1. The diagram shows an upside-down bottle containing two liquids, A ( A =1250 k

Question

1. The diagram shows an upside-down bottle containing two liquids, A (A=1250 kg/m3) and B (B=820 kg/m3). Liquid A fills a volume with a narrow cross-section, A1=14 cm2, and a height of h1=12 cm. Liquid B fills a volume with a wide cross-section, A2=180 cm2, and a height of h2=26 cm.

    A) What is the mass of liquid A?

    B) What is the mass of liquid B?

    C) What is the total volume of fluid in the bottle?

    D) What is the gauge pressure at the bottom of the bottle?

2. Now the fluids are mixed and form an emulsion (like oil and water); the total volume of fluid in the bottle does not change.

    A) What is the density of the mixed fluid? (Hint: Think about your answers to the first three questions.)

    B) What is the gauge pressure at the bottom of the bottle now?

Explanation / Answer

1.

(A) Mass of liquid A (mA) = A*A1*h1 = 1250*14*10^(-4)*0.12 = 0.21 Kg

(B) Mass of liquid B (mB) = B*A2*h2 = 820*180*10^(-4)*0.26 = 3.84 Kg

(C) Total volume of the liquid (V) = A1*h1 + A2*h2  = 14*10^(-4)*0.12 + 180*10^(-4)*0.26 = 4.85*10^(-3) m^3

(D) Gauge pressure at the bottom = A*g*h1 + B*g*h2   = 1250*10*0.12 + 820*10*0.26 = 3632 Pa

2.

(A) Density of the mixed fluid () = total mass of the liquids/total volume = (mA + mB)/V = (0.21+3.84)/4.85*10^(-3)

i, e    = 835 Kg/m^3

(B) Gauge pressure at the bottom = *g*(h1+h2) = 835*10*0.4 = 3340 Pa

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