\(P=4 \ \mathrm{cal} \mathrm{} \mathrm{cm}^{-2} \mathrm{~min}^{-1}=4 \times 4.2=16.8 \mathrm{~J} \mathrm{~cm}^{-2} \mathrm{~min}^{-1} \)
\(E=\frac{h c}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{5500 \times 10^{-10}}=3.6 \times 10^{-19} \mathrm{~J} \)
\(n=\frac{E}{hv}=\frac{16.8}{3.6 \times 10^{-19}}=4.67 \times 10^{19} \)
\(n=4.67 \times 10^{19} \)  per cm² per minute.

The maximum distance the person could see is, x = 1.8 m.
The lens should have a focal length of,
f = –x m = –1.8 m.
It is a concave (or) diverging lens.
The power of the lens is,
\(\\ \\ \\ \\ P=-\cfrac { 1 }{ 1.8m } =-0.56D\)

(a) As the intensity of the unpolarised light falling on the first polaroid is I, the intensity of polarized light emerging from it will be \({ I }_{ o }=\left( \cfrac { 1 }{ 2 } \right) \). Let I' be the intensity of light emerging from the second polaroid.
Malus’ law, I' = I_o cos²θ
Here θ is 90^o as the transmission axes are perpendicular to each other.
Substituting,
\(I'=\left( \cfrac { 1 }{ 2 } \right) { cos }^{ 2 }\left( { 90 }^{ o } \right) =0\left[ \therefore cos\left( { 90 }^{ o } \right) =0 \right] \) 
No light comes out from the second polaroid
(b) Let the first polaroid be P₁ and the second polaroid be P₂. They are oriented at 90^o. The third polaroid P₃ is introduced between them at 45^o. Let ′I be the intensity of light emerging from P₃.
Angle between P₁ and P₃ is 45^o. The intensity of light coming out from P₃ is, I' = I_o cos²θ 
Substituting,
\(I'=\left( \cfrac { 1 }{ 2 } \right) { cos }^{ 2 }\left( { 45 }^{ o } \right) =\left( \cfrac { 1 }{ 2 } \right) \left( \cfrac { 1 }{ \sqrt { 2 } } \right) ^{ 2 }=\cfrac { 1 }{ 4' } ;I'\cfrac { I }{ 4 } \)
Finally, the light has to pass through P₂. Angle between P₃ and P₂ is 45^o. Let I″ is the intensity of light coming out from P₂ I''= I'os² θ
Here, I' is the intensity of polarized light existing between P₃ and P₂. I'= \(\cfrac { 1 }{ 4 } \)
Substituting,
\({ I }^{ n }=\left( \cfrac { 1 }{ 4 } \right) { cos }^{ 2 }\left( { 45 }^{ o } \right) =\left( \cfrac { 1 }{ 4 } \right) \left( \cfrac { 1 }{ \sqrt { 2 } } \right)^2 =\cfrac { 1 }{ 8 } \) 
\(I^{ n }=\cfrac { 1 }{ 8 } \) 

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

\(\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{17.5}{100} \),\(\mathrm{~T}_{1 / 2}=5730 \text { Years } \)
Age of the sample (t) = ?
Age of sample, \(t=\frac{2.303 \times \log \left(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\right) \times \mathrm{T}_{1 / 2}}{0.6931}\)
\(t=\frac{2.303 \times \log \left(\frac{100}{17.5}\right) \times 5730}{0.6931} \)
\(t=\frac{2.303 \times \log (5.714) \times 5730}{0.6931} \)
\(t=\frac{2.303 \times 0.7570 \times 5730}{0.6931} \)
t = 14410 years
t = 1.44 x 10⁴ years

Kinetic energies,
KE_Ai = 0
KE_Af = KE_Bt
It should obey law of conservation of energy
Total Initial KE = Total final KE
\(\mathrm{KE}_{\mathrm{A}_{i}}+\mathrm{KE}_{\mathrm{B}_{i}}=\mathrm{KE}_{\mathrm{A}_{\mathrm{t}}}+\mathrm{KE}_{\mathrm{B}_{\mathrm{t}}} \)
\(0+\mathrm{KE}_{\mathrm{B}_{i}}=\mathrm{KE}_{\mathrm{A}_{\mathrm{f}}}+\mathrm{KE}_{\mathrm{A}_{t}} \)
\(\mathrm{KE}_{\mathrm{B}_{\mathrm{i}}}=2 \mathrm{KE}_{\mathrm{A}_{\mathrm{f}}} \)
Minimum energy required for excite the hydrogen atom is 10.2eV
Hence minimum Kinetic energy of H_B is
\(\mathrm{KE}_{\mathrm{B}_{1}}=2 \times 10.2 \mathrm{eV}=20.4 \mathrm{eV}\)

