Electric potential due to a point charge:
Consider a positive charge q kept fixed at the origin. Let P be a point at distance r from the charge q. This is shown in Figure.

Electrostatic potential at a point P
The electric potential at the point P is
\(V=\int _{ \infty }^{ r }{ \left( -\vec { E } \right) d\vec { r } } =-\int _{ \infty }^{ r }{ \vec { E } } .d\vec { r } \)  ...(1)
Electric field due to positive point charge q is
\(\vec { E } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ { r }^{ 2 } } \hat { r } \)
\(V=\frac { -1 }{ 4\pi { \varepsilon }_{ 0 } } \int _{ \infty }^{ r }{ \frac { q }{ { r }^{ 2 } } \hat { r } .d\vec { r } } \)
The infinitesimal displacement vector, \(d\vec { r } =dr\hat { r } \) and using \(\hat { r } \).\(\hat { r } \) = 1, we have
\(V=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \int _{ \infty }^{ r }{ \frac { q }{ { r }^{ 2 } } \hat { r } .dr\hat { r } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \int _{ \infty }^{ r }{ \frac { q }{ { r }^{ 2 } } dr } } \)
After the integration,
\(V=-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } q{ \left\{ -\frac { 1 }{ r } \right\} }_{ \infty }^{ r }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ r } \)
Hence, the electric potential due to a point charge q at a distance r is
\(V=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ r } \)   ....(2) 

In all the sections, the resistors are connected in parallel.
Section I
\(\frac { 1 }{ { R }_{ { P }_{ 1 } } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ { R }_{ { P }_{ 1 } } } =\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =\frac { 2 }{ 2 } \quad { R }_{ { p }_{ 1 } }=1\Omega \)

Section II
\(\frac { 1 }{ { R }_{ { P }_{ 1 } } } =\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =\frac { 2 }{ 4 } ,\quad \frac { 1 }{ { R }_{ { P }_{ 2 } } } =\frac { 1 }{ 2 } ,{ R }_{ { p }_{ 2 } }=2\Omega \)

Section III
\(\frac { 1 }{ { R }_{ { P }_{ 3 } } } =\frac { 1 }{ 6 } +\frac { 1 }{ 6 } =\frac { 2 }{ 6 } \)
\(\frac { 1 }{ { R }_{ { P }_{ 3 } } } =\frac { 1 }{ 3 } ,{ R }_{ { p }_{ 3 } }=3\Omega \)
Equivalent resistance is given by
R = Rp₁ + Rp₂ + Rp₃
R = 1 Ω + 2 Ω + 3 Ω = 6 Ω
The circuit became,

Equivalent resistance between A and B is

Albert Einstein theoretically proposed the existence of gravitational waves in the year 1915.

Size of the particle is between
1-100 nm - Nano.
Size of the particle is greater
than-100 nm - Bulk.

\(V_{ionisation}=\frac{13.6}{n^2}Z^2 volt\)
\(Z=\sqrt\frac{V\times n^2}{13.6}=\sqrt\frac{122.4 \times I^2}{13.6}=\sqrt{9}=3\)

Lorentz force F = Bqv
Energy w = qV
Energy is equal to kinetic energy,
\(qV=\frac{1}{2}mv^2\)
\(v=\sqrt { \frac {2qV }{ m } } \)
\(\therefore Lorentz \ force \ F= Bq\times \sqrt \frac{2qV}{m}=\sqrt { \frac { 2{ B }^{ 2 }{ q }^{ 3 }V }{ m } } \)

P_av = \(\frac{1}{2}\)V₀I₀cosΦ
\(= \frac{1}{2}\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}cos\times\frac{\pi}{3}\times \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}\)

\(\text {emf } e=8 \mathrm{~V}  \)
\(d I=I_1-I_0=2-(-2)=4 \mathrm{~A}  \)
\(\text {dt }=0.05 \mathrm{~s}  \)
\(L=\frac{-e}{d I / d t}=\frac{-8}{4 / 0.05}  \)
\(=\frac{-8 \times 0.05}{4}=\frac{-0.40}{4}  \)
=-0.1 H
-ve sign indicates that self-induced emf always opposes the current w.r.t. time.

