\(sin{ i }_{ c }=\cfrac { 1 }{ \mu } =\cfrac { 1 }{ 1.5 } =0.667\\ \) 
or i_c= 41.8^o
Deviation = 90^o - i = 90^o - 41.8^o = 48.2^o
This si the maximum attainable deviation in refraction. So, the given data favours total internal reflection.
In reflection, deviation = 180^o - 2i
or 90^o = 180^o - 2i or 2i = 90^o or i = 45^o
or i = 45^o

\(\cfrac { 1 }{ F } =\cfrac { 1 }{ { f }_{ 1 } } +\cfrac { 1 }{ { f }_{ 2 } } \) 
\(\cfrac { 1 }{ 50 } =\cfrac { 1 }{ 20 } +\cfrac { 1 }{ { { f }_{ 2 } } } \quad f=-\cfrac { 100 }{ 3 } cm\) 
\({ P }_{ 2 }=\cfrac { 100 }{ -\frac { 100 }{ 3 } } \) 
D = -3D.

(i) At a minimum deviation \(\mu =\cfrac { sin\left( \frac { A+{ \delta }_{ m } }{ 2 } \right) }{ sin\left( \cfrac { A }{ 2 } \right) } \) 
 Given \({ \delta }_{ m }=A\)
\(\mu =\cfrac { sinA }{ sin\cfrac { A }{ 2 } } =\cfrac { 2sin\cfrac { A }{ 2 } cos\cfrac { A }{ 2 } }{ sin\cfrac { A }{ 2 } } \) 
= \(2cos\cfrac { A }{ 2 } \) 
\(\therefore cos\cfrac { A }{ 2 } =\cfrac { \sqrt { 3 } }{ 2 } \cfrac { A }{ 2 } =30\) A = 6^o 
(iii) \(\mu =\sqrt { 3 } \cfrac { 1 }{ { sini }_{ c } } \Rightarrow { sini }_{ c }=\cfrac { 1 }{ \sqrt { 3 } } \) 
ஃ Angle of incidence > i_c
Total internal reflection takes place.

Diffraction effects are observed for beams of electrons scattered by the crystals.
λ = \(\frac { h }{ p } =\frac { h }{ \sqrt { 2mE_{ k } } } =\frac { h }{ \sqrt { 2meV } } \)
= \(\frac { 6.63\times 10^{ -34 } }{ \sqrt { 2\times 9.1\times 10^{ -31 }\times 1.6\times { 10 }^{ -19 }\times 120 } } \)
λ = 0.112 nm

de Broglie wavelength, \(\lambda =\frac { h }{ \sqrt { 2mE } } \)
Given data 
h = 6.62 x 10 - 34 Js, m = 9.1 x 10 - 31kg
E = 50 KeV = 50 x 1.6 x 10^-19J
\(\lambda =\frac { { 6.62\times 10 }^{ -34 } }{ \sqrt { 2\times 9.1\times { 10 }^{ -31 }\times 8\times { 10 }^{ -15 } } } m\)
\(=\frac { 6.62 }{ \sqrt { 145.6 } } \times { 10 }^{ -11 }m\)
\(=\frac { 6.62 }{ 12.07 } \times { 10 }^{ -11 }m=5.48\times { 10 }^{ -12 }m\)
Wavelength of yellow light,
λ' = 5990 x 10^-10 m = 5.99 x 10^-7 m
Now \(\frac { \lambda }{ \lambda ' } ={ 10 }^{ -5 }\)
Since resolving power is inversely proportional to wavelength, therefore, the resolving power of an electron microscope is 105 times larger than the resolving power of the optical microscope.

(i) The amount of light energy or photon energy incident per metre square per second is called intensity of radiation.
(ii) 
(iii) As per Einstein's equation,
(a) The stopping potential is the same for I₁ and I₂ as they have the same frequency.
(b) The saturation currents are as shown in the figure because of I₁ > I₂ > I₃.

Power of radiation, P =\(\frac { nhv }{ l } \) = Nhv, where N is a number of photons per sec.
or N =\(\frac { P }{ m } \)
= \(\frac { 2.0\times { 10 }^{ -3 } }{ 6.63\times 10^{ -34 }\times 6\times { 10 }^{ 14 } } \)
= 5 x 10¹⁵ photons per second.

