(i) n-type
(ii) \(4\times 10^{ 4 }/m^{ 3 }\) 
(iii) Energy gap decreases with doping
Here,
\({ n }_{ i }=6\times 10^{ 8 }m^{ -3 };{ n }_{ e }=9\times 10^{ 12 }m^{ -3 }\)
\( { n }_{ h }=\frac { { n }_{ i }^{ 2 } }{ { n }_{ e } } =\frac { \left( 6\times 10^{ 8 } \right) ^{ 2 } }{ 9\times 10^{ 12 } } =4\times 10^{ 4 }m^{ -3 }\)
As, after doping, \({ n }_{ e }>{ n }_{ h }\) so the new semiconductors is n-type. Energy gap decreases with doping.

The law of conservation of charge states that the total charge in an isolated system remains constant.
The electric charge can neither be produced nor destroyed. This law has been found to be true for all events as well as for those at nuclear and atomic levels. In other words, there is no exception to the law. Like the law of conservation of energy, the law of conservation of charge is also a universal law.
Examples (i) When a glass rod is rubbed with silk the charges developed on the glass rod and the piece of silk are equal and opposite.
(ii) Charge is conserved in all chemical and nuclear reactions.

For the high frequency
\(\omega =2\pi \times { 10 }^{ -4 }rad\quad s^{ -1 }\)
\({ I }_{ 0 }=\frac { \sqrt { 2 } \times 240 }{ \sqrt { { 10 }^{ 4 }+{ \left( 0.5 \right) }^{ 2 }\times 4{ \pi }^{ 2 }\times { 10 }^{ 8 } } } =1.1\times { 10 }^{ -2 }A\)
Term from R is negligible in the above denominator
\(tan\phi =\frac { 2\pi \times { 10 }^{ 4 }\times 0.5 }{ 100 } =100\pi \)
\(\phi ={ 89 }^{ \circ }{ 48 }^{ \prime }={ 89.8 }^{ \circ}\) which is close to \({ \pi }/{ 2 }\ (i.e.{ 90 }^{ \circ })\)
In a d.c. circuit (after steady state) \(\omega =0\), so here L acts like a pure conductor.

(i) 36 x 10^-2 Am²
(ii) 18 x 10^-2Nm

Alpha particles due to its largest mass

Clearly , the fish can see the outside view of the cone with semi vertical angle
But \(\mu \) = 1.sin i_c
or 1/3 = 1/ sin i_c
or sin i_c = 3/4 = 0.75 
\(\theta\)/2  =i_c = sin^-1 (0.75 ) = 48.6⁰

(i) The trajectory of the beam is a parabola and its K.E changes.
(ii) The trajectory of the beam is circular and its K.E. remains the same. Justify your answer.

q \(\rightarrow\)_______Q
1/2 mv2 = kQq/r
Or, v2 a1/r
Or, r a 1/v²
Or, r' = r/4

Magnetic declination The angle between geographical meridian and magnetic meridian  any place of the earth is known as magnetic  declination \(\left( \alpha \right) \) at that place of the earth. 

The direction of propagation of the electromagnetic wave is perpendicular to both electric field vector E and magnetic field vector B, i.e. In the direction of E x B.
This can be seen by the diagram given below.

Here, electromagnetic wave is along the Z-direction which is given by the cross product of E and B

Since the airplane and the gasoline truck usually have wheels with rubber tires, they are insulated from the ground. further, the service ramps are usually made of concrete and are not necessarily good conductors to the earth. Therefore, inspite of grounding metallic ropes, the airplane and the truck could remain charged.
A spark may jump and ignite the explosive gasoline when the metal nozzle is brought near the airplane. The connection of metal of the airplane and the nozzle of the hose with a wire avoids any unbalances of charge and hence the risk of gasoline.

Yes, all the free electron drift in the same direction.

The maximum electric field that a dielectric medium can withstand without breaking down of its insulating property is called dielectric strength. Its value for vacuum is infinity.

Zero, because one plate has positive charge and the other carriers an equal negative charge.

No, the statement is not true. Infact, at a given distance from the centre of magnet, intensity on axial line is twice the intensity on equatorial line.

This is done in order to compensate for loss of energy due to resistance in L-C oscillatory circuit and hence to produce undamped electromagnetic waves or carriers waves.

With the increase in temperature, the drift velocity of free electrons in a metal conductor decreases due to increase in collision frequency of free electrons with the atom/ions of the conductor.

(i) Power = +2 dioptre.
(ii) Here, we have f = +12 cm, R₁ = +10 cm, R₂ = -15 cm.
Refractive index of air is taken as unity.
We use the lens formula. The sign convention has to be applied for f, R₁ and R₂.
Substituting the values, we have
\(\frac { 1 }{ 12 } =(n-1)\left( \frac { 1 }{ 10 } -\frac { 1 }{ 15 } \right) \)
This gives n = 1.5.
(iii) For a glass lens in air, n₂ = 1.5, n₁ = 1, f = +20 cm. Hence, the lens formula gives
\(\frac { 1 }{ 20 } =0.5\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
For the same glass lens in water, n₂ = 1.5, n₁ = 1.33. Therefore \(\frac { 1.33 }{ f } =(1.5-1.33)\left[ \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right] \)
Combining these two equations, we find f = + 78.2 cm.

