Given above is the concentration vs time plot for a dissociation reaction : $\mathrm{A} \rightarrow \mathrm{nB}$.

Based on the data of the initial phase of the reaction (initial 10 min ), the value of n is $\_\_\_\_$ .

Observe the following reactions at $\mathrm{T}(\mathrm{K})$.

I. $\mathrm{A} \rightarrow$ products.

II. $5 \mathrm{Br}^{-}(\mathrm{aq})+\mathrm{BrO}_3{ }^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 3 \mathrm{Br}_2(\mathrm{aq})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$

Both the reactions are started at 10.00 am . The rates of these reactions at 10.10 am are same. The value of $-\frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta \mathrm{t}}$ at 10.10 am is $2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$. The concentration of A at 10.10 am is $10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$. What is the first order rate constant (in $\mathrm{min}^{-1}$ ) of reaction $I$ ?

$$ \text { Given below are two statements : } $$

In the light of the above statements, choose the correct answer from the options given below

The work functions of two metals $\left(\mathrm{M}_{\mathrm{A}}\right.$ and $\left.\mathrm{M}_{\mathrm{B}}\right)$ are in the $1: 2$ ratio. When these metals are exposed to photons of energy 6 eV , the kinetic energy of liberated electrons of $M_A: M_B$ is in the ratio of $2.642: 1$. The work functions (in eV ) of $M_A$ and $M_B$ are respectively.