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,
\(I_{\mathrm{rms}}=0.707 I_m=0.707 \times 20  \)
\(I_{\mathrm{rms}}=14.14 \mathrm{~A}\)

Relative permeability of air-core solenoid is μ_r = 1
Length of the solenoid, I = 2 m, Number of turns, N = 4000
Diameter of the solenoid, d = 4 x 10^-2m
Area of solenoid, A = \(\frac{\pi d^2}{4}\)
\(A =\frac{3.14 \times\left(4 \times 10^{-2}\right)^2}{4}  \)
\(=12.56 \times 10^{-4} \mathrm{~m}^2\)
Self inductance of a solenoid is, \(L=\frac{\mu_0 \mu_r N^2 A}{l}\)
\(\therefore L  =\frac{4 \pi \times 10^{-7} \times 1 \times(4000)^2 \times 12.56 \times 10^{-4}}{2}  \)
\(=2 \times 3.14 \times 10^{-7} \times 16 \times 10^6 \times 12.56 \times 10^{-4}  \)
\(=1262 \times 10^{-5} \mathrm{H}=12.62 \times 10^{-3} \mathrm{H}  \)
\(\therefore L  =12.62 \mathrm{mH}\)
∴ Self inductance of a solenoid L =12.62 mH

Area of the capacitor = A
Radius = R
Current in the wire connected to the plates I = 5 A
The electric field, between the plates of a parallel plate capacitor,
\(E=\frac{\sigma}{\varepsilon_0}  \)
\(E=\frac{Q}{A \varepsilon_0}\)
Q is the charge accumulated at the positive plate.
The flux of this field, \(\phi_E=\frac{Q}{A \varepsilon_0} \times A=\frac{Q}{\varepsilon_0}\)
Displacement current \(i_d=\varepsilon_0 \frac{d \phi_E}{d t}\)
\(=\varepsilon_0 \frac{d}{d t}\left(\frac{Q}{\varepsilon_0}\right)=i_c\)
\(\therefore \mathrm{i}_{\mathrm{d}}=5 \mathrm{~A} \quad\left(\because\right.\) The current through the capacitor i_c = 5 A)
Displacement current = 5 A

(a) In the figure

1 - Electrostatic force F_e = QE
2 - Weight W = mg
3 - Elastic force F = -kx
4 - Upward force = Normal reaction = N
b) In this figure

1- Electrostatic force F = qE
2 - Weight W = mg
3 - Tension acting along the string is T
(c) The charge is attracted towards the positively charged plate because it is a negative charge.

1 - Force = qE
2 - Downward force F = mg

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 \(\mathrm{R}=12 \Omega \ is, \mathrm{I}=\frac{\mathrm{V}^1}{\mathrm{R}}=\frac{2}{12}=0.17 \mathrm{~A}\)

At T_o = 20^oC, resistance R₀ = 10 \(\Omega \)
\(\alpha\) = 0.004/^oC, At T_o = 100^oC  R₁₀₀ = ? (at boiling point of water)
R_T = R₀[1 + \(\left.(\alpha( T-T_{0}\right))\)]
 R₁₀₀ = 10 [1 + (0.004 x (100 - 20))]
= 10 [1 + 0.32] = 10 x 1.32 = 13.2 \(\Omega \)
∴ Resistance at boiling point of water R₁₀₀ = 13.2 \(\Omega \)
(i.e) R_r = 13.2 \(\Omega \)
Comment : As the temperature increases, the resistance of the wire also increases.