The energy of photon E = \(\frac { hc }{ \lambda } \) Joule
=\(\frac { hc }{ e\lambda } \) eV
=\(\frac { 6.63\times { 10 }^{ -34 }\times 3\times 10^{ 8 } }{ 1.6\times 10^{ -19 }\times 3.3\times 10^{ -7 } } \)eV
= 3.75 eV
Since W_O of M_O is greater than E,
∴ M_O will not give photoemission.

i) X-rays are produced in x-ray tube which is essentially a discharge tube.
ii) A tungsten filament F is heated to incandescence by a battery. As a result, electrons are emitted from it by thermionic emission.

iii) The electrons are accelerated to high speeds by the voltage applied between the filament F and the anode.
iv) The target materials like tungsten, molybdenum are embedded in the face of the solid copper anode.
v) The face of the target is inclined at an angle with respect to the electron beam so that x-rays can leave the tube through its side.
vi) When high-speed electrons strike the target, they are decelerated suddenly and lose their kinetic energy.
vii) As a result, x-ray photons are produced. Since most of the kinetic energy of the bombarding electrons gets converted into heat, targets' made of high-melting-point metals and a cooling system are usually employed.

(i) It is used for personal communication and cellular phones offer voice and data connectivity with high speed.
(ii) Transmission of news across the globe is done within a few seconds.
(iii) Using Internet of Things (UIT), it is made possible to control various devices from a single device. Example: home automation using a mobile phone.
(iv) It enables smart classrooms, online availability of notes, monitoring student activities etc. in the field of education.

(i) This gives the variation of collector current (I_C) with changes in base current (I_B) at constant collector-emitter voltage (V_CE).
(ii) It is seen that a small Ie flows even when I_B is zero. This current is called the common emitter leakage current (I_CEQ) which is due to the flow of minority charge carriers.
Forward current gain:
(i) The ratio of the change in collector current (ΔI_C) to the change in base current (ΔI_B) at constant collector-emitter voltage (V_CE) is called forward current gain(β)
\(\beta ={ \left( \frac { \triangle { I }_{ C } }{ \triangle { I }_{ B } } \right) }_{ { V }_{ CE } }\)
(ii) It is value is very high and it generally ranges from 50 to 200. It depends on the construction of the transistors and will be provided by the manufacturer.

Number of atoms undergoing fission per second
\(\frac { 3\times { 10 }^{ 7 } }{ 200\times L6\times { 10 }^{ -13 } } =0.9375\times { 10 }^{ 18 }\)
= Number of atoms undergoing fission in 24 hours = 0.9375 x 10¹⁸X 24 x 3600 = 0.81 x 10²³.
∴ Mass of uranium undergoing fission
= \(\frac { 235 }{ 6.023\times { 10 }^{ 23 } } \times 0.81\times { 10 }^{ 23 }=31.6g\) 

If N_o is the initial quantity, then the quantity left after n half-lives will be \(N={ N }_{ o }{ \left( \frac { 1 }{ 2 } \right) }^{ n }\)
But \(n=\frac { t }{ T } \) where t is the time of disintegration and T is half-life.
Then \(N={ N }_{ o }{ \left( \frac { 1 }{ 2 } \right) }^{ t/T }\)
Now, N = 25% of N_o = \(\frac { { N }_{ 0 } }{ 4 } \)
\(\frac { { N }_{ o } }{ 4 } ={ N }_{ o }{ \left( \frac { 1 }{ 2 } \right) }^{ t/T }\) ∴ 2^t/T = 4 or 2^t/T = 2²
or \(\frac{t}{T}\) = 2 or t = 2T = 2 x 1600 year.

