10 }^{ -7 }\times 2\pi \times 1.05\times { 10 }^{ -3 } }{ 5.3\times { 10 }^{ -11 } } \) 
=12.4 T

\(Here, \ { I }_{ v }=5.0A, \ v=50Hz\)
\((i) \ { I }_{ 0 }=\sqrt { 2 } { I }_{ v }=1.414\times 5.0=7.07A\)
\((ii) \ { I }_{ m }=\frac { 2 }{ \pi } { I }_{ 0 }=\frac { 2 }{ 3.14 } \times 7.07=4.5A\)
\((iii) \ From \ I={ I }_{ 0 } \ sin \ \omega t={ I }_{ 0 }sin2\pi \ vt\)
\(=7.07sin2\pi \times 50\times \frac { 1 }{ 300 } =\frac { 7.07\sqrt { 3 } }{ 2 } =6.12A\)

Here, v = 6 x 10¹⁴ Hz, E₀ = 2 V m^-1
(i) Average energy density of the electric field
\({ u }_{ E }=\frac { 1 }{ 4 } { \epsilon }_{ 0 }{ E }_{ 0 }^{ 2 }=\frac { 1 }{ 4 } \times \left( 8.85\times { 10 }^{ -12 } \right) \times { 2 }^{ 2 }\)
\(=8.85\times { 10 }^{ -12 }J{ m }^{ -3 }\)
(ii) Average energy density of magnetic field
\({ u }_{ B }=\frac { { B }_{ 0 }^{ 2 } }{ 4{ \mu }_{ 0 } } =\frac { 1 }{ 4 } \frac { { \left( { E }_{ 0 }/c \right) }^{ 2 } }{ { \mu }_{ 0 } } =\frac { 1 }{ 4 } \frac { { E }_{ 0 }^{ 2 } }{ 4{ \mu }_{ 0 }{ c }^{ 2 } } \)
\(=\frac { 1 }{ 4 } \times \frac { { 2 }^{ 2 } }{ \left( 4\pi \times { 10 }^{ -7 } \right) \times { \left( 3\times { 10 }^{ 8 } \right) ^{ 2 } } } \)
\(=8.85\times { 10 }^{ -12 }J{ m }^{ -3 }\)

Here, \(C=0.2\mu F=0.2\times { 10 }^{ -6 }F\)
\(=2\times { 10 }^{ -7 }F,\)
\(\omega =400 \ rad/s, \ { I }_{ rms }=2mA=2\times { 10 }^{ -3 }A\)
\({ V }_{ rms }={ I }_{ rms }\times { X }_{ C }\)
\(={ I }_{ rms }\times \frac { 1 }{ { \omega C } } =\left( 2\times { 10 }^{ -3 } \right) \times \frac { 1 }{ 400\times \left( 2\times { 10 }^{ -7 } \right) } \)
= 25 V.
Displacement current = conduction current
= 2 mA

Since drift velocity \({ v }_{ d }\) and current, I flowing in a conductor are related by the relation:
\({ v }_{ d }=\frac { I }{ neA } ....(1)\)
Also drift velocity in terms of average relaxation time τ is given by\(\)
\({ v }_{ d }=\frac { eE\tau }{ m } .......(2)\)
From (1) and (2), we have
\(\frac { eE\tau }{ m } =\frac { I }{ neA } \)
\(or \ \frac { E }{ I } =\frac { m }{ n{ e }^{ 2 }A\tau }\)
\(or \ \frac { V }{ lI } =\frac { m }{ n{ e }^{ 2 }A\tau }\)
\(or \ \frac { V }{ I } =\frac { ml }{ n{ e }^{ 2 }A\tau }........(3)\)
The R.H.S. of Eq(3) is constant
\(\frac { V }{ I } = \ Constant\)
This is Ohm′s law
\(But \ \frac { V }{ I } =R, \)the resistance of the conductorSo Eq(3) becomes
\(R=\frac { ml }{ n{ e }^{ 2 }A\tau }\)
\(But\quad R=\rho \frac { l }{ A }\)
\(\therefore \ \rho \frac { l }{ A } =\frac { ml }{ n{ e }^{ 2 }A\tau }\)
\( \rho =\frac { m }{ n{ e }^{ 2 }A\tau } \)

(a) E = 1.25V

According to Coulomb's law, the magnitude of force acting between two stationary point charges is given by
\(F=\left( \frac { q_{ 1 }q_{ 2 } }{ 4\pi \varepsilon _{ 0 } } \right) \left( \frac { 1 }{ r^{ 2 } } \right) \)
For give q₁ q₂ , \(F\propto \quad \frac { 1 }{ r^{ 2 } } \)
Slope of \(F-\frac { 1 }{ r^{ 2 } } \), graph depends on q_1, q_{2 is} higher for second pair 

\( \therefore \)  slope of \(F-\frac { 1 }{ r^{ 2 } } \) graph corresponding to second pair
(1\(\mu \)C- 3\(\mu \)C) is greater.higher the magnitude of product of charges q₁ and q₂, higher will be the slope.

\(Here, \ E=648 \ J,I=9,e=?,\)
\(dI=9-0=9A,dt=0.45s\)
\(From \ E=\frac { 1 }{ 2 } { LI }^{ 2 }\)
\(648=\frac { 1 }{ 2 } { L(9) }^{ 2 }; \ L=\frac { 648\times2 }{ 9\times9 } =16H\)
\(As \ e=\frac { LdI }{ dt } \therefore e=\frac { 16(9) }{ 0.45 } =320V\)

Yes, when cell itself is being charged, because terminal potential difference,
\(V=ϵ−(−Ir)=ϵ+Ir.\)

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

\(\frac { d }{ 1+\sqrt { 2 } } ,\frac { \sqrt { 2d } }{ 1+\sqrt { 2 } } \)

When magnet is placed with its north pole pointing south, neutral point is obtained on its axial line. Therefore, at neutral point
\({ B }_{ axia }={ B }_{ 0 }\) 
\(or\quad \frac { { \mu }_{ 0 } }{ 4\pi } \times \frac { 2Md }{ { \left( { d }^{ 2 }-{ l }^{ 2 } \right) }^{ 2 } } ={ B }_{ 0 }\)
\(\\ or \ M=\frac { 4\pi }{ { \mu }_{ 0 } } \times \frac { { B }_{ 0 }{ \left( { d }^{ 2 }-{ l }^{ 2 } \right) }^{ 2 } }{ 2d }\)
\( =\ \frac { 1 }{ { 10 }^{ -7 } } \times \frac { 0.34\times { 10 }^{ -4 }\times { \left( { 0.30 }^{ 2 }-{ 0.15 }^{ 2 } \right) }^{ 2 } }{ 2\times 0.30 }\)
\( =2.582{ Am }^{ 2 }\)
If m is pole strength of the magnet, then
 \(m=\frac { M }{ 2l } =\frac { 2.582 }{ 0.30 } =8.606Am\)