The small region in the vicinity of the junction which is depleted of free charge carriers and has e only immobile ions is called depletion region. The accumulation of negative charges in the p-region and positive charge in the n-region sets up a potential difference across the junction. This acts as a barrier and is called barrier potential V_B.
(ii) 
Working
(a) During positive half cycle of input alternating voltage, the diode is forward biased and a current flows through the load resistor RL and we get an output voltage.
(b) During other negative half cycle of the input alternating voltage, the diode is reverse biased and it does not conduct (under break down region).
Hence, AC voltage can be rectified in the pulsating and unidirectional voltage.

Electric energy consumed per second = 100 J
Heat produced = 80% = \(\frac{80}{100}\times 100\)
= 80 J = \(\frac{80}{4.2}\)
= 19.05 cal

Here, M = 2.5 Am², m = 6.6 g = 6.6 x 10^-3 kg
\(\rho =7.9\times { 10 }^{ 3 }kg/{ m }^{ 3 },\) 
\(V=\frac { m }{ \rho } =\frac { 6.6\times { 10 }^{ -3 } }{ 7.9\times { 10 }^{ 3 } } =0.835\times { 10 }^{ -6 }{ m }^{ 3 }\) 
\(I=\frac { M }{ V } =\frac { 2.5 }{ 0.835\times { 10 }^{ -6 } } =3.0\times { 10 }^{ 6 }{ Am }^{ -1 }\)

 Here, \(N=500, \ A=50{ cm }^{ 2 }=50\times { 10 }^{ -4 }{ m }^{ 2 }\)
\(B=0.14Wb/{ m }^{ 2 },\omega =150rad/s,\)
\(R=5\Omega \quad { R }^{ ' }=10\Omega \quad { i }_{ 0 }=?\)
\({ e }_{ 0 }=NAB\omega =500\times 50\times { 10 }^{ -4 }\times 0.14\times 150\)
\(=52.5volt\)
As total resistance \(=R+{ R }^{ ' }=5+10=15\Omega \)
\(\therefore { \ i }_{ 0 }=\frac { { e }_{ 0 } }{ 15 } =\frac { 52.5 }{ 15 } =3.5A\)

\(1.22\times { 10 }^{ 4 }V\)

(i) As the lens forms a real iamge, it must be a convex lens.
(ii) From the graph, when u = 20 cm , we have v = 20 cm.
For the convex lens forming a real iamge, u is negative and v and f are positive.
U = -20 cm v = +20cm
Using this lens formula,
1/f = 1/v – 1/u = 1/20 – 1/-20 = 1/10 or f = + 10 cm

As \({ f }_{ 0 }\) is the threshold frequency, so \({ { \phi }_{ 0 }=hf }_{ 0 }\)
Using Einstein's photoelectric equation, we have 
\(\frac { 1 }{ 2 } { mv }_{ 1 }^{ 2 }=h\times 2{ f }_{ 0 }-{ hf }_{ 0 }={ hf }_{ 0 }\)
\(and \ { \frac { 1 }{ 2 } }{ mv }_{ 2 }^{ 2 }=h\times 5{ f }_{ 0 }-{ hf }_{ 0 }=4{ hf }_{ 0 }\)
\(\\ \frac { { v }_{ 1 }^{ 2 } }{ { v }_{ 2 }^{ 2 } } =\frac { 1 }{ 4 } \ or \ \frac { { v }_{ 1 } }{ { v }_{ 2 } } =\frac { 1 }{ 2 } \)

(a) For a concave mirror, the focal length (f) is negative.
∴ f < 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴ u < 0
For image distance v, we can write the lens formula as:
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f }\)
\(\frac { 1 }{ v } =\frac { 1 }{ f } -\frac { 1 }{ u } \)  ....(1)
The object lies between f and 2f.
∴ 2f < u < f (∵ u and f are negative)
\(\frac { 1 }{ 2f } >\frac { 1 }{ u } >\frac { 1 }{ f } \)
\(-\frac { 1 }{ 2f } <-\frac { 1 }{ u } <\frac { 1 }{ f } \)
\(\frac { 1 }{ f } -\frac { 1 }{ 2f } <\frac { 1 }{ f } -\frac { 1 }{ u } <0\)  ...(2)
Using equation (1), we get:
\(\frac { 1 }{ 2f } <\frac { 1 }{ v } \)
2f > v
-v > - 2f
Therefore, the image lies beyond 2f.
(b) For a convex mirror, the focal length (f) is positive.
∴ f > 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴ u < 0
For image distance v, we have the mirror formula:
\(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ v } =\frac { 1 }{ f } -\frac { 1 }{ u } \quad \)
Using equation (2), we can conclude that:
\(\frac { 1 }{ v } \) < 0
v > 0
Thus, the image is formed on the back side of the mirror.
Hence, a convex mirror always produces a virtual image, regardless of the object distance.
(c) For a convex mirror, the focal length (f) is positive.
∴ f > 0
When the object is placed on the left side of the mirror, the object distance (u) is negative,
∴ u < 0
For image distance v, we have the mirror formula:
\(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ v } =\frac { 1 }{ f } -\frac { 1 }{ u } \)
But we have u < 0
∴ \(\frac { 1 }{ v } >\frac { 1 }{ f } \)
v < f
Hence, the image formed is diminished and is located between the focus (f) and the pole.
(d) For a concave mirror, the focal length (f) is negative.
∴ f < 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴ u < 0
It is placed between the focus (f) and the pole.
∴ f > u > 0
\(\frac { 1 }{ f } <\frac { 1 }{ u } \)< 0
\(\frac { 1 }{ f } -\frac { 1 }{ u } \)< 0
For image distance v, we have the mirror formula:
\(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ v } =\frac { 1 }{ f } -\frac { 1 }{ u } \)
∴\(\frac { 1 }{ v } \) < 0
v > 0
The image is formed on the right side of the mirror. Hence, it is a virtual image
For u < 0 and v > 0, we can write:
\(\frac { 1 }{ u } >\frac { 1 }{ v } \)
v > u
Magnification, m = \(\frac { u }{ v } \) > 1
Hence, the formed image is enlarged.