(a) The work done to arrange the charges in the corners of the square is independent of the way they are arranged. We can follow any order.
(i) First, the charge +q is brought to the corner P. This requires no work since no charge is already present, W_P = 0
(ii) Work required to bring the charge –q to the corner Q = (-q) x potential at a point Q due to +q located at a point P
W_Q = -q x \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q }{ a } =-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q^{ 2 } }{ a } \)
(iii) Work required to bring the charge +q to the corner R = q x potential at the point R due to charges at the point P and Q.
W_R = q x \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( -\frac { q }{ a } +\frac { q }{ \sqrt { 2 } a } \right) \)
= \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }^{ 2 } }{ a } \left( -1+\frac { 1 }{ \sqrt { 2 } } \right) \)
(iv) Work required to bring the fourth charge –q at the position S = q × potential at the point S due the all the three charges at the point P, Q and R.
W_s = - q x \(\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \left( \frac { q }{ a } +\frac { q }{ a } -\frac { q }{ \sqrt { 2 } a } \right) \)
W_s = \(-\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { q^2 }{ a } \left( 2-\frac { 1 }{ \sqrt { 2 } } \right) \)
(b) Work required to bring the charge q′ to the center of the square = q′ x potential at the center point O due to all the four charges in the four corners.
The potential created by the two +q charges are canceled by the potential created by the -q charges which are located in the opposite corners. Therefore the net electric potential at the center O due to all the charges in the corners is zero.
Hence no work is required to bring any charge to the point O. Physically this implies that if any charge q′ when brought close to O, then it moves to the point O without any external force.

λ = 589 nm = 589 x 10⁻⁹ m
For the film to have strong reflection, the reflected waves should interfere constructively. The least optical path difference introduced by the film should be λ/2. The optical path difference between the waves reflected from the two surfaces of the film is 2μd. Thus, for strong reflection, 2μd = λ/2 [As given in equation 6.145. with n = 1]
Rewriting, \(d=\frac{\lambda}{4 \mu}\) 
Substituting, \(d=\frac{589 \times 10^{9}}{4 \times 1.25}=117.8 \times 10^{-9}\) 
d = 117.8 x 10^-9 = 117.8 nm
For the film to be anti-reflecting, the reflected rays should interfere destructively. The least optical path difference introduced by the film should be λ. The optical path difference between the waves reflected from the two surfaces of the film is 2μd. For strong reflection, 2μd = λ [As given in equation 6.146. with n = 1]
Rewriting, \(d=\cfrac { \lambda }{ 2\mu } \)
Substituting, \(d=\cfrac { 589\times { 10 }^{ 9 } }{ 2\times 1.25 } =235.6\times { 10 }^{ -9 }\) 
d = 235.6 x 10^-9 = 235.6 nm

Diode D₁ and D₄ is reverse biased [open]
Diode D₁ and D₃ are forward biased.
The resistances are in series
R = 1 + 10 + 1 - 12 Ω
Barier Potential, V = 0.7 + 0.7 = 1.4 V (Silicon diode)
Applying Kirchhoff's voltage Law,
0.7 + I(1) + I(10) + 0.7 + I(1) = 3V
12 I = 3 - 1.4
12 I = 1.6
\(I=\frac{1.6}{12}=\mathbf{0 . 1 3 3 A}\)

Given data:
λ = 900 nm = 900 x 10^-9 m
λ ₂ = 750 nm = 750 x 10^-9 m
D = 2 m d = 2 nm = 2 x 10^-3 m
Let n^th order bright fringe of λ₁,
Coincides with (n + 1)^th order bright fringe of λ ₂
\(y_{n}=\frac{n \lambda_{1} D}{d}, Y_{n+1}=\frac{(n+1) \lambda_{2} D}{d} \)
\(\frac{n \lambda_{1} D}{d}=\frac{(n+1) \lambda_{2} D}{d} \)
\(n \lambda_{1}=(n+1) \lambda_{2} \)
\(\frac{n+1}{n}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{900 \times 10^{-9}}{750 \times 10^{-9}}=\frac{18}{15}=\frac{6}{5} \)
\(1+\frac{1}{n} =\frac{6}{5} \)
\(\frac{1}{n} =\frac{6}{5}-1=\frac{6-5}{5} \)
\(\frac{1}{n} =\frac{1}{5} \)
n = 5
n + 1 = 6
5^th bright fringe of  λ ₁ coincides with 6^th bright fringe of λ ₂ in the least distance of y
\(y =\frac{n \lambda_{1} D}{d}=\frac{5 \times 900 \times 10^{-9} \times 2}{2 \times 10^{-3}} \)
\(=4500 \times 10^{-6}=4.5 \times 10^{-3} \)
y = 4.5 mm

