The first refracting ABC forms the image I₁ of the object O. The image I₁ acts as virtual object for the second refracting surface ADC, which forms the real image I as shown in the diagram
For refraction at ABC 
\(\frac { { n }_{ 2 } }{ { v }_{ 1 } } -\frac { { n }_{ 1 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 1 } } \)
For refraction at ADC
\(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ { v }_{ 1 } } =\frac { { n }_{ 1 }-{ n }_{ 2 } }{ { R }_{ 2 } } \)
Adding equation (i) and equation (ii)
 \(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
We know, If  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\) 

(ii) Let initially, I_r, be current flowing in all the three circuits. If frequency of applied AC source is increased, then the change in current will occur in following manner.
(a) AC circuit contanining resistance only where,
V_i = initial frequency of AC source.

There is no effect on current with the increase i frequency.
(b) AC circuit containing inductance only with the increase of frequency of AC soure, inductive reactance increase as, 
\(I=\frac { { V }_{ rms } }{ { X }_{ L } } =\frac { { V }_{ rms } }{ 2\pi vL } \)
\(\Rightarrow \ { X }_{ L }=2\pi vL\)
\(for \ given \ circuit, \ I\propto \frac { 1 }{ V } \)
Current decrease with the increase in frequency.

(c) AC circuit containing capacitor only 
\({ X }_{ C }=\frac { 1 }{ \omega C } =\frac { 1 }{ 2\pi vC }\)
\(Current, \ I=\frac { { V }_{ rms } }{ { X }_{ C } } =\frac { { V }_{ rms } }{ (\frac { 1 }{ 2\pi vC } ) }\)
\( I=2\pi vC{ V }_{ rms }\)
For given circuit, \(I\propto v\)
Current increase with the increase in frequency.

(i) (a) When a capacitor is placed in an external electric field, the free charges present inside the conductor redistribute themselves in such a manner that electric field within the conductor. This happens until a static situation is achieved,i.e. when the two fields cancel each other and the net electrostatic field in the conductor becomes zero.

(b) In contrast to conductors, dielectrics are non-conducting substance, i.e. they have no charge carriers.Thus, in a dielectric, free movement of charges in not possible.It turns out that the external field induces dipole moment by stretching molecules of the dielectric.
The collective effect of all the molecular dipole moments is the net charge on the surface of the dielectric which produces a field that opposes the external field. However, the opposing field is so induced, that does not exactly cancel the extent of the effect depends on the nature of dielectric.

Both polar and non-polar dielectrics develop net dipole moment in the presence of an external field. The dipole moment per unit volume is called polarisation and is denoted by P for linear isotropic dielectrics.
P = X^E
Where, X is constant of proportionality and is called electric susceptibility of the electric slab.
(ii) (a) At point C, inside the shell. Electric field inside a spherical shell is zero.
Thus, the force experienced by charge at centre C will also be zero.
\(\because \)  F_c = qE (E_inside  the shell = 0)
\(\therefore \) F_c = 0
At point A, | F_A | = 2Q \(\left[ \frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } .\frac { 3Q/2 }{ { x }^{ 2 } } \right] \\.\)
\( F \ = \ \frac { { 3Q }^{ 2 } }{ { { 4\pi \varepsilon }_{ 0 } }{ x }^{ 2 } } ,\) away from shell
Electric flux through the shell,
\(\Phi =\frac { 1 }{ { \varepsilon }_{ 0 } } \) x magnitude of charge enclosed by shell
\(=\frac { 1 }{ { \varepsilon }_{ 0 } } \times \frac { Q }{ 2 } \Rightarrow \Phi =\frac { Q }{ { 2\varepsilon }_{ 0 } } \)

Consider that a wire of length l and area of cross-section A be subjected to an electric field of strength E.
If V = Potential difference applied across the end of the wire,
\(E=\frac { V }{ l } or \ V=El\)

Let n = number of free electrons per unit volume of the conductor
\({ v }_{ d }\) = drift velocity of electrons charge flowing through the conductor wire,
q = nAle
Time taken by the electrons to cross the conductor,
\(t=\frac { Distance\quad }{ Velocity } =\frac { 1 }{ { v }_{ d } } \) 
\( Current \ I=\frac { charge }{ time } =\frac { nAle }{ \frac { l }{ { v }_{ d } } } \)

Capacitance of capacitor C₁ is 100 pF.
Capacitance of capacitor C₂ is 200 pF.
Capacitance of capacitor C₃ is 200 pF.
Capacitance of capacitor C₄ is 100 pF.
Supply potential, V = 300 V
Capacitors C₂ and C₃ are connected in series. Let their equivalent capacitance be C' .
\(\therefore \frac{1}{C^{\prime}}=\frac{1}{200}+\frac{1}{200}=\frac{2}{200}\)
C' = 100 pF
Capacitors C₁ and C' are in parallel. Let their equivalent capacitance be 
\(\therefore\) C'' =C' + C₁
= 1000 + 100 =200 pF
C" and C₄ are connected in series. Let their equivalent capacitance be C.
\(\therefore \frac{1}{C}=\frac{1}{C^{\prime \prime}}+\frac{1}{C_{4}}\)
\(=\frac{1}{200}+\frac{1}{100}=\frac{2+1}{200}\)
\(C=\frac{200}{3} \mathrm{pF}\)
Hence, the equivalent capacitance of the circuit is \(\frac{200}{3} \mathrm{pF}\)
Potential difference across C" = V" 
Potential difference across C₄ = V₄
\(\therefore\) Vⁿ + V ₄= V =300V
Charge on C₄ is given by 
Q₄= CV
\(=\frac{200}{3} \times 10^{-12} \times 300\)
= 2 x 10^-8 C
\(\therefore V_{4}=\frac{Q_{4}}{C_{4}}\)
\(=\frac{2 \times 10^{-8}}{100 \times 10^{-12}}=200 \mathrm{~V}\)
\(\therefore\)  Voltage across C₁ is given below
V₁ =V -V₄
= 300- 200 =100 V
Hence, potential difference, V₁, across C₁ is 100 V.
Charge on C1 is given by,
Q₁ = C₁V₁
= 100 x 10^-12 x 100
 = 10^-8 C
C₂ and C₃ having same capacitances have a potential difference of 100 V together. Since C₂ and C₃ are in series,
the potential difference across C₂ and C₃ is given by,
V₂= V₃ = 50 V
Therefore, charge on C₂ is given by,
Q₂ = C₂V₂ 
= 200 x 10^-12 x 50
= 10^-8 C
And charge on C₃ is given by,
Q₃ = C₃V₃
= 200 x 10^-12 x 50 = 10^-8 C
Hence, the equivalent capacitance of the given circuit is \(\frac{200}{3} \mathrm{pF}\)

\(E=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }^{ 2 } } for \ a \ point \ charge\)
Now (volume) charge density
\(p=\frac { Q }{ (\frac { 4\pi }{ 3 } { R }^{ 3 }) } \)
(i) Charge contained within a sphere of radius r (0 < r < R)
\(Q=p\frac { 4\pi }{ 3 } { r }^{ 3 }=Q(\frac { { r }^{ 3 } }{ { R }^{ 3 } } )\)
Electric Field
\(E=\frac { 1 }{ 4\pi { \varepsilon }_{ o } } \frac { { Q }^{ 1 } }{ { r }^{ 2 } } =(\frac { 1 }{ { 4\pi \varepsilon }_{ o } } \frac { Q }{ { R }^{ 3 } } )\)
For r > R electric field = (Electric field due to a point charge Q at the centre
\(=\frac { 1 }{ { 4\pi \varepsilon }_{ o } } \frac { Q }{ { r }^{ 2 } } \)

(i) When θ = 0; ፔ = O and p and E are parallel and the dipole is in a position of stable equilibrium.
(ii) When θ = 180°, ፔ = 0 and p and E are anti-parallel and the dipole is in a position of unstable equilibrium.

