r ∝ A^1/3
Surface Area = 4πr²
Hence, Surface Area ∝ A^2/3

\(\frac{\mathrm{hc}}{\lambda}=\phi+\mathrm{eV}  \)    .....(1)
\(\frac{\mathrm{hc}}{2 \lambda}=\phi+\frac{\mathrm{eV}}{4}\)    .....(2)
multiply (2) eqn by 4
\(\frac{2 h c}{\lambda}=4 \phi+\mathrm{eV}\)   .....(3)
subtract eqn (1) from (3), we get
\(\frac{ h c}{\lambda}=3 \phi \Rightarrow \phi = \frac{ h c}{3\lambda}\)
\(\frac{ h c}{\lambda_o}=\frac{ h c}{3\lambda}\)
⋋_o = 3⋋

\(B=\frac { { \mu }_{ ° }NI a^2}{ { 2(a^2+x^2)}^{ \frac { 3 }{ 2 } } } \)
put a = R
and x = R/2, we get,
\(B=\frac { 8N{ \mu }_{ ° }I }{ { 5 }^{ 3 / 2 }R } \)

L = 20 x 10^-3H. C = 50 x 10^-6 F, R= 40Ω
enf V = 10 sin 340 t
\(\therefore V_0=10 \mathrm{~V}, \omega=340  \)
\(X_1=1 \omega^{\prime}=20 \times 10^3 \times 340  \)
\(=6800 \times 10^{-1}=6.8 \Omega  \)
\(X_C=\frac{1}{C .}  \)
\(=\frac{1}{50 \times 10^{-\alpha} \times 340}=\frac{10^{\circ}}{17000}=\frac{10^{\prime}}{17}=58.823 \Omega   \)
\(Z=\sqrt{R^2+\left(X_6-X_1\right)^2}  \)
\(=\sqrt{(40)^2+(58.82-6.8)^2}  \)
\(=\sqrt{(40)^2+(52.02)^2}  \)
\(=65.62 \Omega\)
The peak current in the circuit is,
\(I_0=\frac{V_0}{Z}=\frac{10}{65.62}  \)
\(\cos 0=\frac{R}{Z}=\frac{40}{65.62} \)
\(\text{Power loss in A.C. circuit }=V_{r m} 1_{r \rightarrow \infty} \cos \phi \)
\(=\frac{1}{2} V_{\mathrm{o}} I_{\mathrm{c}} \cos \phi  \)
\(=\frac{1}{2} \times 10 \times \frac{10}{65.62} \times \frac{40}{65.62}\)
\(\frac{2000}{4305.98}\)
= 0.46 W

Force ∝ charge 
After the separation, the magnitude of charge will be increased. So the force will be more than before.

The combined width of a ruling and a slit is called grating element (e = a + b). Points on successive slits separated by a distance equal to the grating element are called corresponding points.

A transducer is a device that converts information such as pressurc, temperature, sound, picture etc. into an equivalent electrical signal or vice versa.

Band of very large number of closely spaced energy levels in a very small energy range is known as energy band.

The energy gap between the valance band and conduction band is called forbidden energy gap

(i) Alpha ray has high ionizing power.
(ii) α - particle has a large mass and large nuclear cross-section. So it has high ionizing power.

Given, focal length, f = 150 cm = 1.5 m
Equation for power of lens is, \(p=\cfrac { 1 }{ f } \) 
Substituting the values,
\(p=\cfrac { 1 }{ 1.5 } =0.67 D\) 
As the power is positive, it is a converging lens.

The element chlorine is a mixture of 75.77% of  \(_{ 17 }^{ 35 }{ Cl }\) and 24.23% of  \(_{ 17 }^{ 37 }{ Cl }\). So the average atomic mass will be
\(\frac { 75.77 }{ 100 } \times 34.96885u+\frac { 24.23 }{ 100 } \times 36.96593u\)
= 35.453 u
In fact, the chemist uses the average atomic mass or simply called chemical atomic weight (35.453 u for chlorine) of an element. So it must be remembered that the atomic mass which is mentioned in the periodic table is basically averaged atomic mass.

The reciprocal of resistivity of a material is called electrical conductivity \(\left( \sigma \right) \sigma =\cfrac { 1 }{ \rho } \) .
It depends upon number density nature of material relaxation time & temperature
i.e \(\sigma =\cfrac { { ne }^{ 2 }\tau }{ m } \).

