$A$ plot of $ \frac{1}{T} $ vs. $ \ln k $ for a reaction gives the slope $ -1 \times 10^{4} \ K $. The energy of activation for the reaction is (Given $ R = 8.314 \ J \ K^{-1} \ mol^{-1} $)

The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40^{\circ} C$ to $30^{\circ} C$. The activation energy of the reaction is.........$kJ \, mol^{-1}$.
[Take; $R = 8.314 \, J \, mol^{-1} \, K^{-1}$,$\ln(3.555) = 1.268$]

The velocity constant of a reaction at $290 \ K$ was found to be $3.2 \times 10^{-3}$. At $300 \ K$ it will be

Consider the following statements:
$(i)$ Increase in concentration of reactant increases the rate of a zero order reaction
$(ii)$ Rate constant $k$ is equal to $A$ if $E_a = 0$
$(iii)$ Rate constant $k$ is equal to $A$ if $E_a = \infty$
$(iv)$ $\log_e k$ vs $T$ is a straight line
$(v)$ $\log_e k$ vs $1/T$ is a straight line
Which of the following statements are correct?

The rate of a chemical reaction doubles for every $10^\circ \text{C}$ rise in temperature. If the temperature is increased by $50^\circ \text{C}$,the rate of reaction will increase by a factor of ........

Assertion : If the activation energy of a reaction is zero,temperature will have no effect on the rate constant.
Reason : Lower the activation energy,faster is the reaction.