Consider the given figure,

Given, C₁ = 20 \(\mu\)F, C₂ = 30 \(\mu\)F, C₃ = 15 \(\mu\)F
Potential at A, V = 90 V
As we can see that capacitor C₃ is earthed, therefore potential across C₃ will be zero.
since, capacitors C₁,C₂ and C₃ are connected in series therefore
\(\begin{aligned} \frac{1}{C_{\text {eq }}} & =\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \\ \end{aligned}\)
\(\begin{aligned} =\frac{1}{20}+\frac{1}{30}+\frac{1}{15} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad & \frac{1}{C_{\text {eq }}}=\frac{3+2+4}{60}=\frac{9}{60} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad C_{\text {eq }} & =\frac{60}{9}=\frac{20}{3} \mu \mathrm{F} \end{aligned}\)
Since, charge remains same in series combination.
So, \(\begin{aligned} q & =C_{\mathrm{eq}} V=\frac{20}{3} \times 90 \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad q & =600 \mu \mathrm{C} \\ \end{aligned}\)
\(\begin{aligned} & =600 \times 10^{-6} \mathrm{C} \\ \end{aligned}\)
\(\begin{aligned} & =6 \times 10^{-4} \mathrm{C} \end{aligned}\)
\(\therefore\) Potential difference across C₂ = \(\frac{q}{V_2}\)
\(\begin{array}{ll} \Rightarrow & V_2=\frac{q}{C_2} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & V_2=\frac{6 \times 10^{-4}}{30 \times 10^{-6}}=20 \mathrm{~V} \end{array}\)
\(\therefore\) Energy stored in capacitor C₂ is given by
\(\begin{aligned} U & =\frac{1}{2} C_2 V_2^2 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{2} \times 30 \times 10^{-6} \times(20)^2 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{2} \times 30 \times 400 \times 10^{-6} \\ \end{aligned}\)
\(\begin{aligned} & =6 \times 10^{-3} \mathrm{~J} \end{aligned}\)

\(Here,length,2a=2cm=2\times { 10 }^{ -2 }m.\)
  \(\theta ={ 60 }^{ 0 },\tau =8\sqrt { 3 } N-m\)
Charges, Q = 4\(\times\) 10^-9 C, U = ?
As we know that, \(\tau =Q(2a) E\sin { \theta } \)
\(\Longrightarrow \)Electric field,
\(e=\frac { \tau }{ Q(2a)\sin { \theta } } =\frac { 8\sqrt { 3 } }{ 4\times { 10 }^{ -9 }\times 2\times { 10 }^{ -2 }\times \sin { { 60 }^{ 0 } } } N/C\)
\(\\ \therefore Potential\quad energy, U=-\rho E\cos { \theta } =-Q(2a)E\cos { \theta }\)
\( \\ =-4\times { 10 }^{ -9 }+2\times { 10 }^{ -2 }\times \frac { 8\sqrt { 3 } \times \cos { { 60 }^{ 0 } } }{ 4\times { 10 }^{ -9 }\times 2\times { 10 }^{ -2 }\times \sin { { 60 }^{ 0 } } }\)
\( \\ \Longrightarrow U=-\frac { 8\sqrt { 3 } }{ \sqrt { 3 } } =-8J\)​​​​​​

Capacitance, C = 30μF = 30 × 10⁻⁶F
Inductance, L = 27 mH = 27 × 10⁻³ H
Angular frequency is given as:

\({ \omega }_{ r }=\frac { 1 }{ \sqrt { LC } } =\frac { 1 }{ \sqrt { 27\times { 10 }^{ -3 }\times 30\times { 10 }^{ -6 } } } =\frac { { 10 }^{ 4 } }{ 9 } =1.1\times { 10 }^{ 3 }rad/s\)
Hence, the angular frequency of free oscillations of the circuit is 1.11 × 10³ rad/s.

\(Here, \ R=100\Omega , \ { E }_{ v }=220V, \ v=50Hz\)
\((a) \ { I }_{ v }=?, \ { I }_{ v }=\frac { { E }_{ v } }{ R } =\frac { 220 }{ 100 } =2.2A\)
\((b) \ Net \ power \ consumed \ over \ a \ full \ cycle,\)
\(P={ E }_{ v }{ I }_{ v }=220\times 2.2=484W\)

The law of conservation of charge states that the total charge in an isolated system remains constant.
The electric charge can neither be produced nor destroyed. This law has been found to be true for all events as well as for those at nuclear and atomic levels. In other words, there is no exception to the law. Like the law of conservation of energy, the law of conservation of charge is also a universal law.
Examples (i) When a glass rod is rubbed with silk the charges developed on the glass rod and the piece of silk are equal and opposite.
(ii) Charge is conserved in all chemical and nuclear reactions.

Since \({ F }_{ e }=-\frac { { ke }^{ 2 } }{ { r }^{ 2 } } \)
So     \({ ke }^{ 2 }={ -F }_{ e }{ r }^{ 2 }\)
And    \({ F }_{ G }=-G\frac { { m }_{ p }{ m }_{ e } }{ { r }^{ 2 } } \)
So \({ Gm }_{ p }{ m }_{ e }={ -F }_{ G }{ r }^{ 2 }\)
So dimensions of \(\frac { { ke }^{ 2 } }{ { Gm }_{ e }{ m }_{ p } } =\frac { { F }_{ e }{ r }^{ 2 } }{ { F }_{ G }{ r }^{ 2 } } =\frac { { F }_{ e } }{ { F }_{ G } } \)
 \(=\frac { { [MLT }^{ -2 }] }{ { [MLT }^{ -2 }] } =[{ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }]\)
 = No dimensions
The value of \(\frac { { ke }^{ 2 } }{ { Gm }_{ e }{ m }_{ p } } \)
  \(=\frac { 9\times { 10 }^{ 9 }\times (1.6\times { 10 }^{ -19 })^{ 2 } }{ 6.67\times { 10 }^{ -11 }(1.67\times { 10 }^{ -27 })(9.1\times { 10 }^{ -31 }) } \)
  \(=2.9\times { 10 }^{ 39 }\)
From Eq. (1) and (2), we have
  \(\left| \frac { { F }_{ e } }{ { F }_{ G } } \right| =\frac { { ke }^{ 2 } }{ { Gm }_{ p }{ m }_{ e } } =2.9\times { 10 }^{ 39 }\)
The ratio of the two forces shows that electrical forces are enormously stronger than the gravitational forces.

Total emf in the circuit = SV - 4V = 4V|
Total resistance of the circuit = S\({ \Omega }\)
Hence current flowing in the circuit
i = V/R = 4/8  A = 0.5A
Current flowing through the resistors:
Current through 0.5 0, 1.0 0 and 4.5 0 is 0.5 A
Current through 3.0 \({ \Omega }\) is  \(\frac { 1 }{ 6 } \)A
Current through 6.0 \({ \Omega }\) \(\frac { 1 }{ 3 } \)A

From Ohm's law, we have V = IR
\(\Rightarrow\) \(V=I \rho \frac{l}{A}\)   \(\left[ \because \quad R=\rho \frac { l }{ A } \right] \)      .......(i)
When the rod is cut parallel, and rejoined by length, the length of the conductor becomes 2l, whereas the area decrease to \(\frac{A}{2}\). If the current remains the same, the potential changes as
V = \(I\rho \frac { 2l }{ A/2 } =4\times I\rho \frac { l }{ A } \)=4V           [using Eq (i)]
The new potential applied across the metal rod will be four times the original potential (V).

