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

Electric field lines do not pass through a conductor.Hence,  the interior of the conductor is free from the influence of the electric field

Yes, this would happen when phase difference between alternating voltage and alternating current is \({ 90 }^{ \circ }.\) It can happen when the circuit contains pure L or pure C.

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.

(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.

\(4\mu C\)

The following figure shows a Cassegrain telescope consisting of a concave mirror and a convex mirror.

Distance between the objective mirror and the secondary mirror, d = 20 mm
Radius of curvature of the objective mirror, R₁ = 220 mm
Hence, focal length of the objective mirror,  \({ f }_{ 1 }=\frac { { R }_{ 1 } }{ 2 } =110\)
Radius of curvature of the secondary mirror, R₁ = 140 mm
Hence, focal length of the secondary mirror, \({ f }_{ 2 }=\frac { { R }_{ 2 } }{ 2 } =\frac { 140 }{ 2 } \) = 70 mm
The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror.
Hence, the virtual object distance for the secondary mirror, u = f₁ - d
= 110 - 20
= 90 mm
Applying the mirror formula for the secondary mirror, we can calculate image distance (v) as:
\(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ { f }_{ 2 } } \)
\(\frac { 1 }{ v } =\frac { 1 }{ { f }_{ 2 } } -\frac { 1 }{ u } \)
\(\frac { 1 }{ 70 } -\frac { 1 }{ 90 } =\frac { 9-7 }{ 630 } =\frac { 2 }{ 630 } \)
∴ v = \(\frac{630}{2}\) = 315 mm
Hence, the final image will be formed 315 mm away from the secondary mirror.

(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.

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) Niyaz shows the values of courage to tell truth and determination.
(ii) It is based on the principle when a current carrying coil placed in external magnetic field, it develops torque.
(iii) Sensitivity of galvanometer can be increased by 
(a) increasing number of turns in the coil and
(b) by increasing current in the coil.

\(Here,\lambda =300nm=300\times { 10 }^{ -9 }m\)
\(=3\times { 10 }^{ -7 }m,{ V }_{ 0 }=0.54V\)
(i) Energy of the incident photon,
\(E=\frac { hc }{ \lambda } =\frac { \left( 6.63\times { 10 }^{ -34 }Js \right) \times \left( 3\times { 10 }^{ 8 }m/s \right) }{ 3\times { 10 }^{ -7 } } =6.63\times { 10 }^{ -19 }J\)
\(=\frac { 6.63\times { 10 }^{ -19 } }{ 1.6\times { 10 }^{ -19 } } eV=4.14eV\)
(ii) Max. K.E. of emitted photoelectron is
\({ K }_{ max }=e{ V }_{ 0 }=e\times 54V=0.54eV\)
(iii) \({ As\ K }_{ max }=\frac { hc }{ \lambda } -{ \phi }_{ 0 }\ or{ \ \phi }_{ 0 }=\frac { hc }{ \lambda } -{ K }_{ max }\)      
\(=4.14eV-0.54eV=3.6eV\)

Here, N = 200, L = 10mH = 10\(\times\)10^-3H,  I = 4mA = 4\(\times\)10^-3 A;\(\phi \) = ?
Total magnetic flux linked with the coil \(\phi \) = NLI = 200\(\times\)(10\(\times\)10^-3)4\(\times\)10^-3 = 8\(\times\)10^-3 Wb.
Magnetic flux through the cross-section of the coil=magnetic flux linked with each turn
= \(\frac { \phi }{ N } =\frac { 8\times { 10 }^{ -3 } }{ 200 } =4\times { 10 }^{ -5 }Wb\)

Here, I = 3.3 A; B = 0.33 G = 0.33 x 10^-4 T, \(\delta \) = 0^o. Horizontal component of earth's magnetic field H = B cos \(\delta \) = 0.33 x 10^-4 cos 0^o = 0.33 x 10^-4 T. the neutral points will be parallel and above the cable. Let r be the distance of neutral point from the cable. Then
\(H=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2I }{ r } \ or \ r=\frac { { \mu }_{ o } }{ 4\pi } \times \frac { 2I }{ H } \) 
\(r={ 10 }^{ -7 }\times \frac { 2\times 3.3 }{ 0.33\times { 10 }^{ -4 } } =2\times { 10 }^{ -2 }m=2 \ cm\) 
Thus the set of neutral points will lie parallel to cable, and above the cable at distance of 2 cm from cable.

Here, \( (RC={ 10 }^{ 4 }\times{ 10 }^{ -11 }={ 10 }^{ -7 }s\ \frac { 1 }{ { v }_{ c } } =\frac { 1 }{ 50\times{ 10 }^{ 6 } } ={ 2 }\times10^{ -8 }s\), the circuit is good enough for the detection.