A mercury manometer consists of two unequal arms of equal cross-section and length 100 cm and 50 cm. The two open ends are sealed with air in the tube at a pressure of 80 cm 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 10 cm in steady state, find the length of the mercury column risen in the longer tube.

A mercury manometer consists of two unequal arms of equal cross-sectio

1.	A mercury manometer consists of two unequal arms of equal cross-section and length 100 cm and 50 cm. The two open ends are sealed with air in the tube at a pressure of 80 cm 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 10 cm in steady state, find the length of the mercury column risen in the longer tube.
Answer» Solution :For shorter ARM , INITIAL volume of air = A `times` 50 cm When MERCURY is introduced in the tube, the volume of air `=A times 40cm^(3)` Let `P_(1)` be the pressure of the air in shorter arm, then by PV = constant, we have `80 times (A times50)=P_(1) times (A times 40)` `THEREFORE P_(1)=100 cm` For longer arm, Initial volume of air `=A times 100 cm^(3)` Let `l_(0)` be the length of mercury, then volume of air `=A times(100-l_(0))` Let `P_(2)` be the pressure of the arm in the longer air, then by Pascal.s law `P_(A)=P_(B)` `l_(0)+P_(2)=10+P_(1)` `P_(2)=(10+P_(1))-l_(0)=(10+100)-l_(0)=(110-l_(0))` cm Now by PV = constant, we have `(110-l_(0)) times A(100-l_(0))=80 times (A times 100)" (or) "(110-l_(0))(100-l_(0))=8000` `l_(0)^(2)=210l_(0)+3000=0, l_(0)=15.5cm`