From the graph it is clear that $$A$$ and $$B$$ have the same stopping potential and therefore, the same frequency. Also, $$B$$ and $$C$$ have the same intensity.

From question,
$$\eqalign{ & {B_0} = 20nT = 20 \times {10^{ - 9}}\;T \cr & \left( {\because {\text{velocity}}\,{\text{of}}\,{\text{light}}\,{\text{in}}\,{\text{vacuum}}\,C = 3 \times {{10}^8}\;m{s^{ - 1}}} \right) \cr & {{\vec E}_0} = {{\vec B}_0} \times \vec C \cr & \left| {{{\vec E}_0}} \right| = \left| {\vec B} \right|.\left| {\vec C} \right| = 20 \times {10^{ - 9}} \times 3 \times {10^8} \cr & = 6\,V/m \cr} $$

As $$\lambda $$ decreases, $$\nu $$ increases and hence the speed of photoelectron increases. The chances of photo electron to meet the anode increases and hence photo electric current increases.

In amplitude modulation, the amplitude of the high frequency carrier wave made to vary in proportional to the amplitude of audio signal.

Here $$\omega = 2\pi \times 6 \times {10^{14}}$$
$$ \Rightarrow f = 6 \times {10^{14}}Hz$$
Wavelength
$$\lambda = \frac{c}{f} = \frac{{3 \times {{10}^8}}}{{6 \times {{10}^{14}}}} = 0.5 \times {10^{ - 6}}\;m = 5000\mathop A\limits^ \circ $$
Given $$E = \frac{{12375}}{{5000}} = 2.48eV$$
Using $$E = W + e{V_s}$$
$$ \Rightarrow 2.48 = 2 + e{V_s}$$
or $${V_s} = 0.48V$$

From Einstein’s photoelectric equation,
$$\eqalign{ & \frac{{hc}}{{{l_1}}} - f = \frac{1}{2}\;m{\left( {2v} \right)^2}\,......\left( {\text{i}} \right) \cr & {\text{and}}\,\frac{{hc}}{{{1_2}}} - f = \frac{1}{2}m{v^2}\,......\left( {{\text{ii}}} \right) \cr} $$
From eqn. (i) & (ii)
$$\eqalign{ & \Rightarrow \frac{{\frac{{hc}}{{{\lambda _1}}} - \phi }}{{\frac{{hc}}{{{\lambda _2}}} - \phi }} = 4 \Rightarrow \frac{{hc}}{{{\lambda _1}}} - \phi = \frac{{4hc}}{{{\lambda _2}}} - 4\phi \cr & \Rightarrow \frac{{4hc}}{{{\lambda _2}}} - \frac{{hc}}{{{\lambda _1}}} = 3\phi \Rightarrow \phi = \frac{1}{3}hc\left( {\frac{4}{{{\lambda _2}}} - \frac{1}{{{\lambda _1}}}} \right) \cr & = \frac{1}{3} \times 1240\left( {\frac{{4 \times 350 - 540}}{{350 \times 540}}} \right) \cr & = 1.8eV \cr} $$

$$\eqalign{ & {\text{Power,}}\,P = \frac{{nh\nu }}{t} \cr & \Rightarrow \nu = \frac{{P \times t}}{{nh}} \cr & = \frac{{4 \times {{10}^3} \times 1}}{{{{10}^{20}} \times 6.63 \times {{10}^{ - 34}}}} = 6 \times {10^{16}}Hz \cr} $$

The equation of amplitude modulated wave
$$m = \left( {{v_0} + A\cos \omega t} \right)\sin \omega t$$
$$\eqalign{ & = {v_0}\sin {\omega _0}t + A\cos \omega t\sin {\omega _0}t \cr & = {v_0}\sin {\omega _0}t + \frac{A}{2}\left[ {\sin \left( {{\omega _0} - \omega } \right)t + \sin \left( {{\omega _0} + \omega } \right)t} \right] \cr} $$

$$\frac{{hC}}{\lambda } - {\phi _0} = K.{\text{E}} = \frac{{{p^2}}}{{2{m_e}}} = \frac{{{h^2}}}{{2{m_e}\lambda _d^2}}$$
Differentiating on both sides
$$\eqalign{ & \frac{{ - hC}}{{{\lambda ^2}}}d\lambda = \frac{{{h^2}}}{{2{m_e}}}\left( {\frac{{ - 2}}{{{\lambda _d}^3}} \times d{\lambda _d}} \right) \cr & \therefore \frac{{d{\lambda _d}}}{{d\lambda }} \propto \frac{{{\lambda _d}^3}}{{{\lambda ^2}}} \cr} $$

The energy possessed by photons of wavelength
$$550\,nm\,{\text{is}}\,\frac{{1240}}{{550}} = 2.25\,eV$$
The energy possessed by photons of wavelength
$$450\,nm\,{\text{is}}\,\frac{{1240}}{{450}} = 2.76\,eV$$
The energy possessed by photons of wavelength
$$350\,nm\,{\text{is}}\,\frac{{1240}}{{350}} = 3.54\,eV$$
For metal plate $$p$$ :
$${\phi _p} = 2\,eV.$$
All the wavelengths are capable of ejecting electrons. Therefore, the current is maximum. Also as the work function is lowest in $$p,$$ the kinetic energy of ejected electron will be highest and therefore, the stopping potential is highest.
For metal plate $$q$$ :
$${\phi _q} = 2.5\,eV.$$
Photons of wavelength $$550\,nm$$  will not be able to eject electrons and therefore, the current is smaller than $$p.$$ The work function is greater than $$q$$ therefore the stopping potential is lower in comparison to $$p.$$
For metal plate $$r$$ :
$${\phi _r} = 3\,eV.$$
Only wavelength of $$350\,nm$$  will be able to eject electrons and therefore, current is minimum. Also the stopping potential is least.