Give ,carrier wave frequency
\({ f }_{ c }=20MHz=20\times { 10 }^{ 6 }Hz\)
Bandwidth required for modulation is
\({ 2f }_{ m }=3kHz=3\times { 10 }^{ 3 }Hz\)
\(\Rightarrow \ \ { f }_{ m }=\frac { 3\times { 10 }^{ 3 } }{ 2 } =1.5\times { 10 }^{ 3 }Hz\) s
Demodulation by a diode is possible,if the condition 
Thus,\(\frac { 1 }{ { f }_{ c } } =\frac { 1 }{ 20\times { 10 }^{ 6 }Hz } =0.5\times { 10 }^{ -7 }s\) .....(i)
and \(\frac { 1 }{ { f }_{ m } } =\frac { 1 }{ 1.5\times { 10 }^{ 3 }Hz } =0.7\times { 10 }^{ -3 }s\)    ......(ii)
Now,gain through all the option of R and C one by one,we get
(i)  \(RC=1k\Omega \times 0.01\mu F\)
\(\\ ={ 10 }^{ 3 }\Omega \times ({ 0.01 }\times { 10 }^{ -6 }F)={ 10 }^{ -5 }s\)
Here, condition \(\frac { 1 }{ { f }_{ c } } <\) is satisfied  
Hence, it can be demodulated.
(ii) \((RC=1k\Omega \times 0.01\mu F={ 10 }^{ 4 }\Omega \times { 10 }^{ -8 }F={ 10 }^{ -4 }s)\)
Here, condition\(\frac { 1 }{ { f }_{ c } }\) < is satisfied
 Hence, it  can be demodulated
(iii) \(RC=10k\Omega \times 1\mu F={ 10 }^{ 4 }\Omega \times { 10 }^{ -6 }F={ 10 }^{ -2 }s\)
Here, condition\(\frac { 1 }{ { f }_{ c } } >RC,\) so this cannot be demodulated.

The distance travelled by the signal is 5km. Loss suffered in path of transmission = 2dB/km
So, total loss suffered in 5 km = \(-2\times 5=-10dB\)
Total amplifier gain = 10dB + 20dB = 30dB
Overall gain in signal = 30 - 10 = 30dB
According to the question, gain in dB = \(10{ log }_{ 10 }\frac { { P }_{ 0 } }{ { P }_{ 1 } } \)
\(\because \ 20=10{ log }_{ 10 }\frac { { P }_{ 0 } }{ P_{ i } } or\quad { log }_{ 10 }\frac { { P }_{ 0 } }{ P_{ i } } =2\)
Where\({ P }_{ i }=1.01mW\ and\ { P }_{ 0 }\) is the output power.  
\(\because \ \frac { { P }_{ 0 } }{ { P }_{ i } } ={ 10 }^{ 2 }=100\)
\(\\ \Rightarrow { P }_{ 0 }={ p }_{ i }\times 100=1.01\times 100\ or\ { p }_{ 0 }=101mW\)
Thus the output is 101 mW.

In normal adjustments, m = \(|\frac { { f }_{ 0 } }{ { f }_{ e } } |=20\\ \)
Also, length of telescope, f₀ + f_e = 105
20f₀ + f _e = 105
21f _e = 105
f _e = 5 cm
f₀ = 20 f_e = 20 x 5 = 100 cm

As, we know that
Electric field intensity, 
\( E= \ \sqrt { \frac { { Pc\mu }_{ 0 } }{ { 4\pi r }^{ 2 } } }\)
\( i.e.\ E\alpha \ \sqrt { \frac { P }{ { r }^{ 2 } } } \)
As, r is constant in this question.
So,  \(E\alpha \sqrt { P } \)
For two different situations,
\({ \ E }_{ 1 }\alpha \sqrt { { P }_{ 1 } } and \ { E }_{ 2 } \ \alpha \sqrt { { P }_{ 2 } } \)
\(frac { { E }_{ 1 } }{ { E }_{ 2 } } =\left( \frac { { P }_{ 1 } }{ { P }_{ 2 } } \right) ^{ { 1 }/{ 2 } }\Rightarrow \frac { { E }_{ 1 } }{ { E }_{ 2 } } =\left( \frac { 100 }{ 50 } \right) ^{ { 1 }/{ 2 } }=\sqrt { 2 }\)
\(\Rightarrow \ { E }_{ 2 }=\frac { { E }_{ 1 } }{ \sqrt { 2 } }\)
\(As,d { E }_{ 1 }=E\Rightarrow { E }_{ 2 }=\frac { E }{ \sqrt { 2 } } \)

Electric field intensity at Point O due to + Q charge,
\({ E }_{ 1 }=\frac { 1 }{ 4\pi { \varepsilon }_{ o } } \times \frac { Q }{ { \left( { x }_{ 2 } \right) }^{ 2 } } (towards \ B) .....(i)\)
Electric field intensity at Point O due to - Q charge,
\({ E }_{ 2 }=\frac { 1 }{ 4\pi { \varepsilon }_{ o } } \times \frac { Q }{ { \left( { x }_{ 1 } \right) }^{ 2 } } (towards \ B).....(i)\)
\({ \because }\) E₁ and E₂ act along the same direction.
\({ \because }\) Net electric field intensity at point O,
\(E={ E }_{ 1 }+{ E }_{ 2 }=\frac { 1 }{ 4\pi { \varepsilon }_{ o } } \times \frac { Q }{ { \left( { x }_{ 2 } \right) }^{ 2 } } +\frac { 1 }{ 4\pi { \varepsilon }_{ o } } \times \frac { Q }{ { \left( { x }_{ 1 } \right) }^{ 2 } } (towards \ B)\)
\( E=\frac { Q }{ 4\pi { \varepsilon }_{ o } } \left[ \frac { 1 }{ { x }_{ 2 }^{ 2 } } +\frac { 1 }{ { x }_{ 1 }^{ 2 } } \right] \)

