Answer ;
(a) For diagram (a):
According to the question, atmospheric pressure,
P0 = 76 cm of Hg
Gauge pressure is the difference in mercury levels between the two arms.
The gauge pressure is 20 cm of Hg.
We know that the Absolute pressure is equal to the sum of Atmospheric pressure and the Gauge pressure> therefore we get:
= 76 + 20 = 96 cm of Hg
For diagram (b):
Differences in mercury levels between the two arms = –18 cm
Hence, the gauge pressure is –18 cm of Hg.
We know that the Absolute pressure is equal to the sum of Atmospheric pressure and the Gauge pressure> therefore we get:
= 76 cm – 18 cm = 58 cm
(b) We are given that 13.6 cm of water is poured into the right arm of figure (b). It is given that relative density of mercury is 13.6. 1 centimeter of mercury is equivalent to a 13.6 cm column of water.
Suppose that h denotes the difference in the mercury levels of the two arms.
Now, pressure in the right arm PR = Atmospheric pressure + 1 cm of Hg
Therefore, pressure in the right arm PR = 76 + 1 = 77 cm of Hg . . . . . . (a)
The mercury column rises in the left arm,
Thus the pressure in the left limb, PL = 58 + h . . . . . . (b)
Equating equations (a) and (b) we get :
77 = 58 + h
As a result, the difference in mercury levels between the two arms, h = 19 cm, is calculated.