First slide

Fluid statics

Question

An open U-tube contains mercury. When 11.2 cm of water is poured into one of the arms of the tube, how high does the mercury rise in the other arm from its initial level?

Moderate

A

0.56 cm
B

1.35 cm
C

0.41 cm
D

2.32 cm

Solution

Let mercury rise in the other arm= x
Then from pascal's law
$P_{A} = P_{B} \Rightarrow 11.2 \times ρ_{w} g = 2 {xρ}_{Hg} \times g$
$x = \frac{11.2 \times 1}{2 \times 13.6} = 0.41 cm$

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