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
\(\mathbf{W}_{\text {mag }}=-0.8 \mathrm{J}\)

Magnetic field \(\overset { \rightarrow }{ B } ={ 0.500\hat { i } T } \)
Velocity of the particle
\(\hat { v } \) = (1.95 x 10⁵\(\hat { i } \) + 2.00 x 10⁵\(\hat { k } \)) ms^-1
Charge of the proton q = 1.67 x 10^-19 C
Mass of the proton m = 1.67 x 10^-27kg
(a) The force experienced by the proton is \(\overset { \rightarrow }{ F } \) = q(\(\overset { \rightarrow }{ v } \) x \(\overset { \rightarrow }{ B } \) )
= 160 x 10^-19 x  ((1.95 x 10⁵\(\hat { i } \) + 2.00 x 10⁵\(\hat { k } \)) x (0.500 \(\hat { i } \)))
\(\overset { \rightarrow }{ F } \)= 1.60 x 10^-14 \(\hat { j } \)  N
Therefore, from Newton’s second law,
\(\overset { \rightarrow }{ a } =\frac { 1 }{ m } \overset { \rightarrow }{ F } =\frac { 1 }{ 1.67\times { 10 }^{ -27 } } (1.60\times { 10 }^{ -14 })\hat j\)
\(=9.58\times { 10 }^{ 12 }\hat jm{ s }^{ -2 }\)
(b) Trajectory is helical Radius of helical path is
\(R=\frac { { mv }_{ z } }{ \left| q \right| B } =\frac { 1.67\times { 10 }^{ -27 }\times 2.00\times { 10 }^{ 5 } }{ 1.60\times { 10 }^{ -19 }\times 0.500 } \)
= 4.175 x 10^-3m = 4.18mm
Pitch of the helix is the distance travelled along x-axis in a time T, which is P = v_x T
But time, \(T=\frac { 2\pi }{ \omega } =\frac { 2\pi m }{ \left| q \right| B } =\frac { 2\times 3.14\times 1.67\times { 10 }^{ -27 } }{ 1.60\times 1{ 0 }^{ -19 }\times 0.500 } =13.1\times { 10 }^{ -8 }s\)
Hence, pitch of the helix is
\(p={ v }_{ x }T=(1.95\times { 10 }^{ 5 })(13.1\times { 10 }^{ -8 })=25.5\times { 10 }^{ -3 }m=25.5mm\)
The proton experiences appreciable acceleration in the magnetic field, hence the pitch of the helix is almost six times greater than the radius of the helix.

L = \(\frac { 2 }{ \pi } \)H; C = \(\frac { { 10 }^{ -4 } }{ \pi } F\); R = 100Ω
V_RMS = 220 V; f = 50Hz
X_L= 2πfl = 2π x 50 x \(\frac { 2 }{ \pi } \) = 200Ω
X_c = \(\frac { 1 }{ 2\pi fC } =\frac { 1 }{ 2\pi \times 50\times \frac { 10^{ -4 } }{ \pi } } 100 \Omega\)
(i) Impedance, Z = \(\sqrt { { R }^{ 2 }+({ X }_{ L }-{ X }_{ C })^2 } \)
=\(\sqrt { 100^{ 2 }+(200-100)^{ 2 } } \) = 141.4Ω
(ii) Peak value of current,
I_m = \(\frac { { v }_{ m } }{ Z } =\frac { \sqrt { 2 } V_{ RMS } }{ Z } \)
= \(\frac { \sqrt { 2 } \times 220 }{ 141.4 } \) = 2.2 A
(iii) Power factor of the circuit
\(cos\phi =\frac { R }{ Z } =\frac { 100 }{ 141.4 } \)= 0.707
(iv) Power factor at resonance
\(cos\phi =\frac { R }{ Z } =\frac { R }{ R } \) = 1

Figure (a)

