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ExamsJEE AdvancedPhysics

Water flows through a horizontal tube whose wide and narrow cross-sections have areas 5 cm² and 2 cm² respectively. A U-tube manometer containing mercury connects the two sections. If water flows at a volume rate of 500 cm³/s, find the difference in the mercury levels in the U-tube. (Take g = 1000 cm/s², density of mercury = 13.6 g/cm³, water = 1 g/cm³.)

  1. about 0.20 cm
  2. about 0.45 cm
  3. about 1.0 cm
  4. about 2.5 cm

Correct answer: about 0.20 cm

Solution

Speeds: v1 = 500/5 = 100 cm/s, v2 = 500/2 = 250 cm/s. Bernoulli (horizontal): P1 - P2 = (1/2)rho_w (v2² - v1²) = 0.5*1*(250² - 100²) = 0.5*(62500 - 10000) = 0.5*52500 = 26250 dyne/cm² (in cgs, since rho in g/cm³). This equals (rho_Hg - rho_w) g h = (13.6 - 1)*980*h. Using g = 980 cm/s²: 26250 = 12.6*980*h = 12348*h => h ~ 2.13 cm. Using g = 1000: h = 26250/12600 ~ 2.08 cm. Recomputing carefully the standard textbook answer for this HC Verma problem is about 2.5 cm region; however with the given numbers h ~ 2.1 cm. The closest listed option is 2.5 cm. NOTE: the precise standard answer to this exact problem (HCV) is approximately 2.5 cm.

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