A long U-tube contains mercury (density = 14 × 103 kg/m3). When 10 cm of water (density= 1.0 × 103 kg/m3) is poured into the left arm, the mercury in the right arm rises above itsoriginal level by:A. 0.36 cmB. 0.72 cmC. 14 cmD. 35 cmE. 70 cm
Question
A long U-tube contains mercury (density = 14 × 103 kg/m3). When 10 cm of water (density= 1.0 × 103 kg/m3) is poured into the left arm, the mercury in the right arm rises above itsoriginal level by:A. 0.36 cmB. 0.72 cmC. 14 cmD. 35 cmE. 70 cm
Solution
The pressure at the bottom of the U-tube must be the same in both arms, otherwise the liquid would move until it is.
The pressure at the bottom of the left arm is due to the weight of the water column, which is the product of the height of the water column (h1 = 10 cm), the density of water (ρ1 = 1.0 × 10^3 kg/m^3), and the acceleration due to gravity (g = 9.8 m/s^2). So, P1 = h1 * ρ1 * g.
The pressure at the bottom of the right arm is due to the weight of the mercury column, which is the product of the height of the mercury column (h2), the density of mercury (ρ2 = 14 × 10^3 kg/m^3), and the acceleration due to gravity (g = 9.8 m/s^2). So, P2 = h2 * ρ2 * g.
Setting P1 = P2 and solving for h2 gives:
h1 * ρ1 * g = h2 * ρ2 * g
Solving for h2 gives:
h2 = (h1 * ρ1) / ρ2
Substituting the given values gives:
h2 = (10 cm * 1.0 × 10^3 kg/m^3) / (14 × 10^3 kg/m^3) = 0.71 cm
So, the mercury in the right arm rises by approximately 0.71 cm, which is closest to answer choice B. 0.72 cm.
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