λ = 500 nm = 500 x 10^-9 m; m = 4;
θ = 30°; number of lines per cm = ?
Equation for diffraction maximum in grating is, sin θ = Nm λ
Rewriting, \(N=\cfrac { sin\theta }{ m\lambda } \) 
Substituting,
\(N=\frac{0.5}{4 \times 500 \times 10^{-9}}\) 
= 2.5 x 10⁵ m^-1
= 2.5 x 10³ cm^-1

⋋₂ = 600 x 10^-9 m, ⇒ d = 1 x 10^-3 m, D =2m
n⋋ = dsinθ
For first dark fringe
n = 1 For minimum
d sinθ = ⋋, here
CO = Nc (approximately)

From Fig, sinθ \(=\frac{x/2}{D} =x/2D\)
\(sin \theta =\frac{ \lambda}{d} \)
\(\lambda/d=x/2D\)
\(\therefore x =\frac{2D\lambda}{d}=\frac{2 \times 2 \times 600\times 10^{-9}}{ 10^{-3}} \)
\(x=2.4 \times10^{-3}m =2.4\mathrm{~mm} \)

Binding energy BE = [Zm_H + Nm_n - M_A]c²
For helium nucleus, Z = 2, N = A–Z = 4–2 = 2
Mass defect
Δm = [(2 x 1.00785u) + (2 x 1.008665 u) -4.00260u] Δm = 0.03043 u
B.E = 0.03043u x c²
B.E = 0.03043 x 931 MeV = 28.33 MeV
[∵ luc² = 931 MeV]
The binding energy of the \(_{ 2 }^{ 4 }{ He }\) nucleus is 28.33 MeV.

Linear mass density of the conductor is = 0.2 g/m
Mass per unit length \(\frac{M}{l}=0.2 \times 10^{-3} \mathrm{~kg} / \mathrm{m}\)
Magnetic field B = 1T.
Acceleration due to gravity, g = 10 ms^-2
Force \(=\frac{m}{l} \times g\)
= 0.2 x 10^-3 x 10 = 0.2 x 10^-2
F = 2 x 10^-3 N   ....(1)
If the coil is placed in the magnetic field then the force acting on the coil is
F= BIl   ....(2)
From the equation (1) and (2) we get
BIl = 2 x 10^-3
∴ 1 x L x I = 2 x 10^-3
∴ I = 2 x 10^-3 A      [∴ l = 1m]
∴ I = 2mA

Angle, \( \theta=60^{\circ}\)
Peak value of current, \(I_{P_m}=20 \mathrm{~A}\)
(i) Instantaneous value of current at 60^o,
\(i  =I_m \sin \theta  \)
\(=20 \times \sin 60^{\circ}=\frac{20 \times \sqrt{3}}{2}  \)
\(=10 \times 1.732  \)
\(i  =17.32 \mathrm{~A}\)
(ii) Average value of current,
\(I_{\mathrm{av}}=0.637 I_m  \)
\(I_{\mathrm{av}}=0.637 \times 20  \)
\(I_{\mathrm{av}}=12.74 \mathrm{~A}\)
(iii) RMS value of current,
\(I_{\mathrm{rms}}=0.707 I_m=0.707 \times 20  \)
\(I_{\mathrm{rms}}=14.14 \mathrm{~A}\)

Area of a rectangular coil, \(A=6 \times 10^{-4} \mathrm{~m}^2, Resistance \ \mathrm{R}=35 \Omega\)
Number of turns of the coil, N = 3500 turns
Magnetic field, B = 0.4 T
Angle of orientation, \(\theta=\pi-\frac{\pi}{2}=\frac{\pi}{2} \mathrm{rad}=90^{\circ}\)
Induced emf, e=N A B sin θ
\(e=3500 \times 6 \times 10^{-4} \times 0.4 \times \sin 90^{\circ}\)
\(=3500 \times 2.4 \times 10^{-4} \times 1  \)
\(=840 \times 10^{-3} \mathrm{V} \)
\(I=\frac{e}{R}=\frac{1440 \times 10^{-1}}{35}  \)
\(I=24 \times 10^{-3} A\)
Charge, Q = It
\(=24 \times 10^{-3} \times 2  \)
\(=48 \times 10^{-3} \mathrm{C} \)
\(\therefore Charge Q=48 \times 10^{-3} \mathrm{C}\)

