As we know \(V=\frac{k q}{r}\)
\( \therefore V_{1}=\frac{k Q}{r} ; V_{2}=\frac{k Q}{2 r} \)
\(\Rightarrow \frac{V_{1}}{V_{2}}=\frac{2}{1} \)
\(\because V_{1}>V_{2} \)

Since after connecting these two spheres, the potential of each sphere will be equal, i.e.
\( V_{1} =V_{2} \)
\(\therefore \frac{k Q_{1}^{\prime}}{r} =\frac{k Q_{2}^{\prime}}{2 r} \)
\(\therefore Q_{1}^{\prime} =\frac{Q_{2}^{\prime}}{2}\)
Electric charges are conserved.
\( \because \ Q+Q =Q_{1}^{\prime}+Q_{2}^{\prime} \)
\(2 Q =\frac{3 Q_{2}^{\prime}}{2} \Rightarrow Q_{2}^{\prime}=\frac{4}{3} Q\)
Charge following = \(Q_{2}^{\prime}-Q=\frac{4 Q}{3}-Q=\frac{Q}{3}\)

Let us consider a system of three charges q₁, q₂ and q₃ located at r₁, r₂ and r₃ respectively. To bring q, first from infinity to r₁, no work is required. Next we bring q₂ from infinity to r₂. As before, work done in this step is:

\(q_{2} V_{1}\left(r_{2}\right)=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{12}} \ ...(i)\)
The charges q₁ and q₂ produce a potential, which at any point P is given by
\(V_{1,2}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1}}{r_{1 P}}+\frac{q_{2}}{r_{2 P}}\right)...(ii)\)
Work done next in bringing q3 from infinity to the point r₃ is q₃ times V_1,2 at r₃.
\(q_{3} V_{1,2}\left(r_{3}\right)=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1} q_{3}}{r_{13}}+\frac{q_{2} q_{3}}{r_{23}}\right)...(iii)\)
The total work done in assembling the charges at the given locations is obtained by adding the work done in different steps (i) and (iii).
\(U=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1} q_{2}}{r_{12}}+\frac{q_{1} q_{3}}{r_{13}}+\frac{q_{2} q_{3}}{r_{23}}\right)\)

As magnification, m = \(\frac{I}{O}=\frac{v}{u} \Rightarrow I=4 \times\) length of object
\(\Rightarrow \frac{I}{O}=4\Rightarrow \frac{v}{u}=4 \Rightarrow v=4 u\)
Using lens formula, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{(-4 u)}-\frac{1}{(-u)}\)
\( \Rightarrow \frac{1}{f}=-\frac{1}{4 u}+\frac{1}{u} \Rightarrow \frac{1}{20}=\frac{4-1}{4 u}=\frac{3}{4 u} \)
\(\Rightarrow u=\frac{20 \times 3}{4}=15 \mathrm{~cm} \)
⇒ v = 4u = 15 x 4 = 60 cm
Distance of the object, u = 15 cm
Distance of the image, v = 60 cm
The image is on the same side of the object.

The force on charge +q is + q \(\overset { \rightarrow }{ E } \) and on charge -q is -q \(\overset { \rightarrow }{ E } \). These, two parallel forces, acting in the opposite direction, constitute a couple resulting in the torque 't
Magnitude of torque = qE x 2a sin
= 2qaE sin 9
Therefore, \(\overset { \rightarrow }{ t } =\overset { \rightarrow }{ p } \times\overset { \rightarrow }{ E } \)
where \(\overset { \rightarrow }{ p } =2q\overset { \rightarrow }{ a } \)
Two pairs of perpendicular vectors in expression are given below :
(i) \(\overset { \rightarrow }{ t } \) is perpendicular to \(\overset { \rightarrow }{ p } \)
(ii) \(\overset { \rightarrow }{ t } \) is perpendicular to \(\overset { \rightarrow }{ E } \)

Position of minima due to diffraction at a single slit is given by
\( a\sin { \theta =n\lambda } \)
\( \therefore \ \sin { \theta } =\frac { n\lambda }{ a } =\frac { 1\times600\times{ 10 }^{ -9 } }{ 1.0\times{ 10 }^{ -3 } } \)
\( =6\times{ 10 }^{ -4 }m\)
\(For \ small\theta ,\ \sin { \theta } \simeq \ \theta =6\times{ 10 }^{ -4 } \ radian \)
Let y be the distance of first minima on either side of central maxima, then 
\( \theta =\frac { y }{ D } \)
\(or \ y=\theta D=6\times{ 10 }^{ -4 }\times2=12\times{ 10 }^{ -4 }m \)
So the distance between first dark fringes on either side of central maxima is
\(2y=2\times12\times{ 10 }^{ -4 }=24\times{ 10 }^{ -4 }m=2.4mm\)

