The refractive index of a transparent medium decreases with increase in wavelength of the incident light.
We have \(\mu_{g a}=\frac{v_{a}}{v_{g}}=\frac{3 \times 10^{8}}{2 \times 10^{8}}=\frac{3}{2}=1.5\)
\( \therefore \quad \mu_{g a}=\frac{1}{\sin i_{c}} \)
\(\Rightarrow i_{c}=\sin ^{-1}\left(\frac{1}{\mu_{g a}}\right) \)
\(\Rightarrow \quad i_{c}=\sin ^{-1}\left(\frac{2}{3}\right)=41.8\)

Here, \(I_{1}=I_{2}+I_{3}\)      ......(i)
From the loop ADBA,
\(3 I_{1}+3 I_{2}=2-1 ; 3 I_{1}+3 I_{2}=1\)       .....(ii)
From the loop DCBD, 
\(4 I_{3}-3 I_{2}=3-1 ; 4 I_{3}-3 I_{2}=2\)       .....(iii)
\(4\left(I_{1}-I_{2}\right)-3 I_{2}=2 \quad\left(\because I_{3}=I_{1}-I_{2}\right)\)
\(4 I_{1}-7 I_{2}=2\)        .....(iv)
From equations (ii) and (iv), we get
\( 3 I_{1}+3 I_{2} =1 \)
\(4 I_{1}-7 I_{2} =2 \)
\(I_{1} =\frac{13}{33}, \quad I_{2}=-\frac{2}{33}, \ I_{3}=\frac{15}{33}\)

Given, v = 2 x 10⁸ m / s and \(\mu _{ r }\)
The speed of electromagnetic waves in  medium is given by 
v = \(\frac { 1 }{ \sqrt { \mu \varepsilon } } \)
Where, \(\mu \) and \(\varepsilon \) are absolute permeability and absolute permittivity of the medium.
Now,  \(\mu \) = \(\mu _{ 0 }\mu _{ r }\)
and  \(\varepsilon =\varepsilon _{ 0 }\varepsilon _{ r }\)
Eq. (i) becomes, v = \(\frac { 1 }{ \sqrt { \mu _{ 0 }\mu _{ r }\varepsilon _{ 0 }\varepsilon _{ r } } } \)
= \(\frac { 1 }{ \sqrt { \mu _{ 0 }\varepsilon _{ 0 } } } \) x  \(\frac { 1 }{ \sqrt { \mu _{ r }\varepsilon _{ r } } } \)
v = \(\frac { c }{ \sqrt { \mu _{ r }\varepsilon _{ r } } } \)  \(\left[ c=\frac { 1 }{ \sqrt { \mu _{ 0 }\varepsilon _{ 0 } } } \right] \)
On squaring both sides, we get 
\(\varepsilon _{ r }=\frac { c^{ 2 } }{ v^{ 2 }\mu _{ r } } \)
= \(\frac { (3\times10^{ 8 })^{ 2 } }{ (2\times10^{ 8 })^{ 2 }\times1 } \) = 2.25

Given, diameter = 0.24 cm,
radius, \(r=\frac { 0.24\times { 10 }^{ -2 } }{ 2 } \) = 0.12 x 10^-2 m
and Current, I = 10A
\(\therefore\) Current density, \(J=\frac { I }{ A } =\frac { I }{ \pi { { r }^{ 2 } } } \)
\(=\frac{10}{3.14 \times\left(0.12 \times 10^{-2}\right)^2}=2.2 \times 10^6 \mathrm{Am}^{-2}\)

Here, \(C_o=2\mu F=2\times 10^{-6}F, K=5, V=100 volt\)
When dielectric slab is removed, capacity becomes \(1\over K\) times and potential becomes K times.
New potential \(V_o=5\times 100=500V\)
Work required to remove dielectric
\(=U_2-U_1={1\over 2}C_oV_o^2-{1\over 2}CV^2\)
\(={1\over 2}[2\times 10^{-6}(500)^2-10\times 10^{-6}(100)^2]\)
\(={1\over2}(0.5-0.1)=0.2J\) 

Here, r = 1 mm = 10^-3 m;
l = 50 cm = 0.50 m;
R₁ = \(0.31 \ \Omega\), 
t₁ = 25^oC,
R₂ = \(0.51 \ \Omega\),
t₂ = 125^oC
(i) Temperature coefficient of resistance
\(\alpha=\frac{R_2-R_1}{R_1(t_2-t_1)}=\frac{0.51-0.31}{0.31(125-25)}\)
= \(\frac{0.20}{0.31\times100}\) = 6.45 x 10^-3 oC^-1
(ii) Resistance at 0^oC is
\(R_0=\frac{R_1}{1+\alpha t_1}=\frac{0.31}{1+(6.4\times10^{-3})\times 25}=0.267 \ \Omega\)
(ii) Resistivity at 0^oC
\(\rho_0 =\frac{R_0A}{l}=\frac{R_0 \times \pi r^2}{l}=\frac{0.267 \times 3.242 \times (10^{-3})^2}{0.50}\)
  = \(1.68 \times 10^{-6} \ \Omega m\)
Resistivity at 25^oC,
\(\rho_{25}=\rho_0[1+\alpha \times 25]\)
= \(1.68 \times 10^{-6}[1+6.4 \times10^{-3}\times25]\)
=\(1.948 \times 10^{-6} \ \Omega m\)

