The U-tube shown in Fig. 1 has legs of unequal internal diameters dj = 10 mm and

Question

The U-tube shown in Fig. 1 has legs of unequal internal diameters dj = 10 mm and d; = 5 mm. which are partly filled with immiscible liquids of density p1 = 1,800 kg / m3, and P2 = 1,200 kg / m3, respectively, and are open to the atmosphere at the top. If an additional 1.2 cm5 of the second liquid is added to the right-hand leg, h^ will change by an amount increment. Increment(cm) = 6.11 cm If the level hB falls by 0.6 cm, the level he will rise by a distance If hc= 3 cm and hA = 2 cm, then Fig. 1. U-tube with immiscible liquids 5(cm) = 0.15cm hB = 1.667 cm

Explanation / Answer

1. conservation of mass of second liquid or assuming its density is constant we can equate volume added to the change in volume of the liquid then we get 1.2 cc = pi* (0.25^2)*delta,hA solving we get delta,hA = 6.11 cm 2. this part is independent of first part above so u are not adding any liquid in here so hA remains same or volume of second liquid remains same and also volume of second liquid remains same but hB and hC change so for this we can write the decrease in volume due to fall in hB on one limb is compensated by increase in volume due to raise in hC on other limb so we get pi*(0.5^2)*delta,hC = pi*(0.25^2)*delta,hB then we get delta,hC = 0.15 cm 3.applying pressure at the same horizontal levels in the same liquids is same and taking atmospheric pressure as p0 we get p0 + rho1*g*hC = p0 + rho2*g*hA + rho1*g*hB solving for hB gives us hB = 1.667 cm

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