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

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

Potentiometer does not draw any (net) current front-the cell. Voltmeter draws some current from cell, when connected across at, hence measures terminal voltage.

Range of frequency suitable for space wave propagation is 100 MHz to 220 MHz.

No current flows through the galvanometer G when circuit is balanced.

On distributing currents as per Kirchhoff's first rule.
Applying Kirchhoff's second rule
(i) In mesh ABDA,
∴ -I₁R₁ + (I - I₁)R₄ = 0
⇒ I₁R₁ = (I - I₁)R₄    .......(i)
(ii) In mesh BCDB,
-I₁R₂ + (I - I₁)R₃ = 0
I₁R₂ = (I - I₁)R₃         .......(ii)
On dividing Eq. (i) by Eq. (ii), we get
\(\frac { { I }_{ 1 }{ R }_{ 1 } }{ { I }_{ 1 }{ R }_{ 2 } } =\frac { (I-{ I }_{ 1 }){ R }_{ 4 } }{ (I-{ I }_{ 1 }){ R }_{ 3 } } \)
\(\frac { { R }_{ 1 } }{ { R }_{ 2 } } =\frac { { R }_{ 4 } }{ { R }_{ 3 } } \)
This is necessary and required balanced condition of balanced Wheatstone bridge.

Infrared waves have the wavelength between 1 mm to 700 nm.
Application:
(i) In knowing the molecular structure therapy to heat muscular pain.
(ii) In remote control of TV, VCR, etc.

The photocurrent increase linearly with the intensity of incident radiation.

The electric field due to a line charge falls of with distance as 1/r.

The reflected and refracted rays will be perpendicular when \(i={ i }_{ p },\)
where \(\tan { { i }_{ p } } =\mu =\sqrt { 3 } \ \therefore \ { i }_{ p }={ 60 }^{ ° }\)

\(X_L=\omega L=2 \pi v L,\)

(i) Constantan [Cu (60%) + Ni (40%)]
(ii) Manganin [Cu (84%) + Mn (12%) + Ni (4%)]

Input Resistance r_i = \({ \left( \frac { \Delta { V }_{ BE } }{ \Delta { I }_{ B } } \right) }_{ { V }_{ CE } }\)
= \(\frac { \left( 0.82-0.72 \right) V }{ \left( 80-30 \right) \times { 10 }^{ -6 } } \)
= \(\frac{0.10}{50\times{10}^{-6}}\)
= \(\frac{100\times{10}^{3}}{50}\)
= 2000 Ω

Given, velocity of electromagnetic wave in vacum, c = 3 X 10⁸ m / s
Relative electric permeability,\(\varepsilon _{ r }\) = 2
and magnetic permeability, \(\mu _{ r }\) = 1
Since, velocity of electromagnetic wave in a medium can be calculated by
v = \(\frac { 1 }{ \sqrt { \varepsilon _{ 0 }\varepsilon _{ r }\mu _{ 0 }\mu _{ r } } } =\frac { 1 }{ \sqrt { \varepsilon _{ 0 }\mu _{ 0 } } \quad x \quad \sqrt { \mu _{ r }\varepsilon _{ r } } } \)
Where,
\(\frac { 1 }{ \sqrt { \varepsilon _{ 0 }\mu _{ 0 } } } =c \ \Rightarrow \ v \ = \ \frac { c }{ \sqrt { \mu _{ r }\varepsilon _{ r } } } \)     ...(i)
Therefore, v \(=\frac { 3x10^{ 8 } }{ \sqrt { 2x1 } } \Rightarrow \) \(v=\frac { 3 }{ \sqrt { 2 } } \times10^{ 8 }\) m/s.

(a) Coverage range
\(d=\sqrt { 2hR } =\sqrt { 2\times160\times6.4\times{ 10 }^{ 6 } }\)
\(=45255m=45.255\ km\)
(b) Population Covered = \(\rho \times \pi d^{ 2 }\)
\(=1200\times\frac { 22 }{ 7 } \times(45.255){ 10 }^{ 6 }\)
\(=77.24\times{ 10 }^{ 5 }=77.24\ lakh\)
(c) \(d^{ ' }=\sqrt { 2h^{ ' }R } =2d=2\sqrt { 2hr } \)
\(\ or \ { h }^{ ' }=4h=4\times160=640m\)
Increase in height of tower =\(h^{ ' }=h\)
\(=640-160=480m\)

(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 
\(e{ V }_{ 0 }=\frac { hc }{ \lambda } -{ \phi }_{ 0 }\)
\(or\ \lambda =\frac { hc }{ \left( e{ V }_{ 0 }+{ \phi }_{ 0 } \right) }\)
\(or \ \lambda =\frac { \left( 6.63\times { 10 }^{ -34 }Js \right) \times \left( 3\times { 10 }^{ 8 }m/s \right) }{ \left( e\times 0.6V+2.14eV \right) } \)
\(\lambda =\frac { 19.89\times { 10 }^{ -26 }Jm }{ 2.74\times 1.6\times { 10 }^{ -19 }J } =454nm\)

Electric potential at middle point V = V₁ + V₂
\(V={1\over 4\pi\epsilon_o}[{q_1\over r_1}+{q_2\over r_2}]\)
\(=9\times10^9[{10\times 10^{-6}\over 0.01}+{20\times10^{-6}\over 0.01}]\)
\(=27\times 10^6V=27 MV\) 

Focal length of the convex lens, f₁ = 30 cm
Focal length of the concave lens, f₂ = -20 cm
Distance between the two lenses, d = 8.0 cm
(a) When the parallel beam of light is incident on the convex lens first:
According to the lens formula, we have:
\(\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { f }_{ 1 } } \)
Where,
u₁ = Object distance = ∞
v₁ = Image distance
\(\frac { 1 }{ { v }_{ 1 } } =\frac { 1 }{ 30 } -\frac { 1 }{ \infty } =\frac { 1 }{ 30 } \)
∴ v₁ = 30 cm
The image will act as a virtual object for the concave lens.
Applying lens formula to the concave lens, we have:
\(\frac { 1 }{ { v }_{ 2 } } -\frac { 1 }{ { u }_{ 2 } } =\frac { 1 }{ { f }_{ 2 } } \)
Where,
u₂ = Object distance
= (30 - d) = 30 - 8 = 22 cm
v₂= Image distance
\(\frac { 1 }{ { v }_{ 2 } } =\frac { 1 }{ 22 } -\frac { 1 }{ 20 } =\frac { 10-11 }{ 220 } =\frac { -1 }{ 220 } \)
∴ v₂ = -220 cm
The parallel incident beam appears to diverge from a point that is \(\left( 220-\frac { d }{ 2 } =220-4 \right) 216\) cm from the centre of the combination of the two lenses.
(ii) When the parallel beam of light is incident, from the left, on the concave lens first:
According to the lens formula, we have:
\(\frac { 1 }{ { v }_{ 2 } } -\frac { 1 }{ { u }_{ 2 } } =\frac { 1 }{ { f }_{ 2 } } \)
\(\frac { 1 }{ { v }_{ 2 } } =\frac { 1 }{ { f }_{ 2 } } +\frac { 1 }{ { u }_{ 2 } } \)

Where,
u₂ = Object distance = -∞
v₂ = Image distance
\(\frac { 1 }{ { v }_{ 2 } } =\frac { 1 }{ -20 } +\frac { 1 }{ -\infty } =-\frac { 1 }{ 20 } \)
∴ v₂ = -20 cm
The image will act as a real object for the convex lens.
Applying lens formula to the convex lens, we have:
\(\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { f }_{ 1 } } \)
Where,
u₁ = Object distance
= -(20 + d) = -(20 + 8) = -28 cm
v₁ = Image distance
\(\frac { 1 }{ { v }_{ 1 } } =\frac { 1 }{ 30 } +\frac { 1 }{ -28 } =\frac { 14-15 }{ 420 } =\frac { -1 }{ 420 } \)
∴ v₂ = - 420 cm
Hence, the parallel incident beam appear to diverge from a point that is (420 - 4) 416 cm from the left of the centre of the combination of the two lenses
The answer does depend on the side of the combination at which the parallel beam of light is incident. The notion of effective focal length does not seem to be useful for this combination.
(b) Height of the image, h₁ = 1.5 cm
Object distance from the side of the convex lens, u₁ = -40 cm
|u₁| = 40 cm
According to the lens formula:
\(\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { f }_{ 1 } } \)
Where,
v₁ = Image distance
\(\frac { 1 }{ { v }_{ 1 } } =\frac { 1 }{ 30 } +\frac { 1 }{ -40 } =\frac { 4-3 }{ 120 } =\frac { 1 }{ 120 } \)
∴ v₁ = 120 cm
 \(m=\frac { { v }_{ 1 } }{ \left| { u }_{ 1 } \right| } \)
= \(\frac { 120 }{ 40 } =3\)
Hence, the magnification due to the convex lens is 3.
The image formed by the convex lens acts as an object for the concave lens.
According to the lens formula:
\(\frac { 1 }{ { v }_{ 2 } } -\frac { 1 }{ { u }_{ 2 } } =\frac { 1 }{ { f }_{ 2 } } \)
Where,
u₂ = Object distance
= +(120 - 8) = 112 cm.
v₂ = Image distance
\(\frac { 1 }{ { v }_{ 2 } } =\frac { 1 }{ -20 } +\frac { 1 }{ 112 } =\frac { -112+20 }{ 2240 } =\frac { -92 }{ 2240 } \)
∴ v₂ = \(\frac { 2240 }{ 92 } \) cm
Magnification, \({ m }^{ ' }=\left| \frac { { v }_{ 2 } }{ { u }_{ 2 } } \right| \)
\(=\frac { 2240 }{ 92 } \times \frac { 1 }{ 112 } =\frac { 20 }{ 92 } \)
Hence, the magnification due to the concave lens is \(\frac { 20 }{ 92 } \)
The magnification produced by the combination of the two lenses is calculated as:
m x m'
\(=3\times \frac { 20 }{ 92 } =\frac { 60 }{ 92 } =0.652\)
The magnification of the combination is given as:
\(\frac { { h }_{ 2 } }{ { h }_{ 1 } } =0.652\)
h₂ = 0.652 x h₁
Where,
h₁ = Object size = 1.5 cm
h₂ = Size of the image
∴  h₂ = 0.652 x 1.5 = 0.98 cm
Hence, the height of the image is 0.98 cm

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

\(Here,\quad N=200,\quad A=15\times 40{ cm }^{ 2 }=600\times { 10 }^{ -4 }{ m }^{ 2 }\)
\(A=15\times 40{ cm }^{ 2 }=600\times { 10 }^{ -4 }{ m }^{ 2 },\quad v-50r.p.s.,\quad B=0.08Wb/{ m }^{ 2 },\quad e=?\)
\(\\ { e }_{ 0 }=NAB\omega =NAB\left( 2\pi v \right) =\left( 200 \right) \times \left( 6\times { 10 }^{ -2 } \right) \times 0.08\times 2\times \frac { 22 }{ 7 } \times 50=301.7volt\)
In the question, angle are given between the plane of the coil and magnetic field. We have to use angles which normal to the plane of coil makes with the field. Therefore, from e = e₀ sin \(\omega \)t = e₀ sin\(\theta \)
\((a) \ Here \ \theta ={ 90 }^{ \circ }-{ 0 }^{ \circ }={ 90 }^{ \circ }\)
\(\\ \therefore \ e=301.7 \ sin \ { 90 }^{ \circ }=301.7 \ volt\)
\( (b) \ Here, \ \theta ={ 90 }^{ \circ }-{ 60 }^{ \circ }={ 30 }^{ \circ }\)
\( \therefore \ e=301.7sin \ { 30 }^{ \circ }=301.7\times \frac { 1 }{ 2 } =150.85 \ volt\)
\( (c) \ Here, \ \theta ={ 90 }^{ \circ }-{ 90 }^{ \circ }={ 0 }^{ \circ }\)
\(\therefore \ e=301.7sin \ { 0 }^{ \circ }=zero\)