\(y=\overline{A}=\overline{1011}=0100\)
The Boolean expression for NOT gate, \(y=\bar{A}\)

\(\frac{BE}{A}=\frac{\Delta\times931MeV}{V}=\frac{0.042 \times931}{7}\)
= 5.586 = 5.6 MeV

\(\lambda_0  =\frac{h c}{\phi}  \)
\(=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.313 \times 1.6 \times 10^{-19}}  \)
\( =\frac{19.8782400}{5.3} \times 10^{-7}  \)
\(\lambda_0  =3.750 \times 10^{-7} \simeq 3750 \stackrel{o}A\)

d' = 2d, β' = β, D' = ?
W.K.T, Fringe width
\(\beta = \frac{D\lambda}{d} \Rightarrow D' = \frac{Dd'}{d}\)
\(D' = \frac{D2d}{d}=2D\)

Apparent depth = 3 + 5 = 8 cm
Real depth = thickness of the slab = t
n = 1.5
\(n=\frac{Real \ depth}{Apparent \ depth}\)
\(\therefore 1.5=\frac{t}{8}\)
t = 1.5 x 8
t = 12 cm

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

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

The charge + q will be stable between B₁ and B₂ with respect to the displacement.

R_t = R_o [1 + α(T₂ - T₁)
2 = 1[1 + 0.00125(T₂ - 293)
I = 5/4 x 10^-3 (T₂ - 293)
T₂ = 1093 K
T₂ = 820 ^oC

Total power = 15 x 40 + 5 x 100 + 5 x 80 + 1000
= 600 + 500 + 400 + 1000
= 2500 W
P = VI, V = 220 V
\(I=\frac{P}{V}=\frac{2500}{220}=11.363 \ A\)
≃ 12 A

\(P=\frac{V^2}{R}=\frac{240 \times 240}{120}=480 \ W\)

Force on particle towards upper plate B, F^-_y = qE
Vertical acceleration of particle ay=\(\frac{qE}{m}\)
Initial vertical velocity V_y = 0
Time taken by particle between the plates t = \(\frac{L}{V_x}\)
From equation of motion s = ut + \(\frac{1}{2}\) at²
Vertical deflection y = 0 + \(\frac{1}{2}\) a_yt²
\(=0+\frac { 1 }{ 2 } \left( \frac { qE }{ m } \right) { \left( \frac { L }{ { V }_{ x } } \right) }^{ 2 }\)
\(y=\frac { q{ EL }^{ 2 } }{ 2m{ V }_{ x }^{ 2 } } \)

Energy density U\(=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }{ E }_{ 2 }\)
But \(E=\frac { \sigma }{ { \varepsilon }_{ 0 } } =\frac { Q }{ { A\varepsilon }_{ 0 } } \)
\(\therefore U=\frac { 1 }{ 2 } .\frac { { \varepsilon }_{ 0 }{ Q }_{ 2 } }{ { A }^{ 2 }{ \varepsilon }_{ 0 } } \Rightarrow U=\frac { { Q }_{ 2 } }{ 2A^{ 2 } } \)
\(U\alpha \frac { 1 }{ { A }^{ 2 } } \Rightarrow { U }_{ A }>{ U }_{ B }\)

The resistanct is defined as the ratio of potential difference across the given conductor to the current passing through the conductor.
\(R=\cfrac { V }{ I } \)

House hold appliances are always connected in parallel so that even if one is switched off, the other devices could function properly.

(i) The dipole moment of dipole \(|\overset { \rightarrow }{ P } |=q\times 2a\)
Force on -q at A = -q \(\overset { \rightarrow }{ E } \)
Force on + q at B = +q\(\overset { \rightarrow }{ E } \)
Net force on the dipole = q\(\overset { \rightarrow }{ E } \)- q\(\overset { \rightarrow }{ E } \) = 0
(ii) Work done on dipole when it is rotated through 180°
W = ∆U = pE (cos θ₁ - cos θ₂)
= pE (cos 0° - cos180^o)
= pE(1 - (-1)
W = 2pE

When an external electric field is applied, the centers of positive and negative charges are separated by a small distance which induces dipole moment in the direction of the external electric field. Then the dielectric is said to be polarized by an external electric field.

Magnetic field B = 1 T
Mass of the proton, m_p = 1.67 x 10⁻²⁷kg
Charge of the proton, q = 1.60 x 10⁻¹⁹C
\(f=\frac { qB }{ { { 2\pi m }_{ p } } } =\frac { \left( 1.60\times { 10 }^{ -19 } \right) \left( 1 \right) }{ 2\left( 3.14 \right) \left( 1.67\times { 10 }^{ -27 } \right) } \)
= 15.3 x 10⁶ Hz = 15.3 MHz

The magnetic field due to current in the upper semicircle and lower semicircle of the circular coil are equal in magnitude but opposite in direction. Hence, the net magnetic field at the center of the loop (at point O) is zero \(\overset { \rightarrow }{ B } =\overset { \rightarrow }{ 0 } \).