C_p = C_b + C_c
\(C_{P}=6+2=8 \mu \mathrm{F} \)
\(\frac{1}{C_{s}}=\frac{1}{C_a}+\frac{1}{C_p}=\frac{1}{C_d} \)
\(C_{s}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8} \)
\(\therefore C_{s}=\frac{8}{3} \mu \mathrm{F} \)
Total capacitance \(C_{s}=\frac{8}{3} \times 10^{-6} \mathrm{~F} \)
Total Charge, \(Q=C_{s} V=\frac{8}{3} \times 10^{-6} \times 9 \)
\(Q_{a}=24 \mu C\)
(a) Charge on capacity \(Q_{a}=24 \mu C\)         .....(1)
Charge on capacitor \(Q_{b}=24 \times \frac{6}{8}=18 \mu \mathrm{C} \)             ..............(2)
Charge on capacitor  \(Q_{c}=24 \times \frac{2}{8}=6 \mu \mathrm{C} \)              ..............(3)
Charge on capacitor \(Q_{d}=24 \times \frac{8}{8}=24 \mu \mathrm{C} \)              ..............(4)
(b) Potential difference across \(C_{a} \ is\ V_{a}=\frac{Q_{a}}{C_{a}} \)
\(=\frac{24}{8}=3 \mathrm{~V} \)      ...(5)
Potential difference across \(C_{b}\ is \ V_{b}=\frac{Q_{b}}{C_{b}} \)
\(=\frac{18}{6}=3 \mathbf{V}\)       ........(6)
Potential difference across \(C_{c}\ is \ V_{c}=\frac{Q_{c}}{C_{c}}=\frac{6}{2}=3 \mathrm{~V} \)       ......(7)
Potential difference across \(C_{d}\ is \ V_{d}=\frac{Q_{d}}{C_{d}} \)
\(=\frac{24}{8}=3 \mathbf{V} \)          ....(8)
(c) Energy stored in each capacitor \(U=\frac{1}{2} C V^{2}\)
Energy stored in \(\mathrm{C}_{\mathrm{a}} \text { is } U_{a}=\frac{1}{2} C_{a} V_{a}^{2}\)
\(U_{\mathrm{a}}=\frac{1}{2} \times 8 \times 10^{-6} \times 3 \times 3=36 \mu \mathrm{J}\)   ....(9)
Energy stored in \(C_{b}\ is \ U_{b}=\frac{1}{2} C_{b} V_{b}^{2} \)
\(U=\frac{1}{2} \times 6 \times 10^{-6} \times 3 \times 3 \)
\(=27 \mu \mathrm{J} \)         ........(10)
Energy stored in C_e is \(U_{c} =\frac{1}{2} C_{c} V_{c}^{2} \)
\(=\frac{1}{2} \times 2 \times 10^{-6} \times 3 \times 3=9 \mu \mathrm{J} \)          ......(11)
Energy stored in C_d is \(U_{d} =\frac{1}{2} C_{d} V_{d}^{2} \)
\(U_d=\frac{1}{2} \times 8 \times 10^{-6} \times 3 \times 3 \)
\(=36 \times 10^{-6} \mathrm{~J}=36 \mu \mathrm{J} \)      .........(12)