Size of the candle, h = 2.5 cm
Image size = h’
Object distance, u = -27 cm
Radius of curvature of the concave mirror, R = -36 cm
\(f=\frac { R }{ 2 } =-18\)cm
Image distance = v
The image distance can be obtained using the mirror formula:
\(\frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } \)
\(\frac { 1 }{ v } =\frac { 1 }{ f } -\frac { 1 }{ u } \)
\(\frac { 1 }{ -18 } =\frac { 1 }{ -27 } =\frac { -3+2 }{ 54 } =-\frac { 1 }{ 54 } \)
∴ v = -54 cm
Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.
The magnification of the image is given as:
\(m=\frac { { h }^{ ' } }{ h } =-\frac { v }{ u } \)
\(\therefore { h }^{ ' }=-\frac { v }{ u } \times h\)
\(=-\left( \frac { -54 }{ -27 } \right) \times 2.5=-5\)cm
The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and real.
If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.

(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.

There are number of energy losses in a transformer. Copper losses due to Joule's heating produced across the resistance of primary and secondary coils. It can be reduced by using copper wires.

Hysteresis losses due to repeated magnetisation and demagnetization of the core of transformer. It is minimised by using soft iron core, as area of hysteresis loop for soft iron is small and hence energy loss also becomes small. 
Iron losses due to eddy currents produced in soft iron core. It is minimised by using laminated iron core.
Flux losses due to flux leakage or incomplete flux linkage and can be minimised by proper coupling of primary and secondary coils.
Here, N_p = 100, \({ { {N}_{s} }\over{ {N}_{p} } }=100\)
\({\epsilon}_{i}={\epsilon}_{p}=220V,{P}_{I}=1100\ W\)
Np = 100 \(\therefore\) Ns = 10000
\({I}_{p}={{{P}_{I}}\over{{\epsilon}_{p}}}={{1100}\over{220}}=5A\)
\({\epsilon}_{s}={{{N}_{s}}\over{{N}_{p}}}\times{\epsilon}_{r}=100\times 220=22000 \ V\)
\({I}_{s}={{{P}_{0}}\over{{\epsilon}_{s}}}={{1100}\over{22000}}={{1}\over{20}}A\)  \((\because {P}_{0}={P}_{1})\)
P_s = P₀ = P_I = 1100 W

The magnitudes of the electric fields due to the two charges + q and - q are given by
\(E_{+q} = \frac{q}{4\pi\epsilon_0}\frac{1}{r^2+2a^2}\)
\(E_{-q} = \frac{q}{4\pi\epsilon_0}\frac{1}{r^2+2a^2}\)
and both are equal The directions of E_+q and E_-q are as shown in fig.  Clearly, the components normal to the dipole axis cancel away. The components along the dipole axis add up. The total electric field is opposite to \(\hat{p}\) . We have
E = - (E_+q+ E_-q) cos \(\theta\) \(\hat{p}\)
   \(= \frac{-2qa}{4\pi\epsilon_0(r^2+a^2)^\frac{3}{2}} \hat{p}\)
At large diatances (r > > a), this reduces to
\(= \frac{-2qa}{4\pi\epsilon_0(r^3)} \hat{p}\) (r > > a)
Thus, the graph takes the form as shown below:

Electric dipole of charges +q and -q separated by distance 2a is shown in figure. It is placed in uniform electric field at an angle \(\theta\) with it.

Torque on dipole= force x perpendicular distance
 = qE x 2a sin\(\theta\)
 = 2 qa E sin \(\theta\)
 = pEsin \(\theta\)
  \(\overrightarrow\tau = \overrightarrow{p}\times\overrightarrow{E}\) 

\(Here, \ N=100, \ A=0.1{ m }^{ 2 }, \ v=50 \ rpm=\frac { 5 }{ 6 } rps\)
\(\overset { \rightarrow }{ B } =\left( 0.01 \right) { k } T,i.e.,\ B=0.01\ T \ along \ z-axis\)
\(\\ { e }_{ 0 }=NAB\omega =100\times 0.1\times 0.01\times 2\pi \times \frac { 5 }{ 6 } =0.52V\)
(ii) As the e.m.f. generated varies sinusoidally with time, so the average e.m.f. generated in the coil-over one complete revolution is zero.