Path difference = \(\cfrac { xd }{ D } +\left( { n }_{ 2 }-{ n }_{ 1 } \right) t\) 
For P, = 0; x path difference
= (n² - n¹) t = 0.3 t
\(I={ I }_{ o }{ cos }^{ 2 }\cfrac { \phi }{ 2 } \) 
\(\cfrac { I }{ { I }_{ o } } ={ cos }^{ 2 }\cfrac { \phi }{ 2 } ,cos\cfrac { \phi }{ 2 } =\cfrac { \sqrt { 3 } }{ 2 } \) 
\(\cfrac { \phi }{ 2 } =\cfrac { \pi }{ 6 } ,\phi =\cfrac { \pi }{ 3 } \) 
Path difference = \(\cfrac { \lambda }{ 2\pi } \phi =\cfrac { 2\lambda }{ 2\pi } \cfrac { \pi }{ 3 } =\cfrac { \pi }{ 6 } \) 
\(0.3t=5\lambda +\cfrac { \pi }{ 6 } \)

The magnification due to the objective lens
\({ m }_{ o }=\cfrac { { v }_{ o } }{ \left( -{ \mu }_{ o} \right) } \) 
If the object is close to the focus of the objective lens then
u_o = f_o and v = L
(L = distance between two lenses)
\({ m }_{ o }=\cfrac { L }{ { f }_{ o } } \) 
If the final image is at the near point, then magnification due to the eye lens is
\({ m }_{ e }=\left( 1+\cfrac { D }{ { f }_{ e } } \right) \) 
\(M={ m }_{ o }\times { m }_{ e }=\cfrac { L }{ { f }_{ o } } \left( 1+\cfrac { D }{ { f }_{ e } } \right) \) 
The separation between the two lenses is 6.25 cm.

V_BE =0.7V
Input junction is a forward biased.
Since,
V_BE = 0.7V
V_CE = V_BE+ V_CB
V_CB = V_CE-V_BE
To determine V_CB we find I_C
\({ I }_{ C }\cong { I }_{ C }\frac { 2-{ V }_{ BE } }{ { R }_{ 2 } } =\frac { 2-0.7 }{ 1k\Omega } \)
I_C = 1.3mA
V_CE = V_CC - I_C (R₁ + R₂)
= 10 - 1.3mA (10K + 1K)
V_CE = - 4.3 V
V_CE =-4.3V-0.7
V_CB = - 5V

1eV = 1.6 x 10^-19 J
or \(1J=\frac { 1 }{ 1.6\times { 10 }^{ -19 } } eV=6.25\times { 10 }^{ 18 }eV\)
Mass of C-12 atom = 12 amu
= 12 x 931 MeV = 12 x 931 x 1.6 x 10^-13J
\(=\frac { 12\times 931\times 1.6\times { 10 }^{ -13 } }{ (3\times { 10 }^{ 8 })^{ 2 } } =1.99\times { 10 }^{ -26 }kg\)

(i) The smoke detector uses around 0.2 mg of a man-made weak radioactive isotope called americium \((_{ 95 }^{ 241 }{ Am })\)

(ii) This radioactive source is placed between two oppositely charged metal plates and α radiations from \(_{ 95 }^{ 241 }{ Am }\) continuously ionize the nitrogen, oxygen molecules in the air space between the plates
(iii) As a result, there will be a continuous flow of small steady currents in the circuit.
(iv) If smoke enters, the radiation is being absorbed by the smoke particles rather than air molecules.
(v) As a result, the ionization and along with it the current is reduced. This drop-in current is detected by the circuit and the alarm starts.
(vi) The radiation dosage emitted by americium is very much less than the safe level, so it can be considered harmless.

