Fluid Mechanics

JEE Main level, single correct, medium

48 of 192

Question 48 of 192

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 AA and BB are 6 cm26 \mathrm{~cm}^{2} and 3 cm23 \mathrm{~cm}^{2} respectively. The rate of flow will be ____\_\_\_\_ cm3/s\mathrm{cm}^{3} / \mathrm{s}. (take g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^{2})

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Answer: (D) 2003200 \sqrt{3}

Given: h=5h = 5 cm, A1=6A_1 = 6 cm², A2=3A_2 = 3 cm², g=1000g = 1000 cm/s²
By continuity: A1v1=A2v2⇒v2=2v1A_1v_1 = A_2v_2 \Rightarrow v_2 = 2v_1
By Bernoulli's equation: P1−P2=12ρ(v22−v12)P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)
Since P1−P2=ρghP_1 - P_2 = \rho gh:
ρgh=12ρ(4v12−v12)=3ρv122\rho gh = \frac{1}{2}\rho(4v_1^2 - v_1^2) = \frac{3\rho v_1^2}{2}
v1=2gh3=2×1000×53=1003v_1 = \sqrt{\frac{2gh}{3}} = \sqrt{\frac{2 \times 1000 \times 5}{3}} = \frac{100}{\sqrt{3}} cm/s
Rate of flow Q=A1v1=6×1003=2003Q = A_1v_1 = 6 \times \frac{100}{\sqrt{3}} = 200\sqrt{3} cm³/s

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