Here,  P = \(16 \ \Omega\), Q = \(12 \ \Omega\), R = \(9 \ \Omega\), S = \(4 \ \Omega\) 
Let \(9 \ \Omega\) be shunted with resistance x to balance the bridge. Then effective resistance of \(9 \ \Omega\) and x ohm in parallel,
\(R'=\frac{9x}{9 + x}\)
For balanced bridge,
\(\frac{P}{Q}=\frac{R'}{S}\)
\(\therefore \frac{16}{12}=\frac{9x/(9+x)}{4}\)
or  \(\frac{4}{3}=\frac{9x}{4(9+x)}\)
or  144 + 16 x = 27 x
or x = \(13.1 \ \Omega\)

Given current sensitivity
 \({ I }_{ S }=20\mu A{ div }^{ -1 }=20\times { 10 }^{ -6 }A{ div }^{ -1 }\) 
The galvanometer ha 30 divisions, so current for full-scale deflection
,\({ I }_{ g }=30\times 20\times { 10 }^{ -6 }A=6\times { 10 }^{ -4 }A\) 
When the galvanometer is to be converted into an ammeter, thus the value of shunt required
\(S=\frac { { I }_{ g } }{ I-{ I }_{ g } } G=\frac { 6\times { 10 }^{ -7 } }{ 1-6\times { 10 }^{ -4 } } \times 25\)  (in parallel)
\(\\ =0.015\Omega \)
When the galvanometer is to be converted into a voltmeter, then the value of resistance R is given by
\(R=\frac { V }{ I } -{ R }_{ A }\) 
\(=\frac { 1 }{ 1 } -\frac { GS }{ G+S } =1-\frac { 25\times 0.015 }{ 25+0.015 } \)
\(=1-0.015=0.985\Omega \) (in series)

Resistance of the galvanometer coil, G = 15 Ω
Current for which the galvanometer shows full scale deflection,
 = 4 mA = 4 x 10^-3 A
Range of the ammeter is 0, which needs to be converted to 6 A.
Current, I = 6 A
A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it into an ammeter. The value of S is given as:
\(S=\frac{I_{g} G}{I-I_{g}}\)
\(=\frac{4 \times 10^{-3} \times 15}{6-4 \times 10^{-3}}\)
\(S=\frac{6 \times 10^{-2}}{6-0.004}=\frac{0.06}{5.996}\)
\(\approx 0.01 \Omega=10 \mathrm{~m} \Omega\)
Hence, a  10 mΩ shunt resistor is to be connected in parallel with the galvanometer.

(a) A wavefront is defined as the locus of all the particles of a medium vibrating in the same phase at a given instant. The shape of a wavefront depends upon the shape of the source of disturbance and it is normal to the direction of propagation of wave.
A line drawn perpendicular to the plane wavefront gives the direction of propagation of a wave and is called ray of light.
(b) Huygens' Principle and Reflection at a Plane Surface
(c) Position of nth maxima, \(y_n=\frac{n D \lambda}{d}\) (Bright fringe) and position of mth minima,
\(y_m^{\prime}=\left(m-\frac{1}{2}\right) \frac{D \lambda}{d}\) (Dark fringe)
Given, d = 3 mm = 3 \(\times\) 10^-3 m
D = 1 m
and y₄ - y₂' = 5 mm = 5 \(\times\) 10^-3m
Now, \(\frac{4 D \lambda}{d}-\left(2-\frac{1}{2}\right) \frac{D \lambda}{d}=5 \times 10^{-3} \mathrm{~m}\)
\(\Rightarrow\) \(\frac{4 D \lambda}{d}-\frac{3}{2} \frac{D \lambda}{d}=5 \times 10^{-3}\)
\(\begin{aligned} \Rightarrow \left(4-\frac{3}{2}\right) \frac{D \lambda}{d}=5 \times 10^{-3} \end{aligned}\)
\(\begin{aligned} \Rightarrow \frac{5}{2}\left(\frac{(1) \lambda}{3 \times 10^{-3}}\right)=5 \times 10^{-3} \end{aligned}\)
\(\Rightarrow\)  \(\lambda=\frac{3}{2} \times 10^{-6} \mathrm{~m}\)
\(\Rightarrow\) \(\lambda=1.5 \times 10^{-6} \mathrm{~m}\)
\(\Rightarrow\) \(\lambda=1.5 \mu \mathrm{m}\)

ii) use , dQ = \(\lambda \), dl and Gauss' theorem \(\phi=\frac{Q}{\varepsilon_0}\)

If the final image is formed at the distance of distinct visioa, the magnifying power of the telescope is given as
\(m=\frac{f_{o}}{-f_{e}}\left(1+\frac{f_{e}}{D}\right)\)
Astronomical telescope in near point adjustment.

For an astronomical telescope, we will select lens of power 10D for eye piece and lens of power 0.5 D for objective because magnifying power of telescope is \(m=\frac{f_{o}}{f_{e}}\)
Resolving power of the telescope:
Given: \( D_{0}=6 \mathrm{~cm}, \lambda =540 \times 10^{-9} \mathrm{~m} \)
\( \text { R.P. } =\frac{D_{o}}{1.22 \lambda}=\frac{6 \times 10^{-2}}{1.22 \times 540 \times 10^{-9}} \)
\(= 0.9 \times 10^{5} \)

(i) The shunt is a low resistance connected in parallel with the galvanometer. Therefore, the combined resistance is less than that of the shunt.
(ii) Let I be the total current passing through the shunt galvanometer. Let \({ I }_{ g }\),\({ I }_{ s }\) be the currents through galvanometer of resistances G and shunt of resistance S respectively. As current divides itself in the inverse ratio of the resistances, therefore, fraction of current passing through galvanometer
= \(\frac { { I }_{ g } }{ I } =\frac { S }{ G+S } =\frac { 3 }{ 30+3 } =\frac { 1 }{ 11 }\)
\(Fraction \ of \ current \ passing \ through \ shunt=\frac { { I }_{ s } }{ I } =\frac { G }{ G+S } =\frac { 30 }{ 30+3 } =\frac { 10 }{ 11 } \\ \)
(iii) From the above study, we find that the current always divides itself in the inverse ratio of resistances. The major part of current flows through the shunt, which is the path of low resistance. Similarly in life, if different paths are available to reach a destination, then one selects a path of least resistance to reach there. This would save both, time and energy.

(i) The force is due to the interaction between magnetic fields due to currents in two linear parallel conductors.
(ii) Here, \({ l }_{ 1 }=2cm=2\times { 10 }^{ -2 }m;{ l }_{ 2 }=20\times { 10 }^{ -2 }m;{ I }_{ 1 }=2.0A;{ I }_{ 2 }=5.0A;r=4\times { 10 }^{ -2 }m\)
Since action and reaction are equal and opposite so the magnitude of repulsive force on 20 cm long wire = the magnitude of repulsive force on 2 cm long wire
= \(\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2{ I }_{ 1 }{ I }_{ 2 } }{ r } { l }_{ 1 }={ 10 }^{ -7 }\times \frac { 2\times 2\times 5\times (2\times { 10 }^{ -2 }) }{ 4\times { 10 }^{ -2 } }\)
\( ={ 10 }^{ -6 }\)
This study shows that when the current through two parallel conductors flow in the same direction, they attract each other and vice versa. It implies that when the thoughts and actions of two business partners are aligned in the same direction, their behavior is cohesive and they succeed. If there thoughts and actions are opposing, the partnership is likely to collapse. The same thing is true for husband and wife. For a successful family life, the coherence of thoughts and action is a must. 

Since drift velocity \({ v }_{ d }\) and current, I flowing in a conductor are related by the relation:
\({ v }_{ d }=\frac { I }{ neA }.........(1)\) 
\(Also \ drift \ velocity \ in \ terms \ of \ average \ relaxation \ time \ \tau \ is \ given \ by\)
\({ v }_{ d }=\frac { eE\tau }{ m } .........(2)\)
\( From \ (1) \ and \ (2), \ we \ have\)
\( \frac { eE\tau }{ m } =\frac { I }{ neA }\)
\(or\quad \frac { E }{ I } =\frac { m }{ n{ e }^{ 2 }A\tau }\)
\( or \ \frac { V }{ lI } =\frac { m }{ n{ e }^{ 2 }A\tau } \)
\( or \ \frac { V }{ I } =\frac { ml }{ n{ e }^{ 2 }A\tau } ...........(3)\)
\( The \ R.H.S. \ of \ Eq(3) \ is \ constant\)
\( \frac { V }{ I } = \ Constant\)
\( This \ is \ Ohm's \ law\)

The right-hand thumb rule gives the direction of magnetic field which is stated as under:
Curl the palm of your right hand around a circular wire with the fingers, pointing in the direction of the current and the right hand thumb gives the direction of magnetic field.