Circuit Diagram:
Here, 2 cells are in series,
\(\therefore \varepsilon_{\mathrm{tot}} =\varepsilon+\varepsilon=2 \varepsilon  \)
\(\varepsilon_{\mathrm{tot}}  =10 \mathrm{~V}\)
Here 4, 6 and 12 are in parallel
\(\therefore \frac{1}{\mathrm{R}_{\mathrm{p}}}  =\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_1}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}  \)
\(\mathrm{R}_{\mathrm{p}}  =2 \Omega\)
Now, the circuit becomes,
(i) current drawn from the cell (through the circuit) is,
\(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{s}}}=\frac{10}{8+2}=1 \mathrm{~A}\)
Potential drop across the parallel combination of 3 resistors is \(\mathrm{V}^{\prime}=1 R_P=1 \times 2=2 \mathrm{~V}\)
(ii) Current through 8 resistor is I =1 A
\((\because 8 \Omega, 2 \Omega \text { in series) }\)
Current through \(\mathrm{R}=4 \Omega \ is, \mathrm{I}=\frac{\mathrm{V}^{\prime}}{\mathrm{R}}=\frac{2}{4}=0.5 \mathrm{~A}\)
Current through \(\mathrm{R}=6 \Omega \ is, \mathrm{I}=\frac{\mathrm{V}^1}{\mathrm{R}}=\frac{2}{6}=0.33 \mathrm{~A}\)
Current through \(\mathrm{R}=12 \Omega \ is, \mathrm{I}=\frac{\mathrm{V}^1}{\mathrm{R}}=\frac{2}{12}=0.17 \mathrm{~A}\)

R = 10 Ω, I = 5 A, t = 5 minutes = 5 x 60 s
H = I² R t
= 5² x 10 x 5 x 60
= 25 x 10 x 300
= 25 x 3000
= 75000 J (or) 75 kJ

The given values I = 3.93 A, ξ = 12 V,
R = 3 Ω
(a) The terminal voltage of the battery is equal to voltage drop across the resistor
V = IR = 3.93 x 3 = 11.79 V
The internal resistance of the battery,
\(r=\left[ \frac { \xi -V }{ V } \right] R=\left[ \frac { 12-11.79 }{ 11.79 } \right] \times 3=0.05\Omega \)
(b) The power delivered by the battery P = Iξ = 3.93 x 12 = 47.1 W
The power delivered to the resistor = I² R = 46.3 W
The remaining power = (47.1 – 46.3) = 0.8 W is delivered to the internal resistance and cannot be used to do useful work. (it is equal to I² r).

When the water molecules are aligned in the direction of the electric field, it has minimum potential energy. The work done to rotate the dipole from θ = 0° to 90° is equal to the potential energy difference between these two configurations.
W = ΔU = U(90°) - U(0°)
From the equation U =−pE cosθ = −\(\hat p.\hat E\) ,
we write U = − pE cosθ, Next, we calculate the work done to rotate one water molecule from θ = 0° to 90°.
For one water molecule
W = - pE cos90^o + pE cos0^o  = pE
W=  6.3 x 10^-30 x 3 x 10⁵ = 18.9 x 10^-25J
For 10²² water molecules, the total work done is
W_tot = 18.9 x 10^-25 x 10²² = 18.9 x 10^-3J

\(\text { K.E }=81.9 \times 10^{-15} \mathrm{~J} \)
\(\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}=\frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-3} \times 1836 \times 81.9 \times 10^{-15}}} \)
\(\lambda =\mathbf{4 . 0 0} \times 10^{-14} \mathrm{~m} \)

\(v=10^{14} \mathrm{~Hz} ; \mathrm{E}=19.86 \mathrm{~J} \)
\(E=nhv \Rightarrow n=\frac{E}{hv}\)
\(=\frac{19.86}{6.626 \times 10^{-34} \times 10^{14}}=2.99 \times 10^{20} \)
\(\mathrm{n} \simeq 3 \times 10^{20} \)

f_o = 125 cm; f_e = 2 cm; m = ? L = ?; θ_i = ?
(a) Equation for magnification of telescope,
\(m=\cfrac { { f }_{ o } }{ { f }_{ e } } \) 
Substituting, \(m=\cfrac { 125 }{ 2 } =62.5\) 
(b) Equation for approximate length of telescope, L = f_o+ f_e
Substituting, L = 125 + 2 = 127 cm = 1.27 m
(c) Equation for angular magnification,\(m=\cfrac { { \theta }_{ 1 } }{ { \theta }_{ 0 } } \) 
Rewriting, \({ \theta }_{ 1 }=m\times { \theta }_{ 0 }\) 
Substituting,
\({ \theta }_{ i }=62.5\times 1'=62.5'=\cfrac { 62.5 }{ 60 } =1.04^{ o }\) = 1^o2'30''