The fusion reaction is as under:
\(_{ 1 }^{ 2 }{ H+ }_{ 1 }^{ 2 }{ H\rightarrow }_{ 2 }^{ 4 }{ He }+Q\) (energy)
Deuteron contains 2 nucleons. Binding energy per nucleon is 1.1 MeV. Binding energy of each deuteron nucleus is 2 x 1.1 i.e. 2 .2 MeV
Total binding energy before reaction
= 2 x 2.2 MeV = 4.4 MeV
Total binding energy after reaction
= 4 x 7 MeV = 28 MeV
Clearly the energy released is (28 - 4.4) MeV
i.e. 23.6. MeV

Rutherford atom model helps in the calculation of the diameter of the nucleus and also the size of the atom but has the following limitations
(a) This model fails to explain the distribution of electrons around the nucleus and also the stability of the atom.
According to classical electrodynamics,.any accelerated charge emits electromagnetic radiations. Due to emission of radiations, it loses its energy. Hence, it can no longer .sustain the circular motion. The radius of the orbit, therefore, becomes smaller and smaller (undergoes spiral motion) and finally the electron should fall into the nucleus and the atoms should disintegrate. But this does not happen.

(b) According to this model, emission of radiation must be continuous and must give continuous emission spectrum but experimentally we observe only line (discrete) emission spectrum for atoms.

Formula:
Intensity, I = \(\frac{Power\ of\ visible\ light}{Area}\)
\(I=\frac { \frac { 5 }{ 100 } \times 100 }{ 4\pi { (1) }^{ 2 } } =0.4{ W/m }^{ 2 }\)

G = 15Ω, I_g = 2mA = 2 x 10^-3A
1 = 5A
Shunt resistance, S = \(\frac { { I }_{ g }G }{ I-{ I }_{ g } } \)
= \(\frac { 2\times 10^{ -3 }\times 15 }{ 5-2\times 10^{ -3 } } \)
= 0.006Ω
The resistance S = 0.006Ω is connected in parallel with the galvanometer in order to decrease the resistance of the galvanometer.

According to Ampere's circuital law,
\(\oint _{ s }^{ }{ \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }I\quad ...(1)\)
As the current flows across the area bounded by loop S₁, so
\(\oint _{ { s }_{ 1 },s }^{ }{ \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }I\quad ...(2)\)
But the area bounded by S₂ lies in the region between the plates capacitor where no current flows across it.
\(\therefore \oint _{ { s }_{ 1 } }^{ }{ \overrightarrow { B } .\overrightarrow { dl } } =0\)
Consider that loops enclosing S₁ & S₂ are infinitesimally close to each other. Then
\(\oint _{ { s }_{ 1 } }^{ }{ \overrightarrow { B } .\overrightarrow { dl } } =\oint _{ { s }_{ 2 } }^{ }{ \overrightarrow { B } .\overrightarrow { dl } } \)
This equation is inconsistent with equations (2) & (3). To remove this maxwell said that a changing electric field (during charging) between the capacitor plates must induce a magnetic field which in turn must be associated with current I_d.
\({ I }_{ d }={ \varepsilon }_{ 0 }\left( \frac { d{ \phi }_{ E } }{ dt } \right) \) [\(\frac { d{ \phi }_{ E } }{ dt } \) change in electric flux]
The total current must be
I = I_conduction + I_displacement
\({ I }_{ c }={ \varepsilon }_{ 0 }\frac { d{ \phi }_{ E } }{ dt } \)
Hence the generalized from of Ampere's circuital law is
\(\oint _{ s }^{ }{ \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }\left[ { I }_{ c }+{ \varepsilon }_{ 0 }\frac { d{ \phi }_{ E } }{ dt } \right] \)

ፒ = μBsinθ
Magnetic moment m = 3.6 x 10^-2JT^-1
Magnetic field B = 2 x 10^-2 T
Magnetic needle is inclined at an angle θ = 30°
Torque ፒ =?
ፒ = 3.6 x 10^-2 x 2 x 10^-2 sin 30°
t = 3.6 x 10^-4 Nm.

Given:
A = 20cm x 30cm
B = 0.3T
Let θ be the angle made by the field B with the normal to the plane of the coil
(i) Here, θ = 90° - 90° = 0°
So, flux Φ = BA cosθ
= 0.3 x 6 x 10^-2 x cos0°
e = 1.8 x 10^-2 Wb
(ii) Here, θ = 90° - 30° = 60°
Φ = 0.3 x 6 x 10^-2x cos60°
Φ = 0.9 X 10^-2 Wb
(iii) Here, 90° - 0° = 90°
Φ = 0.3 x 6 x 10^-2x cos90°
Φ = 0°

(i) An AC is that current that changes continuously in magnitude and periodically in direction. The value of ac at any instant
I = I sin ωt = I sin 2π⋎t
Where I_o is the peak value of current and angular frequency ω = 2 π⋎ = \(\frac { 2\pi }{ T } \)
(ii) The frequency of the direct current is zero.

For the bridge to be balanced \(\cfrac { \\ P }{ Q } =\cfrac { R }{ S } \)
Since additionally a resistance 'x' is added in series with S, equation (1) can be written as
\(\cfrac { \\ P }{ Q } =\cfrac { R }{ (S+x) } \)
\(\left( S+x \right) =\cfrac { QR }{ P } x=\cfrac { \theta R }{ P } -S\)
\(x=\cfrac { 8\times 12 }{ 7 } -7\)
\(x=\frac{96}{7}-7=6.714 \Omega\)

Potential difference V = 3 V
Current, I = 5 A
By Ohms law, \(R=\cfrac { V }{ I } =\cfrac { 3 }{ 5 } =0.6\Omega \)

(i) q₁ = q₂ = 6.5 x 10^-7C
r = 50 cm = 0.5m
Electrostatic force of repulsion,
\(F=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \)
\(\\ =\frac { 9\times { 10 }^{ 9 }\times { (6.5\times { 10 }^{ -7 }) }^{ 2 } }{ { (0.5) }^{ 2 } } \)
F = 1.521 x 10^-2N
(ii) Now if q₁,q₂ are doubled and r' is halved then F becomes 16 times.
i.e., New force of repulsion, \(F'=16\times \frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \)
F' = 16F
= 16 x 1.521510^-2
F' = 0.24 N.

Let R₁, R₂, R₃be the resistance of the wires 
Then \({ R }_{ 1 }:{ R }_{ 2 };R_{ 3 }=l\cfrac { 1_{ 1 } }{ { r }_{ 1 }^{ 2 } } :\cfrac { { l }_{ 2 } }{ { r }_{ 2 }^{ 2 } } :\cfrac { { l }_{ 3 } }{ { r }_{ 3 }^{ 2 } } =\cfrac { 2 }{ 9 } :\cfrac { 3 }{ 16 } ;\cfrac { 4 }{ 25 } \)
The ratio of the circuits in the three wires must be inverse of the above ratio \(\left[ \because I=\cfrac { V }{ R } i.eI\alpha \cfrac { I }{ R } \right] \)
\(\therefore { I }_{ 1 }:{ I }_{ 2 }:{ I }_{ 3 }=\cfrac { 9 }{ 2 } :\cfrac { 16 }{ 3 } :\cfrac { 25 }{ 4 } =54:64:75\)
As total current = 5A [54 + 64+ 75 = 193]
\(\therefore { I }_{ 1 }=\cfrac { 5\times 54 }{ 193 } =1.40;{ I }_{ 2 }=\cfrac { 5\times 64 }{ 193 } =1.66A\)
\({ I }_{ 3 }=\cfrac { 5\times 75 }{ 193 } =1.94A\)

Area of the plate, A = 25 cm²
= 25 x 10^-4m²
Distance between the plates, d = 2 mm
d = 2 x 10^-3m
Potential difference, V = 12V
Charge, q = CV \(\left[ C=\frac { { \varepsilon }_{ 0 }A }{ d } \right] \)
\(q=\left( \frac { { \varepsilon }_{ 0 }A }{ d } \right) V\)
\(=\frac { 8.85\times { 10 }^{ -12 }\times 25\times { 10 }^{ -4 }\times 12 }{ 2\times { 10s }^{ -3 } } \)
q = 1.33 x 10^-10C

r = 25cm = 25 x 10^-2m
\(\sigma =\frac { 3 }{ \pi } { cm }^{ -2 }\)
\(AS,\quad \sigma =\frac { q }{ A } =\frac { q }{ 4\pi { r }^{ 2 } } \) [A - surface Area of the sphere]
\(q=(4\pi { r }^{ 2 })\sigma \)
\(=4\pi (0.25{ ) }^{ 2 }\times \frac { 3 }{ \pi } \)
= 0.75C.