Given, E₁ = 2.25 V, l₁ = 30 cm
l₂ = 60 cm, E₂ = ?
As, we know that in the case of the potentiometer, the potential gradient remains constant.
So, E ∝  l
\(\frac { E1 }{ E2 } =\frac { l1 }{ l2 } \)      .....(i)
Substituting the given values in Eq.(i), we get
\(\frac { 2.24 }{ E2 } =\frac { 30 }{ 60 } \) 
\(E2=\frac { 2.25X60 }{ 30 } =4.5\ V\)

The amplitude modulated radiowaves having frequency 530 kHz to 1710 kHz (or wavelength between 175 m to 566 m) which are travelling directly following the surface of earth are known as ground waves. The short wave communication over long distance is only possible via sky waves. It is not possible via ground waves because the ground waves can bend round the corners of the objects on earth and hence, their intensity falls with distance. Moreover the ground wave transmission becomes weaker as frequency increases.

The saturation current depends upon the maximum number of photoelectrons emitted per second from the photosensitive surface, which in turn dep[ends upon the intensity of incident light. The saturation current is independent of the anode potential as it cannot change the maximum number of electrons emitted from photo-cathode.

(a) As explained in theory, \(\alpha \)-decay is caused by the quantum mechanical tunnelling of an alpha particle through a repulsive Coulomb barrier. The rate of tunnelling would depend upon the height and width of the barrier. The decay cannot be all at once. And that is the reason why half life of \(_{ 92 }{ { U }^{ 238 } }\) against \(\alpha \)-decay is large.
(b) The possible nuclear reaction is \(_{ 92 }{ { U }^{ 238 } }\rightarrow _{ 92 }{ { U }^{ 237 } }+_{ 0 }{ { n }^{ 1 } }\)
The data shows that \(m\left( _{ 92 }{ { U }^{ 237 } } \right) +m\left( n \right) \) is greater than \(m\left( _{ 92 }{ { U }^{ 238 } } \right) .\) Therefore, the decay is not allowed energetically. Rather, some external energy has to be supplied to separate a neutron from \(_{ 92 }{ { U }^{ 238 } }\).

The instantaneous activity of a radioactive sample is given by \(A={ A }_{ 0 }{ e }^{ -\lambda t }\)
Differentiating both sides w.r.t. t, we get \(\frac { dA }{ dt } ={ A }_{ 0 }{ e }^{ -\lambda t }\left( -\lambda \right) =-\lambda { A }_{ 0 }{ e }^{ -\lambda t }=-\lambda A=-\lambda \left( \lambda N \right) =-{ \lambda }^{ 2 }N\)
Using \(\lambda =\frac { 0.693 }{ T } , \ we \ get \ \frac { dA }{ dt } =-{ \left( \frac { 0.693 }{ T } \right) }^{ 2 }N, \ i.e., \ \frac { dA }{ dt } \propto \frac { 1 }{ { T }^{ 2 } } ,\) which was to be proved.

It happens because there 9 is an electric field over A but no electric field over B as it lies inside a hollow conductor. Third law of motion is not violated because force arises between A and the hollow sphere, the charge B is simply an internal part of the sphere.

In LR circuit, the impedance is given by
\({ Z }_{ L }=\sqrt { { R }^{ 2 }+{ X }_{ L }^{ 2 } } \quad ...(i)\)
In LCR circuit, the impedance is given by
\(Z=\sqrt { { R }^{ 2 }+{ \left( { X }_{ L }-{ X }_{ C } \right) }^{ 2 } } \quad ...(ii)\)
From equations, (i) and (ii), we find that
\(Z<{ Z }_{ L }\)
\(since \ I=\frac { E }{ Z } \)
As Z decreases on connecting a  capacitor in series with LR circuit, hence the current I in the circuit increases.

Number of turns on the circular coil, n = 20
Radius of the coil,r = 10 cm = 0.1 m
Magnetic field strength, B = 0.10 T
Current in the coil, I = 5.0 A
(a) The total torque on the coil is zero because the field is uniform.
(b) The total force on the coil is zero because the field is uniform.
(c) Cross-sectional area of copper coil, A = 10^-5 m²
Number of free electrons per cubic meter in copper, N = 10²⁹ /m³
Charge on the electron, e = 1.6 x 10^-19 C
Magnetic force, F = Bev_d
Where,
v_d = Drift velocity of electrons
\(=\frac{I}{N e A}\)
\(\therefore F=\frac{B e I}{N e A}\)
\(=\frac{0.10 \times 5.0}{10^{29} \times 10^{-5}}=5 \times 10^{-25} \mathrm{~N}\)
Hence, the average force on each electron is 5 x 10^-25 N.