Given, \(f=\frac{2}{3} R, R_{1}=+R, R_{2}=-R\)
∴ Using lens Maker's formula,
\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\( \Rightarrow \frac{3}{2 R}=(\mu-1)\left(\frac{2}{R}\right) \)
\(\Rightarrow \quad \mu-1=\frac{3}{4} \)
\(\Rightarrow \quad \mu=1+\frac{3}{4}=\frac{7}{4} \)

Given, refractive index of the material of the prism, μ  = 1.5
∴ Critical angle for the material,
\(\sin C=\frac{1}{\mu}=\frac{1}{1.5}=2 / 3\)
\(\Rightarrow C=\sin ^{-1}\left(\frac{2}{3}\right) \simeq 42^{\circ}\)
From the ray diagram, it is clear that angle of incidence i = 30°< C.
Therefore, ftie ray incident at the face AC will not suffer total internal reflection and merges out through this face.

Yes, but motion is simple harmonic only when charge - q is not very far from the centre of ring on its axis. Otherwise motion is periodic, but not simple harmonic in nature.

It is independent of the distance. It' s a straight line parallel to x-axis.

We apply Kirchoff's current law in the shown circuit.
At junction B,
i₁​=i_g​+i₃
At junction D,
i_2​+i_g​=i₄​
If current through the galvanometer is zero,
i_g = 0
thus i₁ = i₃
and i₂ = i₄
Applying Kirchoff's voltage law for loop ABDA,
i₁P + i_gG = i₂R
Applying Kirchoff's voltage law for loop BCDB,
i₃Q + i₄S + i_gG
When i_g = 0,
i₁P = i₂R
and i₃Q = i₄S
But i₁ = i₃ and i₂ = i₄,
Therefore \(\frac{P}{Q}=\frac{R}{S}\)

(i) Biot-Savart law can be expressed alternatively as Ampere's circuital law.
(ii) Biot-Savart law is angle dependence law, whereas Coulomb's law in electrostatic is angle independent law.
(iii) The magnetic field due to an infinitely long current carrying wire,
\(\begin{aligned} B=\frac{\mu_0 I}{2 \pi r} \end{aligned}\)
\(\begin{aligned} \therefore \quad B & \propto I \end{aligned}\)
(iv) The magnetic field at a point on the long current carrying wire,
\(\begin{aligned} d B=\frac{\mu_0}{4 \pi} \cdot \frac{I d l \sin \theta}{r^2} \end{aligned}\)
\(\begin{aligned} =\frac{\mu_0}{4 \pi} \cdot \frac{I d l \sin 0^{\circ}}{r^2} \quad \quad \quad{\left[\because \theta=0^{\circ}\right]} \end{aligned}\)
= 0

(i) (a): Gauss's law is applicable for any closed surface. Gauss's law is most useful in situation where the charge distribution has spherical or cylindrical symmetry or is distributed uniformly over the plane.
Whereas electric dipole is a system of two equal and opposite point charges separated by a very small and finite distance.
So both statements are correct.
(ii) (b): According to Gauss's law, the electric flux through the sphere is 
\(\phi=\frac{q_{\mathrm{in}}}{\varepsilon_{0}}=\frac{8.85 \times 10^{-13} \mathrm{C}}{8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}}=0.1 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^{2}\)
(iii) (c) : For uniformly volume charge density,
\(E=\frac{\rho r}{3 \varepsilon_{0}}\)
\(E \propto r\)
(iv) (a): r = 25 ern = 0.25 m \(\sigma=\frac{3}{\pi} \mathrm{C} / \mathrm{m}^{2}\)
As, \(\sigma=\frac{q}{4 \pi r^{2}} \Rightarrow q=4 \pi \times(0.25)^{2} \times \frac{3}{\pi}=0.75 \mathrm{C}\)
(v) (b): The line charge density at a point on a line is the charge per unit length of the line at that point
\(\lambda=\frac{d q}{d L}\)
Thus, the SI unit for \(\lambda \text { is } \mathrm{Cm}^{-1} \text {. }\)

(b) : \(\mu=\frac{\mu_{g}}{\mu_{c}}=\frac{1.5}{1.65}<1\)
\(\text { From } \frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
f becomes negative.
Therefore, the lens behaves as a diverging lens.

(b): If maximum intensity is observed at P then for maximum intensity to be also observed at Q, S₁ and S₂ must have phase difference of 2 m\(\pi\) (where m is an integer).

(a): Magnetic moment \(=\frac{\text { joule }}{\text { tesla }}=\frac{W}{B}=\frac{W}{F / q v}\)
\(=\frac{W q v}{F}=\frac{\left[M L^{2} T^{-2}\right][A T]\left[L T^{-1}\right]}{\left[M L T^{-2}\right]}\)
\(=\left[\mathrm{AL}^{2}\right]=\mathrm{amp} \mathrm{m}^{2}\)