(i) \(mobility=\cfrac { Drift\ velocity }{ electric\ field } \) (or) \(\mu =\cfrac { { V }_{ d } }{ E } \)
(ii) \({ \mu }_{ d }=\cfrac { eE }{ mL } .\tau \) (or) \(\cfrac { { v }_{ d } }{ E } =\left( \cfrac { e }{ mL } \right) .\tau \)
\(\mu =\cfrac { e }{ mL } .\tau \)

(i) The term RMS refers to time-varying sinusoidal currents and voltages and is not used in DC systems.
(ii) The root mean square value of an alternating current is defined as the square root of the mean of the squares of all currents over one cycle. It is denoted by I_RMS. For alternating voltages, the RMS value is given by I_VRM^.
(iii) The alternating current i = I_m sin ωt or i = I_m sin θ, is represented graphically in Figure. The corresponding squared current wave is also shown by the dotted lines.
(iv) The sum of the squares of all currents over one cycle is given by the area of one cycle of the squared wave. Therefore,
I_RMS = \(\sqrt{\frac{Area \ of \ one \ cycle \ of \ squared \ wave}{Base\ length \ of \ one\ cycle}}\) ....(1)
(v) An elementary area of thickness dθ is considered in the first half-cycle of the squared current wave as shown in Figure. Let i² be the mid-ordinate of the element. Area of the element = i2dθ
Area of one cycle of squared.
wave = \(\int _{ 0 }^{ 2\pi }{ { i }^{ 2 }d\theta } \)

= \(\int _{ 0 }^{ 2\pi }{ { { I }^{ 2 } }_{ m } } { sin }^{ 2 }\theta d\theta ={ { I }^{ 2 } }_{ m }\int _{ 0 }^{ 2\pi }{ { sin }^{ 2 }\theta d\theta } \)
= \({ { I }^{ 2 } }_{ m }\int _{ 0 }^{ 2\pi }{ \left[ \frac { 1-cos2\theta }{ 2 } \right] d\theta } \)
since \({ sin }^{ 2 }\theta =\frac { 1-cos2\theta }{ 2 } \)
= \(\frac { { { I }^{ 2 } }_{ m } }{ 2 } \left[ \int _{ 0 }^{ \pi }{ id\theta } =\int _{ 0 }^{ \pi }{ { I }_{ m }cos2\theta d\theta } \right] \)
= \(\frac { { { I }^{ 2 } }_{ m } }{ 2 } { \left[ \theta -\frac { sin2\theta }{ 2 } \right] }_{ 0 }^{ 2\pi }\)
= \(\frac { { { I }^{ 2 } }_{ m } }{ 2 } \left[ \left( 2\pi -\frac { sin2\times 2\pi }{ 2 } \right) -\left( 0-\frac { sin0 }{ 2 } \right) \right] \)
= \(\frac { { { I }^{ 2 } }_{ m } }{ 2 } \times \pi ={ { I }^{ 2 } }_{ m }\pi \)    [∵ sin 0 = sin 4π = 0]
Substituting this in equation (1), we get
\(\sqrt { \frac { { { I }^{ 2 } }_{ m }\pi }{ 2\pi } } =\frac { { { I }^{ 2 } }_{ m } }{ \sqrt { 2 } } \)    [Base length of cone cycle is 2π]
I_RMS = 0.707 I_m

\({ V }_{ S }-{ V }_{ R }=\frac { W }{ q } \)
Work W = 50μJ = 50 x 10^-6 J
Charge q = 2μC = 2 x 10^-6 C
V = V_S - V_R \(=\frac { W }{ q } =\frac { 50\times { 10 }^{ -6 } }{ 2\times { 10 }^{ -6 } } =25V\)
V = 25V

The kinetic energy acquired by the electron is given by 
\(=\frac{1}{2}mv^2=e V\)
Therefore, the speed v of the electron is \(v=\sqrt{\frac{2 eV}{m}}\)
Hence, the de Broglie wavelength of the electron is, \(\lambda=\frac{h}{mv}=\frac{h}{\sqrt{2meV}}\)