Resonant frequency, \({ f }=\frac { 1 }{ 2\pi \sqrt { LC } } \)
On simplifying, we get C = \(\frac { 1 }{ { 4\pi }^{ 2 }{ { f }^{ 2 } }L } \)
i) When frequency = 500 kHz,
\(C=\frac { 1 }{ 4\times { 3.14 }^{ 2 }\times { (500\times { 10 }^{ 3 }) }^{ 2 }\times 150\times { 10 }^{ -6 } } \)
= 676 pF
ii) When frequency = 1500 kHz,
\(C=\frac { 1 }{ 4\times { 3.14 }^{ 2 }\times { (1500\times { 10 }^{ 3 }) }^{ 2 }\times 150\times { 10 }^{ -6 } } \)
= 75 pF
Therefore, the capacitor range is 75 to 676 pF.

λ₁ = 560 nm = 560 x 10^-9m;
λ₁ = 420 nm = 420 x 10^-9m;
D = 1 m; d = 3 mm = 3 x 10^-3m
Here, n and λ are inversely proportional for a given y.
Here, n^th order bright fringe of longer wavelength λ₁ coincides with (n+1)^th order bright fringe of shorter wavelength λ₂.
Equation for n^th bright fringe is, \({ y }_{ n }=n\cfrac { \lambda D }{ d } \) 
Here, \(n\cfrac { { \lambda }_{ 1 }D }{ d } =(n+1)\cfrac { { \lambda }_{ 2 }D }{ d } \) (as λ₁>λ₂)
\({ n\lambda }_{ 1 }=\left( n+1 \right) { \lambda }_{ 2 }\) (or) \(\cfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\cfrac { (n+1) }{ n } ;1+\frac{1}{n}=\frac{\lambda_1}{\lambda_2}\) 
\(1+\cfrac { 1 }{ n } =\cfrac { 560\times { 10 }^{ -9 } }{ 420\times { 10 }^{ -9 } } \) (or) \(1+\cfrac { 1 }{ n } =\cfrac { 4 }{ 3 } \) 
\(\frac{1}{n}=\frac{1}{3}\) (or) n = 3
Thus, the 3^rd bright fringe of λ₁ and the 4^th bright fringe of λ₂ coincide at the least distance y.
The least distance from the central fringe where the bright fringes of the two wavelengths coincide is, \(y_{n}=n \frac{\lambda D}{d}\)  
\(y_{n}=3 \times \frac{560 \times 10^{-9} \times 1}{3 \times 10^{-3}}=560 \times 10^{-6} \mathrm{~m}\) 
\(y_{n}=0.560 \times 10^{-3} \mathrm{~m}=0.560 \mathrm{~mm}\) 

Density \(\rho=2.3 \times 10^{17} \mathrm{kgm}^{-3}\), Mass M = 5.97 x 10²⁴ kg
\(\rho=\frac{M}{V}=\frac{M}{\frac{4}{3}\pi R^3}\)
\(R=\left[\frac{M}{\frac{4}{3} \pi \rho}\right]^{\frac{1}{3}}=\left[\frac{3 \mathrm{M}}{4 \pi \rho}\right]^{\frac{1}{3}}=\left[\frac{3 \times 5.97 \times 10^{24}}{4 \times 3.14 \times 2.3 \times 10^{17}}\right]^{\frac{1}{3}}=\left[0.62 \times 10^{7}\right]^{\frac{1}{3}}\)
R = 183.7 m
R ≈180 m

Charge of an electron q = –1.60 × 10^–19 C ⇒ |q| = 1 60 x 10^-19 C
Magnitude of magnetic field B = 0.500 T
Mass of the electron, m = 9.11 × 10^–31 kg
Radius of the orbit, r = 2.50 mm = 2.50 × 10^–3 m
Speed of the electron,  V = \(q \frac{\mathrm{rB}}{\mathrm{m}}\)
\( v = 1.60 \times 10^{-19} \times\frac{ 2.50 \times 10^{-3} \times 0.500}{9.11 \times 10^{-31}}\)
\(v=2.195 \times 10^8 \mathrm{~m} \mathrm{s} ^{-1}\)