According to principle of superposition of electric fields, E (electric field) at a point due to system of three charges,
E  = \({{1}\over{4\pi{\epsilon}_{0}}}=\left[{{q}\over{{ r }_{ { 1 }^{ p } }^{ 2 }}}\hat{r}_{{1}_{p}} +{{{q}_{2}}\over{{ r }_{ { 2 }^{ p } }^{ 2 }}}\hat{r}_{{2}_{p}}+{{{q}_{3}}\over{{ r }_{ { 3 }^{ p } }^{ 2 }}}\hat{r}_{{3}_{p}} \right]\)
= \({ { q }\over{ 4\pi{\epsilon}_{0} } }\times{ { 9 }\over{ 8 } }\) [ using \({S}_{\infty}={ { a }\over{ 1 -r } }\) ] = \({ { 1 }\over{ 4\pi{\epsilon}_{0} } }.{{9q}\over{8}}{NC}^{-1}\)

In adsorption there is a decrease in residual forces of the surface which evolves as heat. Hence it is exothermic in nature.

The plot showing the variation of electric field and electric potential with distance r due to a point charge q is shown as below:

By Kirchhoff's first law at D,
 \({ I }_{ DC }=1A\)          \([\because { I }_{ Dc }+1=2]\)
    
Along ACDBA,                                                             
\({ V }_{ A }+1+1\times 2-2={ V }_{ B }\)  (V_A = 0)
But,  \({ V }_{ B }=1+2-2=1V\)
\({ V }_{ B }=1V\)                           

According to Gauss' theorem,

\(\int\)_s E. dS \(=\frac { q }{ { \varepsilon }_{ 0 } } \Rightarrow\) E. \(4\pi r^2=\frac { q }{ { \varepsilon }_{ 0 } }\)     
\(\therefore\)   \(E=\frac { q }{ 4\pi { \varepsilon }_{ 0 } r^2 }\)
If a charge q₀ is kept on the surface, then
F = E\(\times\) q₀ \(=\frac { q{ q }_{ 0 } }{ 4\pi { \varepsilon }_{ 0 } r^2 } \) , which is Coulomb's law.

During landing or take off, the tyres of aircraft's get charged due to the friction between tyres and ground. In case, the tyres are slightly conducting, the charge developed on the tyres will not stay on them and it finds its way to the earth.

Here the hollow cylinder is positively charged. We know that the electric field lines appear to come out from the conductor. thus the field lines for a uniformely positive charged hollow cyclinder is shown in the figure.

When a bird is perched on a bare high power line, the circuit does not get complete between the bird and the earth, therefore, nothing happens to the bird.
When a man standing on ground touches the same line, the circuit between the man and the earth gets completed. As a result, he gets a fatal shock.

No, it is not correct to write the unit of electric dipole moment as mC. The symbol mC represents milli-coulomb, i.e. unit of electric charges. IN SI system unit symbols are written in alphabetical order.
∴ Unit of dipole moment is C-m.

Electric field lines do not pass through a conductor.Hence,  the interior of the conductor is free from the influence of the electric field

Dielectric constant of water is much greater than that of mica because of the following reason
(i) water has a symmetrical shape as compared to mica
(ii) water has permanent dipole moment.

SI units of pole strength is ampere metre and SI unit of magnetic moment is ampere metre² .

As \(\pi={q\over \epsilon_o}\) so flux does not depend upon the radius of Gaussian surface, it will remain unchanged.

\(X_C=\frac{1}{\omega C}=\frac{1}{2 \pi \omega v C}\) therefore, when v is tripled

Electric field varies inversely as square of distance from the point charge.

\(R_1=\rho(2l)/(A/2)=4 \rho l/A=4R;\) so increase in resistance = R₁ - R = 4R - R = 3R. Resistivity will not change as it independent of the dimensions of the wire but depends upon nature of the wire and also on the temperature of the wire.

A current loop behaves as a bar magnet because
 (i) one face of current loop behaves as a south pole and the other face as north pole.
(ii)  it possesses a magnetic dipole moment (M = IA) and 
(iii) it experiences a torque in an external magnetic field, which tends to align the axis of the loop along the direction of magnetic field as bar magnet does.