From the mirror equation, Eq., we get \(v=\frac{f u}{u-f}\)
For convex mirror, since \(R=2 \mathrm{~m}, f=1 \mathrm{~m}\). Then for \(u=-39 \mathrm{~m}, v=\frac{(-39) \times 1}{-39-1}=\frac{39}{40} \mathrm{~m}\)
Since the jogger moves at a constant speed of \(5 \mathrm{~m} \mathrm{~s}^{-1}\), after 1 s the position of the image v (for \(u=-39+5=-34)\) is (34 / 35) m.
The shift in the position of image in 1 s is \(\frac{39}{40}-\frac{34}{35}=\frac{1365-1360}{1400}=\frac{5}{1400}=\frac{1}{280} \mathrm{~m}\)
Therefore, the average speed of the image when the jogger is between 39 m and 34 m from the mirror, is (1/280) m s^–1 Similarly, it can be seen that for u = –29 m, –19 m and –9 m, the speed with which the image appears to move is
\(\frac{1}{150} \mathrm{~m} \mathrm{~s}^{-1}, \frac{1}{60} \mathrm{~ms}^{-1} \text { and } \frac{1}{10} \mathrm{~ms}^{-1} \text {, respectively. }\)
Although the jogger has been moving with a constant speed, the speed of his/her image appears to increase substantially as he/she moves closer to the mirror. This phenomenon can be noticed by any person sitting in a stationary car or a bus. In case of moving vehicles, a similar phenomenon could be observed if the vehicle in the rear is moving closer with a constant speed.

(ii) As, \(\frac{F}{L}=\frac{\mu_0}{4\pi}.\frac{2 I_1 I_2}{r}\)
\(I_1=I_2=I A, r=1 m\)
\(\frac{F}{L}=2\times10^{-7} Nm^{-1}\)
(iii) Here, magnetic field due to the current carrying conductor at a distance d from it is
given by
\(B=\frac{\mu_0}{4\pi} \frac{2I}{d}\)
∴  Force on proton,
F = (e) (v) B sin 90°
⇒ F = evB
\(F=ev(\frac{\mu_0}{4\pi}\frac{2I}{d})\)
\(F=\frac{\mu_0}{4\pi}.\frac{2Iev}{d}\)
The proton is directed perpendicular to straight conductor and away from it.

The first refracting ABC forms the image I₁ of the object O. The image I₁ acts as virtual object for the second refracting surface ADC, which forms the real image I as shown in the diagram
For refraction at ABC 
\(\frac { { n }_{ 2 } }{ { v }_{ 1 } } -\frac { { n }_{ 1 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 1 } } \)
For refraction at ADC
\(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ { v }_{ 1 } } =\frac { { n }_{ 1 }-{ n }_{ 2 } }{ { R }_{ 2 } } \)
Adding equation (i) and equation (ii)
 \(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
We know, If  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\) 

Let \(\sigma\) be the surface charge density of capacitor plates.
for mica \(E_1={\sigma\over K_1\epsilon_0}\)
and for fibre \(E_2={\sigma\over K_2\epsilon_o}\)  or \({E_1\over E_2}={K_2\over K_1}\)
As \(E_s=6.4\times 10^6V/m\)
\(E_1={K_2\over K_1}\times E_2={2.5\over 8}\times 6.4\times 10^6=2\times 10^6V/m\)
Maximum voltage or capacitor
\(V=E_1d_1+E_2d_2=2\times 10^3+3.2\times 10^3=5200V\)

\(Here,\lambda =5000\overset { \circ }{ A } =5\times { 10 }^{ -7 }m;\)
Energy of one photon,
\(E=\frac { hc }{ \lambda } =\frac { \left( 6.63\times { 10 }^{ -34 } \right) \left( 3\times { 10 }^{ 8 } \right) }{ 5\times { 10 }^{ -7 } } =3.96\times { 10 }^{ -19 }J\)
A 200 W bulb supplies 200 J of energy per second.Energy emitted by lamp per second as visible light,
\({ E }_{ 1 }=200\times \frac { 10 }{ 100 } =20J{ s }^{ -1 }\)
Number of photons emitted per second as visible light.
 \(N=\frac { { E }_{ 1 } }{ E } =\frac { 20 }{ 3.96\times { 10 }^{ -19 } } =5.05\times { 10 }^{ 19 }\)

Here, e = 0.4 V,
dI = 10-(-10) = 20A, dt = 0.40s; L = ?
From e = L dI/dt, \(L=\frac { e\quad dt }{ dI } =\frac { 0.4\times 0.4 }{ 20 } =8\times { 10 }^{ -3 }H\)