0.38 gauss

As, the value of R is increased, the current flowing in the circuit will decrease. And the potential gradient, i.e. potential drop per unit length also decreases, so that the balance length will increase. Thus, J will shift towards B.

After inserting the dielectric medium, let their capacitances become C₁' and C₂'.
For C₁ , C₁'  = KC                              ...(i)                      
For C₂ , C₂' = \(\frac { { K }_{ 1 }{ \varepsilon }_{ 0 }\left( { A }/{ 2 } \right) }{ d } +\frac { { K }_{ 2 }{ \varepsilon }_{ 0 }\left( { A }/{ 2 } \right) }{ d } \)
C₂ acts as if two capcitors each of area A/2 and separation d are connected in parallel combination
\({ C' }_{ 2 }=\frac { { \varepsilon }_{ 0 }{ A } }{ d } \left( \frac { { K }_{ 1 } }{ 2 } +\frac { { K }_{ 2 } }{ 2 } \right) \)
\(\\ { C' }_{ 2 }=C\left( \frac { { K }_{ 1 }+{ K }_{ 2 } }{ 2 } \right) \left[ \because C=\frac { { \varepsilon }_{ 0 }{ A } }{ d } \right] \)       ...(ii)     
According to the problem, C'₁ = C'2
\(\Rightarrow KC=C\left( \frac { { K }_{ 1 }+{ K }_{ 2 } }{ 2 } \right) \Rightarrow K=\frac { { K }_{ 1 }+{ K }_{ 2 } }{ 2 } \)

The insulating porcelain of the spark plugs accumulates a film of dirt.
The surface dirt is hygroscopic and picks up moisture from the air. Therefore, in humid weather, the insulating porcelain of the plugs becomes quasi-conductor.
This allows an appreciable proportion of the spark to leak across the surface of the plug instead of discharging across the gap.

On introduction of dielectric slab to fill the gap between plates of capacitor completely when capacitor is connected with battery.
(i) The potential difference V between capacitors is same due to connectivity with battery and hence, charge q' becomes K times of original charge as
q' = C'V' = (KC) (V) = K(CV) = Kq
q' = Kq
(ii) Electric field intensity continue to be the same as potential difference and separation between two plates remain unaffected as
\(E=\frac { V }{ d } \)
(iii) The capacitance of capacitor becomes K times of original capacitor.
\(\therefore \ C'=KC=\frac { K{ \varepsilon }_{ 0 }A }{ d } \) 

Since, resonant angular frequency \(\omega t\) is same for both the graphs.
i.e.,   \(({\omega}_{0})_{1}=({\omega}_{0})_{2}\)     \(\left( \because {\omega}_{0}={ {I }\over{\sqrt{LC} } } \right)\)
\(\Rightarrow\) \({{1}\over{\sqrt{{L}_{1}{C}_{1}}}}={{1}\over{\sqrt{{L}_{2}{C}_{2}}}}\Rightarrow{L}_{1}{C}_{1}={L}_{1}{C}_{2}\) or \({ { {L}_{1} }\over{{L}_{2} } }={ { {C}_{2} }\over{{C}_{1} } }\)
In graph(I), current Im1 is greater than graph (II) \({I}_{{m}_{1}}\)is greater than graph (II), \({I}_{{m}_{2}}.\)
i.e.  \({I}_{{m}_{1}}>{I}_{{m}_{2}}\Rightarrow{R}_{1}<{R}_{2}\)
Power factor of circuit represented by graph (I) has power factor as cos \(\phi =R.\) 
\(\because\)    \(Q={{I}\over{R}}\sqrt{{{L}\over{C}}}\)  [ For two graphs ]
i.e. \(Q\infty{{I}\over{R}}\)
As, R₁2     \(\therefore\)  Q₁>Q₂
Quality factor of graph (I) is higher than that of graph (II).

(i) Now, the electric field, E = Cx\(\widehat { i }\) is in X-direction only. So, face with surface normal vector perpendicular to this field would give zero electric flux, i.e. \(\phi\) = E dS \(cos90^{o}\) = 0, through it.
                   
