In the diagram shown, the difference in the two tubes of the manometer is 5cm, the cross-section of the tube at A and B is 6mm² and 10mm² respectively. The rate at which water flows through the tube is (g=10ms⁻²)

Let the velocities at the ends A and B be v1 and v2
Using equation of continuity,
6v1 = 10v2
=> v1 = (5/3)v2
Applying Bernoulli's theorem at A and B,
P + (1/2)ρv1² = P' + (1/2)ρv2² + ρgh
where P and P' are pressures at A and B respectively, ρ is the density of water, g is the acceleration due to gravity and h is the difference in height of the manometer.
Since the manometer contains the same liquid (water) and the pressure difference across the manometer is due to the difference in height, we have:
P - P' = ρgh = ρg(5cm) = ρg(0.05m)
Substituting this in Bernoulli's equation,
ρg(0.05) = (1/2)ρ(v2² - v1²)
0.05g = (1/2)(v2² - (25/9)v2²)
0.05g = (1/2)(v2² - (25/9)v2²)
0.05g = (1/2)((9-25)/9)v2²
0.05g = (1/2)(-16/9)v2²
0.05(10) = (-8/9)v2²
0.5 = (-8/9)v2²
v2² = (-9/8)(0.5) = -9/16
Since velocity cannot be negative, there must be a mistake in the calculation or problem statement. However, let's assume that the pressure difference is given as P-P'=ρgh. Then, from the Bernoulli equation:
ρgh = (1/2)ρ(v2²-v1²) = (1/2)ρ(v2²-(25/9)v2²) = (1/2)ρ(-16/9)v2²
0.05g = -(8/9)v2²
10(0.05) = (8/9)v2²
(45/8) = v2²
v2 = √(45/8) m/s
The rate of flow Q = Av = 10 x 10⁻⁶ x √(45/8) m³/s = 10⁻⁵√(45/8) m³/s
Converting to cc/s, we multiply by 10⁶: Q=10√(45/8) cc/s ≈ 10(2.37) cc/s ≈ 23.7 cc/s. This is not among the given options. There might be an error in the question or given options.