Question

A column of Hg of 10 cm length is contained in the middle of a narrow horizontal 1m long tube which is closed at both ends. Both the halves of the tube contain air at a pressure 76 cm Hg. By what distance will the column of Hg be displaced if the tube is held vertical?

A. 3cm B. 1cm C. 4cm D. 5cm

Answer 1

Answer :- 3 cm

x - Column of Hg by which it will come down

P1 - upper part air pressure

P2 - lower part air pressure

P1 + 10 = P2  --- (1)

Equation of state at upper part is =

76 x 45 x a = P1 x (45 + x) x a

P1 = (76 x 45) / (45 + x) ---- (2)

Equation on the lower part will be =

76 x 45 x a = P2 x (45 - x) x a

P2 = (76 x 45) / (45 - x) ---- (3)

Substituting (2) & (3) in (1)

[(76 x 45) / (45 + x) ] + 10 = [P2 = (76 x 45) / (45 - x)]

$x^{2} + 684 x -45x^{2} =0\\x = 3 cms$