\(\vec {E}\) = 10x\(\hat{i},\)  when x = 2 m
\(\vec {E}\) = 10 x 2 x \(\hat{i}\) = 20\(\hat{i}\)
Since, \(E=\frac{-dV}{dx}\therefore V=+20 V\)

Resistivity ∝ temperature for conductor. so, copper → decreases
Resistivity ∝\(\frac{1}{\text {temperature for semiconductor}}\)
so, germanium → increases.

Total length = 2πr = 2π
∴ Resistance of each segment = 2π/2 = π Ω

ß₁ = 7.2 mm, ß₂ = 8.1 mm, λ₁ = 600 nm, λ₂ =?
Dividing, we get  \(\beta = \frac{D\lambda}{d}(or)\beta∝\lambda\)
\(\frac{\beta_1}{\beta_2}=\frac{\lambda_1}{\lambda_2}\)
\(\therefore, { \lambda }_{ 2 }=\frac{\beta_2\times \lambda_1}{\beta_1}=\cfrac { 8.1\times10^{-3} \times 600 \times10^{-9}}{ 7.2 \times10^{-3}} \)
\(=\cfrac { 9 \times 600\times10^{-9} }{ 8 } =675\times 10^{-9 }\Rightarrow\lambda_2= 675nm\)

Waves of high frequency are called carrier waves. If an audio signal (low frequency) is transmitted as such, it does out after traversing some distance. To reach up larger distance, it is superimposed an high frequency waves called carrier wave

The voltage at which the diode breaks down is called the breakdown voltage

Digital electronics deals with logical operations. The variables are called logical variables

At room temperature, a potential difference equal to the barrier potential is required before a reasonable forward current starts flowing across the diode. This voltage is known as threshold voltage or cut-in voltage or knee voltage.

Electric field is strong enough to break (or) rep tune the covalent bonds in the lattice and there by generating electron - hole pairs. This effect is called Zener effect.

The process in which alternating voltage or alternating current is converted direct voltage or direct current is called rectification.

From right hand rule, the magnetic field by the straight wire is directed into the plane of the square loop perpendicularly and its magnetic flux is decreasing. The decrease in flux is opposed by the current induced in the loop by producing a magnetic field in the same direction as the magnetic field of the wire. Again from right hand rule, for this inward magnetic field, the direction of the induced current in the loop is clockwise.

The total flux emanating from the closed surface S enclosing the dipole is zero. So,
\({ \Phi }_{ B }=\oint { \overset { \rightarrow }{ B } .d\overset { \rightarrow }{ A } } =0\)
Here the integral is taken over closed surface. Since no isolated magnetic pole (called magnetic monopole) exists, this integral is always zero,
\(\oint { \overset { \rightarrow }{ B } .d\overset { \rightarrow }{ A } } =0\)
This is similar to Gauss’s law in electrostatics.

Data on focal lengths is not required, The limit of resolution of the telescope is given by
\(\theta =\cfrac { 1.22\lambda }{ D } =\cfrac { 1.22\times 6\times { 10 }^{ -7 } }{ 0.60 } rad=1.22\times { 10 }^{ -6 }\) 

Here, the value chosen corresponds roughly to the wavelength of yellow light. The transverse separation between the two sources subtends an angle equal to
\(\cfrac { { 10 }^{ 10 }m }{ 9.46\times { 10 }^{ 19 } } ={ 10 }^{ -10 }red\) 
Here, we have used 1 light-year = 9.46 x 10¹⁵ m This angle is much too small compared to e above. The two stars of the binary cannot be resolved by the given telescope.

