\(\frac { P }{ Q } =\frac { R }{ S } \), when the network is balanced
\(\frac { 500 }{ 800 } =\frac { x+400 }{ 1000 } \)
\(x+400=\frac { 5 }{ 8 } \times 1000\)
x + 400 = 625 
x = 625 – 400
x = 225 Ω

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.

Charge q = 50 nC = 50 x 10^-9 C , e = 1.6 x 10^-19 C
No. of electrons \(n=\frac{q}{e}\)
\(\frac { q }{ e } =\frac { 50\times { 10 }^{ -9 } }{ 1.6\times { 10 }^{ -19 } } \)
=  31.25 x 10¹⁰ electrons
To produce 50 nC charge, number of electrons transferred = 31.25 x 10¹⁰ electrons

Case I : Current from a cell is I₁ = 0.9 A
Resistance R₁ = 2\(\Omega \)
Case II : Current flowing I₂ = 0.3 A
Resistance R₂ = 7\(\Omega \)
E = I(R + r) = IR + Ir
\(\therefore r=\frac{E}{I}-R\)
By applying for two cases we get,
\(\frac{E}{0.9}-2=\frac{E}{0.3}-7 \)
\(\frac{E}{0.9}-\frac{E}{0.3}=7+2=-5 \)
\(-\frac{2 E}{0.9}=-5 \)
\(E =\frac{5 \times 0.9}{2}=\frac{4.5}{2}=2.25 V \)
∴ Internal resistance, \(r=\frac{E}{I}-R\)
Where I = 0.3 A, R = 7\(\Omega\), E = 2.25 V
\(\therefore r=\frac{2.25}{0.3}-7=\frac{22.5}{3}-7\)
\(r=7.5-7=0.5 \Omega \)
∴ Internal resistance = 0.5 \(\Omega\)

(i) Wavelength \(\lambda =\frac { c }{ \gamma } =\frac { 3\times { 10 }^{ 8 } }{ 1.5\times { 10 }^{ 10 } } =2\times { 10 }^{ -2 }m\)
(ii) \(B=\frac { E }{ c } =\frac { 36 }{ 3\times { 1 }0^{ 8 } } =12\times { 10 }^{ -8 }T\)
Formula: (or) 1.2 x 10^-7T
Average energy of magnetic field \(\overrightarrow { E } \) \({ U }_{ E }=\frac { 1 }{ 2 } .{ \varepsilon }_{ 0 }{ E }^{ 2 }\)
The average energy density of electric field \(\overrightarrow { B } \) \({ U }_{ E }=\frac { 1 }{ 2{ \mu }_{ 0 } } .{ B }^{ 2 }\)
But E = CB & C² = \(\frac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } } \)
\({ U }_{ E }=\frac { 1 }{ 2 } .{ \varepsilon }_{ 0 }{ E }^{ 2 }=\frac { 1 }{ 2 } .{ \varepsilon }_{ 0 }{ (CB) }^{ 2 }\)
\({ U }_{ E }=\frac { 1 }{ 2 } .{ \varepsilon }_{ 0 }.\frac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } } { B }^{ 2 }=\frac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } } { B }^{ 2 }={ U }_{ B }\)
\(\therefore { U }_{ E }={ U }_{ B }\)