Correct answer is option A

We are given the work functions of Aluminium and Gold as $4.1\mathrm{eV}$ and $5.1\mathrm{eV}$, respectively.

The stopping potential ($V_s$) is related to the frequency ($f$) of the incident light by the equation:

$e{V}_s=h(f-f_0)$

Where $e$ is the charge of an electron, $h$ is Planck's constant, and $f_0$ is the threshold frequency. The threshold frequency is related to the work function ($\varphi$) by:

$\varphi =h{f}_0$

So, we can write the equation for stopping potential as:

$V_s=\frac{h}{e}(f-\frac{\varphi }{h})$

This equation represents a straight line with slope $\frac{h}{e}$. Therefore, the slope of the stopping potential versus frequency plot is the same for both Aluminium and Gold. The ratio of the slopes is:

$\frac{\text{Slope for Gold}}{\text{Slope for Aluminium}}=\frac{\frac{h}{e}}{\frac{h}{e}}=1$

So , the ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is 1.

Correct answer is option A

$\begin{array}{rl}& \frac{1}{2}m{v}^2=hf-h{f}_0\\ \\ & \Rightarrow \frac{1}{2}m{v}_1^2=2h{f}_0-h{f}_0=h{f}_0\\ \\ & \text{ also, }\frac{1}{2}m{v}_2^2=5h{f}_0-h{f}_0=4h{f}_0\end{array}$

taking ratio,

$\frac{v_1^2}{v_2^2}=\frac{1}{4}\Rightarrow \frac{v_1}{v_2}=\frac{1}{2}$

Correct answer is option C

$\varphi =\frac{hc}{\lambda }=\frac{1240}{350}\mathrm{eV}=3.54\mathrm{eV}$

$\therefore$ Only metal B will emit photoelectron.

Correct answer is option C

Wavelength of infra-red $=700\mathrm{nm}$ (minimum)

Wavelength of UV $=100-400\mathrm{nm}$

Since we need $\lambda <5000$ Å

$\Rightarrow$ Only UV would be able to emit photoelectrons.

Correct answer is option C

Statement I is correct as stopping potential is independent of power of light used. Statement II is correct as maximum kinetic energy of photoelectron depends on wavelength of light.