The mercury manometer consists of two unequal arms of equal cross section `1 cm^(2)` and lengths `100cm` and `50cm`. The two open ends are sealed with air in the tube at a pressure of `80cm` of mercury. Some amount of mercury is now introduced in the manometer through the stopcock connected to it. If mercury rises in the shorter tube to a length `10cm` in steady state, find the length of the mercury column risen in the longer tube.

Explanation

Let `P_(1)and P_(2)` and be the pressures in centimestre of mercury in the two arms after introducing mercury in the tube. Suppose the mercury column in the second armto `l_(0)cm`.

Using PV=constant for the shorter arm
(80cm)(50cm)=`P_(1)(50cm-10cm)`
or `P_(1)`=100cm....... (i).
Using PV=constant for the longer arm,
(80cm)(100cm)=`P_(2)(100-l_(0))cm` ....(ii)
From the figure,
`P_(1)=P_(2)+(l_(0)-10)cm.` ,
Thus by (i),
100cm=`P_(2)+(l_(0)-10)cm`.
or `P_(2)=110cm-l_(0)cm`
putting in (ii),
`P_(2)=(110cm-l_(0))(100-l_(0))=8000`
or, `l_(0)^(2)-210l_(0)+3000=0`
or `l_(0)=15.5`