If a metal sheet is irradiated with radiations of frequencies $v_1$ and $v_2$ and the kinetic energies of the emitted photoelectrons are in the ratio $1 : x$, then the threshold frequency of the metal is:

When photons of energy $hv$ fall on a metal plate of work function $W_0$,photoelectrons of maximum kinetic energy $K$ are ejected. If the frequency of the radiation is doubled,the maximum kinetic energy of the ejected photoelectrons will be

Light of frequency $4 \times 10^{14} \,Hz$ is incident on a metal surface of work function $2.14 \,eV$, resulting in photoemission of electrons. The maximum kinetic energy of the emitted electrons is $\left[h=6.63 \times 10^{-34} \,J-s\right]$ (in $\,eV$)

When two monochromatic lights of frequency $v$ and $\frac{v}{2}$ are incident on a photoelectric metal,their stopping potentials are $\frac{V_{s}}{2}$ and $V_{s}$ respectively. The threshold frequency for this metal is:

Given below are two statements:
Statement-$I$: The figure shows the variation of stopping potential $(V_0)$ with frequency $(v)$ for two photosensitive materials $M_1$ and $M_2$. The slope gives the value of $\frac{h}{e}$,where $h$ is Planck's constant and $e$ is the charge of an electron.
Statement-$II$: $M_2$ will emit photoelectrons of greater kinetic energy for incident radiation having the same frequency.
In the light of the above statements,choose the most appropriate answer from the options given below.

All electrons ejected from a metallic surface by incident light of wavelength $400 \,nm$ travelled $1 \,m$ in the direction of a uniform electric field of $2 \,N/C$ and came to rest. The work function of the surface is (in $\,eV$)