(i) Its cost is very high due to scarity of large and flawless calcite crystals
(ii) Due to extraordinary ray passing obliquely through it the emergent ray is always displaced a little to one side.
(iii) The effective field of view is quite limited
(iv) Light emerging out of it is not uniformly plane polarised.

Curves a and b have different intensities but the same stopping potential, so curves 'a' and 'b' have the same frequency but different intensities.

i) The momentum of photon of frequency v is given by,
\(p=\frac { hv }{ c } =\frac { h }{ \lambda } \)
ii) The wavelength of a photon in terms of its momentum is,
\(\lambda =\frac { h }{ p } \)
iii) According to de Broglie, the above equation is completely a general one and this is applicable to material particles as well. Therefore, for a particle of mass m traveling with speed u, the wavelength is given by,
\(\lambda =\frac { h }{ mv } =\frac { h }{ p } \)
iv) This wavelength of the matter waves is known as de Broglie wavelength. This equation relates the wave character (the wavelength \(\lambda\)) and the particle character (the momentum p) through Planck's constant.

(i)  It is the branch of physics incorporated with technology towards the design of circuits using transistors and microchips.
(ii) It depicts the behavior and movement of electrons in a semiconductor, vacuum, or gas.
(iii) Electronics deals with electrical circuits that involve active components such as transistors, diodes, integrated circuits, and sensors, associated with the passive components like resistors, inductors, capacitors, and transformers.

The radius of the nth orbit 
\({ r }_{ n }=\frac { { n }^{ 2 }{ h }^{ 2 }{ \epsilon }_{ 0 } }{ \pi m{ Ze }^{ 2 } } \)
r₁ = 0.53Å
∴ The diameter of H₂ = 2 x r₁
= 0.53 x 2 = 1.06Å

(a) 10⁹ Hz - Radiowaves
Application: For radio & television communication.
(b) 10¹¹ Hz - microwaves
Application: used in Radar for aircraft navigation, (microwave oven for cooking).

Uniform magnetic field B = 0.8 T
Angle of orientation θ = 30°
Torque, ፒ = 0.2 Nm.
(i) We know that torque ፒ = P_mB sinθ
\(0.2=p_\mathrm{m} \times 0.8 \times \sin 30^{\circ} \)
\(0.2=p_\mathrm{m} \times 0.8 \times \frac{1}{2} \)
0.2 = 0.4 p_m
\(p_m=\frac{0.2}{0.4}=0.5 \mathrm{Am}^{2}\)
(ii) Work done by the applied force to move the magnet from stable to unstable position.
\(\mathrm{W} =-p_\mathrm{m}B\left[\cos \theta_{2}-\cos \theta_{1}\right] \)
\(\mathrm{W} =-p_\mathrm{m}B\left(\cos 180^{\circ}-\cos 0^{\circ}\right) \)
\(\mathrm{W} =-p_\mathrm{m}B(-1-1)=2 \mathrm{p_mB} \)
\(\mathrm{W} =2 \times 0.5 \times 0.8=\mathbf{0 . 8} \mathbf{J} \)
Work done by the applied magnetic field are in opposite direction
\(\mathbf{W}_{\text {mag }}=-0.8 \mathrm{J}\)

Since isotopes are singly ionized, they have equal charge which is equal to the charge of an electron, q = - 1.6 x 10^-19 C. Mass of uranium \(_{ 92 }^{ 235 }{ U and _{ 92 }^{ 238 }{ U } }\) are 3.90 x 10^-25 kg and 3.95 x 10^-25 kg respectively. Magnetic field applied, B = 0.500 T. Velocity of the electron is 1.00 x 10⁵ m s^-1, then 
(a) the radius of the path of \(_{ 92 }^{ 235 }{ U }\) is r₂₃₅
\({ r }_{ 235 }=\frac { { m }_{ 235 }v }{ \left| q \right| B } =\frac { 3.90\times { 10 }^{ -25 }\times 1.00\times { 10 }^{ 5 } }{ 1.6\times { 10 }^{ -19 }\times 0.500 } =48.8\times { 10 }^{ -2 }m\)
r₂₃₅ = 48.8cm 
The diameter of the semi-circle due to \(_{ 92 }^{ 235 }{ U\ \ is \ \ { d }_{ 235 }=2{ r }_{ 235 } }\) = 97.6 cm
The radius of the path of _{\(_{ 92 }^{ 238 }{ U\ is\ 2{ r }_{ 238 }\ then}\)}
\({ r }_{ 238 }=\frac { { m }_{ 238 }v }{ \left| q \right| B } =\frac { 3.90\times { 10 }^{ -25 }\times 1.00\times { 10 }^{ 5 } }{ 1.6\times { 10 }^{ -19 }\times 0.500 } =49.4\times { 10 }^{ -2 }m\)
r₂₃₈ = 49.4 cm
The diameter of the semi-circle due to  \(^{ 238 }_{92}{ U\ is \ 2{ r }_{ 238 } \ =98.8 \ cm}\)
Therefore the separation distance between the isotopes is \(\triangle d={ d }_{ 238 }-{ d }_{ 235 }=1.2 \ cm\)
(b) The time taken by each isotope to complete one semi-circular path are
\({ t }_{ 235 }=\frac { \text{ magnitude of the displacement} }{ velocity } \)
\(=\frac { 97.6\times { 10 }^{ -2 } }{ 1.00\times { 10 }^{ 5 } } =9.76\times { 10 }^{ -6 }s=9.76\mu s\)
\({ t }_{ 238 }=\frac { \text{magnitude of the displacement }}{ velocity } \)
\(=\frac { 98.8\times { 10 }^{ -2 } }{ 1.00\times { 10 }^{ 5 } } =9.88\times { 10 }^{ -6 }s=9.88\mu s\) 

