Step 1: Use the Reaction Quotient \( Q_c \)
The reaction quotient \( Q_c \) is given by:
$$ Q_c = \frac{[B][C]}{[A]^2} $$
Substitute the given concentrations:
$$ Q_c = \frac{(2 \times 10^{-3})(2 \times 10^{-3})}{(2 \times 10^{-3})^2} = \frac{4 \times 10^{-6}}{4 \times 10^{-6}} = 1 $$
Step 2: Compare \( Q_c \) with \( K_c \)
The equilibrium constant \( K_c \) is given as \( 4 \times 10^{-3} \).
Since \( Q_c = 1 \) and \( K_c = 4 \times 10^{-3} \), we see that:
$$ Q_c > K_c $$
Step 3: Direction of the Reaction
When \( Q_c > K_c \), the reaction proceeds in the backward direction to achieve equilibrium.
Therefore, the reaction will shift to the left, and there is a tendency for the reaction to go in the backward direction.
Step 4: Conclusion
The correct answer is: Option (3): The reaction has a tendency to go in the backward direction.
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :