According to the superposition principle, the total electric potential at a point is equal to the sum of the potentials due to each charge at that point.
Consider the point at which the total potential zero is located at a distance x from the charge +q as shown in the figure.

The total electric potential at P is zero.
V_tot = \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { q }{ x } -\frac { 2q }{ (9-x) } \right) \)=0
Which gives \(\frac { q }{ x } -\frac { 2q }{ (9-x) } \)
or \(\frac { 1 }{ x } =\frac { 2 }{ (9-x) } \)
Hence, x = 3m

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, \(y_{n}=n \frac{\lambda D}{d}\)  
\(y_{n}=3 \times \frac{560 \times 10^{-9} \times 1}{3 \times 10^{-3}}=560 \times 10^{-6} \mathrm{~m}\) 
\(y_{n}=0.560 \times 10^{-3} \mathrm{~m}=0.560 \mathrm{~mm}\) 

\(\mathrm{n}_{1} \lambda_{1}=\mathrm{n}_{2} \lambda_{2} \)
\(\mathrm{n}_{1}=62 \text { fringes } \)
\(\lambda_{1}=5893 \mathrm{~A}^{0} \)
\(\lambda_{2}=4359 \mathrm{~A}^{0} \)
\(\mathrm{n}_{2}=? \)
\(\mathrm{n}_{2}=\frac{\mathrm{n}_{1} \lambda_{1}}{\lambda_{2}} \)
\(=\frac{5893 \times 62}{4359} \)
= 83.81 = 84 fringes.

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

μ_r = 1 (air core)
Length of two solenoids, \(l_1=l_2=80 \times 10^{-2} \mathrm{~m}\)
Cross-sectional area of two solenoids, \(A_1=A_2=5 \times 10^{-4} \mathrm{~m}^2\)
Number of turns in the first coll, N₁=1200 turns
Number of turns in the second coil, N₂ = 400 turns
\(n_1=\frac{N_1}{l}=\frac{1200}{80 \times 10^{-2}}=15 \times 10^{-2}=1500  \)
\(n_2=\frac{N_2}{l}=\frac{400}{80 \times 10^{-2}}=500\)
Mutual inductance, \(M= μ_0μ_r n_1 n_2 A_2 l\)
\(M  =4 \times 3.14 \times 10^{-7} \times 1 \times 1500 \times 500 \times 5 \times 10^{-4} \times 80 \times 10^{-2}  \)
\(=12.56 \times 10^{-7} \times 75 \times 10^4 \times 5 \times 10^{-4} \times 80 \times 10^{-2}  \)
\(=3.76,800 \times 10^{-9}=0.376 \times 10^{-3} \mathrm{H}=0.376 \mathrm{mH}  \)
\(=0.38 \mathrm{mH}  \)
\(\therefore \text {Mutual inductance } =0.38 \mathrm{mH}\)

Power of the pulse P = 60 x 10^-3 W
Time internal t = 10^-6 S
Energy U = Power x time
U = p x t
= 60 x 10^-3 x 10^-6
U = 60 x 10^-9 J
Lineral momentum, \(\mathrm{P}=\frac{\text { Energy }}{\text { speed }}=\frac{U}{C} \ \)
\(\mathrm{P}=\frac{60 \times 10^{-9}}{3 \times 10^{8}} \)
P = 20 x 10^-17 kg ms^-1

A = 70 x 10^-4m²; N = 600 turns
B = 0.4 Wbm^-2; f = 500 rpm
The instantaneous emf is
ε = ε_msinωt
since \({ \epsilon }_{ m }=N{ \Phi }_{ m }\omega =N(BA)(2\pi f)\)
ε = NBA x 2\(\pi\)f x sinωt
(i) When ωt = 0^o,
ε = ε_m sin0 = 0
(ii) When ωt = 90^o,
ε = ε_msin90^o = NBA x 2\(\pi\)f x 1
= 600 x 0.4 x 70 x 10^-4 x 2 x \(\frac{22}{7}\times(\frac{500}{60})\)
= 88V
(iii) When ωt = 90° – 60° = 30°,
ε = ε_m sin30^o = 88 x \(\frac{1}{2}\) = 44V

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.