(i) The threshold frequency is the frequency of incident light at which kinetic energy of ejected photoelectron is zero.
∴ From fig. threshold frequency,
v₀ = 4.5 x 10¹⁴ Hz
(ii) Work function, W = hv₀
= 6.6 x 10^-34 x 4.5 x 10¹⁴ joule
= \(\frac { 6.6\times { 10 }^{ -34 }\times 4.5\times 10^{ 14 } }{ 1.6\times { 10 }^{ -19 } } \) eV
= 1.85 eV

(a) λ_e =\(\frac { h }{ { p }_{ e } } \) and λ_p=\(\frac { h }{ { p }_{ p } } \), λ_e = λ_p =1.00 nm.
So, \(\frac { { \lambda }_{ e } }{ { \lambda }_{ p } } =\frac { { p }_{ p } }{ { p }_{ e } } =\frac { 1 }{ 1 } \Rightarrow \frac { { p }_{ p } }{ { p }_{ e } } =\frac { 1 }{ 1 } \)  = 1:1
(b) From relation K=\(\frac { 1 }{ 2 } mv^{ 2 }=\frac { { p }^{ 2 } }{ 2m } \)
K_e = \(\\ \frac { { p }_{ e }^{ 2 } }{ 2me } \) and K_p = \(\frac { { p }_{ e }^{ 2 } }{ 2m_{ p } } \)
\(\frac { { K }_{ p } }{ { K }_{ e } } =\frac { { p }_{ p }^{ 2 } }{ 2{ m }_{ p } } \times \frac { 2{ m }_{ e } }{ { p }_{ e }^{ 2 } } =\frac { { m }_{ e } }{ { m }_{ p } } \)
Since m_e <<< m_p, So K_p <<< K_e
\(\frac { { K }_{ p } }{ { K }_{ e } } =\frac { 9.1\times { 10 }^{ -31 } }{ 1.67\times 10^{ -27 } } \)
= 5.4 x 10^-4

de Broglie wavelength associated with a moving charge particle having a KE. 'K can be given as
λ =\(\frac { h }{ p } =\frac { h }{ \sqrt { 2mK } } \)            ...(1)
\(\left[ K=\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { { p }^{ 2 } }{ 2m } \right] \)
The kinetic energy of the electron in any orbit of a hydrogen atom can be given as
K = -E = -\(\left( \frac { 13.6 }{ n^{ 2 } } eV \right) =\frac { 13.6 }{ { n }^{ 2 } } \)         ......(2)
Let K₁ and K₄ be the KE of the electron in ground state and third excited state, where n₁ = 1
showground state and n₂ = 4 shows the third excited state.
Using the concept of equations (1) & (2), we have
\(\frac { { \lambda }_{ 1 } }{ { \lambda }_{ 4 } } =\sqrt { \frac { { K }_{ 4 } }{ { K }_{ 1 } } } =\sqrt { \frac { { n }_{ 1 }^{ 2 } }{ { n }_{ 2 }^{ 2 } } } \)
\(\frac { { \lambda }_{ 1 } }{ { \lambda }_{ 4 } } =\sqrt { \frac { { 1 }^{ 2 } }{ { 4 }^{ 2 } } } =\frac { 1 }{ 4 } \)  ⇒ λ₁ = \(\frac { { \lambda }_{ 4 } }{ 4 } \).
i.e. the wavelength in the ground state will decrease.

\(\frac { 1 }{ \sqrt { V } } \)
\(\lambda =\frac { h }{ \sqrt { 2mqV } } =\frac { h }{ \sqrt { 2mq } } .\frac { 1 }{ \sqrt { V } } \)
This equation represents a straight line of slope \(\frac { h }{ \sqrt { 2mq } } \)
Clearly, the slope is inversely proportional to \(\sqrt{m}\). Since the slope of line A is less than the slope of line B therefore the line. A represents the particle of heavier mass.

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.

v = 2.3 x 10⁶ m/s
r = 0.53Å = 0.53 x 10^-10m
Current, i = \(\frac { e }{ T } =\frac { e }{ \left( \frac { 2\pi r }{ v } \right) } =\frac { ev }{ 2\pi r } \)
i = \(\frac { 1.6\times 10^{ -16 }\times 2.3\times 10^{ 6 } }{ 2\times 3.14\times 0.53\times 10^{ -10 } } \)
∴ Magnetic moment, M = IA
M = I(πr²)
= 1.105 x 10^-3 x 3.14 x (0.53 x 10^-10)²
= 9.75 x 10^-24 Am^-2