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 x 10⁵ N/m². Find (a) the pressure of the gas in the cylinder and (b) the pressure of mercury at the bottom of the U tube.
Answers
(a). Since, the tube is opened from the top, this means, total pressure of the gas will be equal to the pressure due to the mercury column + Atmospheric Pressure.
∴ Pressure of the gas in cylinder = Atmospheric Pressure + Pressure due to the mercury column.
Now, Pressure due to the mercury column = ρg(h₂ - h₁)
where, ρ is the density of the mercury = 13.6 g/cm³,
g is the acceleration due to gravity = 9.8 m/s² = 980 cm/s², h₂ is 8 cm and h₁ is 2 cm.
∴ Total Pressure of the gas = 1 atm + (8 - 2) × 13.6 × 980
= 1.013 × 10⁵ Pa + 6 × 13.6 × 980 dyne/cm²
= 101300 + 7996.8 Pa.
= 181268 Pa.
= 1.8 × 10⁵ Pa.
(b). The pressure of the mercury at the bottom of the U-tube
= The pressure of mercury column on the open side + Atmospheric pressure
= 13.6 × 10⁶/1000 × 9.8 × 0.08 + 1.013 × 10⁵
= 1.12 × 10⁵ Pa.
Hope it helps .
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