First slide

Fluid statics

Question

An open U-tube contains mercury, When 11.2 cm of water is poured into one of 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

When water is poured in the left arm, level of mercury in the right arm rises by x cm.
$Now P_{A} = P_{B}$
$\begin{array}{l} P_{atm} + 11 .2 \times ρ_{ω} \times g = P_{atm} + 2 x \times ρ_{H g} \times g \\ \Rightarrow x = \frac{11 .2 \times 1}{2 \times 13 .6} cm = 0 .41 cm \end{array}$

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