\(Here,\lambda =5890\mathring { A } ,\mu =\frac { 4 }{ 3 } \)
\( d\theta ={ 0.20 }^{ \circ },d\theta '=?\)
\( Angular \ width,d\theta =\frac { \lambda }{ d } (in \ air)\)
\(In \ denser \ medium,\lambda \ is \ reduced \ to \ \lambda ' \ by \ relation\)
\(\ \frac { \lambda }{ \lambda ' } =\mu \)
\( The \ angular \ width \ is \ given \ by \ d\theta '=\frac { \lambda ' }{ d } \ in \ denser \ medium.\)
\( Hence\ \frac { d\theta ' }{ d\theta } =\frac { \lambda '/d }{ \lambda /d } =\frac { \lambda ' }{ \lambda } =\frac { 1 }{ \mu }\)
\(ord d\theta '=\frac { d\theta }{ \mu } =\frac { 0.20 }{ 4/3 } ={ 0.15 }^{ \circ }\)

Kirchhoff's law is based on conservation of charge and the fact that there is no accumulation of charges at a junction.

Reading of voltmeter = 4 V, Resistance of voltmeter  \(=10^{4} \Omega\)
Current drawn by the voltmeter, \(I=\frac{4}{10^{4}}=4 \times 10^{-4} \mathrm{~A}\)
Again, \(I=\frac{E}{X+10^{4}}\)
\( \Rightarrow \left(X+10^{4}\right) I =E \)
\(\therefore X I =E-I \times 10^{4} \)
\(\Rightarrow X =\frac{E}{I}-10^{4}=\frac{120}{4 \times 10^{-4}}-10^{4} \)
\(\therefore X =29 \times 10^{4} \Omega=290 \mathrm{k} \Omega \)

Given, focal length of convex lens,
f₁ = + 25cm = + 0.25 m and focal length of concave  lens, f₂ = - 20. cm = - 0..20.m
Equivalent focal length of convex and concave
\(F=\frac { 1 }{ f_{ 1 } } +\frac { 1 }{ f_{ 2 } } =\frac { 1 }{ 25 } +\frac { 1 }{ -20 } =-\frac { 1 }{ 100 } \)
\(\therefore \) F = -100cm = -1m
Now the power of lens, P =  \(\frac { 1 }{ f } \)
For convex lens P₁ ⁼ \(\frac { 1 }{ f_{ 1 } } =\frac { 1 }{ 0.25 } \)
For convex lens P₂ = \(\frac { 1 }{ f_{ 2 } } =\frac { 1 }{ -0.20 } \)
\(\frac { 1 }{ 0.25 } +\frac { 1 }{ -0.20 } =\frac { 100 }{ 25 } +\frac { 100 }{ -20 } \)
\(=\frac { 400-500 }{ 100 } =\frac { -100 }{ 100 } =-1D\)
Here, the focal length of the combination =100. cm = -1 m
Since the focal length is in negative, so the system will be diverging in nature.

(i) Refractive index  \((\mu )\frac { 1 }{ sin \ C } \)  Where C is the critical angle
(ii) Since, refractive index depends upon the wavelength of light, the critical angle for a given pair of media is different for different wavelenghts (colours ) og light.

Change i distance, x = t\((1-{1\over k})\) 

Given,
The area of plate of the capacitor, A = 6 x 10^-3 m²
Distances between the plates, d = 3mm = 3 x 10^-3 m
Voltage supplied, V = 100V
Capacitance of a parallel plate capacitor is given by, \(C=\frac{\epsilon \times A}{d}\) 
Here,
ε = permittivity of free space = 8.854 x10^-12 N^-1 m ^-2 C^-2
\(C=\frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}=17.81 \times 10^{-12} \mathrm{~F}=17.71 \mathrm{pF}\)
Therefore, each plate of the capacitor is having a charge of
q = VC = 100 x 17.81 x 10^-12 C = 1.771 x 10^-9 C

20; 13.5 cm

\(45\times \)10^-2 J

(a) Both A and R are true and R is the correct explanation of A.

(c) If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. There is no gain or loss of energy, which is consistent with the principle of conservation of energy.

(a): As given in the expression \(\beta=\frac{D \lambda}{d}\), fringe width \(\beta\) is independent of n (position of fringe), hence all the fringes are of same width.