The Lagrangian forthe Kepler problem is given by $$L = \frac{1}{2}\left[ {{{\dot r}^2} + {r^2}{{\dot \theta }^2}} \right] + \frac{\mu }{r}\,\,\,\,\,\,\,\left( {\mu > 0} \right)$$
where, $$\left( {r,\,\theta } \right)$$  denotes the polar coordinates and mass of the particle is unity, then

A mass m is constrained to move on a horizontal frictionless surface. It is set in circular motion with radius r₀ and angular speed ω₀ by an applied force $$\overrightarrow {\bf{F}} $$ communicated through an inextensible thread that passesthrough a hole on the surface as shown in figure given below. Then, this force is suddenly doubled.

The magnitude of the radial velocity of the mass

A. increases till mass falls into hole

B. decreases till mass falls into hole

C. remains constant

D. becomes zero at radius r₁, where 0 < r₁ < r₀

The speed of a particle whose kinetic energy is equal to its rest mass energy, is given by (c is the speed of light in vacuum)

A. $$\frac{c}{3}$$

B. $$\frac{{\sqrt 2 }}{3}c$$

C. $$\frac{c}{2}$$

D. $$\frac{{\sqrt 3 }}{2}c$$

The Hamiltonian corresponding to the Lagrangian $$L = a{{\dot x}^2} + b{{\dot y}^2} - kxy$$     is

A. $$\frac{{{p_x}^2}}{{2a}} + \frac{{{p_y}^2}}{{2b}} + kxy$$

B. $$\frac{{{p_x}^2}}{{4a}} + \frac{{{p_y}^2}}{{4b}} - kxy$$

C. $$\frac{{{p_x}^2}}{{4a}} + \frac{{{p_y}^2}}{{4b}} + kxy$$

D. $$\frac{{{p_x}^2 + {p_y}^2}}{{4ab}} + kxy$$

A particle is moving in an inverse square field. If the total energy of the particle is positive, then trajectory of particle is

A. circular

B. elliptical

C. parabolic

D. hyperbolic