Answer: (2) 2720 cm$^3$/s

Solution

According to question, area of cross-section at $A$, $a_A = 40$ cm$^2$ and at $B$, $a_B = 20$ cm$^2$ Let velocity of liquid flow at $A = V_A$ and at $B = V_B$ Using equation of continuity: $a_A V_A = a_B V_B$ $40 V_A = 20 V_B \Rightarrow 2V_A = V_B$ Now, using Bernoulli’s equation: $P_A + \frac{1}{2}\rho V_A^2 = P_B + \frac{1}{2}\rho V_B^2 \Rightarrow P_A - P_B = \frac{1}{2}\rho(V_B^2 - V_A^2)$ $\Rightarrow \Delta P = \frac{1}{2} \times 1000\left(V_B^2 - \frac{V_B^2}{4}\right) \Rightarrow \Delta P = 500 \times \frac{3V_B^2}{4}$ $\Rightarrow V_B = \sqrt{\frac{(\Delta P) \times 4}{1500}} = \sqrt{\frac{(700) \times 4}{1500}}$ m/s $= 1.37 \times 10^2$ cm/s

Volume flow rate $Q = a_B \times V_B = 20 \times 100 \approx 2732$ cm$^3$/s $\approx 2720$ cm$^3$/s