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Question

In the manometer as shown, 2 immiscible fluids - mercury (ρ=13600 kg/m3) and water (ρ=1000 kg/m3) are used as manometric fluids. The water end is exposed to the atmosphere (100 kPa) and the mercury end is exposed to a gas. At this position, the interface between the fluids is at the bottom most point of the manometer. The value of h is found to be 9.45 m. The height of the mercury column is given to be 75 cm. Find the gauge pressure of the gas. (g=9.8 m/s2)


A
100 kPa
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B
50 kPa
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C
200 kPa
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D
0 kPa
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Solution

The correct option is D 0 kPa
Height of water column =0.75+9.45
=10.2 m
As we know,
Pg is the absolute pressure of the gas
Gauge pressure of the gas =PgPa

Pressures at bottommost point will be the same. Hence, equating pressure at the bottommost point
Pa+ρw×g×(10.2)=Pg+ρm×g×(0.75)
Pa+103×g×(10.2)=Pg+13.6×103×g×0.75
PgPa=0 kPa

Gauge pressure of the gas =PgPa=0 kPa

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