A manometer reads the pressure of a gas in an enclosure as shown in given figure (a) When a pump removes some of the gas, the manometer reads as in figure (b) The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.

**(a)** Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury.

**(b) **How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) is poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).

Asked by Pragya Singh | 1 year ago | 72

**(a)** For diagram (a):

Given, Atmospheric pressure, P_{0} = 76 cm of Hg

The difference between the levels of mercury in the two arms is gauge pressure.

Thus, gauge pressure is 20 cm of Hg.

We know, Absolute pressure = Atmospheric pressure + Gauge pressure

= 76 + 20 = 96 cm of Hg

For diagram (b):

Difference between the levels of mercury in the two arms = –18 cm

Hence, gauge pressure is –18 cm of Hg.

And, Absolute pressure = Atmospheric pressure + Gauge pressure

= 76 cm – 18 cm = 58 cm

**(b)** It is given that 13.6 cm of water is poured into the right arm of figure (b).

We know that relative density of mercury = 13.6

=> A 13.6 cm column of water is equivalent to 1 cm of mercury.

Let, h be the difference in the mercury levels of the two arms.

Now, pressure in the right arm P_{R} = Atmospheric pressure + 1 cm of Hg

= 76 + 1 = 77 cm of Hg . . . . . . (a)

The mercury column rises in the left arm,

thus the pressure in the left limb, P_{L} = 58 + h . . . . . . (b)

Equating equations (a) and (b) we get :

77 = 58 + h

Therefore the difference in the mercury levels of the two arms, h = 19 cm

Answered by Abhisek | 1 year ago**Answer the following questions:-**

**(a)** According to the law of atmospheres density of air decreases with increase in height y as \( ρ=\rho _{0}e^{\dfrac{-y}{y_{0}}}\). Where ρ_{0} =1.25 kg m^{-3} is the density of air at sea level and y_{0 }is a constant. Derive this equation/law considering that the atmosphere and acceleration due to gravity remain constant.

**(b)** A zeppelin of volume 1500 m^{3} is filled with helium and it is lifting a mass of 400 kg. Assuming that the radius of the zeppelin remains constant as it ascends. How high will it rise?[ y_{0} = 8000 m and ρH_{e} =0.18 kg m^{-3}].

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Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside ? Surface tension of mercury at the temperature of the experiment is 0.465 N m^{–1}. Density of mercury = 13.6 × 10^{3} kg m^{–3}

In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius \( 2.0 × 10^{–5} m\) and density \( 1.2 × 10^3 kg m^{–3}\). Take the viscosity of air at the temperature of the experiment to be \( 1.8 × 10^{–5} Pa \). How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

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