Let polaroids P₁ and P₃ be in a crossed positions.
Let the polaroid P₂ make an angle e with the pass axis of polaroid Pr
Let I₁ be the intensity of polarised light emerging our of P₁ Then intensity of light after passing through P₂ will I₂ = I₁cos²θ
Since P₃ and P, are in crossed position, therefore, the angle made by P₂ with P₃ is \(\left( \cfrac { \pi }{ 2 } -\theta \right) \) 
∴ The intensity of light coming out of P₃ is 
\({ I }_{ 3 }={ I }_{ 2 }{ cos }^{ 2 }\left( \cfrac { \pi }{ 2 } -\theta \right) \) 
or \({ I }_{ 3 }={ I }_{ 2 }{ cos }^{ 2 }{ sin }^{ 2 }\theta =\theta ={ I }_{ 1 }\left( \cfrac { 1 }{ 2 } sin20 \right) ^{ 2 }\) 
If I₀ is the intensity of the unpolarised light falling on P_I' then \({ I }_{ 1 }=\cfrac { { I }_{ o } }{ 2 } \) 
\(\therefore { I }_{ 3 }=\cfrac { { I }_{ o } }{ 2 } \left( \cfrac { 1 }{ 2 } sin20 \right) ^{ 2 }\) 
(i) Minimum outcoming intensity is zero.
(ii) Maximum outcoming intensity is received
when \(\theta =\cfrac { \pi }{ 4 } \) 
\(\therefore \left( { I }_{ 3 } \right) _{ max }=\cfrac { { I }_{ o } }{ 2 } \left( \cfrac { 1 }{ 2 } \right) ^{ 2 }=\cfrac { { I }_{ o } }{ 8 } \)

\(\cfrac { 1 }{ -25 } -\cfrac { 1 }{ { u }_{ e } } =\cfrac { 1 }{ 5 } \) 
or \({ u }_{ e }=-\cfrac { 25 }{ 6 } cm\) 
\({ v }_{ o }=20-\cfrac { 25 }{ 6 } =\cfrac { 95 }{ 6 } cm\) 
\(\cfrac { 1 }{ { v }_{ 0 } } =\cfrac { 1 }{ { u }_{ 0 } } =\cfrac { 1 }{ { f }_{ 0 } } \) 
\(\cfrac { 6 }{ 95 } -\cfrac { 1 }{ { u }_{ 0 } } =\cfrac { 1 }{ 2 } ,{ u }_{ 0 }=2.29cm\) 
\(m=-\cfrac { 95/6 }{ 190/83 } \left( 1+\cfrac { 25 }{ 5 } \right) =-41.5\)

α = 0.98
I_Cu = 0.6 μA
\(\alpha =\frac { \beta }{ 1+\beta } \)
(or)
\(0.98=\frac { 1 }{ \frac { 1 }{ \beta } +1 } \)
β = 49
I_CEO = (1 + β)I_CBO
= (1 + 49) x 0.6μA
I_CEO = 30 μA
= (1 + 49) x 0.6 μA
I_CEO = 30 μA
I_C = βI_B+ I_CEO
= 49 x 20 μA + 30 μA
I_C = 1.01 mA

Consider a solenoid of length L having N turns. The diameter of the solenoid is assumed to be much smaller when compared to its length and the coil is wound very closely.

Consider a rectangular loop abcd. Then from Ampere's circuital law,
\(\oint _{ C }^{ }{ \vec { B } .\vec { dl } } \) = μ₀ I_enclosed = μ₀ x (total current enclosed by Amperian loop)
The left hand side of the equation is
\(\oint _{ C }^{ }{ \vec { B } .\vec { dl } } =\int _{ a }^{ b }{ \vec { B } .\vec { dl } } +\int _{ b }^{ c }{ \vec { B } .\vec { dl } } +\int _{ c }^{ d }{ \vec { B } .\vec { dl } } +\int _{ d }^{ a }{ \vec { B } .\vec { dl } } \)
Elemental lengths bc and da are perpendicular to magnetic field.
\(\therefore \int _{ b }^{ c }{ \vec { B } .\vec { dl } } =\int _{ b }^{ c }{ |\vec { B } ||\vec { dl } | } cos 90^o = 0\)
similarly,
\(\int _{ d }^{ a }{ \vec { B } .\vec { dl } } = 0\) 
Since the magnetic field outside the solenoid is zero,
\(\int _{ c }^{ d }{ \vec { B } .\vec { dl } } =0\) and
\(\int _{ a }^{ b }{ \vec { B } .\vec { dl } } =BL \quad \quad (\because \theta =0^o)\)
Let I be the current passing through the solenoid of N turns, then
\(\int _{ a }^{ b }{ \vec { B } .\vec { dl } }  = BL = μ_o NI ⇒ B = μ_o\frac {NI}{L}\)
The number of turns per unit length is given by \(\frac { N }{ L } \) = n, Then
B = \(\mu_o \frac { nLI }{ L } \) = μ₀nI
Since n is a constant for a given solenoid and μ₀ is also constant. For a fixed current I, the magnetic field inside the solenoid is also a constant.