The mercury manometer consists if two unequal arms of equal cross section `1 cm^(2) and lenghths 100cm and 50cm. The two open ends are sealed with air in the tube at a pressure of 80cm of mercurey. 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 lenght 10cm in steady state, find the length of the mercury column risen in the longer tube.

Explanation

Let `p_(1)` and `p_(2)` be the pressures in centimetre of mercury in the two arms ater introducing mercury in the tube. Suppose the mercury column rises in the second arm to` l_(0)cm`.
using `pV= constant for the shorter arm,
`(80cm) (50cm)= p_(1)(50cm-10cm)
or, p_(1)=100cm`. ...(i)` Using pV=constant fir 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`
`p_(2)=100cm-l_(0)cm`.
Putting in (ii),
`(110-l_(0))(100-l_(0))=8000`
or,`l_(0)^(2)-210l_(0)+3000=0`
or,`l_(0)=15.5`.
The requrired length is `15.5 cm. `