Step 1: Understanding the Concept:
A Boolean expression is a tautology if it is true for all possible truth values of its variables. For an implication \(X \implies Y\) to be a tautology, whenever \(X\) is true, \(Y\) must also be true.
Step 2: Detailed Explanation:
Let's test Option (B): \(* = \wedge, \square = \vee\).
The expression is \((p \wedge \sim q) \implies (p \vee q)\).
Truth Table: 
Since the last column is all True, the expression is a tautology for these operators.
If we checked (A): \((p \wedge \sim q) \implies (p \wedge q)\). If \(p=T, q=F\), the LHS is \(T\) but RHS is \(F\), so it's not a tautology.
Step 3: Final Answer:
The operators are \(* = \wedge\) and \(\square = \vee\).
Equivalent statement to (p\(\to\)q) \(\vee\) (r\(\to\)q) will be
The number of values of $r \in\{p, q, \sim p, \sim q\}$ for which $((p \wedge q) \Rightarrow(r \vee q)) \wedge((p \wedge r) \Rightarrow q)$ is a tautology, is :
Among the statements :
\((S1)\) \((( p \vee q ) \Rightarrow r ) \Leftrightarrow( p \Rightarrow r )\)
\((S2)\)\((( p \vee q ) \Rightarrow r ) \Leftrightarrow(( p \Rightarrow r ) \vee( q \Rightarrow r ))\)
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)