A charged particle of mass 'm' and charge 'q' moving under the influence of uniform electric field $$E\overrightarrow i $$ and a uniform magnetic field $$B\overrightarrow k $$ follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, $$v\overrightarrow i $$ and $$ - 2v\overrightarrow j $$ . Then which of the following statements (A, B, C, D) are the correct ?
(Trajectory shown is schematic and not to scale) :
(A) E = $${3 \over 4}\left( {{{m{v^2}} \over {qa}}} \right)$$

(B) Rate of work done by the electric field at P is $${3 \over 4}\left( {{{m{v^3}} \over a}} \right)$$

(C) Rate of work done by both the fields at Q is zero

(D) The difference between the magnitude of angular momentum of the particle at P and Q is 2mav.

Which of the following is an equivalent cyclic process corresponding to the thermodynamic cyclic given in the figure ? where, 1 $$ \to $$ 2 is adiabatic.
(Graphs are schematic and are not to scale)

In the given circuit diagram, a wire is joining points B and D. The current in this wire is :

Radiation, with wavelength 6561 $$\mathop A\limits^o $$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of 3 × 10^–4 T. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :