In finding the electric field using Gauss Law the formula $$\left| {\overrightarrow E } \right| = {{{q_{enc}}} \over {{\varepsilon _0}\left| A \right|}}$$ is applicable. In the formula $${{\varepsilon _0}}$$ is permittivity of free space, A is the area of Gaussian surface and q_enc is charge enclosed by the Gaussian surface. The equation can be used in which of the following situation?

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy T_A eV end de-Broglie wavelength $$\lambda _A$$. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is T_B = (T_A – 1.5) eV. If the de-Broglie wavelength of these photoelectrons $$\lambda _B$$ = 2$$\lambda _A$$, then the work function of metal B is :

The coordinates of centre of mass of a uniform flag shaped lamina (thin flat plate) of mass 4kg. (The coordinates of the same are shown in figure) are :

Consider a uniform rod of mass M = 4m and length $$\ell $$ pivoted about its centre. A mass m moving with velocity v making angle $$\theta = {\pi \over 4}$$ to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is :