Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in vertical tubes is 5 cm and the area of cross-sections at A and B are 6text cm^2$6\text{ cm}^{2}$ and 3text cm^2$3\text{ cm}^{2}$ respectively. The rate of flow will be ____ textcm^3$\text{cm}^{3}$/s. (take g = 10 textm/s^2$\text{m/s}^{2}$)

### Related Formula A_1 V_1 = A_2 V_2$$A_1 V_1 = A_2 V_2$$ P_1 + frac12rho V_1^2 = P_2 + frac12rho V_2^2$$P_1 + \frac{1}{2}\rho V_1^2 = P_2 + \frac{1}{2}\rho V_2^2$$ P_1 - P_2 = rho g h$$P_1 - P_2 = \rho g h$$ ### Core Logic From the continuity equation between A and B: A_AV_A = A_BV_B implies 6V_A = 3V_B implies V_B = 2V_A$$A_{A}V_{A} = A_{B}V_{B} \implies 6V_{A} = 3V_{B} \implies V_{B} = 2V_{A}$$ Applying Bernoulli's equation between A and B for a horizontal pipe: P_A + frac12rho V_A^2 = P_B + frac12rho V_B^2$$P_{A} + \frac{1}{2}\rho V_{A}^{2} = P_{B} + \frac{1}{2}\rho V_{B}^{2}$$ P_A - P_B = frac12rho(V_B^2 - V_A^2)$$P_{A} - P_{B} = \frac{1}{2}\rho(V_{B}^{2} - V_{A}^{2})$$ Since the height difference is h = 5text cm = 0.05text m$h = 5\text{ cm} = 0.05\text{ m}$, the pressure difference is rho gh$\rho gh$. rho g times 0.05 = frac12rho( (2V_A)^2 - V_A^2 )$$\rho g \times 0.05 = \frac{1}{2}\rho( (2V_A)^2 - V_A^2 )$$ g times 0.05 = frac12(3V_A^2)$$g \times 0.05 = \frac{1}{2}(3V_{A}^{2})$$ ### Step 1: Calculate Velocity and Volume Flow Rate Solving for V_A$V_A$: V_A = sqrtfrac2 times 10 times 0.053 = sqrtfrac13 = frac1sqrt3text m/s$$V_{A} = \sqrt{\frac{2 \times 10 \times 0.05}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}\text{ m/s}$$ V_A = frac100sqrt3text cm/s$$V_A = \frac{100}{\sqrt{3}}\text{ cm/s}$$ Volume flow rate Q = A_A V_A$Q = A_{A} V_{A}$: Q = 6text cm^2 times frac100sqrt3text cm/s = frac600sqrt3 = 200sqrt3text cm^3text/s$$Q = 6\text{ cm}^2 \times \frac{100}{\sqrt{3}}\text{ cm/s} = \frac{600}{\sqrt{3}} = 200\sqrt{3}\text{ cm}^{3}\text{/s}$$ ### Pattern Recognition In a horizontal Venturi meter, substituting V_2 = V_1 (A_1/A_2)$V_2 = V_1 (A_1/A_2)$ directly into rho g h = frac12rho (V_2^2 - V_1^2)$\rho g h = \frac{1}{2}\rho (V_2^2 - V_1^2)$ is the standard path. Working in CGS vs MKS units requires careful tracking; convert V_A$V_A$ to cm/s before multiplying by A_A$A_A$ in cm². ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids