Which plot of $\ln k$ vs $\frac{1}{T}$ is consistent with the Arrhenius equation?

The activation energy for a reaction which doubles the rate when the temperature is raised from $298 \ K$ to $308 \ K$ is ........... $kJ \ mol^{-1}$

The rate of a reaction quadruples when the temperature changes from $300 \, K$ to $310 \, K$. The activation energy of this reaction is ........... $kJ \, mol^{-1}$ (Assume activation energy and pre-exponential factor are independent of temperature; $\ln 2 = 0.693$; $R = 8.314 \, J \, mol^{-1} \, K^{-1}$)

An endothermic reaction with high activation energy for the forward reaction is given by the diagram:

The rate of the chemical reaction doubles for an increase of $10 \, K$ in absolute temperature from $298 \, K$. Calculate $E_{a}$.

What is the slope of the straight line for the graph drawn between $\ln k$ and $\frac{1}{T}$,where $k$ is the rate constant of a reaction at temperature $T$?