So, flux would be across only two surfaces.
Magnitude of E at left face,
E_L = Cx = Ca  [x = a at left face]
Magnitude of E at right face,
E_R = Cx = C2a = 2aC   [x = 2a at right face]
Thus, corresponding fluxes are
\({ \phi }_{ L }\) = E_L . dS = E _L dS cos\(\theta\)
= - aC \(\times\) a² = - a³ C  [\(\because\) \(\theta=180^{o}\)]
\({ \phi }_{ R }\) = E_R . dS
= 2aC dS cos\(\theta\)  [ \(\because \theta=0^{o}\)]
= 2aCa² = 2a³C             
Now, net flux through cube
= \({ \phi }_{ L }+{ \phi }_{ R }\)
= - a³ C + 2a³C
= a³ C N-m² C^-1     
(ii) Net charge inside the cube, again, we can use Gauss' law to find total charge q inside the cube.
We have' \(\phi=\frac { q }{ { \varepsilon }_{ 0 } } \)
or q = \(\phi{ \varepsilon }_{ 0 }\)
q = a³ C \({ \varepsilon }_{ 0 }\)

According to Ampere's circuital law, the net field is given by \(B={ \mu }_{ 0 }nI\)
(a) The net magnetic field is given by
\({ B }_{ net }={ B }_{ 2 }-{ B }_{ 1 }-{ \mu }_{ 0{ n }_{ 2 } }I-{ \mu }_{ 0 }{ n }_{ 1 }\quad [\because { I }_{ 2 }={ I }_{ 1 }=I]\)
 \(={ \mu }_{ 0 }I({ n }_{ 2 }-{ n }_{ 1 })\)
The direction is from B to A.
(b) As the magnetic fields due to S₁ is confined solely inside S₁ as the solenoids are assumed to be very long. So, there is no magnetic field outside S₁  due to current in S_1, similarly, there is no field outside S₂ .
B_net = 0

(i) By Ohm's law, current I is given by 
\(I=6V/6\Omega =1A\) 
But, \(I=neA{ v }_{ d }\) 
or \({ v }_{ d }=\frac { I }{ neA } =\frac { 1 }{ { 10 }^{ 29 }\times \left( 1.6\times { 10 }^{ -19 } \right) \times \left( { 10 }^{ -6 } \right) } \)  
\(\Rightarrow \)   \({ v }_{ d }=\frac { 1 }{ 1.6 } \times { 10 }^{ -4 }\quad m/s\)  
Number of electrons in the wire = \(nAl\)
KE of all electrons \(=\left( nAl \right) \times \frac { 1 }{ 2 } { m }_{ e }{ v }^{ 2 }d\) 
\(=\left( { 10 }^{ 29 }\times { 10 }^{ -6 }\times { 10 }^{ -1 } \right) \times \frac { 1 }{ 2 } \times \left( 9.1\times { 10 }^{ -31 } \right) \times { \left( \frac { 1 }{ 1.6 } \times { 10 }^{ -4 } \right) }^{ 2 }\)                                                                    
\(=1.78\times { 10 }^{ -17 }J\) 
(ii) Power loss in wire = I²R =\({ \left( 1 \right) }^{ 2 }\times 6=6J{ s }^{ -1 }\)  
\(=\frac { 1.78\times { 10 }^{ -17 } }{ 6J{ s }^{ -1 } } =0.30\times { 10 }^{ -17 }s\)  
\(={ 10 }^{ -17 }s\) 

(ii) The frequencies of charge particle must be equal to the frequency of AC oscillator. This is known as resonance condition. 
This makes time period of charged particle and oscillator equal. Therefore, the time taken by charge particle to complete half revolution is equal to the time of change the polarity of dees. This facilitate the acceleration of charge particle.
If two frequencies do not match, then instead of acceleration, charged particle may accelerate.

The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.
Mathematically, the induced emf is given by
\(\varepsilon=\frac{-d\phi}{dt}\)
First consider the forward motion from x = 0 to x =2b
The flux \({ \phi }_{ B }\) linked with the section SPQR is
\({ \phi} _{ B }=Blx,o\le x\le b\)
= Blb, b \(\le\) x \(\le\) 2b
The Induced emf is,
\(\varepsilon =\frac { d{ \phi }_{ B } }{ dt } \)
= Blv 0  \(\le\) x \(\le\) b
= 0 b \(\le\) x \(\le\) 2b