Step-by-Step Solution

1. Apply Bernoulli's equation between the two points in the Venturi meter:

   \(P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2\)

2. Since the Venturi meter is horizontal, \(h_1 = h_2\), so the \(\rho g h\) terms cancel out:

   \(P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2\)

3. Rearrange the equation to relate the pressure difference to the velocity difference:

   \(P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2)\)

4. The pressure difference is also related to the difference in water levels in the manometer:

   \(P_1 - P_2 = \rho g \Delta h\), where \(\Delta h = 5\) cm = 0.05 m

5. Equate the two expressions for the pressure difference:

   \(\rho g \Delta h = \frac{1}{2}\rho (v_2^2 - v_1^2)\)

6. Simplify and solve for \(v_2^2 - v_1^2\):

   \(v_2^2 - v_1^2 = 2g \Delta h = 2 \times 10 \times 0.05 = 1\)

7. Apply the continuity equation: \(A_1 v_1 = A_2 v_2\), where \(A_1 = A\) and \(A_2 = a\). Given \(\frac{A}{a} = 2\), so \(A = 2a\).

   \(A v_1 = a v_2 \Rightarrow 2a v_1 = a v_2 \Rightarrow v_2 = 2v_1\)

8. Substitute \(v_2 = 2v_1\) into the equation \(v_2^2 - v_1^2 = 1\):

   \((2v_1)^2 - v_1^2 = 1 \Rightarrow 4v_1^2 - v_1^2 = 1 \Rightarrow 3v_1^2 = 1 \Rightarrow v_1^2 = \frac{1}{3} \Rightarrow v_1 = \frac{1}{\sqrt{3}}\)

9. Calculate the volume flow rate \(Q = A v_1\), where \(A = \sqrt{3}\) m\(^2\):

   \(Q = \sqrt{3} \times \frac{1}{\sqrt{3}} = 1\) m\(^3\)/s