\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \)
\(m=\frac{-v}{u}=-4, f=20 \mathrm{~cm} \text { (Given) } \)
V = 4u
\(\frac{1}{f} =\frac{1}{4 u}-\frac{1}{u} \)
\(=\frac{1-4}{4 u}=\frac{-3}{4 u} \)
\(\frac{1}{f} =\frac{-3}{4 u} \)
4u = -3 x f
\(u=\frac{-3}{4} \times 20=-15 \mathrm{~cm}\)

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)

\(W =\left(V_{b}-V_{a}\right) q\left[where\ V=\frac{K Q}{r}\right] \)
To find V_b :
\(V_b=\frac{kQ}{r_3}+\frac{kQ}{r_4}\)
\(V_{b} =\frac{K \times 10 \times 10^{-6}}{\sqrt{50 \times 10^{-4}}}+\frac{K \times 10 \times 10^{-6}}{\sqrt{250 \times 10^{-4}}} \)
\(V_{b} =\frac{K \times 10^{-5}}{\sqrt{50} \times 10^{-2}}+\frac{K \times 10^{-5}}{\sqrt{250} \times 10^{-2}} \)
\(=K \times 10^{-3}\left[\frac{1}{\sqrt{50}}+\frac{1}{\sqrt{250}}\right] \)
\(=9 \times 10^{9} \times 10^{-3}\left[\frac{1}{\sqrt{50}}+\frac{1}{\sqrt{250}}\right] \)
V_b = 1842002 V
To find V_a :
\(V_a=\frac{kQ}{r_1}+\frac{kQ}{r_2}\)
\(V_{a} =\frac{K \times 10 \times 10^{-6}}{\sqrt{5 \times 10^{-2}}}+\frac{K \times 10 \times 10^{-6}}{\sqrt{15 \times 10^{-2}}} \)
V_a = 2400000 V
\(\therefore W_{D} =\left(V_{b}-V_{a}\right) q=(1842002-2400000)( -2 \times 10^{-6} )\)
W = +1.12J
Positive sign implies that to move the charge - 2 μC external work is required.

Distance of separation r = 1 m
Number of electron in human body = 10²⁸
Charge appeared on my friend and me, q = 1% of charge on 10²⁸ electrons
q \(=10^{28} \times 1 / 100 \)
q = 1.6 x 10⁷ C
Force between us \( \mathrm{F}_{\mathrm{e}} =\mathrm{Kq}^{2} / \mathrm{r}^{2} \)
\(\left.=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{7}\right)^{2}}{(1)^{2}} \quad \text { (as } \mathrm{r}=1 \mathrm{~m}\right)\)
= 9 x 2.56 x 10⁹ x 10¹⁴
F_e = 23.04 x 10²³ N
Mass of each person m₁ = m₂ = m = 60 kg
Weight of each person W = mg = 60 x 9.8
W = 588N
F_e / W = \(\frac{23.04\times10^{23}}{588} = 3.9 \times 10^{21}=(or)F_e= 3.9 \times 10^{21}W\)

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.

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

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''

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

No of turns of small solenoid, N₂ = 200
Radius of small solenoid r = 2 cm = 2 x 10^-2 m
Area of small solenoid, A₂= πr² = 3.14 x 4 x 10^-4 m²
Turn density of long solenoid, \(\frac { { N }_{ 1 } }{ l } =90\times10^2m^{-1}, M=?\)
M = μ₀\(\frac { { N }_{ 1 } }{ l } \)N₂A₂
=4 x 3.14 x 10^-7 x 90 x 10² x 200 x 3.14 x 4 x 10^-4 = 2.84 x 10^-3 H
M = 2.84 mH

For a charged particle moving on a magnetic field, \(\vec { F } \)= q(\(\vec { v } \)x\(\vec { B } \))
The work done by the magnetic field is
\(W=\int { \overset { \rightarrow }{ F } .{ d \vec r } =\int { \overset { \rightarrow }{ F } .\overset { \rightarrow }{ v } dt } } \)
\(W=q\int { \left( \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right) } .\overset { \rightarrow }{ v } dt=0\)
Since \(\overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \) is perpendicular to \(\vec { v } \) and hence \(\left( \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right) .\overset { \rightarrow }{ v } =\overset { \rightarrow }{ 0 } \) 
This means that Lorentz force does no work on the particle. From work-kinetic energy theorem, (Refer section 4.2.6, XI th standard Volume I)
\(\frac { dw }{ dt } =p=0\)
Since \(\overset { \rightarrow }{ F } .\overset { \rightarrow }{ v } =0\Rightarrow \overset { \rightarrow }{ F } \ and \ \overset { \rightarrow }{ v } \) are perpendicular to each other. The angle between Lorentz force and velocity of the charged particle is 90^o. Thus Lorentz force changes the direction of the velocity but not the magnitude of the velocity. Hence Lorentz force does no work and also does not alter kinetic energy of the particle. 

N_P = 460 turns; N_S = 40,000 turns
V_P = 230 V; R_S = 10⁴ Ω
(i) Secondary voltage,
\({ V }_{ s }=\frac { { V }_{ p }{ N }_{ s } }{ { N }_{ p } } =\frac { 230\times 40,000 }{ 460 } \)
= 20,000V
Secondary voltage per turn,\(\frac { { V }_{ S } }{ { N }_{ S } } =\frac { 20,000 }{ 40,000 } =0.5V\)
(ii) Power delivered
= \({ V }_{ s }{ I }_{ s }=\frac { { V }_{ s }^{ 2 } }{ { R }_{ s } } =\frac { 20,000\times 20,000 }{ { 10 }^{ 4 } } =40kW\)

(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.

i) Equivalent emf ξ_eq = 5 V
ii) Equivalent internal resistance,
\({ R }_{ eq }=\frac { r }{ n } =\frac { 0.5 }{ 4 } =0.125\Omega \)
iii) total current, \(I=\frac { \xi }{ { R }_{ 5 }+\frac { r }{ n } } \)
\(I=\frac { 5 }{ 10+0.125 } =\frac { 5 }{ 10.125 } \)
I ≈ 0.5 A 
iv) Potential difference across each cell
V = IR = 0.5 x 10 = 5 V
v) Current from each cell, \(I'=\frac { I }{ n } \)
\(I'=\frac { 0.5 }{ 4 } =0.125A\)