Materials made up of particles sizes of 1-100 made up nm differ in their properties from the materials made up of bulk particles. Thus, nano materials find wide range of applications
Automotive industry :
(i) Lightweight construction
(ii) Painting (fillers, base coat, clear coat)
(iii) Catalysts
(iv) Tires (fillers)
(v) Sensors
(vi) Coatings for wind-screen and car bodies.
Chemical industry :
(i) Fillers for paint systems
(ii) Coating systems based on nanocomposites
(iii) Impregnation of papers
(iv) Switchable adhesives
(v) Magnetic fluids
Engineering :
(i) Wear protection for tools and machines (anti blocking coatings, scratch resistant coatings on plastic parts etc.)
(ii) Lubricant-free bearings.
Electronic industry :
(i) Data memory
(ii) Displays
(iii) Laser diodes
(iv ) Glass fibres
(v) Optical switches
(vi) Filters (lR-blocking)
(vii) Conductive, antistatic coatings.
Construction :
(i) Construction materials
(ii) Thermal insulation
(iii) Flame retardants
(iv) Surface-functionalised building materials for wood, floors, stone, facades, tiles, roof tiles, etc.
(v) Facade coating
(v) Groove mortar.
Medicine :
(i) Drug delivery systems
(ii) Active agents
(iii) Contrast medium
(iv) Medical rapid tests
(v) Prostheses and implants
(vi) Antimicrobial agents and coatings
(vii) Agents in cancer therapy.
Textile/fabrics/non-wovens :
(i) Surface-processed textiles
(ii) smart clothes.
Energy :
(i) Fuel cells
(il) Solar cells
(iii) Batteries
(iv) Capacitors.
Cosmetics :
(i) Sun protection
(ii) Lipsticks
(iii) Skin creams
(iv) Tooth paste.
Food and drinks :
(i) Package materials
(ii) Storage life sensors
(iii) Additives
(iv) Clarification of fruit juices.
Household :
(i) Ceramic coatings for irons
(ii) Orders catalyst
(iii) Cleaner for glass, ceramic, floor, windows.
Sports/Outdoor :
(i) Ski wax
(ii) Antifogging of glasses /goggles
(iii) Antifouling coating for ships /boats
(iv) Reinforced tennis rackets and balls.

Laws of Boolean algebra:
Complement law:

The complement law can be realised as Ā = A
OR laws:

The OR laws can be realised as:

AND law:

The AND laws can be realised as:

The Boolean operations obey the folloWing laws:
Communtative laws:
A+B =B+A
A.B =B.A
A sociate laws:
A + (B + C) = (A + B) + C
A. (B.C) = (A.B).C
D stributive laws:
A (B + C) = AB + BC
A + BC = (A + B) (A + C) 
The above laws are used to simplify complicated expressions and to simplify the logic circuitry.

FuIl wave rectifier :
The positive and negative half cycles of the AC input signal pass through the full wave rectifier circuit and hence it is called the full wave rectifier
Construction:

(i) It consists of two p-n junction diodes, a center-tapped transformer, and a load resistor (R₁)
(ii) The centre is usually taken as the ground or zero voltage reference point.
(iii) Due to the centre tap transformer, the output voltage rectified by each diode is only one-half of the total secondary voltage.
Working:
During positive half cycle :
(i) When the positive half cycle of the ac input signal passes through the circuit, terminal M is positive, G is at zero potential and N is at negative potential.
(ii) This forward biases diode D₁ and reverse biases diode D₂.
(iii) Hence, being forward biased, diode D₁ conducts and current flows along the path MD₁AGC.
During negative half cycle:
(i) When the negative half cycle of the AC input signal passes through the circuit, terminal N becomes positive, C is at zero potential and M is at negative potential.
(ii) This forward biases diode D₂ and reverse biases diode D₁.
(iii) Hence, being forward biased, diode D₂ conducts and current flows along the path ND₂BGC.
(iii) During both positive and negative half cycles of the input signal, the current flows through the load in same direction.

(iv) The output signal corresponding to the input signal is shown in Figure. Though both half cycles of AC input are rectified, the output is still pulsating in nature.
(v) The efficiency (η) of full wave rectifier is twice that of a half wave rectifier and is found to be 81.2 %.

Given, n = 4/3, d = 10 m.
Radius of illumination, \(R=\cfrac { d }{ \sqrt { { n }^{ 2 }-1 } } \) 
\(R=\cfrac { 10 }{ \sqrt { \left( 4/3 \right) ^{ 2 }- } 1 } =\cfrac { 10\times 3 }{ \sqrt { 16-9 } } \) 
\(R=\cfrac { 30 }{ \sqrt { 7 } } =11.32cm\) 
To find the critical angle,
\({ i }_{ c }={ sin }^{ -1 }\left( \cfrac { 1 }{ n } \right) \) 
\({ i }_{ c }={ sin }^{ -1 }\left( \cfrac { 1 }{ 4/3 } \right) ={ sin }^{ -1 }\left( \cfrac { 3 }{ 4 } \right) =48.6^{ o }\)
The total angle of view of the cone is, \({ 2i }_{ c }=2\times { 48.6 }^{ 0 }={ 97.2 }^{ 0 }\)