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Question
The heights of mercury surfaces in the two arms of the manometer shown in figure are 2 cm and 8 cm.
Atmospheric pressure = 1.01 × 105 N−2. Find (a) the pressure of the gas in the cylinder and (b) the pressure of mercury at the bottom of the U tube.
Solution
(a) Given:
Height of the first arm, h1 = 8 cm
Height of the second arm, h2 = 2 cm
Density of mercury, ρHg = 13.6 gm/cc
Atmospheric pressure, pa = 1.01 × 105 N/m2 = 1.01 × 106 dyn/cm2
Now,
Let pg be the pressure of the gas.
If we consider both limbs, then the pressure at the bottom of the tube will be the same.
According to the figure, we have:
pg + ρHg × h2 × g = Pa + ρHg × h1 g
`=>` Pg =Pa +ρHg × g(h1-h2)
=1.01 ×106+13.6 ×980 ×(8-2) dyn/cm2
=(1.01 ×106+13.6 ×980 ×6)dyn/cm2
= 1.09 × 105N/m2
(b) Pressure of the mercury at the bottom of the U-tube:
PHg =Pa + ρHg×h1×g
=1.01×106+13.6×8×980
=1.12×105N/m2
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