A = 60⁰ ,  \(\delta \)_m = 30⁰
i = e = ¾ A = 450
as A + \(\delta \) = i + e
60 + \(\delta \) = 45 +45
or \(\delta \) = 30⁰
Refractive index,
\(\mu \) = sin a + \(\delta \)m /2/sin A/2 = sin 60⁰+30⁰/2/sin 60⁰/2
= sin 45⁰/sin30⁰ = 1\(\surd 2\) 1/2 = \(\surd 2\) = 1.414

B1 = 4 x \({ \mu }_{ 0 }\) I/4\(\pi\) a/2 (sin 45  + sin 45)
= 4 x \({ \mu }_{ 0 }\)  2/4 \(\pi\) x 3  (1 / 1.414  + 1/414 )
= 2\({ \mu }_{ 0 }\)/3\(\pi\)  (1/.414 ) + 1/414 ) T

V = V1 + V2
Q = C1V = 6 x 10^-6  x 2 = 12\(\mu \)C
As C₂ is in series same amount of charge will also flow through it now V₂ = Q/C₂ = (12 x 10^-6) /(12 x 10^-6) = 1 volt
Total Battery voltage, V = 2 + 1 = 3 Volt

Let two point charges q₁ and q₂ are situated at points A and B have position vectors r₁ and r₂.

\(\because\)  AP = r - r₁
and BP = r - r₂
Electric field intensity at point P due to q1,
E₁ = \({{1}\over{4\pi{\epsilon}_{0}}}.{{{q}_{1}}\over{{|BP|}^{3}}}AP\)
Similarly, E₂ = \({{1}\over{4\pi{\epsilon}_{0}}}.{{{q}_{1}}\over{{|AP|}^{3}}}BP\)
\(\therefore\)  Net electric field intensity at point P,
E = E₁ + E₂
= \({{1}\over{4\pi{\epsilon}_{0}}}\left[ {{{q}_{1}}\over{{|r-{r}_{1}|}^{3}}}(r-{r}_{1})+{{{q}_{2}}\over{{|r-{r}_{2}|}^{3}}} (r-{r}_{2}) \right]\).

(i) Heat produced in the circuit is directly proportional to the resistance if a constant current is flowing through a circuit, because H = I²Rt or H \(\propto\) R. It is so in series combination of resistors.
(ii) Heat produced in the circuit is inversely proportional to the resistance if a constant pot. diff. is applied across the circuit, because
\(H=\frac{V^2}{R}t \)
or \(H \propto \frac{1}{R}\)
It is so in parallel combination of resistors.

In photoelectric effect, We can observe that most electrons get scattered into the metal by absorbing a photon.
Thus, all the electrons that absorb a photon does not come out as photoelectron. Only a few comes out of metal whose energy becomes greater than the work functibn of metal.

(i) Capacitors are used in radio circuits for tuning purposes.
(ii) Capacitors are used in power supplies for smoothing the rectified current.

If amplitude of modulating signal \(({ A }_{ m })\)exceeds the amplitude of carrier wave \(({ A }_{ c })\) is the carrier is over modulated \(({ \mu }_{ a }>1)\) .This causes distortion during reception.

(i) Current capacity from the combination of cells is the total current available from the cell. Current capacity increases for the parallel combination of cells.
(ii) The effect e.m.f. of the cells of equal e.m.f. in parallel will be equal to e.m.f. of one cell.

The NOT gate is a device which has only one input and one output i.e. \(\bar { A } \)= Y means Y equals NOT A.
This gate cannot be realised by using diodes. However, it can be realised by making the use of a transistor. This can be seen in the figure given below:

Here, the base B of the transistor is connected to the input A through a resistance R_B and the emitter E is earthed. The collector is connected to 5 V battery. The output Y is the voltage at C with respect to the earth.
The resistors, R_B and R_C are so chosen that, if emitter-base junction is unbiased, the transistor is in cut-off mode and if emitter-base junction is forward biased by 5 V, the transistor is in saturation state.

Here \(C_1=C_2=800pF=8\times 10^{-10}F\)
\(V_1=100V, V_2=0\)
Common potential, V = \({C_1V_1+C_2V_2\over C_1+C_2}\)
\(={8\times 10^{-10}\times 100\over (800+800)10^{-12}}=50V\)
Energy stored, \(u_f={1\over 2}(C_1+C_2)V^2\)
\(={1\over 2}(800+800)10^{-12}(50)^2\)
\(u_f=2\times 10^{-6}J\) 

For v = 6MHz ,\(\lambda =\frac { c }{ v } =\frac { 3\times{ 10 }^{ 8 } }{ 6\times{ 10 }^{ 6 } } =50\ m\)
for  v = 12 MHz,\(\lambda =\frac { c }{ v } =\frac { 3\times{ 10 }^{ 8 } }{ 12\times{ 10 }^{ 6 } } =25\ m\)
Thus the corresponding wavelength band is 25 m to 50 m band.

\(Here,\ r=8cm=8\times { 10 }^{ -2 }m\)
\(\frac { dE }{ dt } =​​{ 10 }^{ 13 }V{ m }^{ -1 }{ s }^{ -1 }\)
Displacement current.
\({ I }_{ D }={ \epsilon }_{ 0 }\frac { { d\phi }_{ E } }{ dt } { { \epsilon }_{ 0 } }\frac { d }{ dt } (EA)\)
\( ={ \epsilon }_{ 0 }A\frac { { d }E }{ dt } ={ { \epsilon }_{ 0 }\pi { r }^{ 2 } }\frac { dE }{ dt } \)
\( =(8.85\times { 10 }^{ -12 })\times 3.14\times ({ 8\times { 10 }^{ -2 }) }^{ 2 }\times { 10 }^{ 13 }\)
\( =1.78\ A\)

\(7.896 \times { 10 }^{ -5 }H;19.74mV\)

Area of the virtual image of each square, A = 6.25 mm²
Area of each square, A₀ = 1 mm²
Hence, the linear magnification of the object can be calculated as:
\(m=\sqrt { \frac { A }{ { A }_{ o } } } \)
\(\sqrt { \frac { 6.25 }{ 1 } } =2.5\)
But m = \(\frac{Image \ distance \ (v)}{Object \ distance \ (u)}\)
∴ v = mu
= 2.5 u ....(1)
Focal length of the magnifying glass, f = 10 cm
According to the lens formula, we have the relation:
\(\frac { 1 }{ { f } } =\frac { 1 }{ { v } } -\frac { 1 }{ { u } } \)
\(\frac { 1 }{ 10 } =\frac { 1 }{ 2.5u } -\frac { 1 }{ u } =\frac { 1 }{ u } \left( \frac { 1 }{ 2.5 } -\frac { 1 }{ 1 } \right) =\frac { 1 }{ u } \left( \frac { 1-2.5 }{ 2.5 } \right) \)
\(\therefore u=-\frac { 1.5\times 10 }{ 2.5 } =-6\)
And v = 2.5u
= 2.5 x 6 = -15 cm
The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.

The focal length f = -15/2 cm = -7.5 cm
(i) The object distance u = -10 cm. Then Eq gives
\(\frac { 1 }{ v } +\frac { 1 }{ 10 } =\frac { 1 }{ -7.5 } \)
or \(v=\frac { 10\times 7.5 }{ -2.5 } =-30\) cm
The image is 30 cm from the mirror on the same side as the object
Also, magnification m = \(\frac { v }{ u } =-\frac { (-30) }{ (-10) } =-3\)
The image is magnified, real and inverted.
(ii) The object distance u = -5 cm. Then from Eq
\(\frac { 1 }{ v } +\frac { 1 }{ -5 } =\frac { 1 }{ -7.5 } \)
or \(v=\frac { 5\times 7.5 }{ (7.5-5) } =15\) cm
This image is formed at 15 cm behind the mirror. It is a virtual image.
Magnification m = 15 \(-\frac { v }{ u } =-\frac { 15 }{ (-5) } =3\)
The image is magnified, virtual and erect.

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  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\) 

i) Mr.Santosh is helping in nature, honest and has concern for the students to create interest in the subject.
ii) Magnetic declination, magnetic inclination and horizontal component of the earth's magnetic field.
iii) It is of the order of 10^-5 T.

Here, two parallel plate capacitors have same charge q and same potential V, so they have equal capacitances as
C = q/V
C₁ = C₂
\({\epsilon_oA_1\over d_1}={\epsilon_oA_2\over d_2}\)
or  \(d_2={A_2\over A_1}d_1\)
Now, \(A_1=100 cm^2, A_2=500cm^2\)
\(d_1=0.5mm=0.05cm\)
\(d_2={500\times 0.05\over 100}=0.25cm=2.5mm\)

Rest mass of electron,
\({ m }_{ 0 }=9.1\times { 10 }^{ -31 }kg\)
\(v=0.6c=0.6,\times 3\times { 10 }^{ 8 }=1.8\times { 10 }^{ 8 }{ ms }^{ -1 }\)
As v is comparable to c, hence mass of the electron in motion will be a relativistic mass. So,
\(m=\frac { { m }_{ 0 } }{ \sqrt { 1-{ v }^{ 2 }/{ c }^{ 2 } } } =\frac { { m }_{ 0 } }{ \sqrt { 1-\frac { { \left( 0.6 \right) }^{ 2 } }{ { c }^{ 2 } } } } =\frac { { m }_{ 0 } }{ 0.8 } \)
De-Broglie wavelength,
\(\lambda =\frac { h }{ mv } =\frac { h }{ \left( { m }_{ 0 }/0.8 \right) \times v } =\frac { h\times 0.8 }{ { m }_{ 0 }v } \)
\(=\frac { \left( 6.6\times { 10 }^{ -34 } \right) \times 0.8 }{ \left( 9.1\times { 10 }^{ -31 } \right) \times \left( 1.8\times { 10 }^{ 8 } \right) }\)
\( =0.322\times { 10 }^{ -11 }m=0.0322\times { 10 }^{ -10 }m\)
\(=0.032\overset { \circ }{ A } \)