For D-region, N = 10⁹ m^-3 Here, v = 60 x 10⁶ Hz
Refractive index of the atmospheric layer is
\(\mu =\left[ 1-\frac { 81.45N }{ { v }^{ 2 } } \right] ^{ 1/2 }=\sqrt { 1-\frac { 81.45\times { 10 }^{ 9 } }{ \left( 60\times { 10 }^{ 6 } \right) ^{ 2 } } } \approx 1\)
Now, \(\mu =\frac { \sin { i } }{ \sin { r } } =1\) or  \(\sin { i } =\sin { r } \) or r = i = 30°

\({\epsilon_oA\over A}={4\pi\epsilon_o R}\)
or  \({\epsilon_o\pi(D^2)\over d4}=4\pi\epsilon_oR\)
\(d={D^2\over 16R}={(0.08)^2\over 16\times 0.10}=4\times 10^{-3}m=4mm\) 

(a) For the cut-off or threshold frequency, the energy h v₀ of the incident radiation must be equal to work function  Φ₀, so that
\({ V }_{ 0 }=\frac { { \phi }_{ 0 } }{ h } =\frac { 2.14eV }{ 6.63\times { 10 }^{ -34 }Js }\)
\(=\frac { 2.14\times 1.6\times { 10 }^{ -19 }J }{ 6.63\times { 10 }^{ -34 }Js } =5.16\times { 10 }^{ 14 }Hz\)
Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.
(b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the potential energy eV₀ by the retarding potential V₀. Einstein’s Photoelectric equation is 
\(e{ V }_{ 0 }=\frac { hc }{ \lambda } -{ \phi }_{ 0 }\)
\(or\ \lambda =\frac { hc }{ \left( e{ V }_{ 0 }+{ \phi }_{ 0 } \right) }\)
\(or \ \lambda =\frac { \left( 6.63\times { 10 }^{ -34 }Js \right) \times \left( 3\times { 10 }^{ 8 }m/s \right) }{ \left( e\times 0.6V+2.14eV \right) } \)
\(\lambda =\frac { 19.89\times { 10 }^{ -26 }Jm }{ 2.74\times 1.6\times { 10 }^{ -19 }J } =454nm\)

\(Here,N=200,A=0.30x0.05=0.015{ m }^{ 2 }\)
\({ B }_{ 1 }=0.20Wb{ m }^{ -2 }e=?\)
\({ B }_{ 2 }=0.15Wb{ m }^{ -2 }dt=0.02s\)
\( e=\frac { -N({ \Phi }_{ 2 }-{ \Phi }_{ 1 }) }{ dt } =\frac { -N({ B }_{ 2 }-{ B }_{ 1 })A }{ dt }\)
\(=\frac { -200(0.15-0.20)0.015 }{ 0.02 } =7.5V\)

Since \(\varepsilon _{ E }=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ { E }^{ 2 } }\) 
and \(\varepsilon _{ B }=\frac { { B }^{ 2 } }{ 2\mu _{ 0 } } \) 
\(\therefore \) Total energy density \(u={ u }_{ E }+{ u }_{ B }=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }_{ 2 }+\frac { 1 }{ 2 } \frac { { B }^{ 2 } }{ { \mu }_{ 0 } } \) 
For a plane em wave, Band E are related as
\({ B }_{ v }=\frac { { E }_{ V } }{ C } \) 
\(u=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }_{ v2 }+\frac { 1 }{ 2\mu _{ 0 } } \left( \frac { { E }_{ v } }{ c } \right) ^{ 2 }\) 
\(={ E^{ 2 } }_{ v }\frac { \left( { \mu }_{ 0 }\varepsilon _{ 0^{ c2 } }+1 \right) }{ 2{ \mu }_{ 0^{ c2 } } } ={ E }^{ 2 }_{ v }\) 
or      \(u=\frac { { E }^{ 2 }_{ v }(1+1) }{ 2{ \mu }_{ 0 }\frac { 1 }{ { \mu }_{ 0 }\varepsilon _{ 0 } } } \)   \(\left[ \therefore { c }^{ 2 }=\frac { 1 }{ { \mu }_{ 0 }\varepsilon _{ 0 } } \right] \) 
\(=\varepsilon _{ 0 }{ E }^{ 2 }_{ v }\) 
So     \(I=\frac { Energy/Time }{ Area } \) 
or
\(I=\frac { Energy \ density/Time }{ Area } \times volume\) 
\( =Energy\quad density\times \frac { length }{ Time } \)
\( =u\times c\)
Using eqn.(i), we get
\(\\ I=\varepsilon _{ 0 }{ E }^{ 2 }_{ v }\)

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) No effect
(ii) Current will decrease
(iii) Current will Increase.

