A manometer tube contains mercury of density $13.6 \times 10^{3} k g m^{- 3}$ . What difference in the levels of mercury in the two arms is indicated by a gauge pressure of $1.03 \times 10^{5} P a$ ?

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Solution

$P = l g h$
$p = l g h$
$h = \frac{p}{l g}$
$= \frac{1.03 \times 10^{5}}{13.6 \times 1000 \times 9.8}$
$= 0.077 m$
$= 7.7 c m$

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A manometer tube contains mercury of density 13.6×103kgm−3. What difference in the levels of mercury in the two arms is indicated by a gauge pressure of 1.03×105Pa?

P=lghp=lghh=plg=1.03×10513.6×1000×9.8=0.077m=7.7cm

A manometer tube contains mercury of density $13.6 \times 10^{3} k g m^{- 3}$ . What difference in the levels of mercury in the two arms is indicated by a gauge pressure of $1.03 \times 10^{5} P a$ ?

$P = l g h$
$p = l g h$
$h = \frac{p}{l g}$
$= \frac{1.03 \times 10^{5}}{13.6 \times 1000 \times 9.8}$
$= 0.077 m$
$= 7.7 c m$