To check which bulb will be fused, the voltage drop across each bulb has to be calculated.
The resistance of a bulb,
\(R=\frac { V^{ 2 } }{ P } =\frac { { (Ratedvoltage) }^{ 2 } }{ Ratedpower } \)
For 20W - 220V bulb,
\({ R }_{ 1 }=\frac { { \left( 220 \right) }^{ 2 } }{ 20 } \Omega =2420\Omega \)
For 100W - 220V bulb,
\({ R }_{ 2 }=\frac { { \left( 220 \right) }^{ 2 } }{ 20 } \Omega =484\Omega \)
Both the bulbs are connected in series. So same current will pass through both the bulbs. The current that passes through the circuit, \(I=\frac { V }{ { R }_{ tot } } \)
R_tot (R₁ + R₂)
R_tot = (484 + 2420) \(\Omega\) = 2904 \(\Omega\)
\(I=\frac { 440V }{ 2904\Omega } \approx 0.151A\)
The voltage drop across the 20W bulb is
\(V_1=IR_1=\frac { 440V }{ 2904 }\times2420 \approx 366.6V\)
The voltage drop across the 100W bulb is
\({ V }_{ 2 }=I{ R }_{ 2 }=\frac { 440 }{ 2904 } 484\approx 73.3A\)
The 20 W bulb will get fused because the voltage across it is more than the voltage rating.

\(\lambda_{1}=3310 Å=3310 \times 10^{-10} \mathrm{~m} ; \mathrm{E}_{1}=3 \times 10^{-19} \mathrm{~J} \)
\(\lambda_{2}=5000 Å=5000 \times 10^{-10} \mathrm{~m} ; \mathrm{E}_{2}=0.972 \times 10^{-19} \mathrm{~J} \)
\(\mathrm{E}=\mathrm{E}_{1}-\mathrm{E}_{2}=2.028 \times 10^{-19} J\)
\(\mathrm{hc}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)=\mathrm{E}_{1}-\mathrm{E}_{2} \)
\(\frac{\mathrm{h} \times 3 \times 10^{8}}{10^{-10}}\left(\frac{1}{3310}-\frac{1}{5000}\right)=2.028 \times 10^{-19} \)
\(\mathrm{~h}=\frac{2.028 \times 10^{-19} \times 10^{-10} \times 3310 \times 5000}{3 \times 10^{8} \times 1690}=6.62 \times 10^{-34} \mathrm{Js} \)
\(\phi_{0} =\frac{\mathrm{hc}}{\lambda}-\mathrm{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{3310 \times 10^{-10}}-3 \times 10^{-19} \)
\(=(6-3) \times 10^{-19}=3 \times 10^{-19} \mathrm{~J} \)
\(\phi_{0} =3 \times 10^{-19} \mathrm{~J} \)
Threshold Wavelength,
\(\lambda_{0}=\frac{\mathrm{hc}}{\phi_{0}}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{3 \times 10^{-19}}=6.62 \times 10^{-7} \mathrm{~m} \)
\(\lambda_{0}=6620 \stackrel {o}{A}\)

V = 81 V
∴ K = eV = 1.6 x 10^-19 x 81 = 1.296 x 10^-17 = 1.3 x 10^-17 J
\({ V }=\sqrt \frac {2K}{m}\sqrt { \cfrac { 2\times 1.3\times { 10 }^{ -17 } }{ 9.1\times { 10 }^{ -31 } } } =5.345 \times 10^6 ms^{-1} \) 
v = 5.345 x 10⁶ m s^-1, K = 1.3 x 10^-17 J

Step 1: AC (1 + B) = AC.1 [OR law-2]
Step 2: AC . 1 = AC [AND law – 2]
Therefore, AC + ABC = AC
Thus the Boolean identity is proved.
Circuit Description

From example, we found that the BE of  \(_{ 2 }^{ 4 }{ He }\) = 28.33 Mev
Binding energy per nucleon = \(\overline{B \cdot E}\) = \(28.33 \mathrm{MeV} / 4 \simeq 7 \mathrm{MeV}\).

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 \(\overset { \rightarrow }{ B } =\overset { \rightarrow }{ 0 } \).

L = 400 x 10^-3 H; I_eff = 6 x 10^-3A
f = 1000 Hz
Inductive reactance, X_L= L\(\omega\) = L x 2\(\pi\)f
= 2 x 3.14 x 1000 x 0.4
= 2512 Ω
Voltage across L,
V = I X _L = 6 x 10^-3 x 2512
V = 15.072 V (RMS)