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

(i) Microwaves are used for aircraft navigation, their frequency range is 10⁹ Hz to 10¹² Hz.
(ii) X-rays are used to study crystal structure, their frequency range is 10¹⁶ Hz to 10²⁰ Hz.

According to question,

(i) Net current in the circuit = 12/6 = 2A
Voltage across the battery,
V_b = 12 - 2 x 2 = 8V
Voltage across the resistance
Vr = IR = 2 x 4 = 8V
(ii) In order to measure the device's voltage for a voltmeter, it must be connected in parallel to that device. This is necessary because device in parallel experiences the same potential difference. An ammeter is connected in series
with the circuit because the purpose of the ammeter is to measure the current through the circuit. Since, the ammeter is a low impedance device. Connecting in parallel with the circuit would cause a short-circuit, damaging, the ammeter of the circuit.

R₁ = 10 cm, R₂ = -15 cm, f = 12 cm
\(\frac { 1 }{ f } =(\mu -1)(\frac { 1 }{ R_{ 1 } } -\frac { 1 }{ R_{ 2 } } )\)
\(\frac { 1 }{ 12 } =(\mu -1)(\frac { 1 }{ 10 } -\frac { 1 }{ 15 } )\)
\( \mu = \frac { 3 }{ 2 } \)

\({20}^{o}\)

\({ B }_{ 1 }=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2Md }{ { \left( { d }^{ 2 }-{ l }^{ 2 } \right) }^{ 2 } } ;along \ SN\) 
where symbols have usual meaning.

There will be no depletion in the galvanometer as the condition of balanced bridge will still hold good.

Transistor is said to be in active state when its emitter-base junction is suitably forward biased and base-collector junction is suitably reverse biased.

Electric field varies inversely as square of distance from the point charge.

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.

Focal length of the objective lens, f_o = 15 m = 15 x 10² cm
Focal length of the eyepiece, f_e = 1.0 cm
(a) The angular magnification of a telescope is given as:
α = \(\frac { { f }_{ o } }{ { f }_{ e } } \) = \(\frac { 15\times { 10 }^{ 2 } }{ 1.0 } \) = 1500
Hence, the angular magnification of the given refracting telescope is 1500.
(b) Diameter of the moon, d = 3.48 x 10⁶ m
Let d' be the diameter of the image of the moon formed by the objective lens.
Radius of the lunar orbit, r₀ = 3.8 x 10⁸ m
The angle subtended by the diameter of the moon is equal to the angle subtended by the image.
\(\frac { d }{ { r }_{ o } } =\frac { { d }^{ ' } }{ { f }_{ o } } \)
\(\frac { 3.48\times { 10 }^{ 6 } }{ 3.8\times { 10 }^{ 8 } } =\frac { { d }^{ ' } }{ 15 } \)
\(\therefore { d }^{ ' }=\frac { 3.48 }{ 3.8 } \times { 10 }^{ -2 }\times 15\)
= 13.74 x 10^-2 m = 13.74 cm
Hence, the diameter of the moon’s image formed by the objective lens is 13.74 cm.

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.

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

(a) Keen observer/helpful/concerned/responsible/respectful towards elders(Any two)
(b) The doctor can trace and observe,the difference between the moment of an appropriate through a normal brain and a brain having tumour in it.

\(V=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2} \times \frac{4 \times 10^{-7} \mathrm{C}}{0.09 \mathrm{~m}}\)
= 4 x 10⁴ V 
(b) W = qV = 2 x 10⁻⁹C x 4 x 10⁴ V
= 8 x 10^–5 J
No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along r and another perpendicular to r. The work done corresponding to the later will be zero.

\(Here, \ C=1.0\mu F={ 10 }^{ -6 }F\)
\(R={ 10 }^{ 4 }ohm, \ { E }_{ 0 }=2.0volt\)
\(During \ charging \ of \ the \ condenser,\)
\(q={ q }_{ 0 }\left( 1-{ e }^{ -t/RC } \right) \ I=-{ I }_{ 0 }{ e }^{ -t/RC }, \ where\)
\({ I }_{ 0 }=\frac { E }{ R } =\frac { 2.0 }{ { 10 }^{ 4 } } =2.0\times { 10 }^{ -4 }amp\)
\(At \ t=0.02s, \ I={ I }_{ 0 }\left( { e }^{ -0.02/{ 10 }^{ 4 }\times { 10 }^{ -6 } } \right) =2\times { 10 }^{ -4 }{ e }^{ -2 }\)
\(=\frac { 2\times { 10 }^{ -4 } }{ { e }^{ 2 } } =\frac { 2\times { 10 }^{ -4 } }{ \left( 2.718 \right) ^{ 2 } } =0.27\times { 10 }^{ -4 }A\)
\(I=27\times { 10 }^{ -6 } \ A=27\mu A\)