Slope of the straight line obtained by plotting $\log_{10} k$ against $\frac{1}{T}$ represents which term?

For reaction $A \rightarrow P$, rate constant $k = 1.5 \times 10^3 \text{ s}^{-1}$ at $27^\circ\text{C}$. If activation energy for the above reaction is $60 \text{ kJ mol}^{-1}$, then the temperature (in $^\circ\text{C}$) at which rate constant $k = 4.5 \times 10^3 \text{ s}^{-1}$ is . . . . . . .

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: $A$ reaction can have zero activation energy.
Reason $R$: The minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to threshold value,is called activation energy.
In the light of the above statements,choose the correct answer from the options given below:

For the following reactions:
$A \xrightarrow{700 \ K}$ Product
$A \xrightarrow[\text{catalyst}]{500 \ K}$ Product
it was found that $E_{a}$ is decreased by $30 \ kJ/mol$ in the presence of a catalyst. If the rate remains unchanged,the activation energy for the catalysed reaction is (Assume pre-exponential factor is same):

Reactant $(A)$ produces two products. If $Ea_2 = 2 Ea_1$,then $K_1$ and $K_2$ are related as:
$A \xrightarrow{K_1} B$,activation energy: $Ea_1$
$A \xrightarrow{K_2} C$,activation energy: $Ea_2$

The rate constant is doubled when temperature increases from $27\,^oC$ to $37\,^oC.$ Activation energy in $kJ$ is