V_p = 12 V; V_s = 240 V
I_s = 50 mA; N_p = 100 turns
\(\frac { { V }_{ s } }{ { V }_{ P } } =\frac { { N }_{ s } }{ { N }_{ p } } =\frac { { I }_{ P } }{ { I }_{ S } } =K\)
Transformation ratio, K = \(\frac{240}{12}=20\)
The number of turns in the secondary
N_S = N_P x K = 100 x 20 = 2000
Primary current,
I_P = K x I_s = 20 x 50 mA = 1 A

Frequency of the electromagnetic waves given, f = 2450 MHz
The corresponding angular frequency is
ω = 2πf = 2 x 3.14 x 2450 x 10⁶
= 15,386 x 10⁶ Hz
= 1.54 x 10¹⁰ s^-1
The required magnetic field, B = \(\frac { { m }_{ e }\omega }{ |q| } \)
Mass of the electron, m_e = 9.11 x 10^-31 kg
Charge of the electron,
q = -1.60 x 10^-19C
⇒ |q| = 1.60 x 10^-19 C
B = \(\frac { (9.11\times { 10 }^{ -31 })(1.54\times 10^{ 10 }) }{ (1.60\times 10^{ -19 }) } \)  = 8.7683 x 10^-2T
B = 0.08768 T
This magnetic field can be easily produced with a permanent magnet. So, electromagnetic waves of frequency 2450 MHz can be used for heating and cooking food because they are strongly absorbed by water molecules.

(a) The capacitance of the capacitor in the presence of dielectric is
C = \(\frac { { \varepsilon }_{ r }{ \varepsilon }_{ 0 }A }{ d } =\frac { 5\times 8.85\times 10^{ -12 }\times 6 \times 10^{-4}}{ 6\times 10^{ -3 } } \)
= 44.25 x 10^-13F = 4.425 pF
The stored charge is
Q = CV = 44.25 x 10^-13 x 10
= 442.5 x 10^-13C = 44.25pC
The stored energy is
\(U=\frac { 1 }{ 2 } \) CV² = \(\frac { 1 }{ 2 } \) x 44.25 x 10^-13 x 100
= 2.21 x 10^-10 J
(b) After the removal of the dielectric, since the battery is already disconnected the total charge will not change. But the potential difference between the plates increases. As a result, the capacitance is decreased.
New capacitance is
C₀=\(\frac { C }{ { \varepsilon }_{ r } } =\frac { 44.25\times 10^{ -12 } }{ 5 } \)
= 0.885 x 10^-12 F = 0.885 pF
The stored charge remains same and 44.25 pC. Hence newly stored energy is
U₀ =\(\frac { { Q }^{ 2 } }{ 2{ C }_{ 0 } } =\frac { { Q }^{ 2 }{ \varepsilon }_{ r } }{ 2C } =\varepsilon _{ r }U\)
= 5 x 2.21 x 10^-10J = 11.05 x 10^-10J
The increased energy is ΔU = (11.05 - 2.21) x 10^-10 J  = 8.84 x 10^-10 J
When the dielectric is removed, it experiences an inward pulling force due to the plates. To remove the dielectric, an external agency has to do work on the dielectric which is stored as additional energy. This is the source for the extra energy 8.84 x 10^–10 J.

\(\frac { P }{ Q } =\frac { R }{ S } \)
\(S=\frac { Q }{ P } \times R\)
\(S=\frac { 1000 }{ 100 } \times 40S=400\Omega \)

Charge Q = Current x Time interval
= I x t
At t = 0 s,  
dq = dI x t
= 5 x 0
dq = 0 C
At t = 2 s,    
dg = dI x t
=5 x 2
dq = 10 C
At t = 5 s,
dg = dl x t
=0 x 5
dq = 0 C
At t= 0 s, dg = 0 C; At t = 2 s, dg = 10 C; At t= 5 s, dg = 0 C.