Question

A U-tube manometer contains mercury of density 13,600kgm-3, and oil.
(i) When both ends of the tube are open to the atmosphere, h1 = 0.17 m and Δh = 1cm. Calculate the density of the oil.
(ii) The open end of the left leg is sealed and is pressurized to a gauge pressure P, which causes the level of the free surface of the oil in the left leg to fall 5mm. Calculate the value of P.

Answer 1

Equate the pressure at a horizontal level

Answer 2

I think density of oil will be 799.893Kgm-3

Answer 3

and i think p=101666.579Nm-2

Answer 4

The pressure of the oil column at the bottom of the oil on the mercury forces the mercury column up delta h cm. \[P=\rho _{hg} g \Delta h = \rho _{oil}g H _{oil}\] Similarly, when the the left side is capped and pressurized so as to force the oil column down 5 mm the Hg column is raised 5mm. The additional pressure is transferred through the oil to the Mercury. the diameters of the tube and oil container are irrelevant as is the height of the mercury column on the left.

Answer 5

So P=5mm of Hg ?

Answer 6

NO Diwakar you must add air pressure too

Answer 7

They didn't ask for absolute pressure but gauge pressure which is the pressure above atmospheric. So the gauge pressure is 5 mm Hg .

Answer 8

yes gleem your right sorry Diwakar your right too