When the left arm of a mercury manometer is connected to a cylinder filled with a gas, the level of the mercury in the right arm rises by 2 cm. If the pressure of the gas in the container 110160 Pa, the atmospheric pressure is __________cm of Hg. ( take $g = 10 m s^{- 2}$ )

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 79
Let $P_{o}$ be atmospheric pressure.
So $(P_{1} - P_{0}) = ρ g h$
$(110160 - P_{0}) = \frac{13.56 g}{{c m}^{3}} \times 980 \frac{c m}{3} \times 2 c m$
$110160 - P_{0} = 26577.6$
$P_{0} = 110160 - 26577.6$
$P_{0} = 83582.4 P a s c a l$
$P_{0} = \frac{83582.4 P_{a}}{1333.22 P a / c m H g}$
$P_{0} = 79 c m o f H g$

Suggest Corrections

Similar questions

Open end mercury manometer is connected to a gas chamber containing gas at a pressure of 1.38 bar. The barometric pressure is 750 mm Hg. Mercury level in the arm connected to a gas chamber stands at 11.3 cm. What is the reading in open arm of the manometer?

In a constant volume gas thermometer, the pressure of the working gas is measured by the difference in the levels of mercury in the two arms of a U-tube connected to the gas at one end. When the bulb is placed at the room temperature 27.0℃, the mercury column in the arm open to atmosphere stands 5.00 cms above the level of mercury in the other arm. When the bulb is placed in a hot liquid, the difference of mercury levels becomes 45.0 cms. Calculate the temperature of the liquid. (Atmospheric pressure = 75.0 cm of mercury.)

One arm of a glass U-tube is open to the atmosphere, whereas the other vertical arm is connected to a gas supply. The difference in levels of mercury (density = 13.6 g cm^-3) between the vertical arms of the U-tube manometer is h = 3 cm. If the atmospheric pressure is 760 mm Hg and the gravitational acceleration is 9.8 m s^-2, then the pressure of the gas is about:

Q. 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.

Q. A manometer is connected to a gas containing bulb. the open arm reads

40.0

cm whereas the arm connected to the bulb reads

15.0

cm. If barometric pressure is

74.0

cm Hg, then what is the pressure of a gas in the bar?

Join BYJU'S Learning Program

When the left arm of a mercury manometer is connected to a cylinder filled with a gas, the level of the mercury in the right arm rises by 2 cm. If the pressure of the gas in the container 110160 Pa, the atmospheric pressure is __________cm of Hg. ( take g=10ms−2)

The correct option is A 79Let Po be atmospheric pressure.So (P1−P0)=ρgh(110160−P0)=13.56gcm3×980cm3×2cm110160−P0=26577.6P0=110160−26577.6 P0=83582.4PascalP0=83582.4Pa1333.22Pa/cmHgP0=79cmofHg

When the left arm of a mercury manometer is connected to a cylinder filled with a gas, the level of the mercury in the right arm rises by 2 cm. If the pressure of the gas in the container 110160 Pa, the atmospheric pressure is __________cm of Hg. ( take $g = 10 m s^{- 2}$ )