In a photoelectric experiment,if the wavelength of incident radiation is reduced from $6000 \ \mathring{A}$ to $4000 \ \mathring{A}$ while keeping the intensity of radiation constant,then:

In a photocell circuit,the stopping potential $V_0$ is a measure of the maximum kinetic energy of the photoelectrons. The following graph shows experimentally measured values of stopping potential versus frequency $\nu$ of incident light. The values of Planck's constant and the work function as determined from the graph are (taking the magnitude of electronic charge to be $e = 1.6 \times 10^{-19} \, C$):

$A$ metal surface is illuminated by light of a given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum kinetic energy of the emitted photoelectrons would become

Which of the following graphs shows the correct variation of maximum kinetic energy $(E)$ of photoelectrons with the intensity of incident radiation $(I)$?

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\frac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the first case is one-fourth that in the second case,the work function of the surface of the material is ($c=$ speed of light,$h=$ Planck's constant).

This question has Statement-$1$ and Statement-$2$. Of the four choices given after the statements,choose the one that best describes the two statements.
Statement-$1$: $A$ metallic surface is irradiated by a monochromatic light of frequency $v > v_0$ (the threshold frequency). The maximum kinetic energy and the stopping potential are $K_{max}$ and $V_0$ respectively. If the frequency incident on the surface is doubled,both the $K_{max}$ and $V_0$ are also doubled.
Statement-$2$: The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.