Parallel : \(C_{P}=C_{0}+C_{0}=2 C_{0}\)
Series : \(\frac{1}{C_{S}} =\frac{1}{C_{0}}+\frac{1}{2 C_{0}} \)
\(=\frac{2+1}{2 C_{0}} \)
\(\frac{1}{C_{S}} =\frac{3}{2} C_{0} \)
\(\therefore C_{s} =\frac{2}{3} C_{0} \)
Figure (b)

\(C_{p 1} =2 C_{0},C_{p 2} =2 C_{0}\)
\(\frac{1}{C_{S}} =\frac{1}{C_{p_{1}}}+\frac{1}{C_{P 2}}=\frac{1}{2 C_{0}}+\frac{1}{2 C_{0}} \)
\(\frac{1+1}{2 C_{0}} =\frac{2}{2 C_{0}}=\frac{1}{C_{0}} \)
\(\therefore C_{s} =C_{0} \)
Figure (c)

\(C_{p} =C_{0}+C_{0}=2 C_{0} \)
\(C_{p} =2 C_{0}+C_{0} \)
\(C_p=3 C_{0} \)
Figure (d)

\(\frac{1}{C_{s 1}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{C_{1}+C_{2}}{C_{1} C_{2}} \)
\(\therefore C_{S 1}=\frac{C_{1} C_{2}}{C_{1}+C_{2}} \)   ....(1)
\(\frac{1}{C_{s 2}}=\frac{1}{C_{3}}+\frac{1}{C_{4}}=\frac{C_{4}+C_{3}}{C_{3} C_{4}} \)
\(\therefore C_{s 2}=\frac{C_{3} C_{4}}{C_{3}+C_{4}} \)    ...(2)
\(C_{p} =C_{s 1}+C_{s 2} \)
\(C_p= \frac{C_{1} C_{2}}{C_{1}+C_{2}}+\frac{C_{3} C_{4}}{C_{3}+C_{4}} \)
\(=\frac{C_{1} C_{2}\left(C_{3}+C_{4}\right)+C_{3} C_{4}\left(C_{1}+C_{2}\right)}{\left(C_{1}+C_{2}\right)\left(C_{3}+C_{4}\right)} \)
\(C_{p} =\frac{\mathbf{C}_{1} C_{2} C_{3}+C_{1} C_{2} C_{4}+C_{3} \mathbf{C}_{4} C_{1}+C_{3} C_{4} C_{2}}{\left(C_{1}+C_{2}\right)\left(C_{3}+C_{4}\right)} \)
The Effective capacitance across PQ and RS is same.
\(C_{PQ}=C_{RS}=C=\frac{C_{1} C_{2} C_{3}+C_{2} C_{3} C_{4}+C_{1} C_{2} C_{4}+C_{1} C_{3} C_{4}}{\left(C_{1}+C_{2}\right)\left(C_{3}+C_{4}\right)}\)
Figure (e): Across PQ

\(\frac{1}{C_{S 1}}=\frac{1}{C_{0}}+\frac{1}{C_{0}}=\frac{2}{C_{0}} \)
\(\therefore C_{S 1}=\frac{C_{0}}{2} \)
\(\frac{1}{C_{S 2}}=\frac{1}{C_{0}}+\frac{1}{C_{0}}=\frac{2}{C_{0}} \)
\(C_{S 2}=\frac{C_{0}}{2} \)
\(C_{P}=C_{S 1}+C_{S 2}=\frac{C_{0}}{2}+\frac{C_{0}}{2} \)
\(C_p=C_{0} \)
Resultant Capacitance

C_R = C_o + C_o
C_R = 2C_o

(a) The capacitance of the capacitor is
C = \(\frac { { \varepsilon }_{ 0 }A }{ d } =\frac { 8.85\times { 10 }^{ -12 }\times 25\times 10^{ -4 } }{ 1\times 10^{ -3 } } \)
= 221.2 x 10^-13 F
C = 22.12 x 10^-12F = 22.12pF
(b) The charge stored in any one of the plates is Q = CV, Then
Q = 22.12 x 10^-12 x 10 = 221.2 x 10^-12C = 221.2 pC.

Thus applying Kirchoff’s second law to the closed loop EACE
I₁R₁ + I₂R₂ + I₃R₃ = ξ
and for the closed loop ABCA
I₄R₄ + I₅R₅ - I₂R₂ = 0

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.